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:
2%
Explanation:
.98 is 98% of one and therefore they are missing 2%
Answer:
2-4 mm height of capillary tube.
Explanation:
Sample should be around 2-4 mm in height.
It should be packed well so that it does not have air packets that caues the lowering of melting point.
If you take greater amount, then there will be needed more heat, resulting a wide range of melting point.
K:
m=155g
M=39g/mol
n = 155g / 39g/mol ≈ 3,97mol
KNO₃:
m=122g
M=101g/mol
n = 122g/101g/mol = 1,21mol
2K + 10KNO₃ ⇒ 6K₂O + N₂
2mol : 10mol
3,97mol : 1,21mol
limiting reagent
KNO₃ is limiting reagent
When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-
Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄
On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.
As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.
So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 =
=74.8 %.