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3 years ago

65.5g of copper reacts with __g of oxygen to produce 81g copper(I) oxide

Chemistry

1 answer:

Nikitich [7]3 years ago

7 0

Answer:

grams O₂(g) ≅ 15.9 g O₂(g) (3 sig. figs.)

Explanation:

2Cu + O₂ => 2CuO

Convert given to moles, solve by equation rxn ratio then convert to grams.

Moles CuO formed = 81.0 g / 81.5 g·mol⁻¹ = 0.994 mol CuO

0.994 mol CuO requires 0.994 mol Cu = 0.994 mol Cu x 64 g·mol⁻¹ = 63.6 g Cu => from this, the amount of O₂ needed = 1/2(0.994 mol) O₂(g) = 0.497 mol O₂(g) = (0.497 mol O₂(g))(32 g·mol⁻¹) = 15.904 g O₂(g) ≅ 15.9 g O₂(g) (3 sig. figs.)

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To what volume (in mL) would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN?

CaHeK987 [17]

Answer:

To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN

Explanation:

Dilution is the reduction of the concentration of a chemical in a solution and consists simply of adding more solvent.

In a dilution the amount of solute does not vary. But as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

In a solution it is fulfilled:

Ci* Vi = Cf* Vf

where:

Ci: initial concentration
Vi: initial volume
Cf: final concentration
Vf: final volume

In this case:

Ci= 1.40 M
Vi= 20 mL
Cf= 0.088 M
Vf= ?

Replacing:

1.40 M* 20 mL= 0.088 M* Vf

Solving:

$Vf=\frac{1.40 M* 20 mL}{0.088 M}$

Vf= 318.18 mL

To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN

7 0

3 years ago

Use the balanced equation given below to solve the problem that follows: Calculate the mass in grams of water produced along wit

saveliy_v [14]

Answer: 1.98 g

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}$

$\text{Moles of} CO_2=\frac{5.0L}{22.4L}=0.22moles$

The balanced given equation is:

$C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)$

According to stoichiometry :

4 moles of $CO_2$ will produce = 2 moles of $H_2O$

Thus 0.22 moles of $CO_2$ will produce= $\frac{2}{4}\times 0.22=0.11moles$ of $H_2O$

Mass of $H_2O=moles\times {\text {Molar mass}}=0.11moles\times 18g/mol=1.98g$

Thus 1.98 g of water is produced along with 5.0 L of $CO_2$ at STP

6 0

2 years ago

Which of the following equations is balanced?

stiv31 [10]

Answer:

A) 4P + 5O₂ → 2P₂O₅

Explanation:

A) 4P + 5O₂ → 2P₂O₅

This equation is balanced. There are four phosphorus and ten oxygen atoms are on both side of equation.

B) 5P + 4O₂ → 2P₄O₅

This equation is not balanced. There are five phosphorus and eight oxygen atoms on left, eight phosphorus ten oxygen on right side of equation.

C) 2P + O₂ → P₂O₅

This equation is not balanced. There are two phosphorus, two oxygen atoms on left and two phosphorus five oxygen on right side of equation.

D) 4P + 2O₂ → 2P₄O₅

This equation is not balanced. There are four phosphorus, four oxygen atoms on left and eight phosphorus ten oxygen on right side of equation.

4 0

3 years ago

A molecule has the empirical formula C4H6O. If its molecular weight is determined to be about 212 g/mol, what is the most likely

krok68 [10]

Answer:

The molecular formula is C12H18O3

Explanation:

Step 1: Data given

The empirical formula is C4H6O

Molecular weight is 212 g/mol

atomic mass of C = 12 g/mol

atomic mass of H = 1 g/mol

atomic mass of O = 16 g/mol

Step 2: Calculate the molar mass of the empirical formula

Molar mass = 4* 12 + 6*1 +16

Molar mass = 70 g/mol

Step 3: Calculate the molecular formula

We have to multiply the empirical formula by n

n = the molecular weight of the empirical formula / the molecular weight of the molecular formula

n = 70 /212 ≈ 3

We have to multiply the empirical formula by 3

3*(C4H6O- = C12H18O3

The molecular formula is C12H18O3

3 0

3 years ago

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How does a molecule differ from an atom?

Eddi Din [679]

Answer:

According to my research A molecule is two or more atoms held together by covalent bonds. An atom is the smallest part of an element. ... A sodium atom has one outer electron, and a carbon atom has four outer electrons.

Explanation:

5 0

3 years ago