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IRINA_888 [86]
3 years ago
6

Which is not true regarding reaction rates? (2 points)

Chemistry
1 answer:
PtichkaEL [24]3 years ago
5 0

Let us consider each statement one by one

a) Catalysts are the substance which alter the reaction pathway where activation energy of reaction is less that the un catalyzed reaction. They are not used up in the reaction . the catalyst is restored in the reaction.

b) As mentioned above, the catalyst speed up reaction by lowering the activation energy

c) During a reaction the reactants are consumed up and products are formed. so there is a decrease in concentration of reactant.

Thus all the above statements are true.

D) the rate of decrease in concentration of reactants depends upon the coefficient of reactant in balanced chemical reaction.

For example

H_{2} (g)  + \frac{1}{2} O_{2} (g) -->    H_{2}O(g)

Here the rate of decrease in concentration of hydrogen will be double the rate of decrease in concentration of oxygen.

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Basically it means that there is a sort of skin on the surface of water where the water molecules hold on tight together. If the conditions are right, they can hold tight enough to support your paper clip. The paperclip is not truly floating, it is being held up by the surface tension.

5 0
3 years ago
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Which of the following is a correct statement about collision theory?
rjkz [21]
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3 years ago
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A 0.1326 g sample of magnesium was burned in an oxygen bomb calorimeter. the total heat capacity of the calorimeter plus water w
Sladkaya [172]

Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol

Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.

W are given a chemical reaction:

Mg(s)+\frac{1}{2}O_2(g)\rightarrow MgO(s)

c=5760J/^oC

\Delta T=0.570^oC

To calculate the enthalpy change, we use the formula:

\Delta H=c\Delta T\\\\\Delta H=5760J/^oC\times 0.570^oC=3283.2J

This is the amount of energy released when 0.1326 grams of sample was burned.

So, energy released when 1 gram of sample was burned is = \frac{3283.2J}{0.1326g}=24760.181J/g

Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

\Delta H=24760.181J/g\times 24g/mol\\\\\Delta H=594244.3J/mol\\\\\Delta H=594.244kJ/mol

4 0
3 years ago
Please help it’s due today I will give brainliest
Thepotemich [5.8K]

1st and 4th options are suitable answers, as these 2 changes are not exactly physical changes as it cant return back to original form and as well as its not cooling, so I feel its 1st and 4th options

7 0
3 years ago
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
3 years ago
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