Answer:
0.500 moles of CO2 has a volume of 11.2 L at STP (option B)
Explanation:
Step 1: Data given
Volume of a gas at STP = 11.2 L
STP: Pressure = 1 atm and temperature = 273 K
Step 2: Calculate volume
p*V= n*R*T
V = (n*R*T)/p
⇒with V = the volume of the gas = TO BE DETERMINED
⇒with n = the number of moles of the gas
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 273 K
⇒with p = the pressure of the gas = 1 atm
A
) 0.250 mole of NH3
V = (0.250 * 0.08206 * 273) / 1
V = 5.6 L
B
) 0.500 mole of CO2
V = (0.500 * 0.08206 * 273) / 1
V = 11.2 L
C
) 0.750 mole of NH3
V = (0.750 * 0.08206 * 273) / 1
V = 16.8 L
D) 1.00 mole of CO2
V = (1.00 * 0.08206* 273) / 1
V = 22.4 L
0.500 moles of CO2 has a volume of 11.2 L at STP (option B)
Answer:
The values of n, p and q are 314, 0.12 and 0.88 respectively.
Explanation:
n is the number of subjects given a placebo. The value of n is 314.
p is the sample proportion of the subjects given a placebo that developed headaches. The value of is 12% = 0.12
q is the sample proportion of the subjects given a placebo that did not develop headaches. The value of q = 1 - p = 1 - 0.12 = 0.88
Answer:
a) kc = 0,25
b) [A] = 0,41 M
c) [A] = <em>0,8 M</em>
[B] =<em>0,2 M</em>
[C] = <em>0,2M</em>
Explanation:
The equilibrium-constant expression is defined as the ratio of the concentration of products over concentration of reactants. Each concentration is raised to the power of their coefficient.
Also, pure solid and liquids are not included in the equilibrium-constant expression because they don't affect the concentration of chemicals in the equilibrium.
If global reaction is:
A(g) + B(g) ⇋ 2 C(g) + D(s)
The kc = ![\frac{[C]^2}{[A][B]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BC%5D%5E2%7D%7B%5BA%5D%5BB%5D%7D)
a) The concentrations of each compound are:
[A] =
= <em>0,4 M</em>
[B] =
= <em>0,1 M</em>
[C] =
= <em>0,1 M</em>
<em>kc = </em>
= 0,25
b) The addition of B and D in the same amount will, in equilibrium, produce these changes:
[A] =
[B] =
[C] = 
0,25 = ![\frac{[0,60+2x]^2}{[1,60-x][0,60-x]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B0%2C60%2B2x%5D%5E2%7D%7B%5B1%2C60-x%5D%5B0%2C60-x%5D%7D)
You will obtain
3,75x² +2,95x +0,12 = 0
Solving
x =-0,74363479081119 → No physical sense
x =-0,043031875855476
Thus, concentration of A is:
= <em>0,41 M</em>
c) When volume is suddenly halved concentrations will be the concentrations in equilibrium over 2L:
[A] =
= <em>0,8 M</em>
[B] =
= <em>0,2 M</em>
[C] =
= <em>0,2M</em>
I hope it helps!
Answer:
C₂ = 0.056 ppm
Explanation:
Given data:
Initial volume = 2.0 mL
Initial concentration = 7.0 ppm
Final volume = 250.0 mL
Final concentration = ?
Solution:
Formula:
C₁V₁ = C₂V₂
C₁ = Initial concentration
V₁ = Initial volume
C₂ = Final concentration
V₂ = Final volume
Now we will put the values in formula.
C₁V₁ = C₂V₂
7.0 ppm × 2.0 mL = C₂ × 250.0 mL
C₂ = 14.0 ppm.mL /250.0 mL
C₂ = 0.056 ppm