Answer:3.6 x 101 or 8.77 x 10-1
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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There would be 79 electrons present in each atom of gold. I hope this helps :)
The correct answer is + 32.5 J because heat is absorbed here by copper not evolved also after calculation we will find the H value is positive, lets calculate it:
H = m C Δt where:
m = 10 g
C = 0.13 J/g.°C
Δ t = final temperature - initial temperature = 50 - 25 = + 25 °C
so H = 10 x 0.13 x (+25) = + 32.5 J
<span>0.001 M Ba(OH)2 has a higher pH</span>
