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3 years ago

Which of these is the BEST definition of heredity?

Chemistry

1 answer:

valkas [14]3 years ago

3 0

Answer:

If you post them I can answer for you :)

Explanation:

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The tilt of the Earth's axis of rotation is responsible for the A) ocean's tides. B) size of the Moon. C) brightness of stars. D

SIZIF [17.4K]

its d earths seasons

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4 years ago

The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning th

Klio2033 [76]

Answer:

The enthalpy of the reaction of the chlorine with ozone is -162.5 kJ/mol.

Explanation:

$ClO(g)+O_3(g)\rightarrow Cl(g)+2O_2(g),\Delta H_{rxn}=-122.8 kJ/mol$ ...(1)

$2O_3(g)\rightarrow 3O_2(g),\Delta H_{rxn}=-285.3 kJ/mol$ ...(2)

$O_3(g)+Cl(g)\rightarrow ClO(g)+O_2(g),\Delta H_{rxn}=x$ ...(3)

The enthalpy of reaction for the reaction of chlorine with ozone can be calculated by using Hess's law:

(2 )- (1) = (3)

$x=-285.3 kJ/mol - (-122.8 kJ/mol)=-162.5 kJ/mol$

The enthalpy of the reaction of the chlorine with ozone is -162.5 kJ/mol.

6 0

3 years ago

Will mark BRAINLIEST question and choices in photos

ohaa [14]

Answer:

what to do in this question

8 0

3 years ago

Observe

zheka24 [161]

Answer:copper , single replacement, zinc sulfate

Explanation:

4 0

3 years ago

Read 2 more answers

Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

Mariulka [41]

Answer: The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

Explanation:

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

Initial: 0.025 0.025

At eqllm: 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

6 0

4 years ago