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3 years ago

what is the mass of carbon dioxide which contain the same number of molecules as are contained in 14 gram of oxygen?

Chemistry

1 answer:

Ratling [72]3 years ago

3 0

Answer:

Mass of CO2 WILL BE ~ 9.33 g

Explanation:

Moles of O2 = 14/18

Let the mass of CO2 be x

Then moles of CO2 will be = x/12

moles of CO2 = moles of O2

x/12 = 14/16

x = 9.33 grams

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In this molecule, what's the formal charge on the central O atom?

olga_2 [115]

The answer you are looking for is C

3 0

3 years ago

Read 2 more answers

2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea

Alex_Xolod [135]

Answer: The concentration of hydrogen gas at equilibrium is 0.0275 M

Explanation:

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Moles of HI = 0.550 moles

Volume of container = 2.00 L

$\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M$

For the given chemical equation:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

Initial: 0.275

At eqllm: 0.275-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$K_c=0.0156$

Putting values in above expression, we get:

$0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

4 0

3 years ago

A 46.2 mL,0.568 M calcium nitrate solution is mixed with 80.5mL of 1.396M calcium nitrate solution.Calculate tge concentration o

Ne4ueva [31]

Answer:

1.09 M

Explanation:

Let's define the equation that will be used to calculate the final concentration of the resultant calcium nitrate solution. In order to calculate it, we need to find the total number of moles of calcium nitrate and divide by the total volume of the resultant solution:

$c=\frac{n}{V}$

This equation firstly helps us find the number of moles of calcium nitrate. Multiplying molarity by volume will yield the moles. Adding the moles from the first component to the second component will provide us with the total number of moles of calcium nitrate:

$n_{Ca(NO_3)_2}=46.2 mL\cdot0.568 M+80.5 mL\cdot1.396 M=138.62 mmol$