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Afina-wow [57]
3 years ago
5

In the decomposition reaction, 1 mole of water (mw = 18.015 g/mol) was produced for every mole of cuo (mw = 79.545 g/mol) produc

ed. given that 3.327 g of cuo was produced during the reaction, how many grams of water were released as water vapor? (number of moles = mass (g) / molar mass (g/mol)). choose the closest answer.
Chemistry
1 answer:
natita [175]3 years ago
3 0

Reactives -> Products

CuO and water are products.

I found this reaction which has CuO and water as products: decomposition of Cu(OH)2.

Cu(OH)2 -> CuO + H2O

Stoichiometry calculus involve the mole proportions you can see in the reaction: When 1 mole of Cu(OH)2 reacts, 1 mole of CuO and 1 mole of H2O are formed.

Considering the molar masses:

Cu(OH)2 = 83.56 g/mol

CuO = 79.545 g/mol

H2O = 18.015 g/mol

Then: When 83.56 g of Cu(OH)2 react, 79.545 g of CuO and 18.015 g H2O are formed.

You should use that numbers in the rule of three:

79.545 g CuO __________18.015 g water

3.327 g CuO__________ x =3.327*18.015 /79.545 g water 

x= 0.7535 g water




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As you move across the period on the periodic table what pattern can be described about the electron configurations?
xeze [42]

Answer:

Elements in the same period have same number of electronic shell and electron is increased by one in every coming element with in same electronic shell.

Explanation:

Consider the second period of periodic table. This period consist of eight elements lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.

Electronic configuration of lithium:

Li₃ = [He] 2s¹

Electronic configuration of beryllium:

Be₄ = [He] 2s²

Electronic configuration of boron:

B₅ = [He] 2s² 2p¹

Electronic configuration of carbon:

C₆ = [He] 2s² 2p²

Electronic configuration of nitrogen:

N₇ = [He] 2s² 2p³

Electronic configuration of oxygen:

O₈ = [He] 2s² 2p⁴

Electronic configuration of fluorine:

F₉ = [He] 2s² 2p⁵

Electronic configuration of neon:

Ne₁₀ = [He] 2s² 2p⁶

All these elements present in same period having same electronic shell and number of electron increased by 1.

4 0
3 years ago
A solid sample is hit with a hammer and breaks into jagged, irregular shards. What type of solid was it? A) molecular solid b) n
DaniilM [7]

I honestly don't know the answer

8 0
3 years ago
What is the correct equilibrium constant expression for equation P2(g)
sertanlavr [38]

Answer:

k = [F2]² [PO]² / [P2] [F2O]²

Explanation:

In a chemical equilibrium, the equilibrium constant expression is written as the ratio between the molar concentration of the products over the molar concentration of the reactants. Each species powered to its reaction coefficient. For the equilibrium:

P2(g) + 2F2O(g) ⇄ 2PO(g) + 2F2(g)

The equilibrium constant, k, is:

k = [F2]² [PO]² / [P2] [F2O]²

6 0
3 years ago
A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
sammy [17]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (S_8) = 0.800 g

M_{solute} = Molar mass of solute (S-8) = 256.52 g/mol

W_{solvent} = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m

  • <u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

or,

\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

Hence, the freezing point of solution is 16.5°C

  • <u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

or,

\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

Hence, the boiling point of solution is 118.2°C

5 0
3 years ago
) Calculate the number of moles of sulfuric acid that is contained in 250 mL of 0.500 M sulfuric acid solution
maksim [4K]

Explanation:

According to the analysis, Molarity is amount mole per volume(1L). the amount in mole would be molarity × volume in litres.

0.500M × (250/1000)L= 0.125moles.

I hope this helps**

3 0
3 years ago
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