which of the following is likely closest to the bond angle(s) in a molecule where the central atom is sp3 hybridized?

The formation of 1-bromobutane involves the following reaction mechanism: 1. Nucleophilic substitution (SN2): Butanol reacts with a bromide ion (Br-) to form 1-bromobutane and a hydroxide ion (OH-). Butanol + Br- → 1-bromobutane + OH-

The formation of 1-butene and di-n-butyl ether are side reactions in this process:
1. Elimination (E2) reaction: Butanol undergoes an elimination reaction to form 1-butene and water.
Butanol → 1-butene + H2O

2. Williamson ether synthesis: Two butanol molecules react with each other in the presence of a base, forming di-n-butyl ether and water.
2 Butanol + Base → di-n-butyl ether + H2O + Base

It's necessary to remove water before weighing the 1-bromobutane because water could alter the measured weight, leading to inaccurate results. Water can also promote side reactions like the formation of di-n-butyl ether, which could further impact the yield and purity of the desired product, 1-bromobutane.

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Answer: Valency of Al in \(Al_2O_3\) is +3.

Explanation:

The capacity of an element to gain or lose electrons is called valency of the atom or element.

The valency of oxygen is -2 is the compound \(Al_2O_3\)  and let the valency of Al is x over here.

Therefore, valency of x will be calculated as follows.

\(2x + \times 3(-2) = 0\) (as the charge on compound is 0)

\(2x - 6 = 0\)

\(x = +3\)

Therefore, valency of Al in \(Al_2O_3\) is +3.

liquid water For example, the limited temperature range of liquid water (0°C–100°C) severely limits its use. Aqueous solutions have both a lower freezing point and a higher boiling point than pure water.

How hot (or energetic) a substance or radiation is can be quantified by a physical value called temperature.

There are three different types of temperature scales: those like the SI scale that are defined in terms of the average translational kinetic energy per freely moving microscopic particle, like an atom, molecule, or electron in a body; those that only rely on strictly macroscopic properties and thermodynamic principles, like Kelvin's original definition; and those that are defined by actual empirical properties of particulates rather than by theoretical principles.

Temperature is gauged using a thermometer. It is calibrated using a variety of temperature scales that historically defined themselves using various reference points and thermometric materials. The most widely used scale is the Celsius scale, previously called as "centigrade."

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Answer:

D. The end of a slinky is moved vertically up and down to simulate a wave.

Explanation:

A Slinky is a compact helical coiled toy that can perform many number of tricks such as stretching it-self and come back in its original shape.

A slinky here is used as a model by a scientist to stimulate a wave or to understand the concept of the wave. To study the nature of waves, slinky is used to create waves by moving vertically up and down that leads to the vibration or disturbance in a medium.

Hence, the correct answer is "D. The end of a slinky is moved vertically up and down to simulate a wave."

Answer:

The answer would be option D.

Explanation:

I took the test and D was correct.

Hope this helps. ^^

If the volume is decreased to 2.6 L and pressure is decreased to 3.2 atm, then temperature become 156.98K.

This problem can be solved by using the concept of Ideal gas equation.

The equation of ideal gas can be expressed as

PV = nRT

where,

P is the pressure of the gas

V is the volume taken

n is the number of moles of the gas

R is the gas constant = 6.314 J/ mol K

T is the temperature

Now from given data, we have the values of

V = 3.2 L

P = 5.5 atm

T = 53°C = 273 + 53 = 326 K

By substituting all the values, we get the value of n

3.2 × 5.5 = n × 6.314 × 326

17.6 = n × 2058.36

n = 17.6 / 2058.36

n = 0.0085 mol

Now, we have to find the temperature

n = 0.0085 mol

V = 2.6 L

P = 3.2 atm

By substituting all the values, we get

3.2 × 2.6 = 0.0085 × 6.314 × T

8.32 = 0.053 T

T = 8.32/ 0.053

T = 156.98 K

Thus, we concluded that if the volume is decreased to 2.6 L and pressure is decreased to 3.2 atm, then temperature become 156.98K.

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Two sodium ions and one phosphate ion are needed to form an ionic compound named sodium phosphate \(Na_{3}PO_{4}\).

These two polyatomic ions require opposite charges that are both equivalent to form a bond. As a result, the one -3 phosphate ion needs to be balanced by three +1 sodium ions.

