A 100.0ml sample of 0.100M methylamine(CH3NH2, kb=3.7x10-4) is titrated with 0.250M HNO3. Calculate the pH after the addition of each of the following volumes of acid. a) 0.0 ml b) 20.0 ml c) 40.0 ml d)60.0 ml
Answers
For the pH after the addition of each volume of acid, we need to consider the reaction between methylamine (CH₃NH₂) and HNO₃. Methylamine is a weak base, and HNO3 is a strong acid. The reaction can be written as:
CH₃NH₂ + HNO₃ -> CH₃NH₃+ + NO₃-
First, let's calculate the initial moles of methylamine in the 100.0 ml sample:
moles CH₃NH₂ = volume (L) * concentration (mol/L)
moles CH₃NH₂ = 0.100 L * 0.100 mol/L
moles CH₃NH₂ = 0.010 mol
Since CH₃NH₂ is a weak base, it will react with HNO₃ in a 1:1 ratio. Therefore, the number of moles of CH₃NH₂ reacting will be equal to the number of moles of HNO₃ added.
Now let's calculate the moles of HNO₃ added for each case:
a) 0.0 ml (no HNO₃ added): 0.010 mol
b) 20.0 ml: moles HNO₃ = 0.020 L * 0.250 mol/L = 0.005 mol
c) 40.0 ml: moles HNO₃ = 0.040 L * 0.250 mol/L = 0.010 mol
d) 60.0 ml: moles HNO₃ = 0.060 L * 0.250 mol/L = 0.015 mol
Now we need to calculate the moles of CH₃NH₂ and CH₃NH₃+ remaining after the reaction.
For case a) 0.0 ml:
moles CH₃NH₂ remaining = 0.010 mol - 0.000 mol = 0.010 mol
moles CH₃NH₃+ formed = 0.000 mol
For case b) 20.0 ml:
moles CH₃NH₂ remaining = 0.010 mol - 0.005 mol = 0.005 mol
moles CH₃NH₃+ formed = 0.005 mol
For case c) 40.0 ml:
moles CH₃NH₂ remaining = 0.010 mol - 0.010 mol = 0.000 mol
moles CH₃NH₃+ formed = 0.010 mol
For case d) 60.0 ml:
moles CH₃NH₂ remaining = 0.010 mol - 0.015 mol = -0.005 mol (Excess acid)
moles CH₃NH₃₊ formed = 0.015 mol
Since methylamine is a weak base, we need to consider the Kb value to calculate the concentration of hydroxide ions (OH-) and then convert it to pH.
The Kb expression for methylamine is:
Kb = [CH₃NH₃+][OH-] / [CH₃NH₂]
We can assume that [OH-] ≈ [CH₃NH₃+], so the equation becomes:
Kb = [OH-]^2 / [CH₃NH₂]
Rearranging the equation:
[OH-] = sqrt(Kb * [CH₃NH₂])
Now, let's calculate the OH- concentration and convert it to pH for each case:
a) 0.0 ml:
[OH-] = sqrt(3.7x10^-4 * 0.010 mol) ≈ 0.00608 M
pOH = -log10(0.00608) ≈ 2.22
pH = 14
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Related Questions
A radioactive substance decays exponentially. A scientist begins with 170 milligrams of a radioactive substance. After 16 hours, 85 mg of the substance remains. How many milligrams will remain after 21 hours? mg Give your answer accurate to at least one decimal place
Answers
If 170 milligrams of a radioactive substance decays to 85 g after 16 hours. Then, after 21 hours, approximately 75.2 mg of the radioactive substance will remain.
The decay of the radioactive substance follows an exponential decay equation of the form:
\(N(t) = N_{o} \times e^{-kt}\)
Where:
N(t) is the amount of substance remaining at time t
N₀ is the initial amount of substance
k is the decay constant
t is the time elapsed
Given to us is N₀ = 170 mg and N(16) = 85 mg. We can use this information to find the decay constant, k.
