Answer:
- According to the law <br> Mass of reactants = mass of product, here <br> `underset(10 g)(CaCO_(3))rarr underset(4.4 g)(CO_(2))+underset(x)(CaO)` <br> Hence, x = 10 g - 4.4 g = 5.6 g <br> Which is mass of CaO.d
- In the first compound <br> Hydrogen = 5.93 % <br> Oxygen = `(100-5.93)% = 94.07 %` <br> In the second compound <br> Hydrogen = 11.2 % <br> Oxygen `= (100-11.2)%=88.8%` <br> In the first compound the number of parts by mass of oxygen that combine with one part by mass of hydrogen `=(94.07)/(5.93)=15.86` parts ...
- (The ratio of Cu combining with fixed weight of oxygen in black and red oxide is 1 : 2 respectively. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.) {Check something more in the above attachment!}
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Explanation:
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The correct answer is approximately 1.95g. First, take (1.022*10^10)/(1.277*10^9). This is the same as time elapsed/half-life. This gives you 8, which is the number of half lives. Then multiply mass (500.3g) by (1/2)^8, resulting in 1.95g, which is the answer.
Its all of the above.
- I took the test and mine is correct.
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
