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Elden [556K]
4 years ago
6

A reaction has ?h?rxn= -129 kj and ?s?rxn= 301 j/k .

Chemistry
1 answer:
Gennadij [26K]4 years ago
6 0
I was thinking that the question would be to find for the operating temperature of the reaction. Change in entropy is equal to the total energy divided by the temperature. Assuming it is isothermal, internal energy would be zero. Therefore, the temperature would be:

T = 129000/301 = 428.57 K
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Which of the following has the highest ionization energy? Rubidium Zirconium Niobium
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Niobium has Highest ionization energy.

<u>Explanation:</u>

The next ionization energy removes an electron from the same electron shell, which increases "ionization energy" due to "increased net charge"of the ion from which the electron is being removed.

Let's compare each of the metals first ionization energies,

Rubidium has its first ionization energy as 403 kJ / mol.

Zirconium has its first ionization energy as 660 KJ / mol.

Niobium has its first ionization energy as 664 KJ / mol.

From the given data above, we can infer that Rubidium has the least Ionization energy whereas Niobium has the highest Ionization energy. We can give the increasing order of the given elements as follows-  

Rubidium < Zirconium < Niobium

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3 years ago
Helium gas collected at a pressure of 0.0045 atm is put into a flexible balloon measuring 350 mL in volume. What volume, in mL,
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3 years ago
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4 years ago
Find the mass of 3.27 x 10^23 molecules of H2SO4. Use 3 significant digits<br> and put the units.
marta [7]

Answer:

Approximately 53.3\; \rm g.

Explanation:

Lookup Avogadro's Number: N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1} (three significant figures.)

Lookup the relative atomic mass of \rm H, \rm S, and \rm O on a modern periodic table:

  • \rm H: 1.008.
  • \rm S: 32.06.
  • \rm O: 15.999.

(For example, the relative atomic mass of \rm H is 1.008 means that the mass of one mole of \rm H\! atoms would be approximately 1.008\! grams on average.)

The question counted the number of \rm H_2SO_4 molecules without using any unit. Avogadro's Number N_{\rm A} helps convert the unit of that count to moles.

Each mole of \rm H_2SO_4 molecules includes exactly (1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23} of these \rm H_2SO_4 \! molecules.

3.27 \times 10^{23} \rm H_2SO_4 molecules would correspond to \displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of \rm H_2SO_4 gives the mass of each mole of \rm H_2SO_4\! molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the mass of approximately 0.541389\; \rm mol of \rm H_2SO_4:

\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}.

(Rounded to three significant figures.)

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3 years ago
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3 years ago
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