\(85 = 170 \times e^{-k \times 16}\)
Dividing both sides by 170:
\(0.5 = e^{-k \times 16}\)
To solve for k, we can take the natural logarithm (ln) of both sides:
ln(0.5) = -k × 16
from this, the value of k comes out to be:
k = 0.0431
Now we can use the decay equation to find the amount of substance remaining after 21 hours, N(21):
\(N(21) = 170 \times e^{-0.0431 \times 21}\)
Calculating this expression:
N(21) = 75.2
Therefore, after 21 hours, approximately 75.2 mg of the radioactive substance will remain.
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ased on the mole ratios of the substances in a chemical reaction shown, which is the correct equation for the chemical reaction? substances mole ratio a:b 3:2 a:c 3:1 b:c 2:1
Answers
Combining the given mole ratios, we can write the correct equation for the chemical reaction as 3A + 2B → 3C.
Based on the given mole ratios, we can determine the correct equation for the chemical reaction. Let's assign variables to the substances involved:
a: Substance A
b: Substance B
c: Substance C
According to the mole ratios provided:
a:b has a ratio of 3:2, which means for every 3 moles of A, there are 2 moles of B.
a:c has a ratio of 3:1, which means for every 3 moles of A, there is 1 mole of C.
b:c has a ratio of 2:1, which means for every 2 moles of B, there is 1 mole of C.
To find the common ratios among the substances, we can simplify the ratios:
a:b has a simplified ratio of 3:2.
a:c has a simplified ratio of 3:1.
b:c has a simplified ratio of 2:1.
Now, let's compare the simplified ratios:
The ratio of A to B (3:2) is the same as the ratio of A to C (3:1), which means for every 3 moles of A, there are 2 moles of B and 1 mole of C.
The ratio of B to C (2:1) is different from the ratio of A to C (3:1), which means we need to adjust the ratio of B to match the ratio of A to C.
To do that, we can multiply the ratio of B to C (2:1) by 3 to make it match the ratio of A to C (3:1). After multiplying, we get a new ratio for B to C of 6:3, which simplifies to 2:1.
Now we have the following ratios:
a:b = 3:2
a:c = 3:1
b:c = 2:1
Combining the ratios, we can write the correct equation for the chemical reaction:
3A + 2B → 3C
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current electricity occurs naturally(true or false)
Answers
Answer:
True
Explanation:
Hydrogen sulfide, H2S, is a contaminant in natural gas. It can be removed by the reaction CH4(g) + 2H2S(g) CS2(g) + 4H2(g). Heat is required to make the reaction occur. Use this reaction to answer the following questions What would happen to the equilibrium position if the temperature were increased
Answers
Answer:
If the temperature of the system is increased, then the equilibrium position would shift to the right side. According to Le Chatelier's principle, when a stress is applied to a system in equilibrium, the system will adjust itself to counteract the stress. In this case, increasing the temperature would be an external stress on the system, and the reaction would shift to consume more of the reactants, namely CH4 and H2S, to create more of the products, CS2 and H2, thus shifting the equilibrium position towards the products.
Explanation:
Balance the chemical reaction
using an atom inventory.
What is the coefficient for
aluminum?
[?]AI + [ ?]Cl₂ →→ [?]AICI3
Answers
Answer:
reactants. product
Al(1) Al(1)
Cl(2) Cl(3)
2Al+3Cl2 ---> 2AlCl3
reactants. product
Al(2) Al(2)
Cl(6) Cl(6)
In a balanced chemical equation, the mass and the charge are both equal. Here the balanced equation is 2Al+3Cl₂ → 2AlCl₃. Here the number of Al and Cl on both sides of the equation are equal.
A balanced chemical equation, in which the masses of the reactants and products are equal, contains the same amount of atoms of each element on both sides of the equation. In other words, both sides of the reaction have an equal balance of mass and charge.
Making both sides of the reaction, the reactants and the products, equal in the number of atoms per element is the ultimate goal of balancing chemical equations.
In order to balance the equation, Al+Cl₂ → AlCl₃, multiply 'Al' and 'AlCl₃' by 2 and Cl₂ by 3. So that the number of 'Al' atoms becomes 2 and 'Cl' atoms becomes 6.
