Answer:
pH = 12.8
Explanation:
HF + NaOH → F⁻ + Na⁺ + H₂O
<em>1 mole of HF reacts with 1 mole of NaOH</em>
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Initial moles of HF and NaOH are:
HF = 0.018L × (0.308mol / L) = 5.544x10⁻³mol HF
NaOH = 0.023L × (0.361mol / L) = 8.303x10⁻³mol NaOH
That means moles of NaOH remains after reaction are:
8.303x10⁻³mol - 5.544x10⁻³mol = <em>2.759x10⁻³moles NaOH</em>
Total volume is 18.0mL + 23.0mL = 41.0mL = 0.0410L
Molar concentration of NaOH is
2.759x10⁻³moles NaOH / 0.0410L = 0.0673M = [OH⁻]
pOH = - log [OH⁻] = 1.17
As pH = 14 - pOH
<em>pH = 12.8</em>
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Answer:
1. n = 0.174mol
2. T= 26.8K
3. P = 1.02atm
4. V = 126.88L
Explanation:
1. P= 2.61atm
V = 1.69L
T = 36.1 °C = 36.1 + 273= 309.1K
R = 0.082atm.L/mol /K
n =?
n = PV / RT = (2.61x1.69)/(0.082x309.1)
n = 0.174mol
2. P = 302 kPa = 302000Pa
101325Pa = 1atm
302000Pa = 302000/101325 = 2.98atm
V = 2382 mL = 2.382L
T =?
n = 3.23 mol
R = 0.082atm.L/mol /K
T= PV /nR = (2.98x2.382)/(3.23x0.082) = 26.8K
3. P =?
V = 0.0250 m³ = 25L
T = 288K
n = 1.08mol
R = 0.082atm.L/mol /K
P = nRT/V = (1.08x0.082x288)/25 = 1.02atm
4. P = 782 torr
760Torr = 1 atm
782 torr = 782/760 = 1.03atm
V =?
T = 303K
n = 5.26 mol
R = 0.082atm.L/mol /K
V = nRT/P
V = (5.26x0.082x303)/1.03 = 126.88L
Answer:
false
in chemical reaction element doesnot change into another
Explanation:
Answer:

Explanation:
Let's rewrite the given word equation in its chemical balanced equation representation:
1. Lead(II) nitride is represented by lead, Pb, in an oxidation state of 2+, while nitride is a typical nitrogen anion with a state 3-. As a result, the lowest common multiple between 2 and 3 is 6, meaning 2 lead cations are needed to balance 3 nitrogen anions:
.
2. Ammonium sulfate consists of an ammonium cation with a 1+ charge and sulfate anion with a 2- charge, two ammonium cations needed:
.
3. Lead(II) sulfate would have one lead cation and one sulfate anion, as they have the same magnitude of charges with opposite signs:
.
4. Ammonium nitride would require three amonium cations to balance the nitride anion:
.
Let's write the balanced equation:

Answer:
For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively
Explanation:
The computation of the mass of each element is given below:
As we know that
A1 mole of ammonium nitrate i.e. 2 mol N, 4 mol H, 3 mol
Now we multiply each of above by the molar masses
For N
= 14.0 g/mol × 2
= 28.0 gN
For H
= 1.0 g/mol × 4
= 4.0 gN
ANd, for O
= 16.0 g/mol × 3
= 48.0 gN
Hence, For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively