6. What is true about Calcium (Ca)?

A. It is a very unreactive metal.

B. It is reactive and is found with nonmetals in nature.

C. naturally uncombined with other elements.

D.It is a very reactive nonmetal.

To graph the function g(x) = f(-x), you can start with the graph of f(x) and then reflect it about the y-axis.

To find the graph of the function g(x) = f(-x), we can start with the graph of the function f(x) and then reflect it about the y-axis.

If the graph of f(x) is symmetric with respect to the y-axis, meaning it is unchanged when reflected, then g(x) = f(-x) will have the same graph as f(x).

However, if the graph of f(x) is not symmetric with respect to the y-axis, then g(x) = f(-x) will be a reflection of f(x) about the y-axis.

In either case, the resulting graph of g(x) = f(-x) will be symmetric with respect to the y-axis.

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Answer:

Explanation:

Consider the general chemical reaction

                                                       \(\mathrm{A} \ \overset{k}{\longrightarrow} \ \mathrm{product}\)  .

If [A] is the concentration of A (reactant) at any time t and n is the reaction order for the whole equation, the rate is then related to the concentration of reactant A with the following differential form of equation

                                               \(Rate \ = \ -\displaystyle\frac{d[\mathrm{A}]}{dt} \ = \ k[\mathrm{A}]^{n}\)  .

where k is the rate constant.

*Note that the differential term \(\displaystyle\frac{d[\mathrm{A}]}{dt}\) has a negative sign to denote that the concentration of A is decreasing over time t.

Since the chemical reaction between tert-butyl bromide and water is given to be a first-order reaction, hence n = 1, and the resulting differential equation becomes

                                        \(Rate \ = \ -\displaystyle\frac{d[\mathrm{A}]}{dt} \ = \ k[\mathrm{A}]^{1} \ = \ k[\mathrm{A}]\)

To solve this first-order linear homogenous differential equation, the method of separation of variables can be used.

\(\-\hspace{1cm} \displaystyle\frac{d[\mathrm{A}]}{dt} \ = -\ k[\mathrm{A}] \\ \\ \-\hspace{0.5cm} \displaystyle\frac{1}{[\mathrm{A}]} \, d[\mathrm{A}] \ = -\ k \, dt \\ \\ \int\ {\displaystyle\frac{1}{[\mathrm{A}]} \, d[\mathrm{A}] \ = \ -\int {k} \, {dt}\)

         \(\ln{[\mathrm{A}]} \ = \ -kt \ + \ C \\ \\ \-\hspace{0.45cm} $[A]$ \ = \ e^{-kt \ + \ C} \\ \\ \-\hspace{0.45cm} $[A]$ \ = \ e^{-kt}e^{C} \ \ \ \ \ \ \ \ \ (e^{a \ + \ b} \ = \ e^{a}e^{b} \ \ \ \mathrm{by \ the \ law \ of \ indices})\)

Since the term \(e^{C}\) is a constant, let \(\alpha \ = \ e^{C}\), hence \([\mathrm{A}] \ = \ \alpha e^{-kt}\) or \(C \ = \ \alpha e^{-kt}\) according to the question. Given that the initial concentration (t = 0) of tert-butyl bromide is 0.2 mol/L and k = 0.0537\(\mathrm{h^{-1}}\), so

\(0.2 \ = \ \alpha e^{-0.0537 \ \times \ 0} \\ \\ 0.2 \ = \ \alpha \ \ \ \ \ (e^{0} \ = \ 1)\)

Therefore, the rate equation is \(C \ = \ 0.2e^{-0.0537t}\).

