Answer: The average atomic mass of the element X is 51.99592 amu
Explanation:
Mass of isotope 1 = 49.94605 amu
% abundance of isotope 1 = 4.350% = 
Mass of isotope 2 = 51.94051 amu.
% abundance of isotope 2 = 83.79% = 
Mass of isotope 3 = 52.94065 amu.
% abundance of isotope 2 = 9.500% = 
Mass of isotope 4 = 53.93888 amu.
% abundance of isotope 2 = 2.360% = 
Formula used for average atomic mass of an element :
Therefore, the average atomic mass of the element X is 51.99592 amu
Answer:
The number of solute particles increases, and the boiling point increases.
Explanation:
- It is known from colligative properties that adding solute to the solvent will cause elevation of boiling point.
- Elevation of boiling point (ΔTb) can be expressed as:
<em>ΔTb = Kb.m,</em>
where, Kb molal boiling point elevation constant.
m is the molal concentration of solute.
- Adding more sodium chloride to the solution:
will increase the number of solute particles and also will increase the molal concentration of NaCl solute.
<em>∵ ΔTb ∝ m.</em>
- So, the boiling point increases.
- Thus, the right choice is:
<em>The number of solute particles increases,</em>
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To be able to compare the result with other experiments it has to be reported in moles.
number of moles = mass / molecular weight
number of moles of Mg(H₂PO₄)₂ = 600 / 218 = 2.75 moles
1 atm = 760mmHg
754.3 mmHg / 760 mmHg * 1atm = 0.99 atm
760 mmHg = 101.3 KPa
754.3 mmHg/ 760mmHg *101.3 KPa = 100.54 KPa
Hope this helps!
Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
