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3 years ago

There are 1.806×10^25 molecules of water. How many moles of water does the sample contain?

Chemistry

1 answer:

Arada [10]3 years ago

4 0

1.806e+25 is the answer to your equation.

Hope this Helped!

;D

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Hydrofluoric acid, hf, has a ka of 6.8 × 10−4. what are [h3o+], [f−], and [oh−] in 0.710 m hf?

STALIN [3.7K]

Answer:

[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.

Explanation:

For a weak acid like HF, the dissociation of HF will be:

HF + H₂O ⇄ H₃O⁺ + F⁻.

[H₃O⁺] = [F⁻].

∵ [H₃O⁺] = √Ka.C,

Ka = 6.8 x 10⁻⁴, C = 0.710 M.

∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.

∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(2.2 x 10⁻²) = 4.55 x 10⁻¹³.

6 0

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

n=2

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= $\frac{30.0}{15.01}$

n=2

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= $\frac{129}{64.5}$

n=2

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= $\frac{577}{278.32}$

n=2

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= $\frac{184.5}{61.5}$

n=3

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

n=1

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

4 0

3 years ago

If measuring experimental results, what can you predict about the work output of a 1200 watt hair dryer? the work output of the

eimsori [14]

Answer: first option, the work output of the hairdryer will be less than the work input.

Explanation:

1) The work output measured in watts is the power of hair dryer measured in joules per second.

2) The hair dryer converts electrical energy from the wall outlet to mechanical and thermal energy: hot wind.

3) Nevertheless, you can never expect a 100% efficiency of the machines: due to friction, some energy is converted into useless energy.

So, efiiviency = power output / power input< 1 ⇒

power output = work output / time

input power = work input / time

⇒ work output / work input < 1

⇒ work output < work input.

Which is the first option: the work output of the hairdryer will be less than the work input

6 0

3 years ago

Read 2 more answers

Directions Now that the lab is complete, it is time to write your lab report. The purpose of this guide is to help you write a c

Elza [17]

Answer:

foot

Explanation:

6 0

2 years ago

D5W is a solution used as an intervenous fluid.It is 5.0% by mass solution of the dextrose (C6H12O6) in water. if the density of

Alona [7]

Answer:

0.29 mol/L

Explanation:

Its density is 1.029 g/ml so in a liter (1000 mL) there is 1029 g of solution, but only 5% is dextrose.

0.05x1029=51.45

So in a liter of D5W solution there is 51.45 g of dextrose.

Dextrose molar mass iss 180.156 g/mol, so in 51.45 g of dextrose there is

51.45/180.156=0.29 mol

In one liter of solution there is 0.29 mol of dextrose, so the molarity of such solution is 0.29 mol/L.

6 0

3 years ago