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Paul [167]
4 years ago
14

Gas y effuses half as fast as o2. what is the molar mass of gas y?

Chemistry
1 answer:
devlian [24]4 years ago
4 0
To solve this question, we will use Graham's law which states that:
(R1 / R2) ^ 2 = M2 / M1 where
R1 and R2 are the rates of effusion and M1 and M2 are the molar masses of the two gases.
From the periodic table, we can calculate the molar mass of O2 as follows:
molar mass of O2 = 2*16 = 32 grams

Therefore we have:
R1 / R2 = Ry / RO2 = 1/2
M1 is My we want to get
M2 is molar mass of O2 = 32 grams
Substitute in the above equation to get the molar mass of y as follows:
(1/2) ^2 = (32/My)
1/4 = 32/My
My = 32*4 = 128

Therefore, molar mass of gas y = 128 grams

You might be interested in
The following equation describes how sodium and chlorine react to produce sodium chloride: 2Na + Cl2 -> 2NaCl Is the equation
DedPeter [7]

Answer:

The equation shows balance.

Explanation:

You can easily count the number of elements on each side.

On the left side of the equation, you have 2 moles of Na and 2 moles of Cl.

On the right side of the equation, you also have 2 moles of Na and 2 moles of Cl because the two elements formed a compound meaning that whatever number is in front of them, both elements will receive the same number.

5 0
4 years ago
Please help. urgent. needs explanation too :( thank you
g100num [7]

Answer:

24) F. 3 mol.......ratio of hydrogen and oxygen is 2:1 , if hydrogen 6 then oxygen 3

25) B. 2.5 moles ......take ration fo 2RbNO3 and O2 which is 2:1 then if RbNO3 has 5 moles these O2 will have 2.5 moles

3 0
3 years ago
The concentration of the KCN solution given in Part A corresponds to a mass percent of 0.473 %. What mass of a 0.473 % KCN solut
tiny-mole [99]
If the Ka of HCN = 5.0 x 10^-10
Since
(Ka) (Ka) - 1 x 10^ -14
then
the Kb of its conjugate base (CN-) = 2.0 X 10^-5

since
pH + pOH = 14
when the pH = 10.00
then
the pOH = 4.00
& the OH-
would then equal 1.0 X 10^-4

NaCN as a base does a hydrolysis in water:
CN- & water --> HCN & OH-
notice that equal amounts of OH- & HCN are formed

Kb = [HCN] [OH-] / [CN-]

2.0 X 10^-5 = [1.0 X 10^-4] [1.0 X 10^-4] / [CN-]

[CN-] =(1.0 X 10^-8) / (2.0 X 10^-5)

[CN-] = (5.0 X 10^-4)

that's 0.00050 Molar
which is 0.00050 moles in each liter of aqueous KCN solution
which is
0.00025 moles KCN in 500. mL of aqueous KCN solution

use molar mass of KCN, to find grams:
(0.00025 moles KCN) (65.12 grams KCN / mole) = 0.01628 grams of KCn

which is 16.3 mg of KCN
& rounded to the 2 sig figs which are showing in the Ka of HCN , "5.0" X 10^-10
your answer would be
16 mg of KCN

sorry even after making a correction in calcs , I don't get one of your answers.
the only way that I could get one of them is to pretend that yours was a 1 sig fig problem,
in which case your 16 mg would round off to 20 mg.
but you have 3 sig figs in "500. ml", & 2 sig figs in both the "pH of 10.00."
& The Ka of HCN = "5.0 x 10^-10."

it does however take 12 mg of NaCN, to make 500. mL of aqueous solution pH of 10.00. the molar mass of NaCN has the smaller molar mass of 49.00 grams per mole.
maybe they meant NaCN, but wrote KCN instead.

I hope i answered this correctly for you.
8 0
4 years ago
How can global climate change be studied?
IgorC [24]
Using physical evidence from multiple different sources, there’s lots of other ways but that’s a starting point
7 0
3 years ago
What is the pH of a solution that has the same number of hydrogen ions as hydroxide ions?
Alexxx [7]

Answer:

7

Explanation:

cuz it is trust me hope it helps though

4 0
2 years ago
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