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labwork [276]
3 years ago
14

A flexible vessel contains 32.00 L of gas at a pressure of 1.59 atm. Under conditions of constant temperature and moles of gas,

what is the volume of the gas when the pressure of the vessel is decreased by a factor of three
Chemistry
1 answer:
HACTEHA [7]3 years ago
4 0

Answer:

When the pressure of the vessel is decreased by a factor of three, the volume will increase by a factor 3. This means the new volume is 96.00 L

Explanation:

Step 1: Data given

Volume of the gas = 32.00 L

Pressure of the gas = 1.59 atm

Temperature and moles of gas stay constant

the pressure of the vessel is decreased by a factor of three

Step 2: Calculate the new volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 1.59 atm

⇒with V1 = the initial volume = 32.00 L

⇒with P2 = the decreased pressure = 1.59 / 3 = 0.53 atm

⇒with V2 = the new volume = TO BE DETERMINED

1.59 atm * 32.00 L = 0.53 atm * V2

V2 = (1.59 * 32.00 L) / 0.53 atm

V2 = 96.00 L

When the pressure of the vessel is decreased by a factor of three, the volume will increase by a factor 3. This means the new volume is 96.00 L

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Answer:

5

Explanation:

To balance the hydrogen atoms, we check the number of hydrogen on the left side, this is equal to the 10 hydrogen atoms we have in the alkanol.

Now, on the right hand side, we can see we only have two hydrogen atoms in the water molecule. Now, to make equal the number of hydrogen atoms on both sides, we simply multiply the number of hydrogen there by 5 to make it 10 too

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2 years ago
4. Magnesium and oxygen undergo a chemical reaction to
erastovalidia [21]

Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

  1. Convert grams of MgO to moles of MgO.
  2. Moles of MgO to moles of O₂
  3. And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

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A 0.610 m aqueous solution of kbr has a total mass of 75.0 g. what masses of solute and solvent are present?
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To determine the masses of each component, we need to know the concentration first in terms of molality which is mol per mass of solution. Assuming the density of the solution is equal to that of water we would find:

molality = 0.160 mol KBr / L ( 1 L / 1 kg solution ) = 0.160 mol KBr / kg solution

mass KBr = 0.160 mol KBr / kg solution (.0750 kg solution ) ( 119 g KBr / mol KBr ) = 1.428 g KBr

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Question 4 (4 points)
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