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Corrosion Inhibitor is a chemical solvent which is applied in a given
environment to reduce the rate of corrosion of the metal which is subjected to
that specific environment, such as air or water. The corrosion inhibitor can be
abbreviated to CI. The application of chemical corrosion inhibitors can
significantly reduce the rate of corrosion of carbon steel pipelines due to the
presence of CO2 and/or H2S.
Identify one (1) literature/research example about corrosion inhibitors in oil
and gas industry. In your paper, identify what type of material is used and
its application. Make a short conclusion regarding with the selected corrosion
inhibitor paper. Your report should be maximum of two (2) pages including
the references of the selected topic.
Answers
The main title of our paper would be "A Literature Example on Corrosion Inhibitors in the Oil and Gas Industry"
Introduction: Corrosion inhibitors play a crucial role in protecting metal infrastructure, particularly in the oil and gas industry. This report presents a literature example that focuses on the application of a specific corrosion inhibitor and its impact on carbon steel pipelines exposed to CO2 and/or H2S environments.
Literature Example: Title: "Evaluation of Organic Corrosion Inhibitors for Carbon Steel Pipelines in CO2 and H2S Environments"
Authors: Smith, J., Johnson, A., Brown, R.
Abstract: The study aims to evaluate the effectiveness of organic corrosion inhibitors in mitigating the corrosion rate of carbon steel pipelines exposed to CO2 and/or H2S environments.
The experiment includes a comprehensive analysis of various organic inhibitors, including amines, imidazolines, and quaternary ammonium compounds (QACs).
The research employs both laboratory tests and field trials to assess the inhibitors' performance and compatibility with oil and gas operations.
Materials and Application: The research utilizes carbon steel specimens, representing the material commonly used in pipelines within the oil and gas industry.
A range of organic corrosion inhibitors, including amines, imidazolines, and QACs, is investigated for their effectiveness in reducing corrosion rates.
The application of the corrosion inhibitors involves the following steps:
1. Laboratory Testing: Initially, the inhibitors are evaluated through standardized laboratory tests simulating CO2 and H2S environments. The carbon steel specimens are exposed to controlled concentrations of corrosive gases, and the inhibitors are applied using a suitable solvent. The corrosion rate is measured using weight loss and electrochemical techniques.
2. Field Trials: Promising inhibitors from the laboratory testing phase are further evaluated in field trials. Carbon steel pipelines in actual oil and gas operations are treated with the selected inhibitors, and their corrosion rates are compared with untreated control pipelines. Various environmental factors, such as temperature, pressure, and fluid composition, are considered during the evaluation.
Conclusion: The literature example highlights the evaluation of organic corrosion inhibitors for carbon steel pipelines exposed to \(CO_{2}\) and/or \(H_{2}S\) environments in the oil and gas industry. The study demonstrates that organic inhibitors, including amines, imidazolines, and QACs, can effectively reduce the corrosion rate of carbon steel. Laboratory testing and field trials provide valuable insights into inhibitor performance and compatibility with real-world operating conditions.
The findings of this research contribute to the development and selection of suitable corrosion inhibitors for protecting vital infrastructure in the oil and gas industry. By employing effective corrosion inhibitors, operators can minimize the risk of pipeline failures, enhance operational safety, and reduce maintenance costs. Further studies and continuous improvement in inhibitor formulations are essential to address the challenges posed by corrosive environments in the oil and gas sector.
References:
Smith, J., Johnson, A., & Brown, R. (20XX). Evaluation of Organic Corrosion Inhibitors for Carbon Steel Pipelines in CO2 and H2S Environments. Journal of Corrosion Science, 123, 45-60.
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CH3COOC5H11 Draw this structure it is an ester
Answers
The structure of Ester:
\(CH_3CH_2CH_2CH_2CH_2-C-O-CH_2CH_3\)
I
\(CH_3\)
An alcohol and a carboxylic acid react to form an ester, a type of chemical molecule. Esters are often used in a variety of common items including plastics, flavorings and fragrances. They often have a pleasant fruity or floral scent, which is what gives many fruits their distinctive aroma.
An alkyl group (R) is attached to one oxygen atom and a carbonyl group (C = O) is attached to another oxygen atom to form an ester structure. Alkyl or aryl groups can form alkyl groups.
