Answer:
Explanation:
x in (-oo:+oo)
2 < (1/2)*x-3 // - (1/2)*x-3
2-((1/2)*x)+3 < 0
(-1/2)*x+2+3 < 0
5-1/2*x < 0 // - 5
-1/2*x < -5 // : -1/2
x > -5/(-1/2)
x > 10
x in (10:+oo)
(10:+oo)
Sr is the limiting reactant.
Given the reaction equation;
2Sr + O2 (g) → 2SrO
2 moles of Sr reacts with 1 mole of O2
2 moles Sr will react with x mole of O2
x = 2 ×1/2
x = 1 mole of O2
Since we have more O2 than required, it is the reactant in excess, hence Sr is the limiting reactant.
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The reduction of alkyne to an alkene in the first step allows the best reagent to be chosen for each subsequent step.
Describe reagents.
A reagent is merely an essential component of a chemical reaction, it should be mentioned. It is an ingredient that speeds up the reaction.
With H2 and Lindlar's catalyst, an alkyne is reduced to alkene as the initial step in this process. Alkene will then be brominated to produce allyl bromide as the next step.
In this instance, the required allyl alcohol will be produced via the reaction of allyl bromide with NaOH.
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i. The new concentration is 0.116 M
ii. The new concentration is 0.029 M
<h3>What is the concentration of the original hydrochloric acid acid?</h3>
The concentration of the original or stock hydrochloric acid is calculated as follows:
- Molarity = Percentage concentration * Density * 1000/Molar mass * 100
Molarity of stock HCl = 36 * 1.18* 1000 /36.5 * 100
Molarity of stock HCl = 11.6 mol/dm
i. Using the dilution formula: M₁V₁ = M₂V₂
M₂ = M₁V₁ /V₂
M₂ = 11.6 * 10/1000
M₂ = 0.116 M
The new concentration = 0.116 M
ii. Using M₁V₁ = M₂V₂
M₂ = M₁V₁ /V₂
M₂ = 0.116 * 5/20
M₂ = 0.029 M
The new concentration = 0.029 M
In conclusion, the new concentrations are found using the dilution formula.
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Answer:
1.2km=1,200,000! hope this helps
