Use the reaction to answer the question.

What volume of beaker contains exactly 2.23x10^-2 mol of nitrogen gas at STP?

nikklg [1K]

Answer:

V = 0.5 L

Explanation:

Given data:

Moles of nitrogen = 2.23×10⁻² mol (0.0223 mol)

Temperature = 273 K

Pressure = 1 atm

Volume = ?

Solution:

PV = nRT

V = nRT / P

V = 0.0223 mol × 0.0821 atm. mol⁻¹. L . k⁻¹ × 273 K / 1 atm

V = 0.5 L

4 0

4 years ago

What is the mass percent of chlorine in magnesium chloride?

finlep [7]

The total mass of magnesium chloride is 95.21g/mol

The mass of chlorine is 35.45g/mol

To find the mass percent of chlorine, divide its molar mass by the total mass of the substance and multiply the result by 100:
(35.45/95.21)(100)= 37.2%

4 0

3 years ago

Given the balanced equation representing a reaction: 2h2o(l) + 571.6 kJ -> 2h2(g) + o2(g). What occurred as a result of this

Ivan

Answer: option D) energy was absorbed and entropy increased.

Explanation:

1) Given balanced equation:

2H₂O (l) + 571.6 kJ → 2 H₂ (g) + O₂(g).

2) Being the energy placed on the side of the reactants means that the energy is used (consumed or absorbed). This is an endothermic reaction.

So, the first part is that energy was absorbed.

3) As for the entropy, it is a measure of the disorder or radomness of the system.

Since, two molecules of liquid water were transformed into three molecules of gas, i.e. more molecules and more kinetic energy, therefore the new state has a greater degree of radomness, is more disordered, and you conclude that the entropy increased.

With that, you have shown that the right option is D) energy was absorbed and increased.

7 0

4 years ago

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

Degger [83]

a) before addition of any KOH :

when we use the Ka equation & Ka = 4 x 10^-8 :

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

= 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

= 9.2 x 10^-5 M

when PH = -㏒[H+]

PH = -㏒(9.2 x 10^-5)

= 4

b)After addition of 25 mL of KOH: this produces a buffer solution

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]

first, we have to get moles of HCIO= molarity * volume

=0.21M * 0.05L

= 0.0105 moles

then, moles of KOH = molarity * volume

= 0.21 * 0.025

=0.00525 moles

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L = 0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

= 0.00525 / 0.075

=0.07 M

and molarity of KCIO = moles KCIO / total volume

= 0.00525 / 0.075

= 0.07 M

and when Ka = 4 x 10^-8

∴Pka =-㏒Ka

= -㏒(4 x 10^-8)

= 7.4

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume

= 0.21 M * 0.05L

= 0.0105 moles

then moles KOH = molarity * volume
= 0.22 M* 0.035 L

=0.0077 moles

∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume

= 8 x 10^-5 / 0.085

= 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

= 0.0077M / 0.085L

= 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

= 0.0105mol / 0.1 L

= 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

= 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO]

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

= 3.8
∴PH = 14- POH

=14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M

V2 is the excess volume added of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

= -㏒0.02

= 1.7

∴PH = 14- POH

= 14- 1.7

= 12.3

8 0

3 years ago

Read 2 more answers

Citric acid (H3 C6 H5 O7) is a product of the fermentation of sucrose (C12 H22 O11) in air.

stiks02 [169]

Answer:

960.6175

Explanation:

The balanced equation shows that for every mole of sucrose used, 2 moles of citric acid are produced.

So 2.5 moles of C12H22O11 will produce 2*2.5 moles of Citric acid.

So 2.5 moles of C23H22O11 will produce 5 moles of Citric Acid.

The next step is to figure out the molar mass of Citric Acid.

There are

Hydrogen: H3 + H5 moles of hydrogen in 1 mole of Citric Acid = 8 mol H

Carbon: 1 mol of Citric Acid has 6 moles of Carbon in it = 6 mol C

Oxygen: 1 mol of Citric Acid has 7 moles of Oxygen it it =7 mol O

Hydrogen has a mass of 1 so Citric Acid has 1 * 8 grams H 8

Carbon has a mass of 12 so Citric Acid has 12 * 6 grams C 72

Oxygen has a mass of 16 so Citric Acid has <u>16 * 7 grams O 112</u>

Total grams for 1 mol Citric Acid 192

The official published mass of Citric Acid is 192.1235 so because I used rounded numbers I'm a little off. It is important for any chemistry student student to know where the 192.1235 came from rather than exactly why I'm a little out. Every periodic table lists different masses for each element. If you want a better number, you will have to consult your own periodic table.

1 mol of Citric Acid = 192 grams

5 mols of Citric Acid = x

Cross multiply

1 * x = 5 * 192

x = 960 grams.

If you want to use the published amount then the answer will be 5 * 192.1235 = 960.6175

8 0

3 years ago