Volume of Cl₂(g) is produced at 1.0 atm and 540.°C=4.5×10^4 L
As per the evenly distributed response
2NaCl (g) ----> 2Na(l)+ Cl2(g)
Calculate the amount of Cl2 that was formed as indicated below:
Moles of Cl2 = 31.0 kg of Na x (1000* 1 * 1 / 1*23* 2)
= 673.9 mol
P is equal to 1.0 atm, and T is equal to 813.15 K
when converted to Kelvin by multiplying by a factor of 273.15.
Using Cl2 as an ideal gas, determine the in the following volume:
volume = nRT/P
= 673.9 * 0.0821 * 813.15/ 1
=4.5×10^4 L
As a result, the volume of Cl2 under the given circumstances =4.5×10^4 L
Learn more about Volume here:
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Answer:
2.29 moles of Cr₂O₃ are produced
Explanation:
This is the reaction:
4 Cr + 3O₂ → 2Cr₂O₃
Ratio for this equation is 4:2, so 4 moles of chromium can produce the half of moles of chromium(III) oxide
4.58 mol of Cr may produce (4.58 .2)/4 = 2.29 moles of Cr₂O₃
68000 = 6.8 * 10000 = 6.8 * 10^4
hope this helps? c;
<em>ANSWER - 6 MOLES OF </em><em>IRON</em>
Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(1)
One moles of Fe2O3 forming 2 moles of Fe
3 moles of Fe2O3 will form 2×3 = 6 moles of iron
Answer: M = 22/ (i x28.948)
Explanation:
Pi = osmotic pressure = 22atm
T = Temperature = 353K
M = Molarity = ?
R = gas constant = 0.082atm.L/mol/K
i = van’t Hoff factor
Pi = iMRT
M= Pi /(iRT) = 22 / ( i x 0.082 x 353)
M = 22/ (i x28.948)
