Answer:
A) 31.22
Explanation:
The reaction of sulfuric acid with NaOH is:
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2H₂O
To solve this problem we need to determine the moles of acid that will react, and, using the chemical equation we can determine the moles of NaOH and the volume that a 0.2389M NaOH solution would require to neutralize it.
<em>Moles H₂SO₄ (Molar mass: 98.08g/mol):</em>
0.9368g * 39.04% = 0.3657g H₂SO₄ * (1mol / 98.08g) =
3.7289x10⁻³moles H₂SO₄
And moles of NaOH that you require to neutralize the acid are:
3.7289x10⁻³moles H₂SO₄ * (2 moles NaOH / 1 mole H₂SO₄) =
7.4578x10⁻³ moles NaOH
Using a 0.2389M NaOH solution:
7.4578x10⁻³ moles NaOH * (1L / 0.2389mol) = 0.03122L = 31.22mL
Right answer is:
<h3>A) 31.22
</h3>
Answer:
(a) rate = 4.82 x 10⁻³s⁻¹ [N2O5]
(b) rate = 1.16 x 10⁻⁴ M/s
(c) rate = 2.32 x 10⁻⁴ M/s
(d) rate = 5.80 x 10⁻⁵ M/s
Explanation:
We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration of N₂O₅, so
(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]
(b) rate = 4.82×10⁻³s⁻¹ x 0.0240 M = 1.16 x 10⁻⁴ M/s
(c) Since the reaction is first order if the concentration of N₂O₅ is double the rate will double too: 2 x 1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s
(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to
1.16 x 10⁻⁴ M/s / 2 = 5.80 x 10⁻⁵ M/s
Answer:
1.28 g
Explanation:
Mass of anhydrous compound/molar mass of anhydrous compound = mass of hydrated compound/ molar mass of hydrated compound
Mass of anhydrous compound = ?
Mass of hydrated compound = 2g
Molar mass of anhydrous compound= 160 g/mol
Molar mass of hydrated compound = 250 g/mol
x/160 = 2/250
250x = 2 ×160
x= 2 × 160/250
x= 1.28 g
Should be <span>during deposition.
</span>
This helps increase variation in the offspring.
