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3 years ago

9

A ball rolling down a hill is what form of energy?

2 answers:

Maksim231197 [3]3 years ago

7 0

Answer:

kinetic I think that's the answer

Luden [163]3 years ago

5 0

Answer:

Kinetic energy

Explanation:

An object that is in motion possesses Kinetic energy. A ball rolling down a hill will possess kinetic energy.

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Which of the following compounds are soluble in aqueous solutions? Select ALL that are soluble.

sattari [20]

<h3>Answer:</h3>

K₂S
CaSO₄ (slightly soluble in water)
Pb(ClO₄)₂

<h3>Explanation:</h3>

Mg(OH)₂ and CoCO₃ are insoluble in water.
Solubility rules are used to determine whether a compound will be soluble or not soluble in aqueous form.
K₂S is soluble in aqueous solutions as all salts of Potassium are soluble in water.

Calcium (ii) sulfate is slightly soluble in water.
Hydroxides are insoluble in water except, KOH, NaOH and NH₄OH.and Barium hydroxide. Therefore, Mg(OH)₂ is an insoluble hydroxide.

CoCO₃ is insoluble since carbonates are insoluble except K₂CO₃, Na₂CO₃ and (NH₄)₂CO₃. .

Pb(ClO₄)₂ is soluble because all perchlorates are soluble.

7 0

3 years ago

Textbook mass 2000 grams volume 4000cm3 density

PIT_PIT [208]

Density = mass/volume = 2000/4000 = 0.5 grams/cm3. Hope this hopes!

8 0

3 years ago

Which answer is INCORRECT not the correct one

Elena L [17]

Explanation:

It's 4

Hope it's help....

<h3>^_^</h3>

6 0

3 years ago

Read 2 more answers

Which physical property can NOT be observed using 5 senses without changing the chemical makeup of the object that you are obser

bonufazy [111]

Ummm i believe it’s C

4 0

3 years ago

The pre-exponential constant and activation energy for the diffusion of iron in cobalt are 1.7 x 10-5 m2/s and 273,300 J/mol, re

Kitty [74]

<u>Answer:</u> The temperature of the system will be 1622 K

<u>Explanation:</u>

The equation relating the pre-exponential factor and activation energy follows:

$\log D=\log D_o-\frac{E_a}{2.303RT}$

where,

D = diffusion coefficient = $2.7\times 10^{-14}m^2/s$

$D_o$ = pre-exponential constant = $1.7\times 10^{-5}m^2/s$

$E_a$ = activation energy of iron in cobalt = 273,300 J/mol

R = Gas constant = 8.314 J/mol.K

T = temperature = ?

Putting values in above equation, we get:

$\log (2.7\times 10^{-14})=\log (1.7\times 10^{-5})-\frac{273,300}{2.303\times 8.314\times T}\\\\T=1622K$

Hence, the temperature of the system will be 1622 K

8 0

3 years ago