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When it comes to equilibrium reactions in chemistry, there are a lot of equilibrium constants that can be used. In the case of solubility, the appropriate one to use is the equilibrium constant of solubility product denotes as Ksp. This is the concentration of products raised to their coefficients. For example,
cC ⇔ aA + bB
Ksp = {[A^a][B^b]}
Now, for the this problem, the reaction is
BaSO₄ ⇔ Ba²⁺ + SO₄²⁻
The reaction is already balanced. Since we don't know the value of Ba²⁺ and SO₄²⁻, let's denote this at x.
1.1 × 10⁻¹⁰ = [x][x] =[x²]
[x] = [Ba²⁺] = [SO₄²⁻] = [BaSO₄] = 1.049 × 10⁻⁵ M
Answer:
66 g of CO₂
Solution:
The Balance Chemical Reaction is as follow,
C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O
Or,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)
Step 1: Find out the limiting reagent as;
According to Equation 1,
56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂
So,
125 g of C₂H₂ will react with = X g of O₂
Solving for X,
X = (125 g × 160 g) ÷ 56.1 g
X = 356.5 g of O₂
It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.
Step 2: Calculate Amount of CO₂ produced as;
According to Equation 1,
160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂
So,
60.0 g of O₂ will produce = X g of CO₂
Solving for X,
X = (60.0 g × 176 g) ÷ 160 g
X = 66 g of CO₂
Answer:sample 3
Explanation:because if only have one kind of substance which makes it pure
Answer:
Aluminum iodide (AlI₃)
Explanation:
The synthesis reaction of aluminum (Al) and iodine (I) can be illustrated as shown below:
Aluminium exhibit trivalent positive ion (Al³⁺)
Iodine exhibit univalent negative ion (I¯)
During reaction, there will be an exchange of ion as shown below:
Al³⁺ + I¯ —> AlI₃
Thus, we can write the balanced equation for the reaction as follow:
Al + I₂ —› AlI₃
There are 2 atoms of I on the left side and 3 atoms on the right side. It can be balance by putting 2 in front of AlI₃ and 3 in front of I₂ as shown below:
Al + 3I₂ —› 2AlI₃
There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Al as shown below:
2Al + 3I₂ —› 2AlI₃
Thus the equation is balanced.
The product on the reaction is aluminum iodide (AlI₃)
