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Lilit [14]
3 years ago
12

For the following reaction, identify the element that was oxidized, the element that was reduced, and the reducing agent. Give a

n explanation for each answer. . . SnCl2 + PbCl4 SnCl4 + PbCl2
Chemistry
1 answer:
Galina-37 [17]3 years ago
3 0
The element oxidized is the element in which the charge becomes more positive after the reaction, hence the reducing agent. The element that is reduced is the element or compound in which the charge becomes more negative after the reaction. Sn becomes +4 from +2. Hence Sn is the reduced element while Pb becomes +2 from +4 hence Pb is the element oxidized and PbCl4 is the reducing agent.
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NaCl + H2SO4 ---&gt; Na2SO4 + HCl<br> Balance the double replacement chemical reaction.
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2NaCl+H2SO4-->Na2SO4+2HCl

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Explain what makes a cirrus cloud.
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the answer is below

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Methyl salicylate is a common active ingredient in liniments such as ben-gay. It is also known as oil of wintergreen. It is made
arlik [135]

Answer: The empirical formula is C_3H_3O.

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C_xH_yO_z+O_2\rightarrow CO_2+H_2O

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

We are given:

Mass of CO_2 = 12.24 g

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Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 12.24 g of carbon dioxide, =\frac{12}{44}\times 12.24=3.338g of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 2.505 g of water, =\frac{2}{18}\times 2.505=0.278g of hydrogen will be contained.

Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671  g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.278}{0.104}=3

For H =\frac{0.278}{0.104}=3

For O =\frac{0.104}{0.104}=1

The ratio of C : H : O = 3: 3: 1

Hence the empirical formula is C_3H_3O.

5 0
3 years ago
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