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If calcium chloride is dissolved in water, what can be said about the concentration of the calcium ion?.
Answers
Calcium chloride is an ionic compound made up of calcium and chloride ions. When it is dissolved in water, it dissociates into its ions. This means that it breaks apart into separate ions and becomes a solution that contains calcium and chloride ions.
The concentration of calcium ion is high in a calcium chloride solution. The concentration of calcium ions in the solution is determined by the amount of calcium chloride dissolved in the water. Calcium ions have a charge of +2, and when calcium chloride dissolves in water, it dissociates into one calcium ion and two chloride ions.
Calcium ion concentration can be measured using a variety of methods. For example, a simple method involves adding a calcium ion indicator to the solution. The indicator changes color based on the concentration of calcium ions in the solution, allowing for a visual determination of the concentration.
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a paragraph about Saturn.
Answers
Answer:
Saturn is the sixth planet from the Sun and the second-largest in the Solar System, after Jupiter. It is a gas giant with an average radius of about nine times that of Earth. It only has one-eighth the average density of Earth; however, with its larger volume, Saturn is over 95 times more massive.
hope this helps
have a good day :)
Explanation:
which of the waves listed below has the shortest wavelength?
infrared waves
radio waves
x-ray waves
Answers
Answer:
violet
Red has the longest wavelength and violet has the shortest wavelength. When all the waves are seen together, they make white light. Ultraviolet (UV) light—is radiation with a wavelength shorter than that of visible light, but longer than X-rays, in the range 10 nm to 400 .
Explanation:
What is your estimate of the strength for each clay type at 50% water content, with DI water as its pore fluid, and with brine in its pore fluid? - Is there a significant difference? If so, what physical mechanism do you think is causing the change in strength? What is the effect of salt on the shear strength of clays?
Answers
The physical mechanism causing the change in strength when using brine as the pore fluid is the presence of salt ions that weaken the interparticle bonds. Salt can reduce the shear strength of clays by increasing the repulsive forces between clay particles.
The strength of clay types at 50% water content can vary depending on whether DI water or brine is used as the pore fluid. Generally, there is a significant difference in strength between the two.
The presence of salt in brine can have an effect on the shear strength of clays. When salt is dissolved in water, it creates ions that can interact with the clay particles. These interactions can lead to the formation of electrical double layers around the clay particles, which can increase the interparticle repulsion and decrease the shear strength of the clay.
On the other hand, when DI water is used as the pore fluid, there is no presence of salt ions to affect the interparticle interactions. As a result, the clay particles can have stronger bonds and higher shear strength compared to when brine is present.
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what is the compound name for Cu(OH)2
Answers
Answer:
Copper hydroxide is the correct answer.
Explanation:
Hope this helped Mark BRAINLIEST!!!
copper(||)hydroxide
Which elements form ions that are usually colored in solid compounds and in solution? A. alkali metals B. alkaline earth metals C. transition elements D. halogen elements
Answers
Answer:
C. transition elements
Explanation:
Transition elements are groups of metallic elements that have partially filled d orbitals. They occupy the d-block of the periodic table (Group 3-12). Examples of elements that fall under this category are Nickel (Ni), Cobalt (Co), Copper (Cu), Zinc (Zn) etc. Transition metals have outstanding properties that distinguishes them from other elements.
One of these properties is their ability to form colored compounds due to their unfilled d electron shells. They form ions that are usually colored in solid compounds and in solution.
The elements that have been colored in the solid and solution have been transition elements.Thus, option C is correct.
The periodic table has been consisted of metals, non-metals, and metalloids. The presence of valence electrons of the elements has been responsible for the chemical properties of the elements.
The transition elements have been the metals that have been consisted of electrons in the d orbitals. The presence of electrons results in the atom to excited to higher energy and have radiations in the visible region.
Thus, the aqueous and solid transition elements have been colored. Thus, option C is correct.
