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pav-90 [236]
2 years ago
15

If you have 0.75 moles of an ideal gas at 87 °C and a pressure of 569 torr, what volume will the gas take up?

Chemistry
1 answer:
vivado [14]2 years ago
5 0

Answer:

29 L

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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Which agents cause both chemical and physical weathering
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Weathering is the process by which rocks are broken down by either physical or chemical means. Physical weathering involves physical factors that cause mechanical break down of rocks while chemical weathering involves chemical reactions between the contents of the rocks and other factors such air (oxygen). Agents of weathering are the factors that cause or enhance both physical and chemical weathering. Some of these factors cause both physical and chemical weathering these includes water and temperature.

8 0
3 years ago
A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec
GalinKa [24]

<u>Answer:</u> The moles of precipitate (lead (II) iodide) produced is 1.57\times 10^{-5} moles

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

  • <u>For lead (II) nitrate:</u>

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol

  • <u>For NaI:</u>

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol

For the given chemical reaction:

Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, 3.14\times 10^{-5} moles of NaI will react with = \frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 3.14\times 10^{-5} moles of NaI will produce = \frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is 1.57\times 10^{-5} moles

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2 years ago
Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the sys
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Answer:

Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.

Explanation:

The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:

Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}.

Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both [\mathrm{SO_2\, (g)}] and [\mathrm{O_2\, (g)}] will increase if the pressure is increased through compression. However, because \rm SO_2\, (g) and \rm O_2\, (g) have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient Q.

As a result, the increase in pressure will have no impact on the value of Q\!. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.

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