The concentration of tert-butyl alcohol after 1 hour is

\(C \ = \ 0.2e^{-0.0537 \ \times \ 1} \\ \\ C \ = \ 0.2e^{-0.0537} \\ \\ C \ = \ 0.190 \ \mathrm{mol/L} \ \ \ (3 \ \ \mathrm{s.f.})\)

Answer:

3 half-lives 7. An isotope of Cesium (Cesium-137) has a half life of 30 years. If 1.0 mg of cesium-137 disintegrates over a period of 90 years, how many mg of cesium-137 would remain? of sodium-25 is 60 seconds? : 3 minutes t=601 = 3 half-lives 5.0mg ~ 2.5 mg + 1.25mg →) 625 mg.

Explanation:

Hope this helps

       where:

      \(N_{0}\) = Initial quantity of the sample = 1.0g

       t = time in years of disintegration = 90 years

      \(t\frac{1}{2}\) = Half life of cesium isotope = 30 years

      N(t) = Quantity of substance after disintegration = ?

N(t) =  \(1 ( \frac{1}{2})^ \frac{90}{30}\)

N(t) =  \(1 (\frac{1}{2})^{3}\)

N(t) = 0.125 grams

Therefore, the quantity of cesium-137 that would remain after disintegrating over a period of 90 years is 0.125 grams.

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The electric current is  0.025 A

Electric current refers back to the go with the flow of energy in an electronic circuit and to the amount of strength flowing through a circuit. it's far measured in amperes (A). the bigger the cost in amperes, the more energy is flowing within the circuit.

Ag+ + e¯ →Ag

1F deposits 107.87 g/mol (molecular mass) of silver

1F = 96500 C

Let, 107.87 g/mol needed = 96500 C

Number of coulombs required to deposit 0.3650 g of silver =(96500/107.87) 0.3650  

Q = 326.5 C

According to Faraday’s law, Q = I x t

I = 326.5 C / (216 x 60 s) = 0.025 A

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Let's start by stating the important given data:

time ln[A]

0

20

To answer this question, we have to use the following formula corresponding to the first-order reaction rate constant:

Using

The name of the salt tells us that:

a) there is just one other element combined with the metal by looking at the suffix of the salt's name.

b) the presence of oxygen in a salt can be indicated by the name of the salt.

a) The name of a salt can tell us that there is just one other element combined with the metal by looking at the suffix of the salt's name. If the salt name ends in "-ide," it indicates that the salt is composed of a metal and a single non-metal element.
For example, sodium chloride (NaCl) and potassium bromide (KBr) are salts where the metal (sodium and potassium) is combined with a single non-metal element (chlorine and bromine, respectively). The "-ide" suffix suggests the presence of only one other element in the salt.

b) The presence of oxygen in a salt can be indicated by the name of the salt. If the salt name includes the element oxygen, it suggests that oxygen is present in the salt compound.
For example, sodium carbonate (Na₂CO₃) and calcium sulfate (CaSO₄) contain the element oxygen in their chemical formulas. The presence of oxygen in the salt is implied by the name and the combination of elements in the compound.

Therefore, the name of salt tells us that there is just one other element combined with the metal and there is oxygen present in the salt

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Answer:48

Explanation:

Answer:  3. 0.125 X 10”23 molecules

Explanation:

Answer:

Mount Etna I believe

Explanation:

Answer:

pressure = 0.61512atm

Explanation:

1 torr = 0.00132 atm

466 torr = unknown atm

unknown = 0.61512 atm

pressure = 0.61512atm

From the calculations;

1. The mole ratio is obtained as 0.4

2. The equation is written as;

5V3+ + 2MnO4- + 2H2O → 2Mn2+ + 5VO2+ + 4H+

A balanced redox reaction must have an equal number of electrons transferred in the oxidation and reduction half-reactions to ensure the overall charge is conserved. This balance is achieved by adjusting the stoichiometric coefficients of the reactants and products.

From the given information;

Number of moles of Vanadium III = 0.132 M * 10/1000 L

= 0.00132 moles

Number of moles of permanganate = 0.02  M * 26.40/1000 L

= 0.000528 moles

The mole ratio is;

0.000528 moles/0.00132 moles

= 0.4

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Answer:

Here is a set of possible answers:

By how much can CO2 emissions be reduced?

