In a diamond, what type of bonds link carbon atoms together to form a network solid?

Calculate the empirical formula for a compound containing 67.6% Hg 10.8% S 21.6% O

pashok25 [27]

Answer:

HgSO₄

Explanation:

% => g => moles => ratio => reduce => empirical ratio

%Hg = 67.6% => 67.6g/201g/mol = 0.34mol

%S = 10.8% => 10.8g/32g/mol = 0.34mol

%O = 21.6% => 21.6g/16g/mol = 1.35mol

Hg:S:O => 0.34:0.34:1.35

Reduce to whole number ratio by dividing by the smaller mole value...

Hg:S:O => 0.34/.34:0.34/.34:1.35/.34 => Empirical Ratio = 1:1:4

∴ Empirical Formula is HgSO₄

3 0

3 years ago

What is the Molarity of a solution of HNO3 if it contains 12.6 moles in a 0.75 L solution? *

Korolek [52]

Answer:

M = 16.8 M

Explanation:

Data: HNO3

moles = 12.6 moles

solution volume = 0.75 L

Molarity is represented by the letter M and is defined as the amount of solute expressed in moles per liter of solution.

$M=\frac{moles}{solution volume}$

The data is replaced in the given equation:

$M=\frac{12.6 mol}{0.75L}=16.8\frac{mol}{L}$

7 0

3 years ago

Read 2 more answers

Suppose 50.0g of silver nitrate is reacted with 50g of hydrochloric acid producing silver chloride and a mixture of other produc

docker41 [41]

Answer:

53.6 grams of silver chloride was produced.

Explanation:

$AgNO_3+HCl+\rightarrow AgCl+HNO_3$

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

Mass of silver nitrate = 50.0 g

Mass of hydrogen chloride = 50.0 g

Mass of silver chloride = x

Mass of nitric acid = 46.4 g

Mass of silver nitrate + Mass of hydrogen chloride =

Mass of silver chloride + Mass of nitric acid

[te]50.0 g+50.0 g=x+46.4 g[/tex]

$x=50.0 g+50.0 g - 46.4 g = 53.6 g$

53.6 grams of silver chloride was produced.

8 0

3 years ago

For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car

Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

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

Na₂CO₃: 1 mole
Ca(NO₃)₂: 1 mole
CaCO₃: 1 mole
NaNO₃: 2 mole

Being the molar mass of the compounds:

Na₂CO₃: 106 g/mole
Ca(NO₃)₂: 164 g/mole
CaCO₃: 100 g/mole
NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

Na₂CO₃: 1 mole* 106 g/mole= 106 g
Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
CaCO₃: 1 mole* 100 g/mole= 100 g
NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of CaCO₃?

$mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}$

mass of CaCO₃= 74.81 grams

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

6 0

3 years ago

1

Arisa [49]

Light energy into chemical energy

8 0

2 years ago