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2 years ago

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Goodwin County has plenty of water. The residents of Goodwin County enjoy fishing, boating, and swimming in Goodwin Lake. The Go

odwin River feeds the lake. Precipitation from the area flows into the Goodwin River. A dam in the lake controls flooding. This section of the lake provides usable water to Goodwin County.
goodwin is a
watershed
collection area
divide

1 answer:

MrRa [10]2 years ago

5 0

Answer:

I am pretty sure it is a watershed

Explanation:

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2 years ago

I need help with this assignment

yuradex [85]

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 $C_{8}H_{8}$ :

molecular weight of carbon=12.04u

molecular weight of hydrogen=1u

C=8×12.01=96.08u

H=1×8=8u

molar mass of $C_{8}H_{8}$ =96.08+8=104.08u

n=104.08÷78.0

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

Molecular formula = n(Empirical formula)

Molecular formula = 1( $C_{8}H_{8}$ )

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

Therefore the molecular formula of a compound is $C_{8}H_{8}$

7. Given:

mass of oxide of nitrogen=108g

mass of nitrogen=4.02g

mass of oxygen=11.48g

moles of nitrogen= $\frac{4.04}{14.01}$ = 0.289 moles

moles of oxygen= $\frac{11.46}{15.999}$ =0.716 moles

We divide through by the lowest molar quantity to give an empirical formula of $N_{2} O{5}$ .

Now the molecular formula is multiple of the empirical formula.

So,

108 = n × (2×14.01 + 5×15.999)

Clearly,n=1, and the molecular formula is $N_{2}O_{5}$ .

8.For the table refer the attached file.

Simplest ratio of elements:

Phosphorus=2

Oxygen=3

We know;

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

molecular formula= 2(Empirical formula)

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

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

Therefore the molecular formula of the compound is $P_{4}O_{6}$

9. For the table refer the attached file.

Simplest ratio of elements:

Carbon=2

Hydrogen=9

Oxygen=2

We know;

Empirical formula = $C_{2} H_{4} O$

Molecular formula = 2(Empirical formula)

Molecular formula =2( $C_{2} H_{4} O$ )

Molecular formula = $C_{4}H_{8} O_{2}$

Therefore the molecular formula of the compound is $C_{4}H_{8} O_{2}$

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