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Anon25 [30]
3 years ago
6

Which of the following is not the same as 2.97 milligrams? 0.0000297 kg, 0.00297 g, 0.297 ch

Chemistry
2 answers:
Travka [436]3 years ago
4 0

the correct answer its  0.002975

erica [24]3 years ago
3 0
<h2>Answer:</h2>

              0.0000297 Kg is not the same as 2.97 mg.

<h2>Proof:</h2>

<u>Converting 2.97 mg into Kg:</u>

As,

                             1 mg is equal to  =  1.0 × 10⁻⁶ Kg

So,

                        2.97 mg will be equal to  =  X Kg

Solving for X,

                      X =  (2.97 mg × 1.0 × 10⁻⁶ Kg) ÷ 1 mg

                      X =  2.97 × 10⁻⁶ Kg

Or,

                      X =  0.00000297 Kg

Hence, 0.0000297 Kg is not the same.

<u>Converting 2.97 mg into g:</u>

As,

                             1 mg is equal to  =  0.001 g

So,

                        2.97 mg will be equal to  =  X g

Solving for X,

                      X =  (2.97 mg × 0.001 g) ÷ 1 mg

                      X =  0.00297 g

Hence, 0.00297 g is the same.

<u>Converting 2.97 mg into cg:</u>

As,

                             1 mg is equal to  =  0.10 cg

So,

                        2.97 mg will be equal to  =  X cg

Solving for X,

                      X =  (2.97 mg × 0.10 cg) ÷ 1 mg

                      X =  0.297 cg

Hence, 0.297 cg is the same.

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Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79
BARSIC [14]

Answer :  The mole fraction and partial pressure of CH_4,CO_2 and He gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of CH_4 = 1.79 mole

Moles of CO_2 = 1.20 mole

Moles of He = 3.71 mole

Now we have to calculate the mole fraction of CH_4,CO_2 and He gases.

\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267

and,

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179

and,

\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554

Thus, the mole fraction of CH_4,CO_2 and He gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of CH_4,CO_2 and He gases.

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 5.78 atm

X_i = mole fraction of gas

p_{CH_4}=X_{CH_4}\times p_T

p_{CH_4}=0.267\times 5.78atm=1.54atm

and,

p_{CO_2}=X_{CO_2}\times p_T

p_{CO_2}=0.179\times 5.78atm=1.03atm

and,

p_{He}=X_{He}\times p_T

p_{He}=0.554\times 5.78atm=3.20atm

Thus, the partial pressure of CH_4,CO_2 and He gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0
3 years ago
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