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3 years ago

What is the volume, in liters, of 6.8 mol of Kr gas at STP?

Chemistry

1 answer:

Virty [35]3 years ago

6 0

Answer:

$\boxed {\boxed {\sf D. \ 152 \ L}}$

Explanation:

Any gas at standard temperature and pressure (STP) has a volume of 22.4 liters per mole or 22.4 L/mol. We can create a proportion with this value.

$\frac {22.4 \ L \ Kr}{1 \ mol \ Kr}$

Multiply both sides of the equation by 6.8 moles of krypton.

$6.8 \ mol \ Kr *\frac {22.4 \ L \ Kr}{1 \ mol \ Kr}$

The units of moles of krypton will cancel.

$6.8 *\frac {22.4 \ L \ Kr}{1 }$

The denominator of 1 can be ignored, so this becomes a simple multiplication problem.

$68 * 22.4 \ L \ Kr$

$152.32 \ L \ Kr$

If we round to the nearest whole number, the 3 in the tenths place tells us to leave the 2 in the ones place.

$152 \ L \ Kr$

6.8 moles of krypton gas at standard temperature and pressure is equal to <u>152 liters</u>.

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How many grams of sodium hydroxide are needed to completely react with 50.0 grams of H2SO4?

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Answer:

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2 years ago

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A pure compound is found to be 40.0% carbon by mass, 6.73% hydrogen by mass, and 53.3% oxygen by mass. determine the empirical f

ratelena [41]

Since there is no weight, I would assume that this is a 100g of pure compound.
Okay so I would be changing the percentage to gram to solve for the mole.
So
40.0g C (1 mol C/12.01 g C) = 3.33 mol C
6.73g H (1 mol H/1.01 g H ) = 6.66 mol H
53.3g O (1 mol O/16.00 g O) = 3.33 mol O

With that, two of our moles is 3.33, so we consider that are our 1, as it is also the lowest. Therefore the empirical formula is CH2O

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2 years ago

Balance chemical reaction of oxalic acid with conc. sulphuric acid

seropon [69]

Answer:

Explanation: since oxalic acid is a weak acid it wont provide a strong acidic medium. So in order to provide a strong acidic medium dilute sulphuric acid is added.

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2 years ago

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

8 0

3 years ago

The diagram below shows part of the rock cycle. (6 points)

Anna35 [415]

Answer:

Igneous Rock

Explanation:

Assuming this is a cycle, the volcanic eruption would lead back to rock B; rocks formed by volcanic eruptions are considered Igneous.

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2 years ago