2. Which is true of all elements?

What is the answer? show proof I will report you if you answer for the points

IceJOKER [234]

Answer:

A

Explanation:

carbon dioxide is released in cellular respiration

3 0

2 years ago

A+common+iv+solution+is+0.9%+saline+(nacl+solution).+what+is+the+osmolarity+of+0.9%+saline+mosmoles/l?+the+molecular+weight+of+n

Vedmedyk [2.9K]

An osmolarity of saline solution is 308 mosmol/L.

m(NaCl) = 9 g; the mass of sodium chloride

V(solution) = 1 L; the volume of the saline solution

n(NaCl) = 9 g ÷ 58.44 g/mol

n(NaCl) = 0.155 mol; the amount of sodium chloride

number of ions = 2

Osmotic concentration (osmolarity) is a measure of how many osmoles of particles of solute it contains per liter.

The osmolarity = n(NaCl) ÷ V(solution) × 2

The osmolarity = 0.154 mol ÷ 1 L × 2

The osmolarity = 0.154 mol/L × 1000 mmol/m × 2

The osmolarity of the saline solution = 308 mosm/L.

More about osmolarity: brainly.com/question/13258879

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8 0

10 months ago

What elements were formed in Nuclear fusion?

Sonja [21]

Answer:

tritium and deuterium are combined and result in the formation of helium

6 0

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= $\frac{216.8}{106.4}$

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

2 years ago

What is the correct label for a solution?

Marina86 [1]

The answer is a. a homogenous mixture

6 0

3 years ago

Read 2 more answers