Answer:
Abundance of 32S is 94.41%
Explanation:
The average atomic mass is defined as the sum of the atomic masses of each isotope times its abundance:
Average atomic mass = ∑ Atomic mass istope*Abundance
For the sulfur:
32.07amu = 31.97207X + 32.97146Y + 33.96786*0.0422 <em>(1)</em>
<em>Where X is abundance of 32S and Y abundance of 33S</em>
Also we can write:
1 = X + Y + 0.0422 <em>(2)</em>
0.9578 - X = Y
Because the sum of the abundances = 1
Replacing (2) in (1):
32.07amu = 31.97207X + 32.97146(0.9578 - X) + 33.96786*0.0422
32.07 = 31.97207X + 31.58006 - 32.97146X + 1.43344
-0.9435 = -0.99939X
0.9441 =X
In percentage, abundance of 32S is 94.41%
B ase from the reaction <span>cacn2 3 h2o → caco3 2 nh3, for every 1 mole of caco3 produced there 2 moles of nh3 being produced. to solved this, we must first convert the caco3 to moles.
mass nh3 = 187 g caco3 (1 mol caco3 / 100 g caco3 ) ( 2 mol nh3 / 1 mol caco3) ( 17 g nh3 / 1 mol nh3)
mass nh3 = 63.58 g nh3 is produced</span>
<u>Answer:</u> The value of
for the given reaction is 1.435
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

Given mass of
= 9.2 g
Molar mass of
= 92 g/mol
Volume of solution = 0.50 L
Putting values in above equation, we get:

For the given chemical equation:

<u>Initial:</u> 0.20
<u>At eqllm:</u> 0.20-x 2x
We are given:
Equilibrium concentration of
= 0.057
Evaluating the value of 'x'

The expression of
for above equation follows:
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
![[NO_2]_{eq}=2x=(2\times 0.143)=0.286M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.143%29%3D0.286M)
![[N_2O_4]_{eq}=0.057M](https://tex.z-dn.net/?f=%5BN_2O_4%5D_%7Beq%7D%3D0.057M)
Putting values in above expression, we get:

Hence, the value of
for the given reaction is 1.435
Answer:
d = 70.5 mm
Explanation:
given,
length of pipe = 305 m
discharge rate = 150 gal/min
pipe diameter = ?
1 gal/min = 6.30902 × 10⁻⁵ m³/s
150 gal/min = 150 × 6.30902 × 10⁻⁵ m³/s
= 9.46 × 10⁻³ m³/s


Q = A V



f = 0.048 from moody chart using P/D = 0.00015

d = 70.5 mm
Diameter of the pipe is equal to 70.5 mm
Ideal solutions obey Raoult's law, which states that:
P_i = x_i*(P_pure)_i
where
P_i is the partial pressure of component i above a solution
x_i is the mole fraction of component i in the solution
(P_pure)_i is the vapor pressure of pure component i
In this case,
P_benzene = 0.59 * 745 torr = 439.6 torr
P_toluene = (1-0.59) * 290 torr = 118.9 torr
The total vapor pressure above the solution is the sum of the vapor pressures of the individual components:
P_total = (439.6 + 118.9) torr = 558.5 torr
Assuming the gas phase also behaves ideally, the partial pressure of each gas in the vapor phase is proportional to its molar concentration, so the mole fraction of toluene in the vapor phase is:
118.9 torr/558.5 torr = 0.213