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
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P1V1=P2V2
(1.00atm)(3.60L)=(2.50atm)V
3.60=(2.50atm)V
3.60/2.50=V
1.44L=V
Answer: 1.284M NH3
Explanation: (12.23 grams)/(17.0 gramms/mole) = 0.7191 moles
Dissolved in 560.0 ml (=0.5600L)
(0.7191 moles)/(0.560L) = 1.284M (4 sig figs)
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Answer:
moles Na = 0.1114 g / 22.9898 g/mol=0.004846
moles Tc = 0.4562g /98.9063 g/mol=0.004612
mass O = 0.8961 - ( 0.1114 + 0.4562)=03285 g
moles O = 0.3285 g/ 15.999 g/mol=0.02053
divide by the smallest
0.02053/ 0.004612 =4.45 => O
0.004846/ 0.004612 = 1.0 => Tc
to get whole numbers multiply by 2
Na2Tc2O 9
Explanation:
Hope it right hope it helps
