<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C
14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O
molar mass of C=12 g/mole
molar mass of H=1 g/mole
0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C
0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H
mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O
molar mass of O=16 g/mole
9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602
C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602
C2.66H2.66O1 is the empirical formula;
to obtain whole numbers multiply by 3
3[C2.66H2.66O1] = C8H8O3
above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu
The empirical formula weight and the molecular formula weight are the same .
Molecular formula is C8H8O3.</span>
According to the balanced chemical equation:
4 HPO₃ + 12 C → 2 H₂ + 12 CO + P₄
4 moles of HPO₃ ---gives---> 12 moles of CO
2.73 moles of HPO₃ ---gives---> ? moles of CO
so number of moles of CO =
= 8.19 moles of CO
Number of molecules of CO = number of moles * Avogadro's number
= 8.19 * (6.022 * 10²³) = 4.93 * 10²⁴ molecules
Answer:
It would appear that the H2 gas is collected in a stoppered flask. Therefore, the volume of the gas is 138 mL minus the 5 mL occupied by the HCl, or 133 mL.
Explanation:
<span>The rate of infusion is 2.1L/19h or 2100mL/19h (as 1L = 100 mL).
To convert 19 hours to minutes we multiply as follows:
19 hours = (19 hours) x (60 minutes/1 hour) = 1140 minutes
So the rate of infusion becomes:
2100mL /1140 min
In order to converted mL to drops (gtt) we multiply the rate of infusion with the drop factor to get the drip rate:
(2100mL/1140min) x (20 gtt/mL) = 36.8 gtt/min</span>
Voltage<span>, </span>also called<span> electromotive force, is a quantitative expression of the potential difference in charge between two points in an electrical field.
So ACTUALLY an "electromotive force", but of your answer choices.
D. Electrical Field Energy
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