Step 1: Change density from g/mL to g/L;
0.807 g/mL = 807 g/L
Step 2: Find Moles of N₂;
As,
Density = Mass / Volume
Or,
Mass = Density × Volume
Putting Values,
Mass = 807 g/L × 1 L
Mass = 807 g
Also,
Moles = Mass / M.mass
Putting values,
Moles = 807 g / 28 g.mol⁻¹
Moles = 28.82 moles
Step 3: Apply Ideal Gas Equation to Find Volume of gas occupied,
As,
P V = n R T
V = n R T / P
Putting Values, remember! don't forget to change temperatue into Kelvin (25 °C + 273 = 298 K)
V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm
V = 704.76 L
Answer:
1.79 mol.
Explanation:
- For the balanced reaction:
<em>2NaCl + F₂ → 2NaF + Cl₂.
</em>
It is clear that 2 mol of NaCl react with 1 mol of F₂ to produce 2 mol of NaF and 1 mol of Cl₂.
- Firstly, we can get the no. of moles of F₂ gas using the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 1.2 atm).
V is the volume of the gas in L (V = 18.3 L).
n is the no. of moles of the gas in mol (n = ??? mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (299 K).
∴ no. of moles of F₂ (n) = PV/RT = (1.2 atm)(18.3 L)/(0.0821 L.atm/mol.K)(299 K) = 0.895 mol.
- Now, we can find the no. of moles of NaCl is needed to react with 0.895 mol of F₂:
<em><u>Using cross multiplication:</u></em>
2 mol of NaCl is needed to react with → 1 mol of F₂, from stichiometry.
??? mol of NaCl is needed to react with → 0.895 mol of F₂.
∴ The no. of moles of NaCl needed = (2 mol)(0.895 mol)/(1 mol) = 1.79 mol.
It’s a or b i think it’s b
Answer:
None of the given options
Explanation:
Let's go case by case:
A. No matter the volume, the concentration of Fe(NO₃)₃ (and thus of [Fe³⁺] as well) is 0.050 M.
B. We can calculate the moles of Fe₂(SO₄)₃:
- 0.020 M * 0.80 L = 0.016 mol Fe₂(SO₄)₃
Given that there are two Fe⁺³ moles per Fe₂(SO₄)₃ mol, in the solution we have 0.032 moles of Fe⁺³. With that information in mind we <u>can calculate [Fe⁺³]</u>:
- 0.032 mol Fe⁺³ / 0.80 L = 0.040 M
C. Analog to case A., the molar concentration of Fe⁺³ is 0.040 M.
D. Similar to cases A and C., [Fe⁺³] = 0.010 M.
Thus none of the given options would have [Fe⁺³] = 0.020 M.
Answer:
I think <em><u>alpha</u></em> and <em><u>beta</u></em> is the answer.
