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3 years ago

15

How do you do these?

1 answer:

Agata [3.3K]3 years ago

6 0

Solve these problems like weighted averages:

The first one:

Multiply the masses (isotope numbers) by the decimal form of the percentage. Add them

0.076 (6) + 0.924 (7) = 6.924

The second one:

0.2 (10) + 0.8 (11) = 10.8

If you think about it, these answers make sense. 6.924 is much closer to 7 than to 6 (since there's a lot more lithium-7 than there is lithium-6). 10.8 is closer to 11 than to 10.

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Which of the following was not a big ideal of chemistry

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Answer:

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Answer:

a) $T_b=590.775k$

b) $W_t=1.08*10^4J$

d) $Q=3.778*10^4J$

d) $\triangle V=4.058*10^4J$

Explanation:

From the question we are told that:

Moles of N2 $n=2.50$

Atmospheric pressure $P=100atm$

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$t = 293k$

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a)

Generally the equation for change in temperature is mathematically given by

$\triangle T=\frac{Q}{N*C_v}$

Where

$C_v=Heat\ Capacity \approx 20.76 J/mol/K$

$T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }$

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$T_b=590.775k$

b)

Generally the equation for ideal gas is mathematically given by

$PV=nRT$

For v double

$T_c=2*590.775k$

$T_c=1181.55k$

Therefore

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$Wbc=10805.7J$

Total Work-done $W_t$

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$W_t=0+1.08*10^4$

$W_t=1.08*10^4J$

c)

Generally the equation for amount of heat added is mathematically given by

$Q=nC_p\triangle T$

$Q=2.20*2907*(1181.55-590.775)\\$

$Q=3.778*10^4J$

d)

Generally the equation for change in internal energy of the gas is mathematically given by

$\triangle V=nC_v \triangle T$

$\triangle V=2.20*20.76*(1181.55-293)k$

$\triangle V=4.058*10^4J$

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<em><u>Concentration is one of the factors that determine the rate of a reaction. Reaction rates increases with increase in the concentration of the reactants, which means they are directly proportional.</u></em>
An increase in the concentration of reactants produces more collisions and thus increasing the rate at which the reaction is taking place. Therefore, <u>Increasing the concentration of a reactant increases the frequency of collisions between reactants and will cause an increase in the rate of reaction.</u>

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