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wariber [46]
3 years ago
15

!!! 15 POINTS!!!

Chemistry
1 answer:
xxTIMURxx [149]3 years ago
7 0

The answer is (B). The volume will decrease.

Ideal gas equation; PV = nRT

hence, V = nRT/P...

with R,T and P being constant and n reducing it becomes obvious that th volume will also reduce.

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What structures do both plant cells and animal cells contain?
JulsSmile [24]

A. Nuclei and Mitochondria

This is the answer

3 0
2 years ago
Read 2 more answers
What is the freezing point of an aqueous solution that boils at 105.0 ∘C? Express your answer using two significant figures.
konstantin123 [22]

Answer:

T°fussion of solution is -18°C

Explanation:

We have to involve two colligative properties to solve this. Let's imagine that the solute is non electrolytic, so i = 1

First of all, we apply boiling point elevation

ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kb =  ebuliloscopic constant

105°C - 100° = 0.512 °C kg/mol  . m . 1

5°C / 0.512 °C mol/kg = m

9.7 mol/kg = m

Now that we have the molality we can apply, the Freezing point depression.

ΔT = Kf . m . i

Kf =  cryoscopic constant

0° - (T°fussion of solution) = 1.86 °C/m  . 9.76 m . 1

- (1.86°C /m . 9.7 m) = T°fussion of solution

- 18°C = T°fussion of solution

5 0
3 years ago
50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH sol
boyakko [2]

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid])     (Eq. 01)

Where  

pKa = -log(Ka)        (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2    (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles       (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get  

pH = 3.35

5 0
3 years ago
5.40 g B determine the number of atoms
sdas [7]
Atomic mass Boron ( B ) = 10.811 u.m.a

10.811 g -------------- 6.02x10²³ atoms
5.40 g ----------------- ?? atoms

5.40 x ( 6.02x10²³) / 10.811 =

3.0069x10²³ atoms
4 0
3 years ago
What is unique about the nitrogen family
ankoles [38]
Nitrogen molecule is diatomic, whereas other elements form tetraatomic molecules.
7 0
3 years ago
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