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3 years ago

12. How many seconds are in 90 years? Show all work.

Chemistry

2 answers:

den301095 [7]3 years ago

7 0

Answer:

The answer would be 2,838,240,000

Explanation: In 1 year there are 31,536,000 and if you multiply that number by 90 you would get 2,838,240,000

Varvara68 [4.7K]3 years ago

3 0

Answer:

2838240000

Explanation:

1 calendar year =

31536000

X 90

2838240000

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3 years ago

Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the f

julia-pushkina [17]

Explanation:

As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.

The chemical species which accept proton are called Brønsted-Lowry base.

(a) $HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-$

$HNO_3$ is Bronsted lowry acid and $NO_3^-$ is its conjugate base.

$H_2O$ is Bronsted lowry base and $H_3O^+$ is its conjugate acid.

(b)

$CN^- + H_2O \rightarrow HCN + OH^-$

$CN^-$ is Bronsted lowry base and HCN is its conjugate acid.

$H_2O$ is Bronsted lowry acid and $OH^-$ is its conjugate base.

(c)

$H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-$

$H_2SO_4$ is Bronsted lowry acid and $HSO_4^-$ is its conjugate base.

Cl^- is Bronsted lowry base and HCl is its conjugate acid.

(d)

$HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O$

$HSO_4^-$ is Bronsted lowry acid and $SO_4^{2-}$ is its conjugate base.

OH^- is Bronsted lowry base and $H_2O$ is its conjugate acid.

(e)

$O_{2-}+H_2O \rightarrow 2OH^-$

$O_{2-}$ is Bronsted lowry base and OH- is its conjugate acid.

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4 years ago

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3 years ago

How many valence electrons does Bromine have?

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3 years ago

Read 2 more answers

What is the molarity of a NaOH solution if 28.2 mL of a 0.355 M H2SO4 solution is required to neutralize a 25.0-mL sample of the

Anni [7]

Answer:

A) 0.801

Explanation:

The reaction that takes place is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

First we calculate the H₂SO₄ moles that reacted:

0.355 M * 28.2 mL = 10.011 mmol H₂SO₄

Now we convert H₂SO₄ moles to NaOH moles:

10.011 mmol H₂SO₄ * $\frac{2mmolNaOH}{1mmolH_2SO_4}$ = 20.022 mmol NaOH

Finally we calculate the molarity of the NaOH solution:

20.022 mmol NaOH / 25.0 mL = 0.801 M

So the answer is option A.

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3 years ago