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1. What is the formula for calculating how much heat is required to change "m" ( amount of mass of material ) by "delta T" ( amount of temperature change ) using the material's "Cp" ( specific heat )?
m = mass, delta T = change in temp, Cp = specific heat

m x delta T x Cp = amount of heat/energy in cal
2. What is the formula for calculating energy?
mass x force = energy in joules
3. How to calculate energy ( cal ) needed to melt an amount of a substance ( g) using the substance's heat of fusion?
Heat of fusion is the amount of heat needed to melt 1g of substance.

If heat of fusion of ice is 80cal/g how much energy is needed to melt 20g of ice?

• 80cal
• 20g ice ------- = 20 x 80 = 1600 cal
• 1g
4. How to calculate density?
• mass
• ------ = density
• volume

• 12g
• ----- = 12g/L
• 1L
5. How to calculate concentration?
• amount of solute
• ------------------- = concentration
• amount of solution

• 5g glucose
• ------------ = 5% w/v glucose
• 100mL solution

• 5/100 = 5%
• g/mL = w/v
6. What is the percentage of a concentration whose solution is not 100mL?
• solute
• ------ x 100
• solution

• 32g
• ------
• 254mL

32/254 = .1259... x100 = 12.598% w/v
7. How many units of solute in one unit of solution?
• 23mL solute
• -------------
• 455mL solution

23/455 = .051mg/mL
8. You need to prepare 250mL of .25% w/v awesome solution. How much awesome will you need?
.25% w/v solution is .25g per 100mL solution.

.25g/100mL = .0025 ( times how many mL? ) 250mL

.25g/100mL x 250 = .625mg awesome and 250mL solution. STAT!
9. How many micrograms ( ug )in one gram? What is a ppm?
1g = 1,000,000ug

1ug = 1ppm ( part per million )
10. How many nanograms ( ng ) in one gram? What is a ppb?
1g = 1,000,000,000ng

1ng = 1ppb ( part per billion )
11. Sanitizer concentration at work should be 100ppm. How many grams, micrograms, and nanograms is in 3L of sanitizer?
• 1g = .000001ug
• 1ppm = 1ug/mL
• 1ug = 1000ng

3L x 1000mL/1L = 3000mL

3000mL x 1ug/1mL = 3000ug ( micrograms )

3000ug x 1ug/1000ng = 3,000,000 ng ( nanograms )

3000ug x 1g/1,000,000ug = .003g
12. What is a deciliter?
1/10 of a liter.

1L = 10dL

1dL = .01L
13. If a blood plasma concentration of sodium is 150mg/dL, how many grams are in 5mL of blood?
150mg per dL. 100mL per dL

150mg per 100mL 150/100 = 1.5mg/mL

1.5mg x 5mL = 7.5mg

or

• 150mg
• --------
• 1dL
• -------
• 100mL

150/100 = 1.5 x 5 = 7.5mg
14. What is a mole? How much does one mole of O ( oxygen ) weigh?
A mole is a unit of measurement for measuring 6.022 x 10^23 atoms of an element.

1 mole of oxygen contains 6.022 x 10^23 atoms.

1 mole of O ( or any other element ) is equal to its atomic mass in grams.

O's atomic mass is 16. So one mole of O weighs 16 grams.
15. How many moles does 2.65g of CaCl2 weigh?
• Ca = 40.98 x 1 mole = 40.08
• Cl = 35.45 x 2 moles = 70.9
• --------
• 110.98g/mol

2.65g x 1mol/110.98g = .0239mol
16. What is molarity?
• moles of solute
• ----------------- = molarity
• liters of solution
17. What is the molarity of 3g of vit.C (C6H8O6) in 200mL of solution?
• C - 12.01 x 6 = 72.06
• H - 1.008 x 8 = 8.064
• O - 16 x 6 = 96
• ---------
• 176.024g/mol

1mol / 176.024g = .00568...mol in 1 gram

.00568... x 3 = .0170...moles in 3 grams

.0170... / 200ml = .08521...M
18. How much acetic acid will you need to prepare 150mL of .2M solution if 1mol of acetic acid weighs 60.052g?
60.052g/mol

.2mol x 60.052g/1mol = 12.0104g in .2mol

12.0104g in 1L solution

12.0104g x 150ml x 1L / 1000mL = 1.80156g
19. How many mol ions in 1mol NaCl?
1mol NaCl = 1mol Na + 1mol Cl = 2mol ions.
20. One L of solution contains .19M glucose and .050M FeSO4. What is the totall molarity of the solution?
FeSO4 breaks down into 1 x Fe and 1x SO4 ( sulphate ), creating two mol ions x .050 = .1M for the FeSO4 + .19M glucose =

.1 + .19 = .29M
21. Solution A has .1M sucrose and .05M glucose.

