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Sauron [17]
2 years ago
11

How many grams of carbon are contained in one mole of

e="C_{3}H_{8}" alt="C_{3}H_{8}" align="absmiddle" class="latex-formula"> ?
Chemistry
1 answer:
NNADVOKAT [17]2 years ago
4 0

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \:36 \:\: g

____________________________________

\large \tt Explanation \: :

The given compound has 3 carbon atoms, so in 1 mole of that compound, there will be 3 moles of carbon atoms.

Mass of each mole of carbon atoms = 12 g

For 3 mole carbon atoms, it will be 12 × 3 = 36 grams

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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Answer:

Explanation:

(NH4)3 PO4 +NaOH arrow Na3PO4 +3NH3 +3H2O

Start by seeing what happens with the Na. You need 3 on the left, so put a 3 in front of NaOH

(NH4)3 PO4 +3NaOH arrow Na3PO4 +3NH3 +3H2O  Next work with the nitrogens. YOu have 3 on the left and 3 on the right, so they are OK. Next Go to the stray oxygens.

You have 3 on left in (NaOH) and three on the right in 3H2O so they are fine as well. The last thing you should look at are hydrogens.

There are 12 + 3 on the left which is 15. There are 9 (in 3NH3) and 6 more in the water. They seem fine.

Why didn't I do something with the PO4^(-3)? The reason is a deliberately stayed away from them and balanced everything else. Since they were untouched with 1 on the left and 1 on the right, they are balanced.

Species      Na        H        O         N       PO4

Left             3          15        3         3          1

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8 0
3 years ago
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A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

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Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

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\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

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3 years ago
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