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Zanzabum
3 years ago
9

If the volume of an ideal gas is 22.5 L when it is at a temperature of 365 K, what will the volume be when the gas cools to 338K

? Be sure to show your work and proper units.
Chemistry
1 answer:
julsineya [31]3 years ago
6 0

Answer:

20.8L = Final volume of the gas

Explanation:

Based on Charles's law, the volume of a gas is directly proportional to the temperature of the gas under pressure constant. The equation is:

\frac{V_1}{T_1} =\frac{V_2}{T_2}

<em>where V is volume and T absolute temperature of 1, initial state and 2, final state.</em>

<em />

If initial volume is 22.5L, initial T = 365K and final temperature 338K:

22.5L / 365K = V₂ / 338K

<h3>20.8L = Final volume of the gas</h3>
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Calculate the volume of 0.10 g of titanium (4.51 g/cm³).
Juli2301 [7.4K]

The volume of titanium with mass of 0. 10g and density of 4. 51 g/cm³ is 0. 02 cm³

<h3>What is volume?</h3>

Volume is known to be equal to the mass divided by the density.

It is written thus:

Volume = Mass / density

<h3>How to calculate the volume</h3>

The volume is calculated using the formula:

Volume = mass ÷ density

Given the mass = 0. 10g

Density = 4.51 g/cm³

Substitute the values into the formula

Volume of titanium = 0. 10 ÷ 4.51 = 0. 02 cm³

Thus, the volume of titanium with mass of 0. 10g and density of 4. 51 g/cm³ is 0. 02 cm³

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2 years ago
Hi can anyone answer these two distance and time questions
Likurg_2 [28]
1 is hours and 2 is meters
5 0
3 years ago
Which of these half-reactions represents reduction?
gogolik [260]

Answer: The half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
  • MnO^{-}_{4} \rightarrow Mn^{2+}

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element  takes place is called reduction-half reaction.

For example, the oxidation state of Cr in Cr_{2}O^{2-}_{7} is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}

Similarly, oxidation state of Mn in MnO^{-}_{4} is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}

Thus, we can conclude that half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
  • MnO^{-}_{4} \rightarrow Mn^{2+}
3 0
3 years ago
When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C
natka813 [3]

Answer:

\Delta_{r}U of the reaction is -6313 kJ/mol

\Delta_{r}H of the reaction is -6312 kJ/mol

Explanation:

Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C

q_{cal}= C \times \Delta T

=5.861 \times 3.596 = 21.076\,kJ

q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol

Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

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3 years ago
A solution of ammonia has a pH of 11.8. What is the concentration of OH– ions in the solution?
melomori [17]
[OH-]: <span>0.0063095734448</span>
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