Answer:
CaS, CaBr₂, VBr₅, and V₂S₅.
Explanation:
- The ionic compound should be neutral; the overall charge of it is equal to zero.
- Binary ionic compound is composed of two different ions.
<u>Ca²⁺ can combined with either Br⁻ or S²⁻ to form binary ionic compounds.</u>
- CaS can be formed via combining Ca²⁺ with S²⁻ to form the neutral binary ionic compound CaS.
- CaBr₂ can be formed via combining 1 mole of Ca²⁺ with 2 moles of Br⁻ to form the neutral binary ionic compound CaBr₂.
<u>V⁵⁺ can combined with either Br⁻ or S²⁻ to form binary ionic compounds.</u>
- V₂S₅ can be formed via combining 2 moles of V⁵⁺ with 5 moles of S²⁻ to form the neutral binary ionic compound V₂S₅.
- VBr₅ can be formed via combining 1 mole of V⁵⁺ with 5 moles of Br⁻ to form the neutral binary ionic compound VBr₅.
<em>So, the empirical formula of four binary ionic compounds that could be formed is: CaS, CaBr₂, VBr₅, and V₂S₅.</em>
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Answer:
T½ = 16hours
Explanation:
Final mass (N) = 10g
Initial mass (No) = 20g
Time (t) = 16hours
T½ = ?
T½ = In2 / λ
But λ = ?
In(N/No) = -λt
In(10/20) = -(λ * 16)
In(0.5) = -16λ
-0.693 = -16λ
λ = 0.693 / 16
λ = 0.0433
Note : λ is known as the disintegration constant
T½ = In2 / λ
T½ = 0.693 / 0.0433
T½ = 16hours
The half-life of the sample is 16hours
FeSO₄*7H₂O(s) = FeSO₄(s) + 7H₂O(g)
M(FeSO₄*7H₂O)=278.0 g/mol
M(FeSO₄)=151.9 g/mol
m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)=m(FeSO₄)/M(FeSO₄)
m(FeSO₄)=M(FeSO₄)m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)
m(FeSO₄)=151.9*100.0/278.0=54.6 g
m(FeSO₄)=54.6 g
Ferromagnesian silicate minerals (i looked it up)
Answer:
104.84 moles
Explanation:
Given data:
Moles of Boron produced = ?
Mass of B₂O₃ = 3650 g
Solution:
Chemical equation:
6K + B₂O₃ → 3K₂O + 2B
Number of moles of B₂O₃:
Number of moles = mass/ molar mass
Number of moles = 3650 g/ 69.63 g/mol
Number of moles = 52.42 mol
Now we will compare the moles of B₂O₃ with B from balance chemical equation:
B₂O₃ : B
1 : 2
52.42 : 2×52.42 = 104.84
Thus from 3650 g of B₂O₃ 104.84 moles of boron will produced.
