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3 years ago

During a change of state, the temperature of a substance _____.

2 answers:

4 0

Hello,

Your questions states:

During a change of state, the temperature of a substance _____?

In which you gave us some choices:

A. decreases if the arrangement of particles in the substance changes.
B. remains constant until the change of state is complete.
C. increases if the kinetic energy of the particles in the substance increases.
D. increases during melting and vaporization and decreases during freezing and condensation.

Your answer would be:

B. remains constant until the change of state is complete.

Your explanation/Reasoning:

It absorbs the energy, then after the phase changes it then increases the temperature all over again.

Have a nice day:)

Hope this helps!

~Rendorforestmusic

BabaBlast [244]3 years ago

3 0

B. Remains constant until the change of state is complete.

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A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base

tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ = 0.00375 mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

$HN_3 + OH- ---> N^-_{3} + H_2O$

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L

Concentration of $HN_3$ = $\dfrac{0.001755}{0.0383}$ = 0.0458 M

Concentration of $N^{-}_3$ = $\dfrac{ 0.001995 }{0.0383}$ = 0.0521 M

GIven that :

Ka = $1.9 x 10^{-5}$

Thus; it's pKa = 4.72

$pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})$

$pH =4.72 + log(1.1376)$

$pH =4.72 + 0.05598$

$pH =4.77598$

pH ≅ 4.80

3 0

3 years ago

What is the percentage strength (w/w) of a solution made by dissolving 62.5 g of potassium chloride in 187.5 ml of water?

kiruha [24]

Percent strength (% w/w) of a solution is defined as the amount of solute present in 100 g of the solution.

Given data:

Mass of the solute, potassium chloride = 62.5 g

Volume of water (solution) = 187.5 ml

We know that the density of water = 1 g/ml

Therefore, the mass corresponding to the given volume of water

= 187.5 ml * 1 g/1 ml = 187.5 g

We have a solution of 62.5 g of potassium chloride in 187.5 g water

Therefore, amount of solute in 100 g of water= 62.5 * 100/187.5 = 33.33

The percentage strength = 33.33 %

8 0

3 years ago

Draw the major product formed when the structure shown below undergoes substitution in ch3ch2oh with heat. (use a wavy bond (sin

muminat

Answer:

Explanation:

The missing image is attached below.

The objective of this question is to draw the major product formed from the diagram attached below.

From the diagram attached, we will see the reaction of a tertiary alkyl halide together with a weak nucleophile (ch3ch2oh) undergoing a nucleophilic substitution (SN₁) mechanism to yield a racemic mixture(i.e., compound that is not optically active but contains an equal amount of dextrorotatory and levorotatory stereoisomers) as a product.