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zzz [600]
3 years ago
9

How to calculate the atomic mass of Cl​

Chemistry
1 answer:
slega [8]3 years ago
8 0
To find the atomic mass of chlorine, the atomic mass of each isotope is multiplied by the relative abundance (the percent abundance in decimal form) and then the individual masses are added together. The atomic mass of chlorine is 35.45 amu.
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Lyrx [107]

Answer:

C.Universal indicator turns orange in

weak acid.

Explanation:

option A and B are incorrect coz universal indicator turns red in strong acid and in strong alkali (base) it turns blue

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3 years ago
Which process releases heat?
borishaifa [10]

Answer:

C

Explanation:

burning gasoline i think pls can i have brainliest if right!

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Not all of the water falls as rain, snow,or sleet evaporates. What happens to the rest of the water?
Bumek [7]
Hi there! All of the water returns to the atmosphere, it's called evapotranspiration and condenses. And then the water falls back to the Earth's surface and starts the cycle all over again. If you don't know what I mean, it's the Water Cycle.
4 0
3 years ago
Sodium hydride racts with excess water to produce
romanna [79]

Answer:

0.96g of sodium hydride

Explanation:

Equation of reaction:

NaH + H20 = NaOH + H2

Mass of hydrogen gas produced (m) = PVM/RT

P = 765torr - 28torr = 737torr = 737/760 = 0.97atm, V = 982mL = 982cm^3, M = 2g/mol, R = 82.057cm^3.atm/gmol.K, T = 28°C = 28 + 273K = 301K

m = (0.97×982×2)/(82.057×301) = 0.08g of hydrogen gas

From the equation of reaction

1 mole (24g) of sodium hydride produced 1 mole (2g) of hydrogen gas

0.08g of hydrogen gas would be produced by (24×0.08)/2 = 0.96g of sodium hydride

8 0
3 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibrium described by the equation N2O4(g)
lora16 [44]

<u>Answer:</u> The value of equilibrium constant is 0.997

<u>Explanation:</u>

We are given:

Percent degree of dissociation = 39 %

Degree of dissociation, \alpha = 0.39

Concentration of N_2O_4, c = \frac{1mol}{1L}=1M

The given chemical equation follows:

                     N_2O_4\rightleftharpoons 2NO_2

<u>Initial:</u>                c             -

<u>At Eqllm:</u>         c-c\alpha      2c\alpha

So, equilibrium concentration of N_2O_4=c-c\alpha =[1-(1\times 0.39)]=0.61M

Equilibrium concentration of NO_2=2c\alpha =[2\times 1\times 0.39]=0.78M

The expression of K_{c} for above equation follows:

K_{c}=\frac{[NO_2]^2}{[N_2O_4]}

Putting values in above equation, we get:

K_{c}=\frac{(0.78)^2}{0.61}\\\\K_{c}=0.997

Hence, the value of equilibrium constant is 0.997

4 0
3 years ago
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