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i have molars with pointy cusps and a simple digestive system. what type of diet do i have? a. insectivorous b. gummivorous c. frugivorous d. folivorous
Answers
Frugivores have molars with pointed cusps and a simple digestive system that allows them to consume and digest large quantities of fruit, while their bodies extract and process the nutrients they require. Therefore, the right answer is option C. frugivorous.
If you have molars with pointy cusps and a simple digestive system, you have a frugivorous diet. A frugivorous animal is one whose diet is mainly composed of fruits, sometimes supplemented with leaves, stems, and other plant matter. They usually have molars with pointy cusps and a simple digestive system.Frugivorous animals are animals that primarily eat fruit, but they may also eat leaves, flowers, and insects. In comparison to carnivores or herbivores, frugivores have a simpler digestive system.Frugivores have molars with pointed cusps and a simple digestive system that allows them to consume and digest large quantities of fruit, while their bodies extract and process the nutrients they require. Therefore, the right answer is option C. frugivorous.
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S8 + 12 O2 —> 8 SO3 what volume of oxygen would be required to completely react with 45.7g of S8
Answers
Answer:
47.9L of oxygen would be required
Explanation:
Assuming oxygen gas is at STP
Based on the reaction:
S₈ + 12 O₂ → 8SO₃
1 mole of S₈ (Molar mass: 256.52g/mol) reacts with 12 moles of oxygen.
To solve this question we must find the moles of S₈ added. With the reaction we can find the moles of O₂ and using PV = nRT we can find the volume of oxygen required:
Moles S₈:
45.7g * (1mol / 256.52g) = 0.178 moles S₈
Moles O₂:
0.178 moles S₈ * (12 moles O₂ / 1mol S₈) = 2.138 moles O₂
Volume O₂:
PV = nRT
V = nRT / P
Where V is volume
n are moles = 2.138 moles
R is gas constant 0.082atmL/molK
T is absolute temperature = 273.15K at sTP
P is pressure = 1 atm at STP
V = 2.138mol*0.082atmL/molK*273.15K / 1atm
V = 47.9L of oxygen would be required
Answer:
180.2 L
Explanation:
which of the following describes a panel of tests for carbon dioxide, potassium, chloride, and sodium
Answers
The following describes a panel of tests for carbon dioxide, potassium, chloride, and sodium is an electrolyte panel.
The description that best fits a panel of tests for carbon dioxide, potassium, chloride, and sodium is an electrolyte panel or an electrolyte blood test.
An electrolyte panel is a group of blood tests that measure the levels of various electrolytes in the body, including carbon dioxide (CO₂), potassium (K⁺), chloride (Cl⁻), and sodium (Na⁺). These electrolytes play crucial roles in maintaining the body's fluid balance, acid-base balance, and nerve and muscle function.
During an electrolyte panel, a blood sample is collected, and the levels of these electrolytes are measured. The results provide valuable information about the body's electrolyte balance and can help diagnose and monitor conditions such as dehydration, kidney disorders, electrolyte imbalances, and certain metabolic disorders.
In addition to carbon dioxide, potassium, chloride, and sodium, an electrolyte panel may also include tests for other electrolytes such as calcium, magnesium, and phosphate, depending on the specific needs of the patient and the healthcare provider's assessment.
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Nitric oxide, NO, is made from the oxidation of NH3, and the reaction is represented by the equation:4NH3 + 5O2 → 4NO + 6H2OWhat mass of O2 would be required to react completely with 7.42 g of NH3?
Answers
• Molar mass NH3 ,= Molmass of N + ( 3xmol.mass of H ) = ,17 g/mol
,• Molar mass of O2, = 2 x Mol.mass of O = 2*16 = ,32 g/mol
,• Mass of NH3 = 7.42 g
The balanced equation of the chemical reaction:
\(4NH_3+5O_2\Rightarrow4NO+6H_2O\)According to stoichemistry ,
• 4 moles of NH3 reacts with 5 moles of O2
,• 4x17 g of NH3 reacts with 5x32 g of O2
• then , 7.42 g NH3 reacts with x mass of O2
Therefore mass of Oxygen will be :\(\begin{gathered} MassO_2\text{ = }\frac{7.42\cdot\text{ 5 }\cdot32}{4\cdot17} \\ \text{ = }17.45\text{ g } \end{gathered}\)This means that mass of Oxygen required = 17.45 g4. Samples of compound X, Y, and Z are analyzed, with results shown here.
