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nydimaria [60]
1 year ago
14

Use condensed electron configurations and Lewis electron-dot symbols to depict the ions formed from each of the following atoms,

and predict the formula of their compound:
(a) Cs and S
Chemistry
1 answer:
Dennis_Churaev [7]1 year ago
6 0

The formula for Cs and S is Cs₂S.

For Cs and S,

Ions formed will be Cs⁺ and S²⁻

As cesium belongs to group 1A and sulfur belongs to group 6A. Therefore, the condensed electronic configuration is -

For Cs is [Xe} 6s¹ and for S is [Ne] 3s²3p⁴.

Cs have 1 valence electron and S have 6 valence electrons.

Hence to attain a stable electronic configuration, both two Cs atoms lose one electron and form Cs⁺ and these two electrons will be accepted by one S atom to form S²⁻.

Therefore the formula for the compound is Cs₂S as Cs donate 1 electron and S will accept 2 electrons from Cs.

The formula is Cs₂S.

To learn more about electronic configuration, visit: brainly.com/question/15051483

#SPJ4

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When the following reaction comes to equilibrium, will the concentrations of the reactants or products be greater? Does the answ
Nutka1998 [239]

Answer:

At equilibrium, the concentration of the reactants will be greater than the concentration of the products. This does not depend on the initial concentrations of the reactants and products.

Explanation:

The value of Kc gives us an idea of the extent of the reaction. A big Kc (Kc > 1) means that in the equilibrium there are more products than reactants, and the opposite happens for a small Kc (Kc < 1). The equilibrium is reached no matter what the initial concentrations are.

The value of the equilibrium constant is relatively SMALL; therefore, the concentration of reactants will be GREATER THAN the concentration of products. This result is INDEPENDENT OF the initial concentration of the reactants and products.

4 0
3 years ago
Identify 2 ways to measure mass​
Likurg_2 [28]

Answer:

The two ways to measure mass are subtraction and taring.

8 0
2 years ago
If a sample of magnesium with a mass of 35.0 grams reacts with 35.0 grams of oxygen how much magnesium oxide will be produced?​
slamgirl [31]

Answer:

58.0 g of MgO

Explanation:

in a perfect world, 70 g, however we don't live in a perfect world

The equation of reaction

2Mg + O₂ --> 2MgO

first find which element is limiting:

35 g x 1 mol/24.3 g of Mg x 2 mol of MgO/ 2 mole of Mg = 1.44 moles of MgO

35 g x 1 mol/32g of Mg x 2 mol of MgO/ 1 mole of O₂ = 2.1875 moles of MgO

This means Mg is the limiting factor, so you will be using this moles to find grams of MgO

1.44 mols of MgO x 40.3 g of MgO/ 1 mol = 58.0 g of MgO

8 0
2 years ago
5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time
11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

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y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0
3 years ago
How can balancing equations support the law of conservation
amm1812

Answer:

detail is given below.

Explanation:

This law was given by French chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

For example:

In given photosynthesis reaction:

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If equation is not balanced,

CO₂ + H₂O  → C₆H₁₂O₆ + O₂

It can not follow the law of conservation of mass because mass is not equal on both side of equation.

4 0
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