Answer:
D. 1.48atm
Explanation:
Van der waals equation is given as:
(P +an²/v²) (v - nb) = nRT
Where;
P = pressure (atm)
V = volume (L)
R = gas constant (0.0821 Latm/molK)
a and b = gas constant specific to each gas
T = temperature (K)
n = number of moles
According to the given information; V = 22.4L, T = 0.00°C (273.15K), R = 0.0821 Latm/molK, a = 6.49L^2-atm/mol^2, b = 0.0562 L/mol, n = 1.5mol
Hence;
(P + 6.49 × 1.5²/22.4²) (22.4 - 1.5×0.0562) = 1.5 × 0.0821 × 273.15
(P + 6.49 × 2.25/501.76) (22.4 - 0.0843) = 33.638
(P + 0.0291) (22.316) = 33.638
22.316P + 0.649 = 33.638
22.316P = 33.638 - 0.649
22.316P = 32.989
P = 32.989/22.316
P = 1.478
P = 1.48atm
Mass = no. of moles x molecular weight
m = n x Mr
m = 2.5 mol x (24 + [16 x 2])
m = 140g
OP already did it - CONGRATS!!
here are the steps 2 get the same ans:
(NH4)2 CO3 has 2x N, 8x H, 1x C and 3x O per molecule
so its molecular mass = 2x14 + 8x1 + 1x12 + 3x16
=28+8+12+48
=96g
of that 96g, 8x1=8g is due to Hydrogen
so by ratio n proportion, 1.00g will have 1x8/96 = 1/12g = 0.083g of H
One meter in front of the source at this location will a sensor detect the highest concentration of mercaptan.
Answer:
The correct answer is 4.58 grams.
Explanation:
Based on the Faraday's law of electrolysis, at the time of electrolysis, the amount of deposited substance is directly equivalent to the concentration of the flow of charge all through the solution. If current, I, is passed for time, t, seconds and w is the concentration of the substance deposited, then w is directly proportional to I*t or w = zIt (Here z refers to the electrochemical equivalent or the amount deposited when 1 C is passed).
For the reaction, n * 96500 C = molar mass
1C = molar mass/n*96500 = Equivalent wt / 96500
w = Equivalent wt / 96500 * I * t
In the given reaction,
Pb + PbO2 + 2HSO4- + 2H+ → 2PbSO4 + 2H2O, n = 2, the current or I drawn is 350 A, for time, t 12.2 seconds.
Now putting the values in the equation we get,
w = 207.19 / 2 * 96500 * 350 * 12.2 ( The molecular weight of Pb is 207.19 and the equivalent weight of Pb is 207.19 / 2)
w = 4.58 gm.
