Answer:
powerthrone lol
Explanation:
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Answer:
After 1326s, the concentration of pyruvic acid fall to 1/64 of its initial concentration.
Explanation:
The first order kinetics reaction is:
ln [A] = ln [A]₀ - kt
<em>Where [A] is concentration after t time, [A]₀ is intial concentration and k is reaction constant.</em>
To convert half-life to k you must use:
t(1/2) = ln 2 / K
221s = ln 2 / K
K = ln 2 / 221s
<h3>K = 3.1364x10⁻³s⁻¹</h3>
If [A] = 1/64, [A]₀ = 1:
ln [A] = ln [A]₀ - kt
ln (1/64) = ln 1 - 3.1364x10⁻³t
4.1588 = 3.1364x10⁻³s⁻¹t
1326s = t
<h3>After 1326s, the concentration of pyruvic acid fall to 1/64 of its initial concentration.</h3>
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Answer:
Gamma rays
Since they have high penetrating power.
<h3>Answer:</h3>
#1. Ca²⁺
# 2. Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)
#3. 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)
<h3>Explanation:</h3>
The question above concerns solubility of salts or ions in water.
The solution given contains Ag+, Ca2+, and Co2+ ions.
- In the first case, when Lithium bromide is added to the solution, there is no white precipitate formed.
- In the second case, the addition of Lithium sulfate results in the formation of a precipitate because of the Ca²⁺ in the solution combined with the SO₃²⁻ from lithium sulfate to form an insoluble CaSO₄.
- The net ionic equation for the reaction is;
Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)
- From the solubility rules, all sulfates are soluble except BaSO₄, CaSO₄, and PbSO₄.
- In the third case, the addition of Lithium phosphate results in the formation of a precipitate because Ag⁺ ions in the solution combine with phosphate ions ( PO₄³⁻) from lithium phosphate to form an insoluble salt, Ag₃PO₄.
- The net ionic equation for the reaction is;
3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)
- According to solubility rules, all phosphates are insoluble in water except Na₃PO₄, K₃PO₄, and (NH₄)₃PO₄.
Answer:
sp3d
Explanation:
The ground state electronic configuration of tin is written as; [Kr] 5s²4d¹⁰5p². Hybridization is a concept used to explain the combination of orbitals of appropriate energy to produce suitable orbitals that could be used for bonding.
In forming the compound Snf5^ -1, we have to hybridize the following orbitals on tin; 5p, 5d and 6s orbitals. This gives us a set of sp3d hybrid hence the answer.