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4 years ago

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A compound of copper and sulfur contains 88.39 g of metal and 44.61 g of nonmetaL How many grams of copper are in 5264 kg of com

pound? How many grams of sulfur?

2 answers:

11Alexandr11 [23.1K]4 years ago

6 0

<u>Answer:</u> The mass of copper and sulfur in given amount of compound will be 3498383.16 g and 1765616.84 g respectively.

<u>Explanation:</u>

We are given:

Mass of metal (copper) = 88.39 g

Mass of non-metal (sulfur ) = 44.61 g

Mass of compound = Mass of metal + Mass of non-metal = (88.39 + 44.61) = 133 g

Mass of compound = 5264 kg = 5264000 g (Conversion factor: 1 kg = 1000 g)

To calculate the mass of copper and sulfur in given amount of compound, we use unitary method:

<u>For Copper:</u>

In 133 g of compound, the mass of copper is 88.39 g

So, in 5264000 g of compound, the mass of copper will be = $\frac{88.39}{133}\times 5264000=3498383.16g$

<u>For Sulfur:</u>

In 133 g of compound, the mass of sulfur is 44.61 g

So, in 5264000 g of compound, the mass of sulfur will be = $\frac{44.61}{133}\times 5264000=1765616.84g$

Hence, the mass of copper and sulfur in given amount of compound will be 3498383.16 g and 1765616.84 g respectively.

Dahasolnce [82]4 years ago

3 0

Answer :

The mass of copper in 5264 kg of compound are, 3498383.158 g

The mass of sulfur in 5264 kg of compound are, 1765616.842 g

Explanation : Given,

Mass of metal = 88.39 g

Mass of non-metal = 44.61 g

Given mass of compound = 5264 kg = 5264000 g (1 kg = 1000 g)

Now we have to calculate the total mass of compound.

Total mass of compound = Mass of metal + Mass of non-metal

Total mass of compound = 88.39 + 44.61 = 133 g

Now we have to calculate the mass of copper.

As, 133 g of compound contains 88.39 g of copper

So, 5264000 g of compound contains $\frac{5264000}{133}\times 88.39=3498383.158g$ of copper

The mass of copper in 5264 kg of compound are, 3498383.158 g

Now we have to calculate the mass of sulfur.

As, 133 g of compound contains 44.61 g of sulfur

So, 5264000 g of compound contains $\frac{5264000}{133}\times 44.61=1765616.842g$ of sulfur

The mass of sulfur in 5264 kg of compound are, 1765616.842 g

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Answer:

ΔHrxn = - 1534.3 J

Explanation:

Given the assumptions and the formula for the change in enthalpy:

ΔHrxn = m x C x ΔT, where

m is the mass of solution given 135.4 g

C is the heat capacity 4.2 J/g .K and,

ΔT is the change in temperature

we have ,

T₁ = ( 18.1 + 273) K = 291.1 K

T₂ = ( 15.4 +273) K = 288.4 K

ΔHrxn = 135.3 g x 4.2 J/gK x ( 288.4 -291.1 ) K = - 1534.3 J

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I need help with this assignment

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The answers for the following sums is given below.

1. $Pd_{2}H_{2}$

2. $C_{2} H_{6}$

3. $C_{2} H_{2} O_{2} Cl_{2}$

4. $C_{3}Cl_{3} N_{3}$

5. $Tl_{2 } C_{4} H_{4}O_{6}$

6. $C_{8}H_{8}$

7. $N_{2}O_{5}$

8. $P_{4}O_{6}$

9. $C_{4}H_{8} O_{2}$

Explanation:

1.Given:

Molar mass=216.8g

Molecular formula=Pd $H_{2}$

we know;

Molecular formula=n(Empirical formula)

molecular weight of palladium(Pd)=106.4u

molecular weight of hydrogen(H)=1u

Molar mass of Pd $H_{2}$ :

Pd=106.4×1=106.4u

H=1×2=2

molar mass of Pd $H_{2}$ =106.4+2=108.4

n= $\frac{216.8}{106.4}$

<u><em>n=2</em></u>

Molecular formula=2(Pd $H_{2}$ )

Molecular formula= $Pd_{2}H_{2}$

Therefore the molecular formula of the compound is $Pd_{2}H_{2}$

2. Given:

Molar mass=30.0g

Molecular formula= $CH_{3}$

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

Molar mass of $CH_{3}$ :

C=12.01 × 1 = 12.01u

H=1 × 3 = 3u

molar mass of $CH_{3}$ =12.01 + 3 =15.01u

n= $\frac{30.0}{15.01}$

<u><em>n=2</em></u>

Molecular formula=2( $CH_{3}$ )

Molecular formula= $C_{2} H_{6}$

Therefore the molecular formula of the compound is $C_{2} H_{6}$

3. Given:

Molar mass=129g

Molecular formula=CHOCl

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

molecular weight of chlorie(Cl)=35.5u

Molar mass of CHOCl:

C=12.01 × 1 = 12.01u

H=1 × 1 = 1u

O=16.00×1=16.00u

Cl=35.5×1=35.5u

molar mass of CHOCl=12.01+1+16.00+35.5=64.5u

n= $\frac{129}{64.5}$

<u><em>n=2</em></u>

Molecular formula=2(CHOCl)

Molecular formula= $C_{2} H_{2} O_{2} Cl_{2}$

Therefore the molecular formula of the compound is $C_{2} H_{2} O_{2} Cl_{2}$

5. Given:

Molar mass=577g

Molecular formula= $TlC_{2} H_{2}O_{3}$

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Thallium(Tl)=204.3u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

Molar mass of $TlC_{2} H_{2}O_{3}$ :

C=12.01 × 2= 24.02u

Tl=204.3×1=204.3u

H=1×2=2u

O=16.00×3=48.00

molar mass of $TlC_{2} H_{2}O_{3}$ =204.3+24.02+1+48.00=278.32u

n= $\frac{577}{278.32}$

<u><em>n=2</em></u>

Molecular formula=2 ( $TlC_{2} H_{2}O_{3}$ )

Molecular formula= $Tl_{2 } C_{4} H_{4}O_{6}$

Therefore the molecular formula of the compound is $Tl_{2 } C_{4} H_{4}O_{6}$

4. Molar mass=184.5g

Molecular formula=CClN

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Nitrogen(N)=14u

molecular weight of chlorine(Cl)=35.5u

Molar mass of CClN:

C=12.01 × 1 = 12.01u

N=1×14=14U

Cl=35.5×1=35.5u

molar mass of CClN=12.01+14+35.5=61.5u

n= $\frac{184.5}{61.5}$

<u><em>n=3</em></u>

Molecular formula=3 (CClN)

Molecular formula= $C_{3}Cl_{3} N_{3}$

Therefore the molecular formula of the compound is $C_{3}Cl_{3} N_{3}$

6. For the table refer the attached file.

Simplest ratio of elements:

Carbon=8

Hydrogen=8

Empirical formula= $C_{8}H_{8}$

Molecular formula = $C_{8}H_{8}$

Molar mass of $C_{8}H_{8}$ :

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molecular weight of hydrogen=1u

C=8×12.01=96.08u

H=1×8=8u

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<em><u>n=1</u></em>

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Molecular formula = $C_{8}H_{8}$

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7. Given:

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Now the molecular formula is multiple of the empirical formula.

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8.For the table refer the attached file.

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Empirical formula= $P_{2} O_{3}$

molecular formula= 2(Empirical formula)

Molecular formula =2( $P_{2} O_{3}$ )

Molecular formula = $P_{4}O_{6}$

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9. For the table refer the attached file.

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Molecular formula = 2(Empirical formula)

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