Answer:
15.3 %
Explanation:
Step 1: Given data
- Mass of the sample (ms): 230 g
- Mass of carbon (mC); 136.6 g
- Mass of hydrogen (mH): 26.4 g
- Mass of nitrogen (mN): 31.8 g
Step 2: Calculate the mass of oxygen (mO)
The mass of the sample is equal to the sum of the masses of all the elements.
ms = mC + mH + mN + mO
mO = ms - mC - mH - mN
mO = 230 g - 136.6 g - 26.4 g - 31.8 g
mO = 35.2 g
Step 3: Calculate the mass percent of oxygen
%O = (mO / ms) × 100% = (35.2 g / 230 g) × 100% = 15.3 %
Yes yes you are totally right
Answer:
378mL
Explanation:
The following data were obtained from the question:
Pressure (P) = 99.19 kPa
Temperature (T) = 28°C
Number of mole (n) = 0.015 mole
Volume (V) =...?
Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:
For Pressure:
101.325 KPa = 1 atm
Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm
For Temperature:
T(K) = T(°C) + 273
T(°C) = 28°C
T(K) = 28°C + 273 = 301K.
Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:
PV = nRT
Pressure (P) = 0.98 atm
Temperature (T) = 301K
Number of mole (n) = 0.015 mole
Gas constant (R) = 0.0821atm.L/Kmol.
Volume (V) =...?
0.98 x V = 0.015 x 0.0821 x 301
Divide both side by 0.98
V = (0.015 x 0.0821 x 301) /0.98
V = 0.378 L
Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:
1L = 1000mL
Therefore, 0.378L = 0.378 x 1000 = 378mL
Therefore, the volume of N2 collected is 378mL
I think it's Chlorine but, not 100% sure. so C.
