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2 years ago

HELP ASAP How many millimeters of mercury pressure does a gas exert at 3.16 atm?

Chemistry

2 answers:

Georgia [21]2 years ago

0 0

The conversion factor is 760 mmHg/atm.

3.1 atm * 760 mmHg/atm = 2356 m

azamat2 years ago

0 0

3.1=2356 is the millimeters amount

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The molar mass of c3h8o2 (four significant figures) is ________.

Luda [366]

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2 years ago

Suppose a layer of oil is on the top of a beaker of water. Water in many oils is slightly soluble, so its concentration is so lo

rodikova [14]

Explanation:

It is given that energy to transfer one water molecule is $2.208 \times 10^{-20}$ J/molecule

As it is known that in 1 mole there are $6.022 \times 10^{23}$ atoms.

So, energy in 1 mole = $2.208 \times 10^{-20} \times 6.022 \times 10^{23}$ J/mol

= 13.3 kJ/mol

As, $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times 8.314 atm L/mol K} \times \frac{T_{2} - T_{1}}{T_{1}T_{2}}$

Putting the given values in the above formula as follows.

$log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times 8.314 atm L/mol K} \times \frac{T_{2} - T_{1}}{T_{1}T_{2}}$

$log \frac{k_{2}}{k_{1}} = \frac{13.3 kJ/mol}{2.303 \times 8.314 atm L/mol K} \times \frac{293 K - 283 K}{283K \times 293 K}$

$log \frac{k_{2}}{k_{1}}$ = 0.08377

$\frac{k_{2}}{k_{1}}$ = 1.213 = $\frac{Concentration_{2}}{Concentration_{1}}$

$Concentration_{2}$ = $1.213 \times Concentration_{1}$

= $1.213 \times 9 \times 10^{-4}$

= $10.915 \times 10^{-4}$ water molecules per oil molecule

Thus, we can conclude that the equilibrium concentration at 293 K, assuming that nothing else changes is $10.915 \times 10^{-4}$ .

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2 years ago

How many molecules of water are needed to completely hydrolyze a polymer that is 11 monomers long?

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10 molecules of water are needed.

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2 years ago

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2 years ago

Read 2 more answers

Salt in crude oil must be removed before the oil undergoes processing in a refinery. The

irina1246 [14]

Answer:

$\large \boxed{0.64 \, \%}$

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

$\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}$

2. Calculate the mass of salt in the salty oil

$\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}$

3. Calculate the mass of salt in the spent water

$\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}$

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g

5. Concentration of salt in washed oil

$\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}$

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2 years ago