How much is it going to cost the city to install the wind farms?

Will families have to pay for part of the wind farms?

What is the out-of-pocket cost of installing solar panels on a house after the assistance has been granted?

Are there any yearly operating costs to use solar panels?

How reliable is solar energy in this area? In an average year, how many months can solar energy be used?

Do we have a backup source of energy if the wind farm or solar panels don’t generate enough electricity?

Explanation:

This is the actucal answer so you might want to change it up

The five questions that can be asked in the town meeting are:

A solar plant is plant which use sun energy to produce electricity, they are installed in areas with ample of sunlight.

A wind farm is the used to produce electricity through wind, they used large fans which rotate by wind and generate power.

Thus, these are the five questions that can be asked in the town meeting.

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Answer:

48 grams

Explanation:

The total mass in grams of 0.75 moles of SO2 is 48g.

HOW TO CALCULATE MASS:

mass of substance = moles × molar mass

According to this question, there are 0.75 moles of SO2.

The molar mass of SO2 = 32 + 16(2) = 64g/mol.

Mass of SO2 = 64g/mol × 0.75mol

Mass of SO2 = 48g

Therefore, the total mass in grams of 0.75 moles of SO2 is 48g.

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Answer:

2.73

Explanation:

2.72815277835894

i used an online converter lol it is much faster. if you'd like a step by step guide comment and ill give you one :)

Answer:

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____________________

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# ꧁❣ RainbowSalt2²2² ࿐

A balloon is filled with 500.0 mL of helium at a temperature of 27 degrees Celsius and 755 mmHg. 3.4 liters is the volume in milliliters will it have when it reaches an altitude where the temperature is -33 degree Celsius and the pressure is 0.65 atm.

A measurement of three-dimensional space is volume. It is frequently expressed in numerical form using SI-derived units or different imperial or US-standard units (such the gallon, quart, and cubic inch). Volume and length (cubed) have a symbiotic relationship.

The volume much a container is typically thought of as its capacity, not as the amount of space it takes up. In other words, the volume is the volume of fluid (liquid or gas) that the container may hold.

PV/T = constant

(755 x 500) / (27 + 273) = (494 x V) / (-33 + 273)

V = 3396 ml = 3.4 liters

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The enthalpy of combustion for 1kg of urea is -1223525.84 J/mol.

Urea is a compound that is used in fertilizers and in some plastics.The enthalpy of combustion for urea is the amount of energy that is released when urea is burned. In order to calculate the enthalpy of combustion for 1kg of urea, we need to use the information that is provided to us in the question. Let us start by writing down the balanced equation for the combustion of urea: CO(NH2)2 + 3/2 O2 → CO2 + 2H2O + N2
The balanced equation shows that 1 mole of urea reacts with 1.5 moles of oxygen gas to produce 1 mole of carbon dioxide, 2 moles of water, and 1 mole of nitrogen gas. The enthalpy change for this reaction is equal to the amount of energy that is released when 1 mole of urea is burned.
The heat of combustion (ΔHc) of urea is -632.6 kJ/mol. This means that 632.6 kJ of energy is released when 1 mole of urea is burned. We know that 12g of urea raised the temperature of water by 30°C. We can use this information to calculate the amount of energy that was released when 12g of urea was burned.
The specific heat capacity of water is 4.18 J/g°C. This means that it takes 4.18 J of energy to raise the temperature of 1 gram of water by 1°C. Therefore, it takes 4.18 x 1000 = 4180 J of energy to raise the temperature of 1 kg of water by 1°C.
We know that 12g of urea raised the temperature of water by 30°C. Therefore, the amount of energy that was released when 12g of urea was burned is:
Energy = mass x specific heat capacity x temperature change
Energy = 0.012 kg x 4180 J/kg°C x 30°C
Energy = 1497.6 J
We can now use this information to calculate the enthalpy of combustion for 1kg of urea:
Enthalpy of combustion = energy released / moles of urea burned
Enthalpy of combustion = 1497.6 J / (0.012 kg / 60.06 g/mol)
Enthalpy of combustion = - 1223525.84 J/mol
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5.0 moles of calcium will produce 5.0 moles of calcium carbonate.

