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MaRussiya [10]
3 years ago
15

O2 + ch4 co2 + h2o what would be the coefficient of the diatomic oxygen molecule in the balanced reaction?

Chemistry
2 answers:
dlinn [17]3 years ago
4 0
2 is the coefficient.
ValentinkaMS [17]3 years ago
4 0

Answer:

<h2>The coefficient of the diatomic oxygen has to be 2.</h2>

Explanation:

The balanced reaction should be like this: 2O_{2} + CH_{4} = CO_{2} + 2H_{2}O

So, as you can see, balance mean that molecules in the reaction has to be equal, the amount of element involved has to be the same, due to the Conservation of the Mass Law.

So, Hydrogen and Oxygen are not balanced in the given expression, that's why we need to add those coefficients.

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A bottle contains 3.100 ml of a liquid. the total mass of the bottle and the liquid together is 6.300 g. the mass of the empty b
Olenka [21]
The answer is <span>a. 0.665 g/m</span>
7 0
3 years ago
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If 34.7 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ could be for
Luba_88 [7]

The  number  of grams   of Ag2SO4  that could be formed  is   31.8  grams



    <u><em> calculation</em></u>

Balanced   equation is  as below

2 AgNO3 (aq)  + H2SO4(aq)  →  Ag2SO4 (s)   +2 HNO3 (aq)


  • Find  the  moles  of  each reactant by use  of  mole= mass/molar mass  formula

that is  moles of  AgNO3= 34.7 g / 169.87  g/mol= 0.204 moles

             moles of  H2SO4 =  28.6  g/98  g/mol  =0.292  moles

  • use the  mole  ratio to determine the moles of  Ag2SO4

   that is;

  •    the mole ratio of  AgNo3 : Ag2SO4 is  2:1 therefore  the  moles of Ag2SO4=  0.204  x1/2=0.102 moles

  • The moles  ratio of H2SO4  : Ag2SO4  is  1:1  therefore  the moles of Ag2SO4 = 0.292  moles

 

  •      AgNO3  is the limiting reagent therefore  the moles of   Ag2SO4 = 0.102  moles

<h3>     finally  find  the mass  of Ag2SO4  by use of    mass=mole  x molar mass  formula</h3>

that  is  0.102   moles  x  311.8  g/mol= 31.8 grams

3 0
3 years ago
The following table lists molecular weight data for a polypropylene material. Molecular Weight Range (g/mol) xi wi
nikklg [1K]

Answer:

a) the number-average molecular weight is 32,400 g/mol

b) the weight-average molecular weight is 36,320 g/mol

c) the degree of polymerization is 770

Explanation:

Given the data in the question;

Molecular                  Mean        Number      Weight      Number        Weight

weight                                                                             average         average

(g/mol)                    Mi(g/mol)         xi                wi          xiMi                  wiMi

8,000-16,000            12,000        0.07           0.03         840                360

16,000-24,000          20,000       0.15           0.09         3000              1800

24,000-32,000         28,000       0.26           0.21          7280              5880

32,000-40,000         36,000       0.27           0.27         9720               9720

40,000-48,000         44,000       0.18            0.28         7920              12320

48,000-56,000         52,000       0.07           0.12          3640              6240

∑                                                                                Mn=32,400     Mw=36,320

so;

a)  the number-average molecular weight

Mn = ∑Mixi

so from the table above; summation of Row Mixi

Mn = ∑Mixi = 32,400

Therefore, the number-average molecular weight is 32,400 g/mol

b) the weight-average molecular weight

Mw = ∑Miwi

so from the table above; summation of Row Miwi

Mw = ∑Miwi = 36,320

Therefore, the weight-average molecular weight is 36,320 g/mol

c) the degree of polymerization

the degree of polymerization of polypropylene can be determined using number-average molecular and repeat unit molecular weight.

now, for polypropylene { CH₂ = CH - CH₃ }

the repeat unit consist of 3 carbon atoms and 6 hydrogen atoms

given that;

Atomic weight of Carbon mC = 12.01 g/mol and

Atomic weight of  Hydrogen mH = 1.008 g/mol

now we find the repeat unit molecular weight of polypropylene

m = nCmC + nHmH

where n is the number of repeat of atoms

so we substitute

m = ( 3 × 12.01) + ( 6 × 1.008)

m = 36.03 + 6.048

m = 42.078 g/mol

now we calculate the degree of polymerization;

DP = Mn / m

so we substitute

DP = 32,400 / 42.078

DP = 769.9985 ≈ 770

Therefore,  the degree of polymerization is 770

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