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Answer:
The mass of N a H C O 3 present is 2.431 g
Explanation:
The sample contains 57.2 % N a H C O 3 by mass.
To find the mass of N a H C O 3 in the sample, we need to find what the equivalent of 57.2 %.
Mass of N a H C O 3 = Percentage Composition * Mass of sample
Mass of N a H C O 3 = 57.2 / 100 * 4.25
Mass of N a H C O 3 = 2.431 g
The mass of N a H C O 3 present is 2.431 g
The differential pressure is
797.9 - 715.7 = 82.2 torr = 138815.25 Pa
The height of the mercury in the open-end arm is calculated using
ΔP = ρgh
The density of mercury is 13560 kg/m3
The height of the mercury is
138815.25 Pa = 13560 kg/m3 (9.81 m/s2) h
h = 1.04 m
Answer:
The molar mass would double
Explanation:
This would double the molar mass. Even though this question is about a liquid but we can use an example we see a lot in elements like oxygen, flourine, nitrogen and others whose atomic attractions make them exist as diatomic gases. This is why their molar masses are twice their relative atomic masses (in grams). The same thing would happen here. The molar mass of A would have to double.
Answer:
Well why didn't you do it? Have a great day and I am so sorry for wasting your points I don't need them. I hope someone helps!! ;)
Explanation:
