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Sloan [31]
3 years ago
14

How do you solve this problem

Chemistry
1 answer:
ad-work [718]3 years ago
3 0

Answer: Volume of gas in the stomach, V = 0.0318L or 31.8mL

Explanation:

The number of moles of oxygen will remain constant even though the liquid oxygen will undergo a change of state to gaseous inside the person's stomach due to an increase in temperature.

<em>Number of moles of oxygen gas = mass/molar mass</em>

molar mass of oxygen gas = 32 g/mol

mass of oxygen gas = density * volume

mass of oxygen gas = 1.149 g/ml * 0.035 ml

mass of oxygen gas = 0.040215 g

Number of moles of oxygen gas = 0.0402 g/(32 g/mol)

Number of moles of oxygen gas = 0.00125 moles

<em>Using the ideal gas equation, PV=nRT</em>

where P = 1.0 atm, V = ?, n = 0.00125 moles, R = 0.082 L*atm/K*mol, T = (37 + 273)K = 310 K

<em>V = nRT/P</em>

V = (0.00125moles) * (0.082 L*atm/K*mol) * (310 K) / 1 atm

V = 0.0318L or 31.8mL

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Consider two equal-volume balloons under the same conditions of temperature and pressure. One contains helium, and the other con
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Answer:

1. Number of gas particles (atoms or molecules)

2. Number of moles of gas

3. Average kinetic energy

Explanation:

Since the two gas has the same volume and are under the same conditions of temperature and pressure,

Then:

1. They have the same number of mole because 1 mole of any gas at stp occupies 22.4L. Now both gas will occupy the same volume because they have the same number of mole

2. Since they have the same number of mole, then they both contain the same number of molecules as explained by Avogadro's hypothesis which states that at the same temperature and pressure, 1 mole of any substance contains 6.02x10^23 molecules or atoms.

3. Being under the same conditions of temperature and pressure, they both have the same average kinetic energy. The kinetic energy of gas is directly proportional to the temperature. Now that both gas are under same temperature, their average kinetic energy are the same.

5 0
3 years ago
What happens when you put magnesium carbonate and diluted water
Rina8888 [55]

Answer:

Magnesium carbonate doesn't dissolve in water, only acid, where it will effervesce (bubble).

Explanation:

 An insoluble metal carbonate reacts with a dilute acid to form a soluble salt. Magnesium carbonate, a white solid, and dilute sulfuric acid react to produce magnesium sulfate. Colourless magnesium sulfate heptahydrate crystals are obtained from this solution.

8 0
3 years ago
CH3-CHCl-CH2-CH2-CH2CHCl-CH3 +concentrated KCN
Phantasy [73]
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8 0
4 years ago
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What mass of solute must be used to prepare 500ml of 0.100M aqueous sodium borate Na2B4O7 from solid hydrated sodium borate Na2B
motikmotik

0.000132 g of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O)

Explanation:

First we need to find the number of moles of sodium borate (Na₂B₄O₇) in the solution:

molar concentration = number of moles / volume (L)

number of moles = molar concentration × volume (L)

number of moles of Na₂B₄O₇ = 0.1 × 0.5 = 0.05 moles

We know now that we need 0.05 moles of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O) to make the solution.

Now to find the mass of hydrated sodium borate we use the following formula:

number of moles = mass / molar weight

mass =  number of moles × molar weight

mass of hydrated sodium borate = 0.05 / 381 = 0.000132 g

Learn more about:

molar concentration

brainly.com/question/14106518

#learnwithBrainly

4 0
3 years ago
A solution contains 0.021 M Cl and 0.017 M I. A solution containing copper (I) ions is added to selectively precipitate one of t
lidiya [134]

<u>Answer:</u> Copper (I) iodide will precipitate first.

<u>Explanation:</u>

We are given:

K_{sp} of CuCl = 1.0\times 10^{-6}

K_{sp} of CuI = 5.1\times 10^{-12}

Concentration of Cl^-\text{ ion}=0.021M

Concentration of I^-\text{ ion}=0.017M

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

  • <u>For CuCl:</u>

K_{sp}=[Cu^+][Cl^-]

Putting values in above equation, we get:

1.0\times 10^{-6}=[Cu^+]\times 0.021

[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M

Concentration of copper (I) ion = 4.76\times 10^{-5}M

  • <u>For CuI:</u>

K_{sp}=[Cu^+][I^-]

Putting values in above equation, we get:

5.1\times 10^{-12}=[Cu^+]\times 0.017

[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M

Concentration of copper (I) ion = 3.00\times 10^{-10}M

For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.

7 0
3 years ago
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