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2 years ago

How many grams of CO2 are in 3.6 mol of CO2?

Chemistry

2 answers:

Trava [24]2 years ago

7 0

Explanation:

The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles CO2, or 44.0095 grams.

I hope you help:)

hammer [34]2 years ago

6 0

There are 158.4 grams of CO2 in 3.6 mol of CO2.

<h3>HOW TO CALCULATE MASS?</h3>

The mass of a substance can be calculated by multiplying the number of moles of the substance by its molar mass. That is;

mass of CO2 = no. of moles × molar mass

According to this question, there are 3.6 moles of CO2.

mass of CO2 = 3.6 moles × 44g/mol

mass of CO2 = 158.4g.

Therefore, there are 158.4 grams of CO2 in 3.6 mol of CO2

Learn more about mass at: brainly.com/question/15959704

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What is the mass percent of oxygen in sodium bicarbonate (NaHCO3)?

Alenkasestr [34]

The answer is 57.14%.

First we need to calculate molar mass of NaHCO3. Molar mass is mass of 1 mole of a substance. It is the sum of relative atomic masses, which are masses of atoms of the elements.

Relative atomic mass of Na is 22.99 g
Relative atomic mass of H is 1 g
Relative atomic mass of C is 12.01 g
Relative atomic mass of O is 16 g.

Molar mass of NaHCO3 is:
22.99 g + 1 g + 12.01 g + 3 · 16 g = 84 g

Now, mass of oxygen in NaHCO3 is:
3 · 16 g = 48 g

mass percent of oxygen in NaHCO3:
48 g ÷ 84 g · 100% = 57.14%

Therefore, the mass percent of oxygen in sodium bicarbonate is 57.14%.

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The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

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