Solution B has .05M sucrose and .05M glucose.

Will osmosis occur, in which direction?

Will dialysis occure, in which direction?
• Solution A = .15M
• Solution B = .1M

B has more solution ( water ) than A so osmosis will occur, from B to A.

A has more solute than B so dialysis will occur, from A to B.
22. What is an equivalent?
An equivalent is: the amount ( in moles ) of ions that has the same total charge as a mole of hydrogen ions ( H+).

So, one mole of K+ ions would be "one equivalent", since K+ has the same charge as H+ you need the same amount to have an equivalent.

One mole of sulphate ( S ) ion has a 2- charge. So one mole of S is equal to two equivalents.
23. How many equivalents of phosphate ions are in a solution that contains .31mol phosphate ions?
Phosphate is PO4 and is 3-, so one mole of phosphate is 3 equivalents.

.31 mol x 3 eq = .93 Eq
24. A solution has 2g carbonate ions (CO32-), how many equivalents do we have?
• C - 12.01
• O - 16 x 3 - 48

12.01 + 48 = 60.01g/mol

• 2g x 1mol / 60.01g = .0166...mol/g
• x2 grams
• .0333... mol in solution
• x2 charge
• .0666...Eq
25. What is mEq?
mEq is milliequivalent

• 1mEq = .001Eq
• 1Eq = 1000mEq
26. A solution contains 25mEq of citrate ions ( C6H5O73- ), how many grams of citrate does the solution have?
• C - 12.01 x 6 = 72.06
• H - 1.001 x 5 = 5.005
• O - 16 x 7 = 112
• --------
• 189.065g/mol

Solution = 1L

• 189.065g/mol
• 1mol = 3Eq
• 189.065g / 3Eq= 63.0216...g/Eq
• 1Eq = .001mEq

.063021666g/mEq x 25mEq = 1.575541667g in the solution.

Or 1.6g :)
27. A solution has 2.5g of sulfate ions in 100mL. What is the concentration of this solution in mEq/L?
sulfate is SO42-

• S - 32.06
• O - 16 x 4 - 64
• --------
• 96.06g/mol

1mol / 96.06g = .01041016mol/g

2.5g x .01041016mol = .0260254mol / 100mL

100ml x 10 = 1000mL = 1L

.0260254 x 10 = .260254007M ( mol/L )

.260254007M x 2 ( charge ) = .520508015Eq

.520508015Eq x 1000 = 520.5080158mEq

or 520.508mEq
28. A concentration of sodium ions (NaCl in an intravenous solution is 130mEq/L. What is the mass of sodium in 250mL of this solution?
Na = 22.99g / mol / Eq

Since Na has 1+ charge, 1Eq = 1mol

130mEq / 1L = .13Eq / 1L

.13Eq / 1000ml ( 1L ) = .00013Eq / 1mL

.00013Eq x 250mL = .0325Eq

1Eq = 22.99g

.0325Eq x 22.99g = .747175g in solution
29. If you add 50mL of water to 25mL of .9% w/v NaCl, what will the conentration of the diluted solution be?
C - concentration, V - volume

C1 x V1 = C2 x V2

***ignore the percentage signs here***

.9 x 25mL = C2 x 75mL

.9 x 25 = 22.5 , C2 x 75 = 22.5

22.5 / 75 = .3, C2 = .3

.3%
30. 30mL of a solution has 100ppm of iron, how much water do you need to add if you want a solution of 20ppm?
C1 - 100ppm, V1 - 30ml

100 x 30 = 3000

20ppm x V2 = 3000

3000 / 20 = 150mL
31. You have a bottle of 2M NaI. You need to use this solution to prepare 100mL of .12M NaI. How much of the 2M NaI solution do you need and how much water should you add?
2M NaI = 2mole / 1L

C2 - .12M, V2 - 100ml

C1 - 2M, V1 - ?

.12M x 100mL = 12

2M x ? = 12 , 12 / 2 = 6

V1 = 6mL

Initial volume is only 6mL, you would need to add 94mL to get to final volume.

## Card Set Information

 Author: Ghoelix ID: 102528 Filename: conversion factors Updated: 2011-09-19 01:22:03 Tags: conversion factors chemistry chem 32 Folders: Description: chem 32 conversion factors Show Answers:

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