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kotegsom [21]
3 years ago
7

Consider the following intermediate chemical equations.

Chemistry
2 answers:
mixas84 [53]3 years ago
8 0

Answer:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).

Explanation:

  • We have two equations:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

2H₂O(g) → 2H₂O(l)

  • To add the two equations: we omit H₂O(g)  that is formed by 2 moles in the product side of the first equation and consumed by 2 moles from the reactants side in the second equation

  • So, the overall chemical equation is obtained by combining these intermediate equations is:

<em>CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).</em>

<em></em>

Vadim26 [7]3 years ago
4 0

Answer:  The reaction is Exothermic

CH4(g) + 2O2(g) > CO2(g) + 2H2O(g) + energy

Explanation:

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Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,
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Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

6 0
3 years ago
Consider the following first order decomposition reaction with a half-life of 65 seconds at a given temperature:
katrin [286]

Answer:

130

Explanation:

This is because that 3atm of N2O4 is used up for the 6atm of NO2, so 1 atm N2O4 is left. Resulting in In(1/4).

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Which of the following have a neutral charge
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4 years ago
PH is 7.45. Calculate value of [H3O+] and [OH-]
lana66690 [7]

Answer:

The answer is (H30+) =3,55e-8M and (OH-)=2,82e-7M

Explanation:

We use the formulas:

pH= - log(H30+)  and Kwater=(H30+)x(OH-)

pH= - log(H30+)  ----< (H30+)= antilog- pH=antilog- 7,45=3,55E-8M

Kwater=(H30+)x(OH-)

(OH-)=Kwater/(H30+)= 1,00e-14/3,55e-8 = 2,82e-7

8 0
3 years ago
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