The average atomic weight is, from the name itself, the average weight of all its naturally occurring isotopes. All you have to do is multiple the abundance of each isotope with its individual mass, then add them altogether.
Mass = (0.10*55)+(0.15*56)+(.75*57)
<em>Mass = 56.65 amu</em>
Answer: 72.93 litres
Explanation:
Given that:
Volume of gas (V) = ?
Temperature (T) = 24.0°C
Convert 24.0°C to Kelvin by adding 273
(24.0°C + 273 = 297K)
Pressure (P) = 1.003 atm
Number of moles (n) = 3 moles
Molar gas constant (R) is a constant with a value of 0.0821 atm L K-1 mol-1
Then, apply ideal gas equation
pV = nRT
1.003 atm x V = 3.00 moles x 0.0821 atm L K-1 mol-1 x 297K
1.003 atm•V = 73.15 atm•L
Divide both sides by 1.003 atm
1.003 atm•V/1.003 atm = 73.15 atm•L/1.003 atm
V = 72.93 L
Thus, the volume of the gas is 72.93 litres
Answer:
1.327 g Ag₂CrO₄
Explanation:
The reaction that takes place is:
- 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)
First we need to <em>identify the limiting reactant</em>:
We have:
- 0.20 M * 50.0 mL = 10 mmol of AgNO₃
- 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄
If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.
Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:
- 4 mmol K₂CrO₄ * = 1326.92 mg Ag₂CrO₄
- 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