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3 years ago

Why it is advisided to freshly prepare the ferrous sulphate solution for experiment

1 answer:

5 0

Here is a site my buddie has to help you. Well co-owner..
https://www.quora.com/Why-is-fresly-prepared-FeSO4-required-for-the-ring-test

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The structural formula for CF4 is

nevsk [136]

Answer:

PubChem CID 6393

Structure

Find Similar Structures

Chemical Safety

Compressed Gas

Laboratory Chemical Safety Summary (LCSS) Datasheet

Molecular Formula CF4

Molecular Weight

88.004 g/mol

Explanation:

5 0

3 years ago

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What is the mole fraction of NaOH in an aqueous solution that

astraxan [27]

Answer:

The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

Explanation:

We are given that

Aqueous solution that contains 22.9% NaOH by mass means

22.9 g NaOH in 100 g solution.

Mass of NaOH(WB)=22.9 g

Mass of water =100-22.9=77.1

Na=23

O=16

H=1.01

Molar mass of NaOH(MB)=23+16+1.01=40.01

Number of moles = $\frac{Given\;mass}{Molar\;mass}$

Using the formula

Number of moles of NaOH $(n_B)=\frac{W_B}{M_B}=\frac{22.9}{40.01}$

$n_B=0.572moles$

Molar mass of water=16+2(1.01)=18.02g

Number of moles of water $(n_A)=\frac{77.1}{18.02}$

$n_A=4.279 moles$

Now, mole fraction of NaOH

= $\frac{n_B}{n_B+n_A}$

$=\frac{4.279}{0.572+4.279}$

=0.882

Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

4 0

3 years ago

Ammonia and oxygen react to form nitrogen monoxide and water, like this:

olchik [2.2K]

4NH3+5O2 <=>4NO + 6H2O

Using the definition of Kp, we have
Kp=(Pno^4*Ph2o^6)/(Pnh3^4*Po2^5)
where Pno=partial pressure of NO, etc.

The numerical value for a given temperature can be evaluated when the actual partial pressures are known.

6 0

3 years ago

calculate the (m/v) of nacl of a solution made by diluting 25.0 ml of a 7.50 (m/v)% NaCl solution to a total volume of 75.00 ml

postnew [5]

Answer:

Explanation:

8 0

3 years ago

Read 2 more answers

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