Answer:
d. Copper (II) sulfate
Explanation:
Given data:
Mass of Al = 1.25 g
Mass of CuSO₄ = 3.28 g
What is limiting reactant = ?
Solution:
Chemical equation:
2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 1.25 g/ 27 g/mol
Number of moles = 0.05 mol
Number of moles of CuSO₄:
Number of moles = mass/molar mass
Number of moles = 3.28 g/ 159.6 g/mol
Number of moles = 0.02 mol
now we will compare the moles of reactant with product.
Al : Al₂ (SO₄)₃
2 : 1
0.05 : 1/2×0.05=0.025 mol
Al : Cu
2 : 3
0.05 : 3/2×0.05 = 0.075 mol
CuSO₄ : Al₂ (SO₄)₃
3 : 1
0.02 : 1/3×0.02=0.007 mol
CuSO₄ : Cu
3 : 3
0.02 : 0.02
Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.
Answer:
2 H2(g) + O2(g) → 2 H2O(ℓ) ΔH = −570 kJ
Explanation:
Accuracy is the closeness of a measurement to its true value. Precision is the closeness of more than one measurements to each other. The measurements of Lamant is both accurate and precise while the scale of Colin is the opposite. Kirvin is precise but he is not accurate.
Answer:
Because Iodine is a non-polar but water is polar therefore it can not have a hydrogen bond or any permanent dipole-dipole interactions
If it’s (5.93 x 10^3) x (2.3 x 10^-2)
The answe is:
= 13.63 x 10^1
Round to the nearest two significant figures.
= 14 x 10^1
Making sure it’s in proper scientific notation.
[= 1.4 x 10^2]