A chemical compound known as an ionic compound in chemistry is one that contains ions held together by the electrostatic forces known as ionic bonding. Despite having both positively and negatively charged ions, or cations and anions, the compound is overall neutral.

Thus in sodium phosphate sodium is the cation and phosphate is the anionic.

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For the following reaction, 4.27 grams of hydrogen gas are allowed to react with 11.0 grams of ethylene (C2H4). The maximum amount of ethane (C2H6) that can be formed is 11.79 g. The formula for the limiting reagent is C2H4. The amount of the excess reagent remains after the reaction is complete is is 3.48 g.

The balanced chemical reaction for the given scenario is given below: C2H4 + H2 → C2H6Ethylene (C2H4) and hydrogen (H2) react to give ethane (C2H6).

The balanced chemical equation provides the stoichiometry of the reaction. The stoichiometric ratio can be used to determine the moles of reactants and products involved in the reaction. Since the stoichiometric ratio is in terms of moles, we need to convert the mass of the given reactants to moles.

Let's start by calculating the number of moles of hydrogen and ethylene.

Number of moles of hydrogen (H2) = mass / molar mass = 4.27 g / 2.016 g/mol = 2.12 mol

Number of moles of ethylene (C2H4) = mass / molar mass = 11.0 g / 28.054 g/mol = 0.392 mol

According to the stoichiometry of the reaction, 1 mole of ethylene (C2H4) reacts with 1 mole of hydrogen (H2) to form 1 mole of ethane (C2H6).

Since ethylene (C2H4) is present in less quantity as compared to hydrogen (H2), ethylene (C2H4) will be the limiting reagent in this reaction. Therefore, the maximum amount of ethane (C2H6) that can be formed is obtained by using the number of moles of limiting reagent i.e. ethylene.

Number of moles of ethane (C2H6) = number of moles of limiting reagent = 0.392 mol

Mass of ethane (C2H6) = number of moles × molar mass = 0.392 mol × 30.070 g/mol = 11.79 g

Thus, the maximum amount of ethane (C2H6) that can be formed is 11.79 g.

The limiting reagent is the reactant that is consumed completely during the reaction. Since ethylene (C2H4) is the limiting reagent in this reaction, its formula is C2H4.

Excess reagent is the reactant that remains unconsumed after the reaction is complete. We have already determined that ethylene (C2H4) is the limiting reagent in this reaction, which means that hydrogen (H2) is the excess reagent.

Hence, the amount of the excess reagent i.e. hydrogen (H2) remaining after the reaction is complete is obtained by subtracting the amount of hydrogen used in the reaction from the amount of hydrogen available.

Number of moles of hydrogen used = number of moles of limiting reagent = 0.392 mol

Number of moles of hydrogen available = 2.12 mol

Number of moles of hydrogen remaining = 2.12 mol - 0.392 mol = 1.73 mol

Mass of hydrogen remaining = number of moles × molar mass = 1.73 mol × 2.016 g/mol = 3.48 g

Therefore, the amount of the excess reagent (hydrogen) that remains after the reaction is complete is 3.48 g (approx).

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Phytoremediation is a cost-effective and environmentally friendly method of reducing pollutant concentrations and restoring contaminated ecosystems.

What is Pollutants?

Pollutants are substances or agents that contaminate the environment and have harmful effects on living organisms, natural resources, or the climate. Pollutants can be released into the air, water, or soil from natural sources or human activities such as industrial processes, transportation, agriculture, and waste disposal. Some common examples of pollutants include greenhouse gases, particulate matter, ozone, nitrogen oxides, sulfur dioxide, lead, mercury, pesticides, and plastic waste.

The method of using cattails in swamps to absorb chemical pollutants is called phytoremediation. Phytoremediation is a type of bioremediation that uses plants to remove, detoxify, or sequester contaminants from soil, water, or air. In this process, plants absorb contaminants through their roots or take them up from the air and store them in their tissues or metabolize them into less harmful forms. Cattails are particularly effective at removing organic pollutants such as pesticides, herbicides, and petroleum products, as well as heavy metals like lead, cadmium, and arsenic.

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In general, the formation of a chemical bond lowers the potential energy of a chemical system, leading to a more stable arrangement.

