Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

Unit 2 Chem > Chapter Tests > Flashcards

Chapter Tests Flashcards

You may prefer our related Brainscape-certified flashcards:
Periodic Table flashcards Chemistry 101 flashcards AP® Chemistry flashcards Organic Chemistry 101 flashcards
1
Q

0.0019 L is equivalent to:

A

1.9 mL

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

100 mg is equivalent to:

A

10–4 kg

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

78 mg L–1 is equivalent to:

A

78 ppm

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

4.0 ppm is equivalent to:

A

4.0 μg mL–1

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

4.5%(w/w) KF is equivalent to

A

0.0077 mol L–1

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

When would you use a quantitative analysis ?

A

when determining the sodium content in mineral water

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Which best describes a electrical conductivity analytical method?

A

The measurement of combined level of dissolved salts

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Atomic absorption spectroscopy is suitable for the detection of many metals. A solution of an unknown metal is compared with solutions of known concentration. The most likely unit of concentration used in such analyses would be

A

ppm (Parts per million)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Aluminium can be produced by the reaction between alumina, Al2O3, and carbon. The equation for the reaction is
2Al2O3(l) + 3C(s) → 3CO2(g) + 4Al(l)
What would result in the production of 0.8 mol of aluminium?

A

0.4 mol of Al2O3 and 0.6 mol of C

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Calculate the mass of water that is produced when 2.80 g of propane, C3H8, is completely burned in air.

A

C3H8 + 5O2 → 3CO2 + 4H2O

n(C3H8) = m/M = 2.80/44.0 = 0.0636 mol Therefore,

n(H2O) = 4 × 0.0636 = 0.254 mol and m(H2O)
= n × M =
0.254 × 18.0 = 4.57 g.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

What mass of water will be produced if 25.0 g of hydrogen gas, H2, and 100.0 g of oxygen gas, O2, are mixed and ignited?

A

2H2(g) + O2(g) → 2H2O(l)

n(H2) = m/M = 
25.0/2.0 = 13 mol and
n(O2) = m/M = 100.0/32.0 = 3.13 mol

H2 is in excess (3.13 mol of O2 reacts with 2 × 3.13 = 6.26 mol of H2).
Therefore, O2 is the limiting reactant.

n(H2O) = 2 × n(O2) =
m/M = 2 × 3.13 = 6.26 mol
m(H2O) = n × M =
6.26 × 18.0 = 113 g

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

A standard solution is a solution that has

A

an accurately known formula.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Primary standard must

A

Be readily obtainable in pure form, have a known formula and not absorb water from the atmosphere

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

To prepare a standard solution, the set of apparatus normally used is

A

A wash bottle, an electronic balance and a volumetric flask.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

A titre is a volume measured using a

A

Burette

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

An aliquot is a volume measured using a

A

Pipette

17
Q

A 25.0 mL sample of vinegar was diluted to 200.0 mL in a volumetric flask.
The ethanoic acid, CH3COOH, in a 25.00 mL aliquot of this diluted solution required the addition of 23.75 mL of 0.147 M sodium hydroxide for a complete reaction. The reaction that occurred was
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

(Determine the molarity (M) (concentration in mol L–1) of ethanoic acid in the diluted vinegar
solution.)

A
  • n(NaOH) = c × V = 0.147 × 0.023 75 = 0.003 49 mol -
  • n(CH3COOH) in 20.00 mL aliquot = 0.003 49 mol

-c(CH3COOH) in 20.00 mL aliquot =
n/V = 0.003 49/0.025 00 = 0.140 M

18
Q

A 25.0 mL sample of vinegar was diluted to 200.0 mL in a volumetric flask.
The ethanoic acid, CH3COOH, in a 25.00 mL aliquot of this diluted solution required the addition of 23.75 mL of 0.147 M sodium hydroxide for a complete reaction. The reaction that occurred was
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

(Determine the molarity (concentration in mol L–1) of ethanoic acid in the undiluted vinegar solution.)

A

concentration(CH3COOH) in 250 mL volumetric flask = 0.140 M

-Since the original vinegar solution was diluted by a factor of 200.0/25.0 = 8
8 c(CH3COOH) in the undiluted vinegar = 
8 × 0.140 = 1.12 M.
19
Q

A 25.0 mL sample of vinegar was diluted to 200.0 mL in a volumetric flask.
The ethanoic acid, CH3COOH, in a 25.00 mL aliquot of this diluted solution required the addition of 23.75 mL of 0.147 M sodium hydroxide for a complete reaction. The reaction that occurred was
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

(Express the concentration of ethanoic acid in the undiluted vinegar as %(m/v).)

A

-1.12 M means that there is 1.12 mol of CH3COOH in 1 L of solution.

-Since m = n × M 60 = Mass of CH3COOH
= 1.12 × 60.0 = 67.2 g of CH3COOH in 1 L of solution and 6.72 g of CH3COOH in 100 mL of solution.

  • The concentration = 6.72%(m/v).
20
Q

(Na2CO3.xH2O). A 0.300 g of this sample reacts completely with 20.0 mL of 0.250 M hydrochloric acid according to the equation
Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)

(Determine the percentage by mass of sodium carbonate, Na2CO3, in the washing powder.)

A

n(HCl) = c × V
= 0.250 × 0.0200 = 0.005 00 mol

n(Na2CO3) = 0.5 × n(HCl) =
0.002 50 mol m(Na2CO3) =
n × M = 0.0250 × 106.0 = 0.265 g %Na2CO3 =

(0.265/0.300) × 100 = 88.3%

21
Q

The results of an analysis using methyl orange as an indicator follows.
Concentration of sulfuric acid, H2SO4 = 0.095 60 M Aliquot of sodium hydroxide, NaOH = 25.00 mL
Initial volume of sulfuric acid, H2SO4 = 2.32 mL

(Write a balanced equation with states for the reaction)

A

H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

22
Q

The results of an analysis using methyl orange as an indicator follows.
Concentration of sulfuric acid, H2SO4 = 0.095 60 M Aliquot of sodium hydroxide, NaOH = 25.00 mL
Initial volume of sulfuric acid, H2SO4 = 2.32 mL

(Describe the function of the methyl orange indicator)

A

to detect the end point

23
Q

The results of an analysis using methyl orange as an indicator follows.
Concentration of sulfuric acid, H2SO4 = 0.095 60 M Aliquot of sodium hydroxide, NaOH = 25.00 mL
Initial volume of sulfuric acid, H2SO4 = 2.32 mL

(Determine the volume of the titre, in mL, of H2SO4).

A

V(H2SO4) =

21.48 – 2.32 = 19.16 mL

24
Q

The results of an analysis using methyl orange as an indicator follows.
Concentration of sulfuric acid, H2SO4 = 0.095 60 M Aliquot of sodium hydroxide, NaOH = 25.00 mL
Initial volume of sulfuric acid, H2SO4 = 2.32 mL

(Calculate the amount in moles of sulfuric acid used in the titration)

A

n(H2SO4)=c×V=

  1. 09560×0.01916=
  2. 001832mol
25
Q

The results of an analysis using methyl orange as an indicator follows.
Concentration of sulfuric acid, H2SO4 = 0.095 60 M Aliquot of sodium hydroxide, NaOH = 25.00 mL
Initial volume of sulfuric acid, H2SO4 = 2.32 mL

(Calculate the concentration in molL–1 of the sodium hydroxide solution)

A 
n(NaOH) = 2 × n(H2SO4) = 
2 × 0.001 832 = 0.003 664 mol 
c(NaOH) = n/V = 
0.003 664/0.025 00 = 
0.1466 M

Unit 2 Chem (13 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions