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3 years ago

7. A 26.4-ml sample of ethylene gas, C2H4, has a pres-sure of 2.50 atm at 2.5°C. If the

Chemistry

1 answer:

never [62]3 years ago

6 0

Answer: 1.87 atm

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

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

where,

$P_1$ = initial pressure of gas = 2.50 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 26.4 ml

$V_2$ = final volume of gas = 36.2 ml

$T_1$ = initial temperature of gas = $2.5^oC=273+2.5=275.5K$

$T_2$ = final temperature of gas = $10^oC=273+10=283K$

Now put all the given values in the above equation, we get:

$\frac{2.50\times 26.4}{275.5}=\frac{P_2\times 36.2}{283}$

$P_2=1.87atm$

The new pressure is 1.87 atm by using combined gas law.

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iris [78.8K]

The answer is D, wavelength

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3 years ago

Cobalt-60 is a strong gamma emitter that has a half- life of 5.26 yr. The co balt-60 in a radiotherapy unit must be replaced whe

Alenkasestr [34]

<u>Answer:</u> The sample of Cobalt-60 isotope must be replaced in January 2027

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5.26 years

Putting values in above equation, we get:

$k=\frac{0.693}{5.26yrs}=0.132yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.132yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 75) = 25 grams

Putting values in above equation, we get:

$0.132=\frac{2.303}{t}\log\frac{100}{25}\\\\t=10.5yrs$

The original sample was purchased in June 2016

As, June is the 6th month of the year, which means the time period will be $2016+\frac{6}{12}=2016.5$

Adding the time in the original time period, we get:

$2016.5+10.5=2027$

Hence, the sample of Cobalt-60 isotope must be replaced in January 2027

3 0

3 years ago

Which of the following best explains why bonds form between atoms?

ANEK [815]

Answer:

Explanation:

I believe the answer is D because atoms are always seeking to fill up their outer electron shell/valence shell and want to gain a full octet.

6 0

3 years ago

Why is steam distillation prefered over fractional distillation for the extraction of essencial oils?

frez [133]

Answer:

Because , the advantage of steam distillation over simple distillation is that the lower boiling point reduces decomposition of temperature-sensitive compounds. Steam distillation is useful for the purification of organic compounds, although vacuum distillation is more common

Explanation:

7 0

3 years ago

Use the atomic mass of indium to calculate the relative abundance of indium-113.

ASHA 777 [7]

The relative abundance of indium-113 is 4%.

The isotopes are species of the same element having the same atomic number but a different mass number.

The elements occurring in nature exist as multiple isotopes.

When we take into account the existence of these isotopes and their relative abundance (percent), the average atomic mass of that element can be computed, which is given by the following formula,

Average atomic Mass= (%age of isotope 1) x (Mass of isotope 1) + (%age of isotope 2) x (Mass of isotope 2)/100

Indium exists in the form of Indium-113 and Indium-115.

The mass of Indium-113 is 112.90 u.

The mass of Indium-115 is 114.90 u.

The average atomic mass of Indium is 114.82 u.

Let the %age of isotope 1(Indium-113) be X.

Then, the %age of isotope 2(Indium-115) would be 100-X.

Applying the values in the formula,

Average atomic mass = 112.90X+114.90(100-X)

114.82 = 112.90X+114.90(100-X)

On solving the above equation, the value of X comes out to be 4%.

Thus, the relative abundance/%age abundance of Indium-113 is 4%.

To know more about "Average Atomic Mass", refer to the following link:

brainly.com/question/13753702?referrer=searchResults

#SPJ4

7 0

1 year ago