.What is the KE of a baseball that has a mass of 100 kg and is traveling at 5 m/s?

1250joules

Explanation:

The kinetic energy is energy in motion and it's formula is

KE =1/2MV^2

M represent mass which is 100Kg.

V represent velocity which is 5m/s

Therefore, KE = 1/2×100×5× 5

= 50 × 25

KE= 1250joules

answer : the correct answer is 2) hno₃ and naoh .

arrhenius concept of acid and base :

acid :

according to arrhenius acid are those which release h⁺ or hydronium ion (h₃o⁺) in aqueous solution .

example : hcl +h₂o ↔ h₃o⁺ + cl⁻

hcl is giving h₃o⁺ ion in water , hence it is an arrhenius acid .

base :

according to arrhenius , bases are those which releases oh⁻ ions in aqueous solution .

example : koh (aq) ↔ k⁺ + oh⁻

since koh is producing oh⁻ ions hence , it is an arrhenius base .

checking the options :

1) ch₃oh and naoh :

ch₃oh is known as methanol , an organic compound , which is neutral compound . the dissociation of ch₃oh is negligible . it is very very weak acid (ka = 3.2 * 10 ⁻¹² ) .

naoh being ionic gives na⁺ and oh⁻ ions in aqueous solution as :

naoh ↔ na⁺ + oh⁻ . hence it is an arrhenius base

it can be concluded that ch₃oh and naoh is not arrhenius acid and base.

2) hno₃ and naoh :

hno₃ is an ionic compound which gives ions h⁺ and no₃⁻ ion in aqueous solution as:

hno₃ + h₂o ↔ h₃o⁺ + no₃⁻

since it is giving h⁺ ion . hno₃ is arrhenius acid .

naoh being ionic gives na⁺ and oh⁻ ions in aqueous solution as :

naoh ↔ na⁺ + oh⁻

. hence it is an arrhenius base .

it can be said that hno₃ and naoh are arrhenius acid and base.

3) ch₃oh and hcl

as discussed ch₃oh is neutral compound .

while hcl in aqueous solution dissociates to give h⁺ and cl⁻ ions as :

hcl + h₂o→ h₃o⁺ + cl⁻ . so hcl is arrhenius acid .

hence , ch₃oh and hcl are not arrhenius acid and base.

4) hno₃ and hcl :

hno₃ as discussed gives h⁺ in aqueous solution , while hcl also gives h⁺ in aqueous solution .

hence both hno₃ and hcl are arrhenius acid they are not acid and base .

so the arrhenius acid and base are 2) hno₃ and naoh.