Stored energy depends upon the relative position of various parts of a system.

In bond formation, atoms lose their potential energy to attain stability and heat is evolved. The bond formation is an exothermic reaction.

Atoms bond together to form compounds to attain lower energies than they possess as individual atoms.

A quantity of energy is equal to the difference between the energies of the bonded atoms and the energies of the broken atoms.

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Answer:

so I got it wrong the first time but you should pick two points on the line such as 4,3 and 9,5. Then use the equation y^2 - y^1/ x^2 - X^1. You would get 2/5 which is D.

Explanation

The expected total YAN after the additions are completed would be:
- Tank 13: 50.5 mg/L
- Tank 7: 90 mg/L
- Tank 23: 163.4 mg/L

To determine the amount in mg/L for each addition for each YAN source, we will use the given information for each tank:

For Tank 13:
- DAP addition:

3 lbs/1000 gals
- Fermaid K addition:

2.5 lbs/1000 gals

To calculate the amount in mg/L for each addition, we need to convert the pounds to milligrams and consider the volume of the tank. Let's assume the volume of the tank is 1000 gallons:

- DAP addition:

3 lbs/1000 gals = 3000 mg/1000 gals = 3 mg/L
- Fermaid K addition:

2.5 lbs/1000 gals = 2500 mg/1000 gals = 2.5 mg/L

For Tank 7:
- DAP addition:

2.5 lbs/1000 gals
- Fermaid A addition:

1.5 lbs/1000 gals

Using the same conversion and assuming the volume of the tank is 1000 gallons:

- DAP addition:

2.5 lbs/1000 gals = 2500 mg/1000 gals = 2.5 mg/L
- Fermaid A addition:

1.5 lbs/1000 gals = 1500 mg/1000 gals = 1.5 mg/L

For Tank 23:
- DAP addition:

3.5 lbs/1000 gals
- Fermaid O addition:

2.5 lbs/1000 gals
- Fermaid K addition:

1.4 lbs/1000 gals

Again, assuming the volume of the tank is 1000 gallons:

- DAP addition:

3.5 lbs/1000 gals = 3500 mg/1000 gals = 3.5 mg/L
- Fermaid O addition:

2.5 lbs/1000 gals = 2500 mg/1000 gals = 2.5 mg/L
- Fermaid K addition:

1.4 lbs/1000 gals = 1400 mg/1000 gals = 1.4 mg/L

Moving on to determining the increase in YAN associated with each addition:

For Tank 13:
- DAP addition: 3 mg/L
- Fermaid K addition: 2.5 mg/L

For Tank 7:
- DAP addition:

2.5 mg/L
- Fermaid A addition:

1.5 mg/L

For Tank 23:
- DAP addition:

3.5 mg/L
- Fermaid O addition:

2.5 mg/L
- Fermaid K addition:

1.4 mg/L

Finally, let's calculate the expected total YAN after the additions are completed:

For Tank 13:
Initial YAN level: 45 mg/L
Total YAN = Initial YAN + YAN increase from additions
Total YAN = 45 mg/L + (3 mg/L + 2.5 mg/L) = 50.5 mg/L

For Tank 7:
Initial YAN level: 86 mg/L
Total YAN = Initial YAN + YAN increase from additions
Total YAN = 86 mg/L + (2.5 mg/L + 1.5 mg/L) = 90 mg/L

For Tank 23:
Initial YAN level: 156 mg/L
Total YAN = Initial YAN + YAN increase from additions
Total YAN = 156 mg/L + (3.5 mg/L + 2.5 mg/L + 1.4 mg/L) = 163.4 mg/L

So, the expected total YAN after the additions are completed would be:
- Tank 13: 50.5 mg/L
- Tank 7: 90 mg/L
- Tank 23: 163.4 mg/L

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In the long version of the Periodic table, the elements Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Caesium (Cs), and Francium (Fr) collectively make up one group. Because their oxides and hydroxides are soluble in water and produce a potent alkaline (basic) solution, these metals are collectively referred to as alkali metals.

Alkali metals' distinctive flame colors—red, yellow, violet, crimson, and blue for Li, Na, K, Rb, and Cs, respectively—are qualitative markers of the contemporary analytical techniques used to calculate the quantities of alkali-metal salts in aqueous solution.

Most other metals are harder than alkali metals. The most reactive elements in this category are cesium and francium.

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Answer

Ethanol and Ethylene Glycol

Procedure

Hydrogen bonding occurs when hydrogen forms a molecular bond with a highly electronegative element such as Oxygen, Nitrogen, and Fluorine. Based on the structure of the molecules we can see that ethane does not contain the previously mentioned elements, therefore it will not form hydrogen bonds. Dimethyl ether has an oxygen atom located in the middle of the molecule, making it difficult to form a bond with other dimethyl ether molecules.

Lastly, Ethanol and Ethylene Glycol possess OH groups which are free to interact with similar groups via hydrogen bonding. Additionally, these last compounds exhibit higher boiling points, which can indicate a stronger intermolecular bonding, which is a characteristic of hydrogen bonding.

Answer:

The answer would be

C. The number of neutrons

You're welcome <3

Answer:

Explanation:

A material through which electrons can move easily is referred to as a conductor. Conductors have a high electrical conductivity, which means that electrons can flow through them with relative ease. This is due to the presence of free electrons in the material, which are not tightly bound to individual atoms. Examples of common conductors include metals such as copper, aluminum, and gold.

Answer:

A. 15

Explanation:

From the equation we see that 6 water particles are used to make 1 particle of the hexose.

That means, that to make 1 mole of the hexose, we need 6 moles of water (as a mole is an unit of count of particles).

To make 2.5moles of the hexose, we need 2.5*6 moles of water, so 15 moles of water.

The activation energy of the reaction is approximately 71.46 kJ/mol, rounded to 2 decimal places.

To find the activation energy (Ea) of the reaction, we can use the Arrhenius equation, which relates the rate constant (k) to the temperature (T) and activation energy. The Arrhenius equation is given by:

k = A * e^(-Ea/RT)

Where:

k is the rate constant

A is the pre-exponential factor (frequency factor)

Ea is the activation energy

R is the gas constant (8.314 J/(mol·K))

T is the temperature in Kelvin

We have two sets of data:

At 600 K, k1 = 3.4 m^(-1)s^(-1)

At 750 K, k2 = 31.0 m^(-1)s^(-1)

Taking the natural logarithm (ln) of both sides of the Arrhenius equation, we can rearrange the equation to solve for the activation energy:

ln(k) = ln(A) - (Ea/RT)

We can create two equations using the given data:

ln(k1) = ln(A) - (Ea/(R * 600))

ln(k2) = ln(A) - (Ea/(R * 750))

Subtracting the second equation from the first eliminates the ln(A) term:

ln(k1) - ln(k2) = (Ea/R) * ((1/600) - (1/750))

Simplifying further:

ln(k1/k2) = (Ea/R) * ((750 - 600) / (600 * 750))

Now we can solve for Ea:

Ea = (R * (ln(k1/k2))) / ((750 - 600) / (600 * 750))

Using the given values and the appropriate units:

Ea = (8.314 J/(mol·K) * ln(3.4/31.0)) / ((750 - 600) / (600 * 750))

Converting the units from J to kJ:

Ea = (8.314 × 10^(-3) kJ/(mol·K) * ln(3.4/31.0)) / ((750 - 600) / (600 * 750))

Ea ≈ 71.46 kJ/mol

Therefore, the activation energy of the reaction is approximately 71.46 kJ/mol, rounded to 2 decimal places.

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The lipoproteins LDL and HDL are thought to influence heart disease risk because LDL is contributing to the formation of plaques and atherosclerosis, while HDL helps remove cholesterol from the arteries.

LDL (low-density lipoprotein) and HDL (high-density lipoprotein) are two types of lipoproteins involved in the transport of cholesterol in the bloodstream. They play a significant role in influencing heart disease risk due to their distinct functions and effects on cholesterol metabolism.

1. LDL: it is often referred to as "bad cholesterol" because it carries cholesterol from the liver to peripheral tissues, including arterial walls. High levels of LDL in the bloodstream can lead to the deposition of cholesterol in arterial walls, contributing to the formation of plaques. These plaques can narrow the arteries and restrict blood flow, a condition known as atherosclerosis. If a plaque ruptures, it can trigger the formation of blood clots, potentially leading to a heart attack or stroke.

2. HDL: HDL, on the other hand, is often referred to as "good cholesterol" due to its beneficial effects on heart health. HDL transports cholesterol from peripheral tissues, including arterial walls, back to the liver for metabolism and elimination. It acts as a scavenger, removing excess cholesterol from the arteries and transporting it away, which can help prevent the buildup of plaques. HDL also possesses anti-inflammatory and antioxidant properties, which further contribute to its protective effect against heart disease.

The balance between LDL and HDL levels in the bloodstream is an essential factor in assessing heart disease risk. High levels of LDL and low levels of HDL are associated with an increased risk of developing atherosclerosis and heart disease. Conversely, higher levels of HDL and lower levels of LDL are generally considered favorable for heart health.

LDL and HDL lipoproteins influence heart disease risk because LDL is associated with the deposition of cholesterol in arterial walls, contributing to atherosclerosis, while HDL helps remove cholesterol from the arteries and has a protective effect. Maintaining a healthy balance between LDL and HDL levels is crucial for reducing the risk of heart disease.

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Answer:Conservation laws are fundamental to our understanding of the physical world in that they describe which processes can or cannot occur in nature.

Explanation: An important function of conservation laws is that they make it possible to predict the macroscopic behaviour of a system without having to consider the microscopic details of the course of a physical process or chemical reaction.

Number of (NH4)2SO4 molecules = 2.16 × 10^25 molecules / 2 = 1.08 × 10^25 (NH4)2SO4 molecules. There are approximately 1.08 × 10^25 molecules of ammonium sulfate in 35.8 moles of (NH4)2SO4.

To determine the number of molecules in 35.8 moles of ammonium sulfate ((NH4)2SO4), we need to use Avogadro's number, which states that there are 6.022 × 10^23 particles (atoms, molecules, or ions) in one mole of any substance.

The molecular formula of ammonium sulfate indicates that there are two ammonium ions (NH4+) and one sulfate ion (SO4^2-) in each molecule.

To calculate the number of molecules in 35.8 moles of (NH4)2SO4, we can follow these steps:

Determine the number of moles of (NH4)2SO4 by multiplying the given value by Avogadro's number:

Number of moles = 35.8 moles × (6.022 × 10^23 molecules/mole) = 2.16 × 10^25 molecules

Since each molecule of (NH4)2SO4 contains one ammonium ion and one sulfate ion, the number of molecules can be divided by two to obtain the number of (NH4)2SO4 molecules:

Number of (NH4)2SO4 molecules = 2.16 × 10^25 molecules / 2 = 1.08 × 10^25 (NH4)2SO4 molecules

Therefore, there are approximately 1.08 × 10^25 molecules of ammonium sulfate in 35.8 moles of (NH4)2SO4.

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The balanced reduction half reaction is:

8H+ + 5e- + \(MnO_4\)- → \(Mn^2\)+ + \(_4H_2O\)

1. Identify the elements undergoing oxidation and reduction in the given reaction:

  - \(MN^2\)+ is being oxidized to \(MN^4\)+.

  - \(H_2SO_3\) is being reduced to \(HNO_2\).

2. Write the half-reactions for each process:

  Oxidation half-reaction: \(MN^2\) + → \(MN^4\) + + 2e-

  Reduction half-reaction: \(H_2SO_3\) + 2H+ + 2e- → \(HNO_2\) + \(H_2O\)

3. Balance the number of atoms in each half-reaction:

  Oxidation half-reaction: \(MN^2\)+ → \(MN^4\)+ + 2e-

  Reduction half-reaction: \(_2H_2SO_3\) + 4H+ + 4e- → \(_2HNO_2\) + \(_2H_2O\)

4. Balance the number of hydrogen atoms by adding H+ ions to the side lacking hydrogen:

  Oxidation half-reaction: \(MN^2\)+ → \(MN^4\)+ + 2e-

  Reduction half-reaction: \(_2H_2SO_3\) + 4H+ + 4e- → \(_ 2HNO_2\) + \(_2H_2O\)

5. Balance the number of oxygen atoms by adding H2O molecules to the side lacking oxygen:

  Oxidation half-reaction: \(MN^2\)+ → \(MN^4\)+ + 2e-

  Reduction half-reaction: \(_2H_2SO_3\) + 4H+ + 4e- →\(_ 2HNO_2\) + \(_2H_2O\)

