Answer:
Final concentrations:
Cu²⁺ = 0
Al³⁺ = 3.13 mmol/L = 84.51 mg/L
Cu = 4.7 mmol/L = 300 mg/L
Al = 0.57 mmol/L = 15.49 mg/L
Explanation:
2Al (s) + 3Cu²⁺ (aq) → 2Al³⁺ (aq) + 3Cu (s)
Al: 27 g/mol ∴ 100 mg = 3.7 mmol
Cu: 63.5 g/mol ∴ 300 mg = 4.7 mmol
3 mol Cu²⁺ _______ 2 mol Al
4.7 mmol Cu²⁺ _____ x
x = 3.13 mmol Al
4.7 mmol of Cu²⁺ will be consumed.
3.13 mmol of Al will be consumed.
4.7 mmol of Cu will be produced.
3.13 mmol of Al³⁺ will be produced.
0.57 mmol of Al will remain.
Two or more different elements
H2o means water and co3 is co-signs
Answer:
The correct answer is 8.79 × 10⁻² M.
Explanation:
Based on the given information, the mass of NaI given is 4.11 grams. The molecular mass of NaI is 149.89 gram per mole. The moles of NaI can be determined by using the formula,
No. of moles of NaI = Weight of NaI/ Molecular mass
= 4.11 / 149.89
= 0.027420
The vol. of the solution given is 312 ml or 0.312 L
The molarity can be determined by using the formula,
Molarity = No. of moles/ Volume of the solution in L
= 0.027420/0.312
= 0.0879 M or 8.79 × 10⁻² M
Answer : The new volume of the air is, 6.83 L
Explanation :
Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
![\frac{V_1}{T_1}=\frac{V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BV_1%7D%7BT_1%7D%3D%5Cfrac%7BV_2%7D%7BT_2%7D)
where,
are the initial volume and temperature of the gas.
are the final volume and temperature of the gas.
We are given:
![V_1=5.00L\\T_1=0^oC=(0+273)K=273K\\V_2=?\\T_2=100^oC=(100+273)K=373K](https://tex.z-dn.net/?f=V_1%3D5.00L%5C%5CT_1%3D0%5EoC%3D%280%2B273%29K%3D273K%5C%5CV_2%3D%3F%5C%5CT_2%3D100%5EoC%3D%28100%2B273%29K%3D373K)
Putting values in above equation, we get:
![\frac{5.00L}{273K}=\frac{V_2}{373K}\\\\V_2=6.83L](https://tex.z-dn.net/?f=%5Cfrac%7B5.00L%7D%7B273K%7D%3D%5Cfrac%7BV_2%7D%7B373K%7D%5C%5C%5C%5CV_2%3D6.83L)
Therefore, the new volume of the air is, 6.83 L