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3 years ago

1. How many milliliters of 10.0 M HNO 3 are needed to prepare 0.350 L of 0.400 M solution?

Chemistry

1 answer:

tatyana61 [14]3 years ago

5 0

<h3>Answer:</h3>

14 milliliters

<h3>Explanation:</h3>

We are given;

10.0 M HNO₃

Prepared solution;

Volume of solution as 0.350 L
Molarity as 0.40 M

We are required to determine the initial volume of HNO₃

We are going to use the dilution formula;
The dilution formula is;

M₁V₁ = M₂V₂

Rearranging the formula;

V₁ = M₂V₂ ÷ M₁

=(0.40 M × 0.350 L) ÷ 10.0 M

= 0.014 L

But, 1 L = 1000 mL

Therefore,

Volume = 14 mL

Thus, the volume of 10.0 M HNO₃ is 14 mL

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In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol $t_{1/2}$ .

(a) For first order reaction, rate constant and half life time are related to each other as follows:

$k=\frac{0.6932}{t_{1/2}}=\frac{0.6932}{10.5 min}=0.066 min^{-1}$

Thus, rate constant of the reaction is $0.066 min^{-1}$ .

(b) Rate equation for first order reaction is as follows:

$k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}$

now, 75% of the compound is decomposed, if initial concentration $[A_{0} ]$ is 100 then concentration at time t $[A_{t} ]$ will be 100-75=25.

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$0.066 min^{-1}=\frac{2.303}{t}log\frac{100}{25}=\frac{2.303}{t}(0.6020)$

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$t=\frac{2.303\times 0.6020}{0.066 min^{-1}}=21 min$

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4 years ago

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3 years ago

A mixture containing 20 mole % butane, 35 mole % pentane and rest

notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

$Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB$

Whereas $y$ accounts for the fractions at the outlet distillate and $x$ for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

$0.9=\frac{y_bD}{z_bF}$

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

$y_bD=0.9*z_bF=0.9*0.2*100mol=18mol$

The total distillate flow:

$y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol$

And the total bottoms flow:

$F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol$

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

$x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025$

Then, the molar fraction of pentane and hexane:

$x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422$

$x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553$

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

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