6. Balance the charge on both sides of the equation by adding electrons:

  Oxidation half-reaction: \(MN^2\)+ → \(MN^4\)+ + 2e-

  Reduction half-reaction: \(_2H_2SO_3\) + 4H+ + 4e- → \(_2HNO_2\) + \(_2H_2O\)

7. Multiply each half-reaction by the appropriate factor to equalize the number of electrons transferred:

  Oxidation half-reaction: \(2MN^2\)+ → \(2MN^4\)+ + 4e-

  Reduction half-reaction: \(_2H_2SO_3\) + 4H+ + 4e- → \(_2HNO_2\) + \(_2H_2O\)

8. Finally, combine the half-reactions and cancel out any common terms:

  2\(MN^2\)+ + \(_2H_2SO_3\) + 4H+ + 4e- → \(2MN^4\)+ + \(_2HNO_2\) + \(_2H_2O\)

9. Simplify the equation by dividing through by 2:

  \(MN^2\)+ + \(H_2SO_3\) + 2H+ + 2e- → \(MN^4\)+ + \(HNO_2\) + \(H_2O\)

Therefore, the balanced reduction half-reaction is:

8H+ + 5e- + \(MnO_4\)- → \(MN^2\)+ + \(4H_2O\)

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Reducing half-reaction:\(MN^2+ + 4H^+ + 2e^- → MnO2 + 2H2O\)

To write the balanced reduction half-reaction, we need to identify the species that undergoes reduction, which is the one that gains electrons. In this case,\(MN^2+\)is reduced to \(MnO2\).

To balance the reduction half-reaction, we first balance the atoms of all elements except hydrogen and oxygen. Then, we balance the oxygen atoms by adding \(H2O\) to the side that lacks oxygen. Finally, we balance the hydrogen atoms by adding H^+ to the opposite side. We also add electrons to balance the charge. In this case, the balanced reduction half-reaction requires 2 electrons.

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The presence of nonbonding electron pairs, also known as lone pairs or nonbonding electron domains, can have an effect on the shape of molecules. These lone pairs influence the molecular geometry by exerting electron repulsion and affecting the arrangement of atoms and bonding pairs.

Molecules that commonly exhibit the influence of nonbonding electron pairs on molecular shape include:

Water (H2O): In water, the two lone pairs of electrons on the oxygen atom affect the molecular shape, leading to a bent or V-shaped geometry.

Ammonia (NH3): Ammonia has one lone pair of electrons on the nitrogen atom, which leads to a pyramidal shape.

Nitrogen trifluoride (NF3): NF3 has one lone pair of electrons on the central nitrogen atom, resulting in a trigonal pyramidal shape.

Carbon dioxide (CO2): Although carbon dioxide does not possess any lone pairs on the carbon atom, the presence of two double bonds results in a linear molecular shape.

Sulfur hexafluoride (SF6): The six lone pairs of electrons on the sulfur atom in SF6 cause electronic repulsion, resulting in an octahedral shape.

These are just a few examples, but there are many molecules where nonbonding electron pairs influence the overall molecular shape. The presence and arrangement of these lone pairs affect the bond angles and distortions from ideal geometries in molecules, ultimately determining their three-dimensional shapes.

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Answer:

true

Explanation:

Volcanos is rock and there's where the carbon stored at.

To produce 180 millilitres of a saline solution of the same concentration, we need 1.2 milligrams of salt.

It is given that a ratio of 0.5 milligrams of salt is present in 75 millilitres of solution. Let us assign a variable x to the amount of salt present in 180 millilitres of saline solution.

To maintain the same concentration we can use the method of cross-multiplication:

The following equation can be used to determine the value of x:

0.5 mg / 75 ml = x mg / 180 ml

( 0.5 × 180) / 75 = x mg

x mg = 90 / 75

Hence x is approximately equal to 1.2 mg.

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Answer:

5 moles of Fe(II) are required to react completely with the 5 moles of Sulphur

Explanation:

The balanced equation in this question is

Iron + Sulfur = Iron(II) Sulfide

Fe (II) + S --> Fe(II)S

Thus one mole of sulfur reacts with one mole of Fe(II)

Hence, 5 moles of Fe(II) are required to react completely with the 5 moles of Sulphur