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mixas84 [53]
3 years ago
8

1/2 - 3/7=??????????????????????​

Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
6 0
1/2 - 3/7 = 7/14 - 6/14 = 1/14
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What is the percent composition of N H S O in (NH4)2SO4
Lena [83]

Answer:

The percent composition is 21% N, 6% H, 24% S and 49% O.

Explanation:

1st) The molar mass of (NH4)2SO4 is 132g/mol, and it represents the 100% of the mass composition.

In 1 mole of (NH4)2SO4, there are:

- 2 moles of N.

- 8 moles of H.

- 1 mole of S.

- 4 moles of O.

2nd) It is necessary to calculate the mass of each element, multiplying its molar mass by the number of moles:

- 2 moles of N (14g/mol) = 28g

- 8 moles of H (1g/mol) = 8g

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3rd) With a mathematical rule of three we can calculate the percent composition of each element in the molecule of (NH4)2SO4:

\begin{gathered} \text{ Nitrogen:} \\ 132g-100\% \\ 28g-x=\frac{28g*100\%}{132g} \\ x=21\% \end{gathered}\begin{gathered} \text{ Hydrogen:} \\ 132g-100\operatorname{\%} \\ 8g-x=\frac{8g*100\operatorname{\%}}{132g} \\ x=6\% \end{gathered}\begin{gathered} \text{ Sulfur:} \\ 132g-100\operatorname{\%} \\ 32g-x=\frac{32g*100\operatorname{\%}}{132g} \\ x=24\% \end{gathered}\begin{gathered} \text{ Oxygen:} \\ 100\%-21\%-6\%-24\%=49\% \\  \\  \end{gathered}

In this case, we can calculate the percent composition of Oxygen by subtracting the other percentages, since the total must be 100%.

So, the percent composition is 21% N, 6% H, 24% S and 49% O.

8 0
1 year ago
When having an equal number of protons and electrons, what does that do to the overall charge of the element? Why do you think t
gladu [14]
It means it’s neutrally charged! Protons are positive while electrons are negative, so they balance each other out. Neutrons are neutral so they don’t weigh in.
4 0
3 years ago
Limestone stalactites and stalagmites are formed in caves by the following reaction: Ca2 (aq) 2HCO−3(aq)→CaCO3(s) CO2(g) H2O(l)
Vesna [10]

Answer : The value of \Delta E for this reaction is 36.18 kJ

Explanation :

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As per first law of thermodynamic,

\Delta E=q+w

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q = heat added or rejected by the system

w = work done

As we are given that:

q = 38.65 kJ

w = -2.47 kJ (system work done on surrounding)

Now put all the given values in the above expression, we get:

\Delta E=38.65kJ+(-2.47kJ)

\Delta E=36.18kJ

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Ammonia is the compound.
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3 years ago
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