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anzhelika [568]

3 years ago

10

The theoretical yield of Baso, is 58.35 g. If 44.34 g of BaSO4 are produced from the reaction shown above, what is the percent y

ield of Ba50_?
BaCl2(aq)+Na2SO4(aq)—>2NaCl(aq)+BaSO4(s)

a85.17%
b75.99%
c31.67%
d52.03%

2 answers:

8090 [49]3 years ago

8 0

Answer:75.99

Explanation:saw it on quizlet

Wewaii [24]3 years ago

5 0

Answer:

75.999%

Explanation:

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What is the net charge of a nitrogen atom with 8 electrons

DaniilM [7]

Answer:

– 1

Explanation:

From the question given above, we obtained the following:

Electron = 8

Net charge of nitrogen =.?

Nitrogen has atomic number of 7. This also means that nitrogen has 7 proton because atomic number of an element is the equal to number of protons in the atom of the element.

Thus, we can obtain the net charge of nitrogen with 8 electrons by calculating the difference between the protons and electrons of the nitrogen atom. This can be obtained as follow:

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3 years ago

Electrochemistry might be considered a subspecialty of which of the following areas of chemistry

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3 0

3 years ago

First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta

Klio2033 [76]

Rate equation for first order reaction is as follows:

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}$

Here, k is rate constant of the reaction, t is time of the reaction, $A_{0}$ is initial concentration and $A_{t}$ is concentration at time t.

The rate constant of the reaction is $0.1 day^{-1}$ .

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days$

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days$

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c) Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days$

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

5 0

3 years ago