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3 years ago

In an internal combustion engine, air at atmostpheric

Physics

1 answer:

notsponge [240]3 years ago

8 0

Answer:

1302 K or 1029 C

Explanation:

Air at atmospheric pressure has pressure of 1 atm

20 C = 20 + 273 = 293 K

Assume ideal gas, according to the ideal gas law:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

Where P1, V1 and T1 are the pressure, volume and temperature of the gas before the compression and P2, V2 and T2 are the pressure, volume and temperature of the gas after the compression

$T_2 = T_1\frac{P_2V_2}{P_1V_1}$

Since the gas is compressed to 1/9 of its original volume, V2/V1 = 1/9:

$T_2 = 293\frac{40}{9} = 1302 K$ or 1029 C

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liberstina [14]

Answer : The maximum concentration of silver ion is $5\times 10^{-12}m$

Solution : Given,

$K_{sp}$ for AgBr = $5\times 10^{-13}$

Concentration of NaBr solution = 0.1 m

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$NaBr(aq)\rightleftharpoons Na^++Br^-$

The concentration of NaBr solution is 0.1 m that means,

$[Na^+]=[Br^-]=0.1m$

The equilibrium reaction for AgBr is,

$AgBr\rightleftharpoons Ag^++Br^-$

At equilibrium s s

The expression for solubility product constant for AgBr is,

$K_{sp}=[Ag^+][Br^-]$

The concentration of $Ag^+$ = s

The concentration of $Br^-$ = 0.1 + s

Now put all the given values in $K_{sp}$ expression, we get

$5\times 10^{-13}=(s)(0.1+s)$

By rearranging the terms, we get the value of 's'

$s=5\times 10^{-12}m$

Therefore, the maximum concentration of silver ion is $5\times 10^{-12}m$ .

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anzhelika [568]

Answer:

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Explanation:

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telo118 [61]

Answer:

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