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2 years ago

3. Ethanol has a density of 0.800 g/mL. a. What is the mass of 225 mL of ethanol?

Chemistry

1 answer:

dedylja [7]2 years ago

3 0

Since the ethanol has a density of 0.800 g/mL, the mass of the ethanol is 180 grams.

Given the following data:

Volume of ethanol = 225 mL

Density of lead ball = 0.800 g/mL.

To find the mass of the ethanol;

Density can be defined as mass all over the volume of an object.

Mathematically, the density of a substance is given by the formula;

$Density = \frac{Mass}{Volume}$

Making mass the subject of formula, we have;

$Mass = Density \times Volume$

Substituting the given parameters into the formula, we have;

Mass of ethanol = 180 grams.

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2 years ago

How many moles in 28.0 grams of Oxygen

anastassius [24]

Answer:

1.) 28.0 grams of oxygen28 grams (1 mole/16 grams per mole)=1.75 moles oxygen2.)5.0 moles of Iron5 moles(55.845 grams/1 mole)=279.225 grams Iron3.) 452 grams Argon452 grams(1 mole/39.948 grams)=11.315 moles Argon4.) 16.5 moles Hydrogen16.5 moles(1.01 grams/1 mole)=16.665 grams Hydrogen

Explanation:

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2 years ago

In the lab you weigh out 76.02 g of Iron (Fe). How many moles of Iron do you have in the sample. (Your answer must have a unit..

son4ous [18]

Answer:

We have 1.361 moles in the sample

Explanation:

Mass of iron = 76.02g

Molar mass of iron = 55.845 g/ mole ( This we can find in the periodic table, and menas that 1 mole of iron has a mass of 55.845 g).

To calculate the number of moles we will use following formula:

moles (n) = mass / molar mass

moles iron = 76.02g / 55.845 g/ mole

moles iron = 1.36127 moles

To use the correct number of significant digits we use the following rule for multiplication and division :

⇒ the number with the least number of significant figures decides the number of significant digits.

⇒76.02 has 4 digits ( 2 after the comma) and 55.845 has 5 digits (3 after the comma).

⇒ this means 1.361 moles

We have 1.361 moles in the sample

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2 years ago

One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c

Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of $\rm O_2\; (g)$ is larger than that of $\rm H_2\; (g)$ (by a factor of about $16$ .) Therefore, the mass of the $\rm O_2\; (g)$ sample is significantly larger than that of the $\rm H_2\; (g)$ sample.

Explanation:

The $\rm O_2\; (g)$ and the $\rm H_2\; (g)$ sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles ( $\rm O_2\; (g)$ and $\rm H_2\; (g)$ molecules, respectively.) That is:

$n(\mathrm{O_2}) = n(\mathrm{H}_2)$ .

Note that the mass of a gas $m$ is different from the number of gas particles $n$ in it. In particular, if all particles in this gas have a molar mass of $M$ , then:

$m = n \cdot M$ .

In other words,

$m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2})$ .
$m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2})$ .

The ratio between the mass of the $\rm O_2\; (g)$ and that of the $\rm H_2\; (g)$ sample would be:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}$ .

Since $n(\mathrm{O_2}) = n(\mathrm{H}_2)$ by Avogadro's Law:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ .

Look up relative atomic mass data on a modern periodic table:

$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Therefore:

$M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}$ .
$M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}$ .

Verify whether $\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ :

Left-hand side: $\displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1$ .
Right-hand side: $\displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9$ .

Note that the mass of the $\rm H_2\; (g)$ sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

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3 years ago

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Allisa [31]

Answer:

They are both planets made out of gas!

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2 years ago

Read 2 more answers