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2 years ago

A student wanted to make 11.0g of copper chloride. The equation for this reaction is CuCO3 + 2HCl = CuCl2 + H2O + CO2

1 answer:

7 0

The reaction for formation of CuCl2 is as follows ;
CuCO3 + 2HCl —> CuCl2 + H2O + CO2
The stoichiometry of CuCO3 to CuCl2 is 1:1
Mass of CuCl2 to be produced - 11.0 g
Number of moles of CuCl2 -11.0 g / 134.5 g/mol = 0.082 mol
According to stoichiometry the number of CuCO3 moles required = 0.082 mol
Therefore mass of CuCO3 required = 0.082 mol x 123.5 g/mol = 10.13 g

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The Lewis structure for a chlorate ion, ClO3-, should show ____ single bond(s), ____ double bond(s), and ____ lone pair(s).

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1 single bond, 2 double bond 2 lone pairs

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Which of the following accurately represents the relationship between ceramic and metal?

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The atoms in ceramic materials are held together by a chemical bond. The two most common chemical bonds for ceramic materials are covalent and ionic. For metals, the chemical bond is called the metallic bond. The bonding of atoms together is much stronger in covalent and ionic bonding than in metallic.

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2 years ago

Jim collects a sample of beach sand during a class trip. After a close inspection of the sample he classifies it as a __________

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Jim collects a sample of beach sand during a class trip. After a close inspection of the sample he classifies it as a ____________ mixture. A) heterogeneous

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A 8.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of

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Answer:

C3H6.

Explanation:

Data obtained from the question:

Mass of the compound = 8g

Mass of CO2 = 24.01g

Mass of H2O = 13.10g

Next, we shall determine the mass of C, H and O present in the compound. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of C in compound = Mass of C/Molar Mass of CO2 x 24.01

=> 12/44 x 24.01 = 6.5g

Mass of H in the compound = Mass of H/Molar Mass of H2O x 13.1

=> 2x1/18 x 13.1 = 1.5g

Mass of O in the compound = Mass of compound – (mass of C + Mass of H)

=> 8 – (6.5 + 1.5) = 0

Next, we shall determine the empirical formula of the compound. This is illustrated below:

C = 6.5g

H = 1.

Divide by their molar mass

C = 6.5/12 = 0.54

H = 1.4/1 = 1.

Divide by the smallest

C = 0.54/0.54 = 1

H = 1/0.54 = 2

Therefore, the empirical formula is CH2

Finally, we shall determine the molecular formula as follow:

The molecular formula of a compound is a multiple of the empirical formula.

Molecular formula = [CH2]n

[CH2]n = 44

[12 + (2x1)]n = 44

14n = 44

Divide both side by 14

n = 44/14

n = 3

Molecular formula = [CH2]n = [CH2]3 = C3H6

Therefore, the molecular formula of the compound is C3H6

4 0

2 years ago

For the reaction PCl5(g) <--> PCl3(g) Cl2(g) at equilibrium, which statement correctly describes the effects of increasing

xenn [34]

The given question is incomplete. The complete question is :

For the reaction $PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$ at equilibrium, which statement correctly describes the effects of increasing pressure and adding $PCl_5$ , respectively

a) Increasing pressure causes shift to reactants, adding $PCl_5$ causes shift to products.

b) Increasing pressure causes shift to products ,adding $PCl_5$ causes shift to reactants.

c) Increasing pressure causes shift to products, adding $PCl_5$ causes shift to products.

d) Increasing pressure causes shift to reactants,adding $PCl_5$ causes shift to reactants

Answer: Increasing pressure causes shift to reactants, adding $PCl_5$ causes shift to products.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

a) If the pressure is increased, the volume will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. As the number of moles of gas molecules is lesser at the reactant side. So, the equilibrium will shift in the left direction. i.e. towards reactants.

b) If $PCl_5$ is added, the equilibrium will shift in the direction where $PCl_5$ is decreasing. So, the equilibrium will shift in the right direction. i.e. towards products.

8 0

2 years ago