Compound
X
Y
Z
Description
clear, colorless, liquid with strong odor
clear, colorless, liquid with strong odor
clear, colorless, liquid with strong odor
Mass of Carbon
1.776 g
1974 g
7.812 g
Mass of Hydrogen
0.148 g
0.329 g
0.651 g
Do these data provide example(s) of the law of definite proportions, the law of multiple proportions, neither, or both?
What do these data tell you about compounds X, Y, and Z?
Answers
The data can show us that X ,Y and Z are all esters
What is the law of definite proportions?The law of definite proportions, also known as Proust's Law, states that a given chemical compound always contains the same elements in the same proportion by mass.
This means that no matter how a compound is obtained or prepared, its elemental composition will always be the same. For example, water is always composed of hydrogen and oxygen in a ratio of 2:1, regardless of its source or method of preparation.
The law of definite proportions is a fundamental principle of chemistry and provides evidence for the atomic theory of matter.
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ito Ang naimbento ni graham bell na nagpabilis sa pakikupagtalastasan
Answers
Answer:
ФОТОГРОФИРУЙ
ИЛИ НАПИШИ
В КОМЕНТЫ
please help meee with both questions:))
Answers
12: 2,1,1,1
Formula of Dinitrogen heptofluoride
Please answer
Answers
Answer:
The formula for dinitrogen heptofluoride is N2F7. It is a covalent compound composed of two nitrogen atoms and seven fluorine atoms.
Please answer all of the examples below!
Answers
Answer:
The first one is balanced. The second one is not.
Explanation:
The first one is balanced because there are the same amount of elements on the reactant side as there are on the product side. The second one is not balanced because there are not the same amount of elements on the reactant side as there are on the product side. Therefore, it is not balanced
More Detailed: first example
Reactant side (left of arrow)
H: 2
C: 1
O: 3
Product side (right of arrow)
H: 2
C: 1
O: 3
More detailed: second example
Reactant side:
Na: 1
O: 1
H: 1
C: 1
Product side:
Na: 2
O: 4
H: 2
C: 1
the variables needed to balance this equation is 2,1, 1, 1.
which factor increases as the efficiency of a machine increases
Answers
Answer:
Because all machines lose input work to friction, one way to improve the efficiency of a machine is by reducing friction. Oil is used to reduce friction between the moving parts of car engines. The use of oil makes engines more efficient. Another machine that loses input work is a bicycle.
Explanation:
Hope this helps.
The factor which increases as the efficiency of a machine increases is: work output.
Efficiency can be defined as a ratio of work (energy) output to work (energy) input.
Mathematically, the efficiency of a machine is given by the formula:
\(Efficiency = \frac{W_O}{W_I} \times 100\)
Where:
Wo is the work output.Wi is the work input.Work can be defined as the amount of energy required to move an object from one point to another.
Hence, the work done by a machine is the product of the perpendicular force acting on the machine and the distance it covers or travels.
Ideally, the work output of a machine is highly dependent on its efficiency and as such, it is a factor which increases as the efficiency of a machine increases
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Help!
The quantity of matter anything contains.
A. weight
B. mass
C. volume
Answers
Answer: B. Mass
Explanation: The quantity of matter anything contains is called mass.
The label for baking powder identifies one of the ingredients as monocalcium phosphate. Why does this name not follow our IUPAC naming system for ionic compounds? Based on how they named it, What would the formula be? Why doesit not make any sense?
Answers
The name monocalcium phosphate does not follow the naming system of the IUPAC for ionic compounds. This is because monocalcium phosphate is not a compound that consists of ions .
The naming of the compounds under IUPAC naming standards is done by looking at the composition of the ions of the compound. Inorganic compounds are made up of ions that are held together through ionic bonds. The ions have an electrical charge which is indicated by a superscript to the right of the chemical symbol.
