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1 year ago

Convert from grams to moles for 36g CO2

Chemistry

1 answer:

podryga [215]1 year ago

3 0

Answer:

0.82 moles CO₂

Explanation:

To convert from grams to moles, you need to find the molar mass of CO₂. This conversion is a ratio stating the amount of grams CO₂ per 1 mole. The molar mass is calculated using the molar mass of each element (in CO₂) from the periodic table.

Molar Mass (Carbon): 12.011 g/mol

Molar Mass (Oxygen): 15.998 g/mol

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

36 grams CO₂ 1 mole
---------------------- x -------------------------- = 0.82 moles CO₂
44.007 grams

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#2: A gas at unknown pressure and a volume of 66 cm has its pressure changed to 150 kPa. The new volume

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Answer:

Explanation:

The original pressure can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

where

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V1 is the initial volume

V2 is the final volume

Since we're finding the original pressure

$P_1 = \frac{P_2V_2}{V_1} \\$

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2 years ago

2. How many moles are in 8.30 x 1023 molecules of H2O?

daser333 [38]

Answer:

<h2>1.38 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

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N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

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$n = \frac{8.30 \times {10}^{23} }{6.02 \times {10}^{23} } = \frac{8.30}{6.02} \\ = 1.3787 37...$

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2 years ago

Give the direction of the reaction, if K >> 1. Give the direction of the reaction, if K >> 1. The forward reaction i

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Answer:

A. for K>>1 you can say that the reaction is nearly irreversible so the forward direction is favored. (Products formation)

B. When the temperature rises the equilibrium is going to change but to know how is going to change you have to take into account the kind of reaction. For endothermic reactions (the reverse reaction is favored) and for exothermic reactions (the forward reaction is favored)

Explanation:

A. The equilibrium constant K is defined as

$K=\frac{Products}{reagents}$

In any case

aA +Bb equilibrium Cd +dD

where K is:

$K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$

[] is molar concentration.

If K>>> 1 it means that the molar concentration of products is a lot bigger that the molar concentration of reagents, so the forward reaction is favored.

B. The relation between K and temperature is given by the Van't Hoff equation

$ln(\frac{K_{1}}{K_{2}})=\frac{-delta H^{o}}{R}*(\frac{1}{T_{1}}-\frac{1}{T_{2}})$

Where: H is reaction enthalpy, R is the gas constant and T temperature.

Clearing the equation for $K_{2}$ we get:

$K_{2}=\frac{K_{1}}{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

Here we can study two cases: when delta $H^{o}$ is positive (exothermic reactions) and when is negative (endothermic reactions)

For exothermic reactions when we increase the temperature the denominator in the equation would have a negative exponent so $K_{2}$ is greater that $K_{1}$ and the forward reaction is favored.

When we have an endothermic reaction we will have a positive exponent so $K_{2}$ will be less than $K_{1}$ the forward reactions is not favored.

${e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

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