All categories

3 years ago

The larger the energy difference between reactants and products (ΔE), the slower the rate of reaction. true or false

Chemistry

1 answer:

Nikitich [7]3 years ago

8 0

Answer: true

Explanation:

In chemical reactions, an energy barrier exists between reactants and products. The magnitude of this energy barrier determines the rate of reaction. A lesser energy barrier implies that reactants are converted to products faster since the energy required is not too much. On the other hand, a large energy difference between reactants and products will lead to a slow reaction with very poor yield of products if any.

You might be interested in

What protect the contents of the nucleus

densk [106]

Answer:

The nucleus contains all of the genetic material for a eukaryotic cell, but this genetic material needs to be protected.

Explanation:

Hopes this helps

7 0

3 years ago

Read 2 more answers

Caffeine, a stimulant in coffee and tea, has a molar mass of 194.19 g/mol and a mass percentage composition of 49.48% C, 5.19% H

lozanna [386]

Answer : The molecular formula of a caffeine is, $C_8H_{10}N_4O_2$

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.48g}{12g/mole}=4.12moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.19g}{1g/mole}=5.19moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{28.85g}{14g/mole}=2.06moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{16.48g}{16g/mole}=1.03moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{4.12}{1.03}=4$

For H = $\frac{5.19}{1.03}=5.03\approx 5$

For N = $\frac{2.06}{1.03}=2$

For O = $\frac{1.03}{1.03}=1$

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_4H_5N_2O_1=C_4H_5N_2O$

The empirical formula weight = 4(12) + 5(1) + 2(14) + 16 = 97 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{194.19}{97}=2$

Molecular formula = $(C_4H_5N_2O)_n=(C_4H_5N_2O)_2=C_8H_{10}N_4O_2$

Therefore, the molecular of the caffeine is, $C_8H_{10}N_4O_2$

5 0

3 years ago

Use stoichiometry to calculate the number of moles produced of Iron (lll) oxide given three moles of water.

Tju [1.3M]

The balanced chemical equation is:
2Fe+3H2O=Fe2O3+3H2
We are given the amount of water to be used in the reaction. This will be the starting point for the calculations.

3 mol H2O ( 1 mol Fe2O3 / 3 mol H2O ) = 1 mol Fe2O3

Therefore, the correct answer is option B.

6 0

3 years ago

Which of the following best helps to account for the fact that the F- ion is smaller than the O2- ion?

katrin [286]

Answer: Option (B) is the correct answer.

Explanation:

Both oxygen and fluorine are period 2 elements and when we move across a period then there occurs a decrease in atomic size of the atoms. Hence, the atomic radius of a neutral fluorine atom is smaller than a neutral oxygen atom.

Moreover, atomic number of fluorine is 9 and it has higher nuclear charge due to which it will cause more attraction of electrons. As a result, size of a fluorine ion will be smaller.

On the other hand, size of oxygen atom is larger and has small nuclear charge due to which attraction of electrons by its nucleus will not be strong enough. Hence, the size of $O^{2-}$ will be larger.

Thus, we can conclude that the statement $F^{-}$ has a larger nuclear charge than $O^{2-}$ has, is correct for the fact that the $F^{-}$ ion is smaller than the $O^{2-}$ ion.