You will have divided 50200 by 1000000 which 0.0502
Answer:
Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M
Explanation:
4A + 3B ------> C + 2D
In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s
The amount of A that has reacted at the end of 3 seconds will be
0.08 × 3 = 0.24 M
Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.
From the chemical reaction,
4 moles of A gives 1 mole of C
0.24 M of reacted A will form (0.24 × 1)/4 M of C
Amount of C formed at the end of the 3s interval = 0.06 M
If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.
If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M
Answer:
a) 12/323
b) 8/233
Explanation:
a) The probability of a red ball being drawn is 12/38, or in a simplified fraction, 6/19. To find the probability that 3 are red you would multiply the probability of the fraction for each, except subtracting one from the total each time as the drawn is done without replacement. This is done as follows: 6/19 × 6/18 × 6/17= 12/323
b) The probability of drawing a blue ball is 8/38, or 4/19. To find that the first one is blue and the rest are red, the equation is done as follows: 4/19 × 6/18 × 6/17 = 8/233
(hopefully I did this right)
First, let's compute the number of moles in the system assuming ideal gas behavior.
PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
Solving for n,
n = 2.176 moles
At standard conditions, the standard molar volume is 22.4 L/mol. Thus,
Standard volume = 22.4 L/mol * 2.176 mol =<em> 48.74 L</em>
