Answer:
The answer to your question is P = 1.64 atm
Explanation:
Data
Volume = 2.5 x 10⁷ L
Temperature = 22°C
Pressure = ?
Moles = 1.7 x 10⁶
R = 0.082 atm L/ mol°K
Process
1.- Convert temperature to °K
Temperature = 22 + 273
= 295°K
2.- Use the Ideal gas law to solve this problem
PV = nRT
- Solve for P
P = nRT / V
- Substitution
P = (1.7 x 10⁶)(0.082)(295) / 2.5 x 10⁷
- Simplification
P = 41123000 / 2.5 x 10⁷
- Result
P = 1.64 atm
410g Ag
2.3*10^24 atoms
1 molcule Ag- 6.02g*10^3
The balanced chemical equation would be as follows:
<span>K2PtCl4(aq) + 2NH3(aq) --> Pt(NH3)2Cl2(s) + 2KCl(aq)
We are given the amount of </span>K2PtCl4 to be used in the reaction. This will be the starting point for our calculations. We do as follows:
65 g K2PtCl4 ( 1 mol / 415.09 g ) ( 1 mol Pt(NH3)2Cl2 / 1 mol K2PtCl ) ( 300.051 g / 1 mol ) = 46.99 g Pt(NH3)2Cl produced
Answer:
Solution's mass = 200.055 g
[PbSO₄] = 275 ppm
Explanation:
Solute mass = 0.055 g of lead(II) sulfate
Solvent mass = 200 g of water
Solution mass = Solvent mass + Solution mass
0.055 g + 200 g = 200.055 g
ppm = μg of solute / g of solution
We convert the mass of solute from g to μg
0.055 g . 1×10⁶ μg/ 1g = 5.5×10⁴μg
5.5×10⁴μg / 200.055 g = 275 ppm
ppm can also be determined as mg of solute / kg of solution
It is important that the relation is 1×10⁻⁶
Let's verify: 0.055 g = 55 mg
200.055 g = 0.200055 kg
55 mg / 0.200055 kg = 275 ppm
