1. you need a periodic table and find the atomic mass of Cu (copper), S (sulfur) and O (oxygen). The atomic mass is the number in the box that corresponds to the element and have several decimal places.
2. atomic mass of
Cu = 63.546
S = 32.065
O = 15.9994
3. Then according to the formula of the compound, you add as many time the atomic mass of each element as subindex in the formula and add all the values together to calculate the molecular mass of the compound in grams.
4. 63.546 g + 32.065 g + ( 4 x <span>15.9994) = 159.609 g
5. this value </span><span>159.609 g is the mass in grams of one mol of CuSO4
6 the problem is asking not for the mass of one mole but the mass of 3.65 moles of CuSO4
7 then you have the multiply the value of one mol by the number of moles that the problem is asking you
8. </span><span>159.609 g x 3.65 = 582.571 g
</span>
9 the answer to the problem will be
"there are 582.571 g of CuSO4 in 3.65 moles of CuSO4"
Answer:
1.13 × 10⁶ g
Explanation:
Let's consider the reduction of aluminum (III) from Al₂O₃ to pure aluminum.
Al³⁺ + 3 e⁻ → Al
We can establish the following relations:
- 1 Ampere = 1 Coulomb / second
- The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
- 1 mole of Al is produced when 3 moles of electrons circulate
- The molar mass of Al is 26.98 g/mol.
The mass of aluminum produced under these conditions is:
Answer:
k = -0.0525 s⁻¹
Explanation:
The equaiton for a first order reaction is stated below:
ln[A]=−kt+ln[A]₀.
[A] = 5.50 x 10⁻³ M
[A]₀ = 7.60 x 10⁻² M
t = 85.0 - 35.0 = 50.0 s
The rate constant is represented by k and can be calculated substituting the values given above:
k = (ln[A]₀ - ln[A])/t
k = (ln5.50 x 10⁻³ M - ln7.60 x 10⁻² M)/50.0s
k = -0.0525 s⁻¹
0.737 KL would be the answer, pls give brainliest.
