Answer:
The answer to your question is:
Explanation:
Data
moles H=?
moles of N = 0.0969
moles of NH₃=?
N₂ (g) + 3 H₂ (g) ⇒ 2NH₃ (g)
Process
1.- Set a rule of three to calculate the moles of hydrogen
1 mol of nitrogen ------------- 3 moles of hydrogen
0.0969 moles of N ---------- x
x = (0.0969 x 3) / 1
x = 0.2907 moles of hydrogen
2.- Set a rule of three to calculate the moles of ammonia
1 mol of nitrogen -------------- 2 moles of ammonia
0.0969 mol of N -------------- x
x = (0.0969 x 2) / 1
x = 0.1938 moles of ammonia
1. 5 electrons.
- Therefore, the 3d subshells blanks will be like this:
- ↑ ↑ ↑ ↑ ↑
2. 6 electrons.
- The 3d subshells blanks will be:
- ↑↓ ↑ ↑ ↑ ↑
3. 7 electrons.
- The 3d subshells blanks will be:
- ↑↓ ↑↓ ↑ ↑ ↑
Hope you could understand.
If you have any query, feel free to ask.
The answer is 62.00 g/mol.
Solution:
Knowing that the freezing point of water is 0°C, temperature change Δt is
Δt = 0C - (-1.23°C) = 1.23°C
Since the van 't Hoff factor i is essentially 1 for non-electrolytes dissolved in water, we calculate for the number of moles x of the compound dissolved from the equation
Δt = i Kf m
1.23°C = (1) (1.86°C kg mol-1) (x / 0.105 kg)
x = 0.069435 mol
Therefore, the molar mass of the solute is
molar mass = 4.305g / 0.069435mol = 62.00 g/mol
Answer:
37.1°C.
Explanation:
- Firstly, we need to calculate the amount of heat (Q) released through this reaction:
<em>∵ ΔHsoln = Q/n</em>
no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.
<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>
∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.
Q = m.c.ΔT,
where, Q is the amount of heat released to water (Q = 2781.87 J).
m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).
c is the specific heat capacity of water (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).
∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)
∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.
<em>∴ final temperature = 25°C + 12.1 = 37.1°C.</em>
