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Kaylis [27]
3 years ago
6

Which one will diffuse faster Ammonia or carbon dioxide Why?​

Chemistry
2 answers:
Law Incorporation [45]3 years ago
8 0

jcp-fnoa-kzk join here

suter [353]3 years ago
5 0
The heavier the gas is(in terms of its molecular weight), the slower it is. Hence, ammonia is lighter and from Grahams law of effusion, it then means that it will diffuse faster than CO2, which is heavier.
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What is used to contain an acid spill and what are the characteristics of them?​
Bingel [31]
Is that chemistry?
Cause I'm kinda confused about how to help you with this question
6 0
2 years ago
Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found
Igoryamba

Answer:

The concentrations are :

[HAsc^-]=0.000702 M

[Asc^{2-}]=5.92\times 10^{-8} M

The pH of the solution is 3.15.

Explanation:

H_2Asc\rightleftharpoons HAs^-+H^+         K_{a1}=1.0\times 10^{-5}

Initial

c                0              0

Equilibrium

c-x                x          x

K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}

1.0\times 10^{-5}=\frac{x\times x}{(c-x)}

1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}

Solving for x:

x = 0.000702 M

[HAsc^-]=0.000702 M

HAsc^-\rightleftharpoons As^{2-}+H^+        K_{a2}=5\times 10^{-12}

Initially

x                0          0

At equilibrium ;

(x - y)            y         y

K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}

5\times 10^{-12}=\frac{y\times y}{(x-y)}

5\times 10^{-12}=\frac{y^2}{(x-y)}

Putting value of x = 0.000702 M

5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}

y=5.92\times 10^{-8} M

[Asc^{2-}]=5.92\times 10^{-8} M

Total concentration of [H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M

The pH of the solution :

pH=-\log[H^+]

pH=-\log[7.0206\times 10^{-4} M}=3.15

7 0
3 years ago
It has more protons than Cl but less than K
timama [110]

Answer: Argon (Ar), which has 18 protons.

8 0
3 years ago
Which atom is a carbon atom?<br><br> A<br> B<br> C
Illusion [34]
B
Hope I helped :) shajs
3 0
3 years ago
How many grams of NaCl (molecular weight = 58.4 g mole-1) would you dissolve in water to make a total volume of 500 mL of soluti
kolezko [41]

Given :

Volume , V = 500 mL .

Molarity , M = 0.5 M .

Molecular mass of NaCl is 58.4\ g/mole .

To Find :

How many grams of NaCl is required .

Solution :

Let , NaCl required is x gram .

Molarity is given by :

M=\dfrac{\text{Number of moles}}{\text{Volume (in liters) }}\\\\M=\dfrac{m}{M\times V}\\\\0.5=\dfrac{x}{58.4\times 0.5}\\\\x=0.5^2\times 58.4\ g\\\\x=14.6\ g

Hence , this is the required solution.

6 0
2 years ago
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