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Ilia_Sergeevich [38]
3 years ago
14

Be sure to answer all parts.The combustion of a 40.0−g gaseous mixture of H2 and CH4 releases 3766 kJ of heat. Calculate the amo

unt of each gas in grams. Assume that the product of the combustion of H2 is H2O(l) and that the products of the combustion of CH4 are CO2(g) and H2O(l).
H2 =
CH4=
Chemistry
1 answer:
Svetach [21]3 years ago
7 0

Answer:H2=11.4g

CH4=28.6g

Explanation:The complete combustion of the two gases can be represented by a balanced reaction below

1. CH4 +2O2___CO2+2H2O

2.2H2+O2___2H2O

Combining the two we have CH4 +2H2+3O2___

CO2+4H2O

Since the mixture contains 40gof CH4 and 2, therefore 20g of CH4 and 8g of H2 combines.

Calculated from their molecular Mass i.e CH4 12+4×2)=20 and 2H2= 2×2×2=8g

Mass of CH4=20/28×40=28.6g

2H2=8/28×40=11.4g

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Compounds A and B react to form compounds C and D according to the equation: aA + bB → cC + dD. Under which conditions will the
Ipatiy [6.2K]

Answer: A. The reaction takes place in one step.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction.

Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.

Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.

aA=bB\rightarrow cC+dD

Rate=k[A]^a[B]^b

k= rate constant

a= order with respect to A

b = order with respect to B

5 0
3 years ago
Read 2 more answers
PLZ HELP FAST
astra-53 [7]
#1
The correct answer is A.)

Because the animal has sperm that means that the offspring will be different. This is because at conception, the two parents sex cells combine their DNA to make a new creature.

#2
The correct answer is B.)

Genetic variation produces different offspring, and some are more adapted to others. The ones that survive pass on their genes. This is know as "Survival of the Fittest". Asexually organisms are exactly like their parents, so if they have any type of genetic defaults, then their offspring will get them too. 

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3 0
3 years ago
There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
100 points
Mariana [72]

Answer:

Where is the puzzle!??

I surely can help you

5 0
3 years ago
Read 2 more answers
Draw a structural formula for the alkene you would use to prepare the alcohol shown by hydroboration/oxidation.
Nastasia [14]

Answer:

See explanation

Explanation:

The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.

This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).

In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.

Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.

Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.

5 0
3 years ago
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