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ANTONII [103]
3 years ago
12

Round this number to two significant figures. 634

Chemistry
2 answers:
Aneli [31]3 years ago
7 0
The answer to this is 630
loris [4]3 years ago
4 0

Answer : The correct answer is 630.

Explanation :

The following rules are used to round off a number to the required number of significant figures:

(1) If the rightmost digit to be removed is more than 5, the preceding number is increased by one.

(2) If the rightmost digit to be removed is less than 5, the preceding number is not changed.

(3) If the rightmost digit to be removed is 5, then the preceding number is not changed if it is an even number but it is increased by one if it is an odd number.

As we are given, 634

In the given number, there are 3 significant figures. Now we have to convert it into 2 significant figures.

According to the rules, round off the given measurement in 2 significant figures as 630 because last number i.e 4 is less than number 5 so, last digit become 0.

Therefore, the correct answer is 630.

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<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{c}

For a general chemical reaction:

aA+bB\rightleftharpoons cC+dD

The expression for K_{eq} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)

The expression of K_c for above equation is:

K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}

We are given:

[H_2S]_{eq}=0.671M

[O_2]_{eq}=0.587M

K_c=1.35

Putting values in above expression, we get:

1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}

[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M

Hence, the equilibrium concentration of water is 0.597 M

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3 years ago
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Answer:

Explanation:

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<u>The formula to calculate half-life: </u>

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