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Nady [450]
3 years ago
15

Which mixture is classified as a solution?

Chemistry
1 answer:
svlad2 [7]3 years ago
5 0
Salt water is considered to be a solution
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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
HC2H3O2 (aq) + H2O (l) ⇔ C2H3O2- (aq) + H3O+ (aq) Ka = 1.8 x 10-5
marin [14]

The concentration of [H3O⁺]=2.86 x 10⁻⁶ M

<h3>Further explanation</h3>

In general, the weak acid ionization reaction  

HA (aq) ---> H⁺ (aq) + A⁻ (aq)  

Ka's value  

\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}

Reaction

HC₂H₃O₂ (aq) + H₂O (l) ⇔  (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵

\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}

[H₃O⁺]=2.86 x 10⁻⁶ M

5 0
3 years ago
Determine the specific heat ofmaterial if a 12g sample absorbed 48j as it was heated from 20-40
devlian [24]

Answer:

c =0.2 J/g.°C

Explanation:

Given data:

Specific heat of material = ?

Mass of sample = 12 g

Heat absorbed = 48 J

Initial temperature = 20°C

Final temperature = 40°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =  40°C -20°C

ΔT =  20°C

48 J = 12 g×c×20°C

48 J =240 g.°C×c

c = 48 J/240 g.°C

c =0.2 J/g.°C

6 0
2 years ago
The density of a material is a/an ___________.
Nitella [24]

The density of a material is an intensive property.

<h3>What is intensive property?</h3>

An intensive property of matter is one that does not change with the amount of matter. It is a bulk property, which means that it is a physical property that is independent of sample size or mass. An extensive property, on the other hand, is one that is affected by sample size.

<h3>What factors influence an intensive property?</h3>

Intensive properties are those that are determined solely by the characteristics of the material and not by its quantity - for example, density, temperature, refractive index, color, and pressure. Intensive properties are not additive, which means their value does not change when the amount of material is changed.

Learn more about the intensive property here:-

brainly.com/question/24909279

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5 0
1 year ago
Please help me with these two questions
ANEK [815]
What is the question?
8 0
2 years ago
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