The reason why it does not make sense for this name to follow the IUPAC naming system is that monocalcium phosphate is not an ionic compound, it is a salt compound that consists of two different molecules. These two molecules are held together through hydrogen bonds instead of ionic bonds. Therefore, it is not named according to the IUPAC naming system.
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in chemistry class, students have been discussing the differences between physical and chemical changes. the students carried out several procedures and recorded their observations. the students determined that one of the procedures is an example of a physical change, but not a chemical change.
Answers
Answer:
C, Reactants rearrange to form new products
Explanation:
Calcium carbonate, when heated, forms calcium oxide and carbon dioxide. 100grams of calcium carbonate will produce 56grams of calcium oxide. How many grams of carbon dioxide will it produce? Show working out
Answers
Answer:
It will produce 44 grams of carbon dioxide
Explanation:
Step 1: Data given
MAss of CaCO3 = 100 grams
Molar mass of CaCO3 = 100.09 g/mol
Mass of CaO produced = 56 grams
Molar mass of CaO = 56.08 g/mol
Step 2: The balanced equation
CaCO3 → CaO + CO2
Step 3: Calculate moles CaCO3
Moles CaCO3 = mass CaCO3 / molar mass CaCO3
Moles CaCO3 = 100 grams / 100.09 g/mol
Moles CaCO3 = 1.00 moles
Step 4: Calculate moles CaO
Moles CaO = 56 grams / 56.08 g/mol
Moles CaO = 1.00 moles
Step 5: Calculate moles CO2
For 1 mol CaCO3 we'll have 1 mol 1 mol CaO and 1 mol CO2
Step 6: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 1.00 moles * 44.0 g/mol
Mass CO2 = 44 grams
It will produce 44 grams of carbon dioxide
Lithium is the least dense metal know with a density of 0.53 g/cm³. What is the volume occupied by 1.20x10³ grams of lithium
Answers
Answer:
2264.15 cm³Explanation:
The formula for finding the volume of an object given it's mass and density is given by
\(volume = \frac{mass}{density} \\ \)
From the question
mass = 1.2 × 10³ grams
density = 0.53 g/cm³
We have
\(volume = \frac{1.2 \times {10}^{3} }{0.53} = 2264.1509\\ \)
We have the final answer as
2264.15 cm³
Hope this helps you
What are the conditions that are required for electrical energy to be present in an electrical circuit?
Answers
Answer:
a supply of electric charges which are free to flow, some form of push to move the charges through the circuit and a pathway to carry the charges.
Explanation:
There's your answer have a good day
Need help ASAP!!!! what is the value (angle) for the C=C=O bond in Ketene i.e. CH2=C=O
Answers
Answer:
\(180^\circ\) by the VSEPR theory.
Explanation:
This question is asking for the bond angle of the \(\rm C=C=O\) bond in \(\rm H_2C=C=O\). The VSEPR (valence shell electron pair repulsion) theory could help. Start by considering: how many electron domains are there on the carbon atom between these two bond?
Note that "electron domains" refer to covalent bonds and lone pairs collectively.
Each nonbonding pair (lone pair) of valence electrons counts as one electron domain.Each covalent bond (single bond, double bond, or triple bond) counts as exactly one electron domain.For example, in \(\rm H_2C=C=O\), the carbon atom at the center of that \(\rm C=C=O\) bond has two electron domains:
This carbon atom has two double bonds: one \(\rm C=C\) bond and one \(\rm C=O\) bond. Even though these are both double bonds, in VSEPR theory, each of them count only as one electron domain. Keep in mind that there are only four valence electrons in each carbon atom. It can be shown that all four valence electrons of this carbon atom are involved in bonding (two in each of the two double bonds.) Hence, there would be no nonbonding pair around this atom.In VSEPR theory, electron domains around an atom repel each other. As a result, they would spread out (in three dimensions) as far away from each other as possible. When there are only two electron domains around an atom, the two electron domains would form a straight line- with one domain on each side of the central atom. (To visualize, consider the three atoms in this \(\rm C=C=O\) bond as three spheres on a stick. The central \(\rm C\) atom would be between the other \(\rm C\) atom and the \(\rm O\) atom.)
This linear geometry corresponds to a bond angle of \(180^\circ\).