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Ede4ka [16]
2 years ago
5

A chemist found a 60.00 gram sample containing copper and oxygen. Given the following percent composition, what is the mass of c

opper in the sample?
Copper- 57.08 %
Oxygen- 42.9296


choose an answer:
O 0.5708 g Cu
o 25.02 g Cu
34.25 g cu
63.5 g Cu
Chemistry
1 answer:
Strike441 [17]2 years ago
6 0
The answer should 34.25g of Cu.
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Which does not affect global wind belts
vekshin1

Answer:

b

Explanation:

its local

5 0
3 years ago
A sample of 1.000 g of a compound containing carbon and hydrogen reacts with oxygen at elevated temperature to yield 0.692 g H₂O
ollegr [7]

Answer :

(a) 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) There is no other element present in the compound.

Explanation :

(a) Now we have to determine the masses of C and H in the sample.

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of Carbon and hydrogen respectively.

We are given:

Mass of CO_2=3.381g

Mass of H_2O=0.692g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.381 g of carbon dioxide, \frac{12}{44}\times 3.381=0.922g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.692 g of water, \frac{2}{18}\times 0.692=0.0769g of hydrogen will be contained.

Thus, 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) Now we have to determine the compound contain any other elements or not.

Mass carbon + Mass of hydrogen = 0.922 g + 0.0769 g = 0.999 g ≈ 1 g

This means that there is no other element present in the compound.

3 0
3 years ago
(5.625 + 8.15) x 2.34 + 3.2
dalvyx [7]

The answer is 35.4335

Hope this helped! (Plz mark me brainliest!)

3 0
3 years ago
Read 2 more answers
If 15.00 mL of 0.0100 M Ca(IO3)2 solution are mixed with 0.500 g KI, what is the theoretical yield (in grams) of I2?
n200080 [17]

The theoretical yield of I2 in the reaction would be 0.23 g

<h3>Theoretical yield</h3>

This refers to the stoichiometric yield of a reaction.

From the equation of the reaction:

Ca(IO3)2 + 10 KI + 12 HCl → 6 I2 + CaCl2 + 10 KCl + 6 H2O

The mole ratio of Ca(IO3)2 and I2 is 1: 6

Mole of 15.00 mL, 0.0100 M Ca(IO3)2 = 15/1000 x 0.0100

                                                              = 0.00015 mole

Equivalent mole of I2 = 0.00015 x 6

                                      = 0.009 mole

mass of 0.0009 I2 = 0.0009 x 253.809

                                = 0.23 g

More on stoichiometric calculations can be found here: brainly.com/question/6907332

8 0
3 years ago
A gas sample has a temperature of 22c with an unknown volume. The same gas has a volume of 456 mL when the temperature is 86c wi
Mkey [24]

The initial volume is 116.65 mL

<u>Explanation:</u>

<u />

Given:

Temperature, T₁ = 22°C

T₂ = 86°C

Volume, V₂ = 456 m

V₁ = ?

According to Charle's law:

\frac{V_1}{T_1} = \frac{V_2}{T_2}

Substituting the values:

\frac{V_1}{22} =\frac{456}{86} \\\\V_1 = \frac{456 X 22}{86} \\\\V_1 = 116.65 mL

Therefore, the initial volume is 116.65 mL

3 0
3 years ago
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