Answer:
3rd statment
Explanation:
ray 1 and 2 are same vertical line
Based on the solubility observations, barium & aluminum could be distinguished by the addition of sodium chloride to the solutions.
<h3>What happens when NaCl is added to a solution?</h3>
- The ionic link that held sodium and chloride ions together is broken when water molecules force the ions apart.
- The sodium and chloride atoms are encircled by water molecules after the salt compounds are separated. After that, the salt dissolves and forms a homogenous solution.
- In order to keep patients from dehydrating, sodium chloride, an important nutrient, is employed in healthcare. It is employed as a spice to improve flavor and as a food preservative. Additionally, sodium chloride is employed in the production of polymers and other goods. Additionally, it is used to de-ice sidewalks and roadways.
- Adding water to sodium chloride results in a physical change because no new product is created.
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Answer:
a) T
b) T
c) F
d) F
e) T
f) T
g) T
h) F
I) F
j) F
k) F
l) F
Explanation:
The w/v concentration is obtained from, mass/volume. Hence;
%w/v= 50/1000= 5%
In the %w/w we have;
25g/100 g = 25% w/w
In combustion reaction, energy is given out hence it is exothermic.
Neutralization reaction yields a salt and water
% by mass of carbon is obtained from;
8× 12/114 × 100 = 84.1%
All the ionic substances mentioned have very low solubility in water.
One mole of a substance contains the Avogadro's number of each atom in the compound.
There are two iron atoms so one mole contains 2× 55.85 g of iron.
Some sulphates such as BaSO4 are insoluble in water.
Halides are soluble in water hence NaI is soluble in water.
The equation does not balance with the given coefficients because the number of atoms of each element on both sides differ.
The equation represents a decomposition of calcium carbonate as written.
Answer:
[HI] = 0.7126 M
Explanation:
Step 1: Data given
Kc = 54.3
Temperature = 703 K
Initial concentration of H2 and I2 = 0.453 M
Step 2: the balanced equation
H2 + I2 ⇆ 2HI
Step 3: The initial concentration
[H2] = 0.453 M
[I2] = 0.453 M
[HI] = 0 M
Step 4: The concentration at equilibrium
[H2] = 0.453 - X
[I2] = 0.453 - X
[HI] = 2X
Step 5: Calculate Kc
Kc = [Hi]² / [H2][I2]
54.3 = 4x² / (0.453 - X(0.453-X)
X = 0.3563
[H2] = 0.453 - 0.3563 = 0.0967 M
[I2] = 0.453 - 0.3563 = 0.0967 M
[HI] = 2X = 2*0.3563 = 0.7126 M