2 Na + 2 H2O --> 2 NaOH + H2
Answer:
Yes
Explanation:
A small part of the correspondence is due to the relationship between temperature and the solubility of carbon dioxide in the surface ocean, but the majority of the correspondence is consistent with a feedback between carbon dioxide and climate.
Initial dimension of the pool:
Length (L)= 50.0 m
Width (W)= 25.0 m
Depth (D) = 2.0 m
Initial Volume of the pool (V1) = L*W*D = 50.0 * 25.0 *2.0 = 2500 m3
When the depth is lowered by 3 cm i.e. 0.03 m, the new depth becomes: 2.0 - 0.03 = 1.97 m
The new volume of the pool (V2) = 50.0*25.0*1.97 = 2462.5 m3
Volume of water to be pumped out = V1-V2 = 2500-2462.5 = 35.5 m3
Now, 1 m3 = 1000 L
therefore, 35.5 m3 == 35.5 *10^3 L
It is given that:
3.80 L of water is pumped out in 1 sec
Therefore, 35.5*10^3 L will be pumped out in:
= 1 s * 35.5*10^3 L/3.80 L = 9.34*10^3 sec
Answer: technically it is chemical and physical change
Explanation: chemical bc it’s burning and physical bc it’s going from a log to ashes
Answer:
4N2H4(2) + 2N204 (1) --> 6N2 (g) + 8H20 (g), AHO-2098 kJ
Explanation:
2N2H4(2) + N204 (1) --> 3N2 (g) + 4H20 (g),AHO-1049 kJ
when 6 mol of nitrogen are formed?
The ratio of Nitrogen in the previous therrmochemical equation to when 6 mol f nitrogen are formed is 2; 6/3 = 2.
So to get the new thermoochemical equation, multiply all parameters in the given equation by 2.
We have;
4N2H4(2) + 2N204 (1) --> 6N2 (g) + 8H20 (g), AHO-2098 kJ