Ca + CO2 → CaCO3

Therefore, for every 1 mole of calcium, 1 mole of calcium carbonate will be produced.

Given: 5.0 moles of calcium

Therefore, 5.0 moles of calcium carbonate will be produced.

The reaction of calcium and carbon dioxide produces calcium carbonate. For every 1 mole of calcium, 1 mole of calcium carbonate will be produced. In this case, 5.0 moles of calcium will result in the production of 5.0 moles of calcium carbonate.

This reaction is used in a variety of applications, such as the production of cement, lime, and mortar. Calcium carbonate is also used in the production of medicines and supplements, as well as in the food industry to increase calcium levels in foods. Calcium carbonate is also used in water treatment to remove heavy metals and other contaminants. In conclusion, 5.0 moles of calcium will produce 5.0 moles of calcium carbonate.

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answer:when an herbivore eats meat and a carnivore eats the herbivore energy from the eatin meat is passed indirectly to the carnivore.

Answer:

Primary alcohols can be oxidized to form aldehydes and carboxylic acids; secondary alcohols can be oxidized to give ketones. Tertiary alcohols, in contrast, cannot be oxidized without breaking the molecule's C–C bonds.

Taking into account the reaction stoichiometry, 128.4 grams of C are required to produce 10.7 moles of methane.

In first place, the balanced reaction is:

C + 2 H₂ → CH₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The following rule of three can be applied: If by reaction stoichiometry 1 mole of CH₄ is produced by 12 grams of C, 10.7 moles of CH₄ are produced by how much mass of C?

mass of C= (10.7 moles of CH₄×12 grams of C)÷1 mole of CH₄

mass of C= 128.4 grams

Finally, 128.4 grams of C are required.

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The true molecular formula of the compound is C₂ClF₅

The number of mole can be obtained as by using the ideal gas equation as illustrated below:

n = PV / RT

n = (0.978 × 0.428) / (0.0821 × 297.3)

n = 0.017 mole

Molar mass = mass / mole

Molar mass = 2.650 / 0.017

Molar mass = 155.88 g/mol

Divide by their molar mass

C = 15.5 / 12 = 1.292

Cl = 23 / 35.5 = 0.648

F = 61.5 / 19 = 3.237

Divide by the smallest

C = 1.292 / 0.648 = 2

Cl = 0.648 / 0.648 = 1

F = 3.237 / 0.648 = 5

Thus, the empirical formula of the compound is C₂ClF₅

Molecular formula = empirical × n = molar mass

[C₂ClF₅]ₙ = 118.084

[(12×2) + 35.5 + (19×5)]ₙ = 155.88

154.5n = 155.88

Divide both side by 154.5

n = 155.88 / 154.5

n = 1

Molecular formula = [C₂ClF₅]ₙ

Molecular formula = [C₂ClF₅]1

Molecular formula = C₂ClF₅

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This technique has been used to identify the presence of gases such as oxygen, methane, and carbon dioxide in the atmospheres of exoplanets.  

i) To estimate the frequency of the violet (leftmost) emission, we can use the equation v = c/λ, where v is frequency, c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the emission in meters. The wavelength of the violet emission is 400 nm or 400 x 10^-9 m, so the frequency can be calculated as v = (3.00 x 10^8 m/s) / (400 x 10^-9 m) = 7.50 x 10^14 Hz.

ii) To estimate the energy of the violet emission, we can use the equation E = hv, where E is energy, h is Planck's constant (6.63 x 10^-34 Js), and v is frequency in Hz. Substituting the frequency calculated in part (i), we get E = (6.63 x 10^-34 Js) x (7.50 x 10^14 Hz) = 4.97 x 10^-19 J.

