Answer:
1. 0.097 s
2. 0.420 M
Explanation:
To solve both questions we'll use the formula:
[A]ₓ = [A]₀ - kt
Where [A]ₓ is the concentration of A at a given time; and [A]₀ is the initial concentration.
1) We input the data given by the problem:
0.167 M = 0.700 M - 5.48 M/s * t
And solve for t:
t = 0.097 s
2) We input the new data:
[A]ₓ = 0.500 M - 0.361 M/s * 0.220 s
and solve for [A]ₓ:
[A]ₓ = 0.420 M
Answer:
SnH4
Explanation:
Because the melting point of SnH4 is higher than CH4
Answer:
NH₃ and H⁺
Explanation:
The dissociation of NH₄Cl will lead to two ions , i.e. , NH⁴⁺ and Cl⁻ .
Hence , the reaction is -
NH₄Cl → NH⁴⁺ + Cl⁻
We can conclude that NH₄Cl can be formed from the ions , NH⁴⁺ and Cl⁻
Hence ,
According to the reaction ,
NH⁴⁺ + H₂O ⇆ NH₃ + H₃O⁺
Therefore ,
The reactant that will lead to the formation of NH₄Cl are - NH₃ and H⁺
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.
Learn more about molarity at: brainly.com/question/17138838
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Answer:
204.8g
Explanation:
The number of moles of a substance is related to its mass and molecular mass as follows:
mole (n) = mass (m) ÷ molar mass (MM)
According to this question, 3.50 moles of sodium chloride (NaCl) is added to a food.
Molar mass of NaCl = 23 + 35.5
= 58.5g/mol
Using mole = mass/molar mass
Mass = molar mass × mole
Mass = 58.5g/mol × 3.5mol
Mass = 204.75
Mass = 204.8grams.
Therefore, 204.8grams of NaCl or common salt was added to the food.
