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olga55 [171]
3 years ago
7

Why do we not know yet if we live in multiverse

Chemistry
1 answer:
kherson [118]3 years ago
4 0

The people of he world believe and they don't believe.

I,myself, think we do land don't live in multiverse.

Why: Because in science I hear that there are many universes out there that need to b explored.

Not: Because if we did, don't you think that Astronauts and the U.S. would try and find different universes.

I hope I helped you!:)

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Put the level of organization in order from smallest to largest a. Organelles, cellsorgans, tissuesorgan systems b. Organs, tiss
Lesechka [4]

Answer: My guess is D i hope this is right

Explanation:

8 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=PdH_{2}

       we know;

Molecular formula=n(Empirical formula)

molecular weight of palladium(Pd)=106.4u

molecular weight of hydrogen(H)=1u

Molar mass of PdH_{2}:

Pd=106.4×1=106.4u

H=1×2=2

molar mass of PdH_{2}=106.4+2=108.4

n=\frac{216.8}{106.4}

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

Molecular formula=2(PdH_{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}

4 0
2 years ago
The pKb of the base cyclohexamine, C6H11NH2, is 3.36. What is the pKa of the conjugate acid, C6H11NH3
bonufazy [111]

Answer:

10.64

Explanation:

Let's consider the basic reaction of cyclohexamine, C₆H₁₁NH₂.

C₆H₁₁NH₂(aq) + H₂O(l) ⇄ C₆H₁₁NH₃⁺(aq) + OH⁻     pKb = 3.36

C₆H₁₁NH₃⁺ is its conjugate acid, since it donates H⁺ to form C₆H₁₁NH₂. C₆H₁₁NH₃⁺ acid reaction is as follows:

C₆H₁₁NH₃⁺(aq) + H₂O(l) ⇄ C₆H₁₁NH₂(aq) + H₃O⁺(aq)   pKa

We can find the pKa of C₆H₁₁NH₃⁺ using the following expression.

pKa + pKb = 14.00

pKa = 14.00 - pKb = 14.00 - 3.36 = 10.64

6 0
3 years ago
What are some insulators and conductors of electricity? Name 2 of each
hammer [34]

Answer:

insulator:wood,rubber

conductor:aluminium,copper

4 0
2 years ago
What is the [OH-] of a substance that has a pH of 11?
Deffense [45]

Answer:

0.001 M OH-

Explanation:

[OH-] = 10^-pOH, so

pOH + pH = 14 and 14 - pH = pOH

14 - 11 = 3

[OH⁻] = 10⁻³ ; [OH-] = 0.001 M OH-

6 0
2 years ago
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