b. The spectral lines are produced by the electrons within the atoms of this element, which can absorb or emit specific amounts of energy to move between different energy levels. These energy transitions result in the emission or absorption of photons with specific wavelengths and frequencies, giving rise to the observed emission spectrum.

c. The violet emission line represents the photon with the most energy since it has the shortest wavelength (400 nm) and highest frequency (7.50 x 10^14 Hz) among the lines shown. This highest energy does not necessarily represent the energy of the valence electrons, but rather corresponds to the specific energy transitions occurring within the atoms of the element.

d. Emission spectra can be used to determine the gases present in the atmosphere of a far-away planet by analyzing the specific wavelengths of the emitted or absorbed light from the planet. Each gas has a unique emission or absorption spectrum, allowing scientists to identify the gases present in the planet's atmosphere.

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Answer:

Wavelength

Explanation:

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The chemical formula for copper(II) sulfate is CuSO4.

Copper(II) sulfate is a chemical compound that is made up of copper, sulfur, and oxygen. It has the chemical formula CuSO4 and is commonly referred to as "blue vitriol" or "bluestone." Copper(II) sulfate can be prepared by reacting copper oxide or copper metal with sulfuric acid. It is a blue-colored crystalline solid that dissolves easily in water.

Copper(II) sulfate has many uses in industry and can be used as a fungicide, herbicide, pesticide, and in the manufacture of other chemicals. It is also commonly used in schools and laboratories as a reagent in chemical reactions and experiments.

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Answer:

distillation

Explanation:

i did the assignment

Answer:

distillation

Explanation:

Q = Ksp = 1 × 10^(-12).

Substituting the values into the Nernst equation, we have:

0.799 V = E° - (RT/2F) * ln(1 × 10^(-12))

Now, solving for E°:

E° = 0.799 V + (RT/2F) * ln(1 × 10^(-12))

The value of R is the ideal gas constant, T is the temperature in Kelvin, and F is the Faraday constant.

To find the standard reduction potential at 25°C for the half-reaction Ag2CrO4(s) + 2e¯ → 2Ag(s) + CrO4²-(aq), we can use the Nernst equation, which relates the standard reduction potential (E°) to the equilibrium constant (K) and the reaction quotient (Q).

The Nernst equation is given as follows:

E = E° - (RT/nF) * ln(Q)

Given information:

Ksp = 1 × 10^(-12)

E = +0.799 V (standard reduction potential of Ag+ to Ag)

Since the reaction involves the dissolution of Ag2CrO4(s), the reaction quotient Q can be expressed as [Ag+]²/[CrO4²-].

Since the stoichiometry of the reaction is 2:1 for Ag2CrO4 to Ag+, we can say that [Ag+]² = Ksp.

Therefore, Q = Ksp = 1 × 10^(-12).

Substituting the values into the Nernst equation, we have:

0.799 V = E° - (RT/2F) * ln(1 × 10^(-12))

Now, solving for E°:

E° = 0.799 V + (RT/2F) * ln(1 × 10^(-12))

The value of R is the ideal gas constant, T is the temperature in Kelvin, and F is the Faraday constant.

Please note that without specific values for temperature (T) and the ideal gas constant (R), the exact standard reduction potential at 25°C cannot be determined.

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Answer:

no se englesh

ajajajaja

The isoelectric point (pI) is the pH at which an amino acid or protein.

Functional groups of amino acids in enzymes present have the ability to readily ionize. other amino acids which have ionizable side chains. These include arginine, aspartic acid, cysteine, glutamic acid, histidine, lysine and tyrosine.

There are basically four different classes of amino acids determined by different side chains: (1) non-polar and neutral, (2) polar and neutral, (3) acidic and polar, (4) basic and polar. Principles of Polarity: The greater the electronegativity difference between atoms in a bond, the more polar the bond.

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