Answer: 8.28g Na
Explanation: use ideal gas law
PV= nRT
Solve for moles of Cl2
n= PV/ RT
Substitute:
= 1 atm x 4.0 L / 0.08205 L.atm/ mol. K x 273 K
= 0.18 moles Cl2
Do stoichiometry to solve for m of Na
2 Na + Cl2 => 2 NaCl2
=0.18 moles Cl2 x 2 mol Na/ 1 mol Cl2 x 23g Na / 1 mol Na
= 8.28 g Na.
Answer:
0.07172 L = 7.172 mL.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.
</em>
where, P is the pressure of the gas in atm (P = 1.0 atm, Standard P).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 3.2 x 10⁻³ mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 273 K, Standard T).
<em>∴ V = nRT/P =</em> (3.2 x 10⁻³ mol)(0.0821 L.atm/mol.K)(273 K)/(1.0 atm) = <em>0.07172 L = 7.172 mL.</em>
I believe the answer is a
I believe the correct answer from the choices listed above is the second option. The pair of elements that is most likely to chemically combine and form ionic bonds would be <span>lithium and chlorine. Lithium is metal and chlorine is nonmetal which as a compound forms ionic bonds. Hope this answers the question.</span>
<h3><u>Answer</u>;</h3>
≈ 4.95 g/L
<h3><u>Explanation;</u></h3>
The molar mass of KCl = 74.5 g/mole
Therefore; 0.140 moles will be equivalent to ;
= 0.140 moles × 74.5 g/mole
= 10.43 g
Concentration in g/L
= mass in g/volume in L
= 10.43/2.1
= 4.9667
<h3> <u> ≈ 4.95 g/L</u></h3>
