The molar mass if carbon is 12g/mol
PH scale is from 1 to 14 and indicates how acidic or basic a solution is. To find pH or pOH we need to know the H⁺ ion concentration or OH⁻ concentration.
pH can be calculated using the following equation;
pH = -log[H⁺]
the H⁺ concentration of the given acid is 1.0 x 10⁻⁴ M. substituting this we can find the pH
pH = -log[1x10⁻⁴]
pH = 4
answer is 1) 4
Answer:
b) The dehydrated sample absorbed moisture after heating
Explanation:
a) Strong initial heating caused some of the hydrate sample to splatter out.
This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).
b) The dehydrated sample absorbed moisture after heating.
Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.
c) The amount of the hydrate sample used was too small.
It will create some errors but they do not create a difference of 13% difference as stated in the problem.
d) The crucible was not heated to constant mass before use.
Here the error is small.
e) Excess heating caused the dehydrated sample to decompose.
Usually the inorganic compounds are stable in the temperature range of this kind of experiments. If you have an organic compound which retain water molecules you may decompose the sample forming volatile compounds which will leave crucible so the error will be quite high.
Answer:
ΔH°C2H2Cl4(l) = -333,36 kJ/mol
ΔH°r₂ = -35,14 kJ/mol
Explanation:
The ΔH°r of the first reaction is:
ΔH°r = -385,76 kJ/mol = (ΔH°C2H2Cl4(l) + ΔH°H2(g)) - (ΔH°C2H4(g) + 2ΔHCl2 (g))
ΔH°H2(g) = 0 kJ/mol
ΔH°C2H4(g) = 52,4 kJ/mol
Δ°HCl2 (g) = 0 kJ/mol
Replacing:
ΔH°C2H2Cl4(l) = -385,76 kJ/mol + 52,4 kJ/mol = <em>-333,36 kJ/mol</em>
The standard heat of the second reaction is:
ΔH°r₂ = ΔH°C2HCl3(l) + ΔH°HCl(g) - ΔH°C2H2Cl4(l)
Where:
ΔH°C2HCl3(l) = -276,2 kJ/mol; ΔH°HCl(g) = -92,3 kJ/mol; ΔH°C2H2Cl4(l) = -333,36 kJ/mol
Replacing:
ΔH°r₂ = -276,2 kJ/mol -92,3 kJ/mol + 333,36 kJ/mol
<em>ΔH°r₂ = -35,14 kJ/mol</em>
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I hope it helps!
1. The formula of the hydrate is FePO₄•4H₂O
2. The name of the hydrate is Iron(iii) phosphate tetrahydrate.
1. Determination of the formula of the hydrate
- H₂O = 32.3%
- FePO₄ = 67.7%
- Formula =?
The formula of the hydrate can be obtained as follow:
Divide by their molar mass
H₂O = 32.3 / 18 = 1.794
FePO₄ = 67.7 / 151 = 0.448
Divide by the smallest
H₂O = 1.794 / 0.448 = 4
FePO₄ = 0.448 / 0.448 = 1
Therefore, the formula of the hydrate is FePO₄•4H₂O
The name of the hydrate is Iron(iii) phosphate tetrahydrate
Learn more about empirical formula: brainly.com/question/16143068