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arsen [322]
3 years ago
14

Chlorine has two isotopes, Cl-35 and Cl-37 with the average atomic mass 35.5. what is the percentage composition of Cl-35 and Cl

-37 respectively?
A 25% and 75%
B 75% and 25%
C 65% and 35%
D 35% and 65%​
Chemistry
1 answer:
Novosadov [1.4K]3 years ago
4 0

Answer:

75% of Cl-35 and 25% of Cl-37

Explanation:

Equation to find the mass number of an element. repeat the brackets for however many isotopes it has.

\frac{(abundance \:  \times  \:  mr) \: +  \: (abundance \:  \times  \:  mr)}{100}

\frac{(75 \times 35) + (25 \times 37)}{100}  = 35.5

You can try guess it because 35.5 is closer to 35 than 37 so it has a higher abundance of Cl-35

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Adrenal medulla include various hormones as secreted hormones from there.  Fight or flight is also initiate by the hormones secreted from adrenal gland that are epinephrine, nor epinephrine.

Thus, the correct answer is - adrenal gland.

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In a generic chemical reaction involving reactants A and B and products C and D, aA + bB → cC + dD, the standard enthalpy ΔH∘rxn
velikii [3]

Answer:

A. ΔH∘rxn =  21.9 KJ/mol

B. ΔH∘rxn = 103 KJ/mol

C. C2H5OH + 3O2 → 2CO2 + 3H2O

Explanation:

A.

The standard reaction equation is given as:

aA + bB → cC + dD

Its standard enthalpy is given as:

ΔH∘rxn = cΔH∘f(C) + dΔH∘f(D) − aΔH∘f(A) − bΔH∘f(B)

Reaction given to us is:

H2O(l) + CCl4(l) → COCl2(g) + 2HCl(g)

So, its standard enthalpy will be:

ΔH∘rxn = (1)ΔH∘f(CoCl2 (g)) + (2)ΔH∘f(HCl(g)) − (1)ΔH∘f(H2O(l)) − (1)ΔH∘f(CCl4(l))

using the values from table:

ΔH∘rxn = - 218.8 KJ/mol + (2)(- 92.3 KJ/mol) - (- 285.8 KJ/mol) - (- 139.5 KJ/mol)

<u>ΔH∘rxn =  21.9 KJ/mol</u>

<u></u>

B.

Reaction given to us is:

2A + B ⇌ 2C + 2D

So, its standard enthalpy will be:

ΔH∘rxn = (2)ΔH∘f(C) + (2)ΔH∘f(D) − (2)ΔH∘f(A) − (1)ΔH∘f(B)

using the values from table:

ΔH∘rxn = (2)181 KJ/mol + (2)(- 523 KJ/mol) - (2)(- 225 KJ/mol) - (- 337 KJ/mol)

<u>ΔH∘rxn = 103 KJ/mol</u>

<u></u>

C.

Balanced equation for combustion of ethanol is:

<u>C2H5OH + 3O2 → 2CO2 + 3H2O</u>

3 0
3 years ago
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