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3 years ago

15

If the average speed of a helium atom at a certain temperature is 1230. m/s, what is the speed in miles per hour? (1mi =1609m)

1 answer:

Darya [45]3 years ago

6 0

In order to do conversion on these types of problems. <span>
1.You first have to determine the relationship and the current unit and the target unit which it will be converted into.</span> <span>
2.In this case feet to meters and seconds to minutes</span> <span>
3.Determine the conversion rate: </span>1mi =1609 m<span /> and 1 h = 3600 sec; <span>
4.You first calculate: 1230 m x </span>1 mi/1609 m = 0.764 mi/s
<span>5.<span>Next, is the time which is 0.764 mi/s x 3600s/1hr = </span></span><span>2752.02 mi/h
</span>

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vfiekz [6]

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thank you

<em><u>I </u></em><em><u>hope</u></em><em><u> </u></em><em><u>this </u></em><em><u>is </u></em><em><u>helpful</u></em>

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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:

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

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$=5.861 \times 3.596 = 21.076\,kJ$

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

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

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

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

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

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

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

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

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3 years ago