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.
The correct answer is a. This is because the pH of a solution is defined as -log10(concentration of H+ ions). An inverse logarithmic scale such as this means that a solution with a lower concentration of H+ ions will have a higher pH than one with a higher concentration. Therefore we know that the pH of the second sample will be higher than the first.
Since the logarithmic scale has the base 10, a change by 1 on the scale is a consequence of multiplication/division of the H+ concentration by a factor of 10. As the scale is inverse, this means that a decrease of concentration by factor 1000 is equivalent to increasing the pH by (1000/10) = 3.
Delta E = Ef - Ei
E = energy , h = plank constant , v = frequency
h= 6.626 * 10 ^-34 j*s , T = 10 ^ 12 , v = 74 * 10 ^12 Hz , Hz = s^-1
E = ( 6.626 * 10^ -34 j*s) ( 74 * 10 ^ 12 s^ -1 ) = 4.90 * 10 ^ -20 J
Delta E = Ef - Ei
-4.90 * 10 ^ -20 J = -2.18 * 10 ^ -18J ( 1/4 ^2 - 1/x ^2)
0.0225 = 0.0625 - ( 1/x ^ 2)
0.225 - 0.0625 = - 1/ x ^ 2
- 0.0400 = - 1/x ^2 = -1 / - 0.0400 = x^2
25 = x^2
x = 5
Out of the five major mass extinction, one is Cretaceous-Tertiary extinction (K-T extinction). It happened 66 million years ago marked by the end of the Cretinous period.
Answer:
B?
Explanation:
In the example, the amount of hydrogen is 202,650 x 0.025 / 293.15 x 8.314472 = 2.078 moles. Use the mass of the hydrogen gas to calculate the gas moles directly; divide the hydrogen weight by its molar mass of 2 g/mole. For example, 250 grams (g) of the hydrogen gas corresponds to 250 g / 2 g/mole = 125 moles.
