Answer:
CO = zero
CO2 =1 bar
O2 = 2.02 bar
Explanation:
We are given
initial pressure of CO = 1bar
total pressure = 3.52 bar
so initial pressure of O2 = 3.52 - 1 = 2.52 bar
the reaction is
2CO + O2 → 2CO2
using the unitary method
2 moles of CO2 → 1 mole of O2
1 bar of CO → 
(required)
but we have more oxygen present , that means CO is the limiting reagent
- Final pressure of CO will be zero as it is the limiting reagent so it will be consumed completely
- 1 bar of CO →
of CO2 - 2.52 bar O2 (initially) - 0.5 bar (reacted) = 2.02bar O2
The balance chemical equation is:
NaCH₃COO + HCl → NaCl + HCH₃COO
Make
The k is the proportionality constant of the reaction. Graphically, this is the slope of the graph. Since the graph is linear, then there is only 1 value of k. To calculate this, choose two random points in the line. Suppose we use (0.15,10) and (0.30,20), calculate for the slope.
Slope = k = (10 - 20)/(0.15 - 0.30) = 66.67 mL CO₂/g CaCO₃
Answer:
0.521 moles still present in the container.
Explanation:
It is possible to answer this question by using the general gas law, that is:
PV = nRT
<em>Where P represents pressure of the gas, v its volume, n moles, R gas constant law and T absolute temperature (21.7°C + 273.15 = 294.85K)</em>
Replacing with values of the initial conditions of the container, its volume is:
V = nRT / P
V = 2.00mol*0.082atmL/molK*294.85K / 3.75atm
V = 12.9L
When some gas is released, absolute temperature is 28.1°C + 273.15 = 301.25K, the pressure is 0.998atm and <em>the volume of the container still constant. </em>Again, using general gas law:
PV / RT = n
0.998atm*12.9L / 0.082atmL/molK*301.25K = n
0.521 moles = n
<h3>0.521 moles still present in the container.</h3>
<em />
Answer:
1.8 × 10⁻¹⁶ mol
Explanation:
(a) Calculate the solubility of the Sr₃(PO₄)₂
Let s = the solubility of Sr₃(PO₄)₂.
The equation for the equilibrium is
Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹
1.2 + 3s 2s
![K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} = 1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BSr%24%5E%7B2%2B%7D%24%5D%24%5E%7B3%7D%24%5BPO%24_%7B4%7D%5E%7B3-%7D%24%5D%24%5E%7B2%7D%24%7D%20%3D%20%281.2%20%2B%203s%29%5E%7B3%7D%5Ctimes%20%282s%29%5E%7B2%7D%20%3D%20%201.0%20%5Ctimes%2010%5E%7B-31%7D%5C%5C%5Ctext%7BAssume%20%7D%203s%20%5Cll%201.2%5C%5C1.2%5E%7B3%7D%20%5Ctimes%204s%5E%7B2%7D%20%3D%201.0%20%5Ctimes%2010%5E%7B-31%7D%5C%5C6.91s%5E%7B2%7D%20%3D%201.0%20%5Ctimes%2010%5E%7B-31%7D%5C%5Cs%5E%7B2%7D%20%3D%20%5Cdfrac%7B1.0%20%5Ctimes%2010%5E%7B-31%7D%7D%7B6.91%7D%20%3D%201.45%20%5Ctimes%2010%5E%7B-32%7D%5C%5C%5C%5Cs%20%3D%20%5Csqrt%7B%201.45%20%5Ctimes%2010%5E%7B-32%7D%7D%20%3D%201.20%20%5Ctimes%2010%5E%7B-16%7D%20%5Ctext%7B%20mol%2FL%7D%5C%5C)
(b) Concentration of PO₄³⁻
[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹
(c) Moles of PO₄³⁻
Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol
