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xxTIMURxx [149]
2 years ago
15

HIII PLEASE HELP BRUUH

Chemistry
1 answer:
leonid [27]2 years ago
4 0

1. 2Al(s)+6HCl(aq)⇒2AlCl₃(aq)+3H₂(g)

2. 2AgNO₃ (aq) + Cu (s)⇒Cu(NO₃)₂ (aq) + 2Ag (s)

3. 2C₃H₈O(l) + 9O₂(g) ⇒ 6CO₂(g) + 8H₂O(g)

<h3>Further explanation</h3>

There are several reactions that can occur in a chemical reaction: single replacement, double replacement, synthesis, decomposition or combustion, etc.

1.Al(s)+HCl(aq)⇒AlCl₃(aq)+H₂(g)

type : single replacement

balance :

2Al(s)+6HCl(aq)⇒2AlCl₃(aq)+3H₂(g)

2. AgNO₃ (aq) + Cu (s) ⇒ Cu(NO₃)₂ (aq) + Ag (s)

type : single replacement

balance :

2AgNO₃ (aq) + Cu (s)⇒Cu(NO₃)₂ (aq) + 2Ag (s)

3. C₃H₈O + O₂ ⇒ CO₂ + H₂O

type : combustion of alcohol

balance :

2C₃H₈O(l) + 9O₂(g) ⇒ 6CO₂(g) + 8H₂O(g)

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Each step in the following process has a yield of 70.0%. CH4+4Cl2⟶CCl4+4HCl CCl4+2HF⟶CCl2F2+2HCl The CCl4 formed in the first st
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n(HCl)=1.96 mol

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3 years ago
When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H
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Answer: The empirical formula is CH_2 and molecular formula is C_4H_8

Explanation:

We are given:

Mass of CO_2 = 18.95 g

Mass of H_2O= 7.759 g

Molar mass of carbon dioxide = 44 g/mol

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 18.59 g of carbon dioxide, =\frac{12}{44}\times 18.59=5.07g of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 7.759 g of water, =\frac{2}{18}\times 7.759=0.862g of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles

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

For C =\frac{0.422}{0.422}=1

For H =\frac{0.862}{0.422}=2

The ratio of C : H = 1: 2

Hence the empirical formula is CH_2.

The empirical weight of CH_2 = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4

The molecular formula will be=4\times CH_2=C_4H_8

6 0
2 years ago
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