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3 years ago

How is the mass of carbon related to the mole?

Chemistry

1 answer:

Anuta_ua [19.1K]3 years ago

4 0

Answer:

Simply put, you can go from moles to grams and vice versa by using the mass of 1 mole of that substance, i.e its molar mass. For example, the molar mass of carbon is 12.011 g/mol. This means that 1 mole of carbon, or 6.022⋅1023 atoms of carbon, weigh 12.011 g.

Explanation:

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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

Which of the following is not a use for line graphs? *

kati45 [8]

Answer:

Explanation:

because nothing is changing, so what are you gonna track with the line graph

6 0

3 years ago

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Covalent solutes are considered non-electrolytes. What does this mean for the conductivity of the solution? A) Non-electrolytes

iogann1982 [59]

Electrolytes are those which dissociates in solution and produces ions.

Ions can carry current,so Electrolytes conduct electiricity.

And non electrolytes are those which do not dissociate in solution and doesnt produce ions.

Since non electrolytes do not produce ions they cannot conduct electricity.

<u>Hence the right option is:</u>

B) Non-electrolytes dissolve and do not dissociate in water providing no charged ions to conduct electricity.

5 0

2 years ago

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Is methane used as fule<br> gas true or false<br>

Hitman42 [59]

Answer:

its true methane is used as a fuel gas

Explanation:

4 0

2 years ago

If 4.50 l of water vapor at 50.2 °c and 0.121 atm reacts with excess iron, how many grams of iron(iii) oxide will be produced?

Flura [38]

When the balanced equation for this reaction is:
2Fe + 3H2O → Fe2O3 + 3H2

and according to the vapour pressure formula:
PV= nRT
when we have P is the vapor pressure of H2O= 0.121 atm
and V is the volume of H2O = 4.5 L
and T in Kelvin = 52.5 +273 = 325.5 K
R= 0.08205 atm-L/g mol-K
So we can get n H2O
So, by substitution:
n H2O = PV/RT
= (0.121*4.5)/(0.08205 * 325.5) = 0.02038 gmol
n Fe2O3 = 0.02038 * (1Fe2O3/ 3H2O) = 0.00679 gmol
Note: we get (1FeO3/3H2O) ratio from the balanced equation.
we can get the Mass of Fe2O3 from this formula:
Mass = number of moles * molecular weight
when we have a molecular weight of Fe2O3 = 159.7
= 0.00679 * 159.7 = 1.084 g
∴ 1.084 gm of Fe2O3 will produced

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3 years ago

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