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artcher [175]
3 years ago
13

How many significant figures the measurement have 56.0g 0.0004m 1003ml 0.0350s

Chemistry
1 answer:
Setler [38]3 years ago
7 0
In order to do this, we have to first know the significant figure rules.
<span>Rule #1: All non-zero digits are significant. (1234)
 Rule #2: Zeros in front of a number are not significant. (0.093)
 Rule #3: Zeros between non-zero digits are significant. (78309)
 Rule #4: Zeros at the end of a number are significant if there is a decimal point in the number. (0.05470)

So by going by the rules, 56.0g has three sig figs, because there is a decimal point.
0.0004m only has one sig fig, according to Rule #2.
1003ml has 4 sig figs, because the zeroes are wedged in the two sig fig numbers.
And lastly, 0.0350s has 3 sig figs because the number after a decimal point counts.

</span>
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2NH4Cl(s)+Ba(OH)2⋅8H2O(s)→2NH3(aq)+BaCl2(aq)+10H2O(l) The ΔH for this reaction is 54.8 kJ . How much energy would be absorbed if
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1) Chemical equation

<span>2NH4Cl(s)+Ba(OH)2⋅8H2O(s)→2NH3(aq)+BaCl2(aq)+10H2O(l)

2) Stoichiometric ratios

2 mol NH4Cl(s) : 54.8 KJ

3) Convert 24.7 g of NH4Cl into number of moles, using the molar mass

molar mass of NH4Cl = 14 g/mol + 4*1 g/mol + 35.5 g/mol = 53.5 g/mol

number of moles = mass in grams / molar mass

number of moles = 24.7 g / 53.5 g/mol = 0.462 moles

4) Use proportions:

2 moles NH4Cl / 54.8 kJ = 0.462 moles / x

=> x = 0.462 moles * 54.8 kJ / 2 moles = 12.7 kJ

Answer: 12.7 kJ</span>
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When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr
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<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

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\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of N_2O_4 = 9.2 g

Molar mass of N_2O_4 = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M

For the given chemical equation:

                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

We are given:

Equilibrium concentration of N_2O_4 = 0.057

Evaluating the value of 'x'

\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

[NO_2]_{eq}=2x=(2\times 0.143)=0.286M

[N_2O_4]_{eq}=0.057M

Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

6 0
3 years ago
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