Answer:
C3 H6 Cl 3
Explanation:
C -24.2%
H - 4.0%
Cl - (100-24.2 - 4.0)=73.8 %
We can take 100g of the substance, then we have
C -24.2 g
H - 4.0 g
Cl - 73.8 g
Find the moles of these elements
C -24.2 g/12.0 g/mol =2.0 mol
H - 4.0 g/1.0 g/mol = 4. 0 mol
Cl - 73.8 g/ 35.5 g/mol = 2.1 mol
Ratio of these elements gives simplest formula of the substance
C : H : Cl = 2 : 4 : 2 = 1 : 2 : 1
CH2Cl
Molar mass (CH2Cl) = 1*12.0 +2*1.0 + 1*35.5 = 49.5 g/mol
Real molar mass = 150 g/mol
real molar mass/ Molar mass (CH2Cl) = 150 /49.5=3
So, Real formula should be C3 H6 Cl 3.
Answer:
The empirical formula of the hydrocarbon is CH.
Explanation:
The following data were obtained from the question:
Mass of hydrocarbon = 2.9 mg
Mass of CO2 = 9.803 mg
Mass of H2O = 2.006 mg
Next, we shall determine the mass of carbon (C) and hydrogen (H) in the compound since hydrocarbon contains carbon and hydrogen only.
This is illustrated below:
Molar mass of CO2 = 12 + (16x2) = 12 + 32 = 44 g/mol
Mass of CO2 = 9.803 mg
Mass of C in the compound =?
Mass of C in the compound =
12/44 x 9.803
= 2.674 mg
Molar mass of H2O = (2x1) + 16 = 2 + 16 = 18 g/mol
Mass of H2O = 2.006 mg
Mass of H in the compound =
2/18 x 2.006
= 0.223 mg
Finally, we shall determine the empirical formula of the hydrocarbon as follow:
Carbon (C) = 2.674 mg
Hydrogen (H) = 0.223 mg
Divide by their molar mass
C = 2.674 /12 = 0.223
H = 0.223 / 1 = 0.223
Divide both side by the the smallest
C = 0.223/0.223 = 1
H = 0.223/0.223 = 1
Therefore, the empirical formula of the hydrocarbon is CH.
Answer:
The concentration of the chemist's silver(I) nitrate solution is 0.897 mmol/L
Explanation:
<u>Step 1: </u>Data given
Number of moles AgNO3 = 269 micromol = 269 * 10^-6 mol
Volume of AgNO3 = 300 mL = 0.3 L
Molar mass of AgNO3 = 169.87 g/mol
<u>Step 2: </u>Calculate molarity of AgNO3
Molarity = number of moles per volume (in Liters)
Molarity AgNO3 = 269 *10^-6 mol / 0.3 L
Molarity AgNO3 = 8.97 * 10^-4 M
8.97 *10^-4 mol/L = 897 micromol/L = 0.897 mmol/L
The concentration of the chemist's silver(I) nitrate solution is 0.897 mmol/L
Answer:
A. 0.01 M
Explanation:
Molar mass (KCl) = 39.1 + 35.5 = 74.6 g/mol
0.37 g KCl * 1 mol/74.6 g = 0.004960 mol
500 mL = 0.5 L
Molarity = mol solute/L solution = 0.004960 mol/0.5 L = 0.01 M
Answer:
the temprtaur of the gassssss
Explanation: