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Katyanochek1 [597]

4 years ago

15

Mercury(II) oxide (HgO) decomposes to form mercury (Hg) and oxygen (O2). The balanced chemical equation is shown below. 2HgO mc0

20-1.jpg 2Hg + O2 The molar mass of HgO is 216.59 g/mol. The molar mass of O2 is 32.00 g/mol. How many moles of HgO are needed to produce 250.0 g of O2?

2 answers:

Nostrana [21]4 years ago

6 0

Answer:

moles of HgO needed to produce 250.0 g of O2 = 15.63 moles

Explanation:

The chemical reaction is:

$2HgO\rightarrow 2Hg + O2$

Amount of O2 produced = 250.0 g

Molar mass of HgO = 216.59 g/mol

Molar mass of O2 = 32.00 g/mol

Moles of O2 produced = $\frac{Mass\ of\ O2}{Molar\ Mass} = \frac{250.0g}{32 g/mol } = 7.813 moles$

Based on the reaction stoichiometry:

2 moles of HgO produces 1 mole of O2

Therefore, moles of HgO needed to produce 7.813 moles of O2 is:

= $\frac{7.813\ O2 * 2\ HgO}{1\ O2} = 15.63 moles$

Alecsey [184]4 years ago

4 0

2HgO = 2Hg + O₂

n(HgO)=2n(O₂)=2m(O₂)/M(O₂)

m(HgO)=n(HgO)M(HgO)=2m(O₂)M(HgO)/M(O₂)

m(HgO)=2*250.0*216.59/32.00 = 3384.219 g ≈ 3 kg 384 g

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Answer:

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Explanation:

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8 0

3 years ago

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There are two naturally occurring isotopes of boron. 10 B has a mass of 10.0129 u. 11 B has a mass of 11.0093 u. Determine the a

Vanyuwa [196]

<h2>Natural Abundance for 10B is 19.60%</h2>

Explanation:

The natural isotopic abundance of 10B is 19.60%.

The natural isotopic abundance of 11B is 80.40%.

The isotopic masses of boron are 10.0129 u and 11.009 u respectively.

For calculation of abundance of both the isotopes -

Supposing it was 50/50, the average mass would be 10.5, so to increase the mass we need a more percentage of 11.

Determining it as an equation -

10x + 11y= 10.8

x+y=1 (ratio)

10x + 10y = 10

By taking the denominator away from the numerator

we get;

y = 0.8

x + y = 1

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To get percentages we need to multiply it by 100

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5 0

3 years ago

How much force is required to lift a rock a vertical distance of 8 m if 80 j of work were done

yanalaym [24]

we have

work done (W)= force(F) × displacemen(s)

or, 80= F× 8

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3 years ago

In the gas-phase reaction 2 A + B ⇌ 3 C + 2 D, it was found that, when 1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and all

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Answer:

(i)The mole fractions are :

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$B=\frac{1.4}{4.3} \\=0.3256$

$C=\frac{0.9}{4.3} \\=0.2093$

$D=\frac{1.6}{4.3} \\=0.3721$

(ii) $K=0.4508$

(iii)ΔG = 1.974kJ

Explanation:

The given equation is :

$2A+B$ ⇄ $3C+2D$

Let $\alpha$ be the number of moles dissociated per mole of $B$

Thus ,

<em>The initial number of moles of :</em>

$A=1$

$B=2$

$C=0$

$D=1$

$2A\\(1-2\alpha)$ + $B\\2(1-\alpha)$ ⇄ $3C\\(3\alpha)$ + $2D\\(1+2\alpha)$

And finally the number of moles of $C[tex] is 0.9Thus ,[tex]3\alpha=0.9\\\alpha=0.3[tex]The final number of moles of:[tex]A = 1-2\alpha=1-2*0.3=0.4mol[tex] [tex]B=2(1-\alpha)=2(1-0.3)=1.4mol[tex][tex]D=1+2\alpha=1+2*0.3=1.6mol[tex]Thus , total number of moles are : 0.4+1.4+0.9+1.6=4.3(i)The mole fractions are : [tex]A=\frac{0.4}{4.3} \\=0.0930$

$B=\frac{1.4}{4.3} \\=0.3256$

$C=\frac{0.9}{4.3} \\=0.2093$

$D=\frac{1.6}{4.3} \\=0.3721$

(ii)

$K=\frac{(P_C^3)(P_D^2)}{(P_A^2)(P_B)}$

Where ,

$P_A,P_B,P_C,P_D$ are the partial pressures of A,B,C,D respectively.

Total pressure = 1 bar .

∴

<em> $P_A= 0.0930*1=0.0930$ </em>

<em> $P_B= 0.3256*1=0.3256$ </em>

<em> $P_C= 0.2093*1=0.2093$ </em>

<em> $P_D= 0.3721*1=0.3721$ </em>

$K=\frac{0.2093^3*0.3721^2}{0.0930^2*0.3256} \\K=0.4508$

(iii)

Δ $G=-RTlnK\\$

ΔG = $-8.314*(273+25)*ln(0.4508)\\=1973.96J\\=1.974kJ$

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Explanation:

Sorry if wrong

7 0

3 years ago

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