All categories

3 years ago

Explain how changes in temperature affect the particle motion of a substance.

1 answer:

8 0

When the temperature increases, warms, or heats up the molecules become excited. Their movement becomes faster and they tend to move far apart from other molecules.

You might be interested in

In aqueous solution amino acids are rarely found in the neutral, unionized form.

True [87]

a. True.

There is always an equilibrium of the type

NH₃⁺CHRCOOH ⇌ NH₃⁺CHRCOO⁻ ⇌ NH₂CHRCOO⁻

The compound is <em>always in an ionized form</em>.

There are no unionized NH₂CHRCOOH molecules in the solution.

3 0

3 years ago

Which reaction below represents a balanced, double replacement chemical reaction? A) C2H5OH + 3O2 → 2CO2 + 3H2O B) Rb2O + Cu(C2H

guajiro [1.7K]

Option B. <span>Rb2O + Cu(C2H3O2)2 → 2RbC2H3O2 + CuO is the correct answer. HOPE IT HELPS</span>

8 0

3 years ago

Read 2 more answers

In which of the following reactions will Kc = Kp? Group of answer choices N2(g) + 3 H2(g) 2 NH3(g) N2O4(g) 2NO2(g) CO(g) + 2 H2(

Artist 52 [7]

Answer:

H₂(g) +I₂(g) ⟶ 2HI(g)

Explanation:

Kc =Kₚ when the number of moles of gaseous products equals the number of moles of gaseous reactants.

The HI reaction has two moles of gas on each side of the reaction arrow.

K = (Products)ⁿ/(Reactants)ⁿ = (Products/Reactants)ⁿ

Thus, if n is the same for products and reactants, you will get the same number whether you use concentrations or pressures, and Kc = Kₚ

8 0

3 years ago

Now let us refine our model by noting that there is a second source of Na+ ions in the cell: NaI. Suppose the outside of the cel

MArishka [77]

Answer:

$E=55mV$

Explanation:

Hello,

Considering the given information, the new concentration of Na+inside will be (13.6 + 4) mM = 17.6 mM, and outside will be (150 + 0.04) mM = 150.04 mM due to the previous specified conditions.

Now, the recalculation of the Nerst potential at the supposed temperature of 25 °C (which is modifiable) is done via:

$E=\frac{RT}{zF} ln(\frac{C_{Na^+,outside}}{C_{Na^+,outside}} )\\\\E=\frac{8.314\frac{J}{mol*K}*298K}{1*9.65x10^4\frac{C}{mol} } *ln(\frac{150.04}{17.6} )=0.055V*\frac{1x10^3mV}{1V} \\E=55mV$

Best regards.

8 0

3 years ago

The hydrolysis of esters in base is called ________. a. the hunsdiecker reaction b. transesterification c. saponification d. the

KengaRu [80]

The hydrolysis of esters in base is called saponification .

So, option C is correct one.

The saponification is the process that involves conversion of fats , oils , lipids into soap and water in the presence of alkaline medium. Saponification is the process of making soap.

During the saponification process, the mixture has an acidity, which tells that it's not safe for usage. After the saponification process is complete, the pH should be a base.The process of formation of carboxylic salt and water by hydrolysis of ester in base is called saponification.

learn more about saponification

brainly.com/question/2263502

#SPJ4

7 0

2 years ago