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2 years ago

What causes pressure inside a tire

Chemistry

1 answer:

earnstyle [38]2 years ago

7 0

Answer: Changes In Temperature

Explanation:

Causes your tires to lose pressure as the air inside them expands and contracts, triggering the “Check Tire Pressure”

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Which of the following reactions will proceed as drawn below? (Yes or No)

Crazy boy [7]

Answer:

i wish i could help but t know either so i wish you luck for an answer once again truly sorry don'

Explanation:

i am sorry but i do not know :(

5 0

3 years ago

(2)Answer asappppp pls I neee help

pav-90 [236]

Answer:

It’s correct

Explanation:

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2 years ago

What is the concentration of H+ in 0.0025 M HClO4? What is the pH of the solution? What is the OH− concentration in the solution

alexandr402 [8]

Answer:

A. The concentration of H+ is 0.0025 M

B. The pH is 2.6

C. The concentration of OH- is 3.98x10^-12 M

Explanation:

We'll begin by writing the balanced

dissociation equation of HClO4. This is illustrated below:

HClO4 —> H+ + ClO4-

A. Determination of the concentration of H+ in 0.0025 M HClO4. This is illustrated below:

From the balanced equation above,

1 mole of HClO4 produced 1 mole of H+.

Therefore, 0.0025 M of HClO4 will also produce 0.0025 M of H+.

The concentration of H+ is 0.0025 M

B. Determination of the pH.

The pH of the solution can be obtained as follow:

The concentration of H+, [H+]

= 0.0025 M

pH =?

pH = - log [H+]

pH = - log 0.0025

pH = 2.6

C. Determination of the concentration of OH-

To obtain the concentration of OH-, we must first calculate the pOH of the solution. This is illustrated below:

pH + pOH = 14

pH = 2.6

pOH =?

pH + pOH = 14

2.6 + pOH = 14

Collect like terms

pOH = 14 - 2.6

pOH = 11.4

Now, we can calculate the concentration of the OH- as follow:

pOH = - Log [OH-]

pOH = 11.4

11.4 = - Log [OH-]

- 11.4 = log [OH-]

[OH-] = anti log (- 11.4)

[OH-] = 3.98x10^-12 M

8 0

3 years ago

True or false when water melts energy goes out

SVETLANKA909090 [29]

I think the answer is yes

6 0

3 years ago

Read 2 more answers

Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2

mina [271]

<u>Answer:</u> The amount of energy released per gram of $B_5H_9$ is -71.92 kJ

<u>Explanation:</u>

For the given chemical reaction:

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]$

Taking the standard enthalpy of formation:

$\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol$