It’s c I just did this so yea c
Answer:
8.33 hours
Explanation:
In order to solve this problem, we must apply Graham's law of diffusion in gases. Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its vapour density. For two gases we can write;
R1/R2=√d2/d1
Where;
R1= rate of diffusion of hydrogen
R2= rate diffusion of unknown gas
d1= vapour density of hydrogen
d2= vapour density of the unknown gas
Volume of hydrogen gas = 360cm^3
Time taken for hydrogen gas to diffuse= 1 hour =3600 secs
R1 = 360 cm^3/3600 secs = 0.1 cm^3 s-1
Vapour density of unknown gas = 25
Vapour density of hydrogen = 1
Substituting values,
0.1/R2 = √25/1
0.1/R2 = 5/1
5R2 = 0.1 × 1
R2 = 0.1/5
R2= 0.02 cm^3s-1
Volume of unknown gas = 600cm^3
Time taken for unknown gas to diffuse= volume of unknown gas/ rate of diffusion of unknown gas
Time taken for unknown gas to diffuse= 600/0.02
Time= 30,000 seconds or 8.33 hours
3 moles of NaOH reacts with 1 mole of phosphoric acid.
<h3>What is the moles ratio of the NaOH and Phosphoric acid reaction?</h3>
The moles ratio of the reaction between NaOH and Phosphoric acid is given by the equation of the reaction below:

Based on the equation of the reaction, 3 moles of NaOH reacts with 1 mole of phosphoric acid.
Learn more about mole ratio at: brainly.com/question/19099163
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Answer:
(−ΔH°1)+12(−ΔH°2)+(ΔH°3)+(ΔH°4)
I hope this helps you out!! :)
Answer:
True
Explanation:
The oil drop experiment was carried out by Robert Millikan and Harvey Fletcher in 1909 to determine the charge of an electron. By balancing downward gravitation force with upward drag and electric force, they suspended small charged droplets of oil between two metal electrodes.
The charge over an oil droplet was often an integral value of e, was determined by changing the intensity of the electric field.
So, the given statement is true