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3 years ago

What is the mass, in grams, of 9.16×10^24 molecules of methanol (CH3OH)?

Chemistry

1 answer:

Aleks04 [339]3 years ago

3 0

Answer:

The mass in grams is 487.5 for methanol.

Explanation:

We must calcuate the grams of a compound through the molar mass of it.

Molar mass of methanol: 32.04 g/mol

We know that 1 mol has NA molecules and they weigh 32.04 g so, this is the rule of three we must make:

6.02x10²³ molecules _____ 32.04 g

9.16x10²⁴ molecules _____ (9.16x10²⁴ . 32.04) / 6.02x10²³ = 487.5 g

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Carbon dioxide, water vapor, and oxygen pass through ---------------

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The answer is B) stomata.

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Identify each of the following sets of quantum numbers as allowed or not allowed in the hydrogen atom.

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Answer:

See explanation.

Explanation:

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In this case, according to the given information, it turns out possible for us to firstly recall the electron configuration of hydrogen:

$1s^1$

To realize that the principal quantum number is 1, the angular is 0 as well as the magnetic one; therefore we infer that all the given n's are not allowed, just l=0 is allowed as well as ml=0 yet the rest, are not allowed.

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3 years ago

The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa

Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of $MgCl_2$ (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of $MgCl_2$ (solute) = 3.37 g

First we have to calculate the mass of solute.

$\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2$

$\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g$

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg$

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

$\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%$

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

$\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%$

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

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3 years ago

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Answer:

Explanation:

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3 years ago

From the graph of Density vs. Concentration, created in Graph 1, what was the relationship between the concentration of the suga

USPshnik [31]

The graph is not given in the question, so, the required graph is attached below:

Answer:

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