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2 years ago

If 42 grams of carbon in 52 grams of oxygen are used how many grams of CO2 will be produced. I need to show work.

Chemistry

1 answer:

UNO [17]2 years ago

4 0

Answer:

How many moles of oxygen gas are required to make 8.33 moles of carbon dioxide? ... be used to produce 1.99 grams of water. 1.99 mg H2O X. 1mol H2O. 18.0g X ... c. If the reaction produces 5.3 mg of carbon dioxide how many grams of water ... X. 25mol O2. 2mol C8H18. X. 32.0g O2. 1mol O2. = 4.80 x 103g O2. Answer ...

Explanation:

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Agree or disagree? Explain (2 ideas, 2 examples)

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Answer:

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3 years ago

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2 years ago

Which of the following describes what occurs as a gas becomes a liquid?

lora16 [44]

As the gas cools it condenses and becomes a liquid its atoms also become smaller

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3 years ago

An aqueous solution has a volume of 2.0 L and contains 36.0g of glucose. If the Molar Mass of glucose is 180g/mol, what is the m

choli [55]

Answer:

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2 years ago

You have a 2.0 mL sample of acetic acid (molar mass 60.05 g/mol) of unknown concentration. You titrate it to its endpoint with 2

Katyanochek1 [597]

<u>Answer:</u> The mass of acetic acid used is 0.12 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $CH_3COOH$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=2.0mL\\n_2=1\\M_2=0.1M\\V_2=20.0mL$

Putting values in above equation, we get:

$1\times M_1\times 2.0=1\times 0.1\times 20.0\\\\M_1=\frac{1\times 0.1\times 20.0}{1\times 2.0}=1M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of acetic acid = ? g

Molar mass of acetic acid = 60.05 g/mol

Molarity of solution = 1 M

Volume of the solution = 2.0 mL

Putting values in above equation, we get:

$1mol/L=\frac{\text{Mass of acetic acid}\times 1000}{60.05g/mol\times 2.0}\\\\\text{Mass of acetic acid}=\frac{1\times 60.05\times 2}{1000}=0.12g$

Hence, the mass of acetic acid used is 0.12 grams

5 0

3 years ago