test 3 review

solubility of a solution

- "like dissolves like" - If solute is polar and solvent is polar, then it will dissolve and a solution will form - If solute is polar and solvent is non-polar then Solution will not form

Concentration conversion factor for (m/m) (mass/mass)

- 10% (m/m) KCl solution > 10 g of KCl in 100. g of KCl solution > 10 g KCl/100. g of solution

Concentration conversion factor for (v/v) (volume/volume)

- 12% (v/v) ethanol solution > 12 mL of ethanol in 100. mL of ethanol solution > 12 mL ethanol/100. mL solution

STP (standard temperature and pressure)

- 273 K and 1 atm - 1 atm = 760 mmHg

Concentration conversion factor for (m/v) (mass/volume)

- 5% (m/v) glucose solution > 5 g of glucose in 100. mL of glucose solution > 5 g glucose/ 100. mL solution

Molarity (M)

- 6.0 M HCl solution > 6.0 moles of HCl in 1 L of HCl solution > 6.0 moles HCl/ 1L solution

endothermic reaction

- A reaction in which energy/heat is absorbed - The energy of the product is greater than the energy of the reactants - heat is in the reactant - one way to write it is that kJ is with the reactant - or that its delta H is a positive

exothermic reaction

- A reaction in which energy/heat is released - The energy of the product is less than the energy of the reactants - heat is a product - one way to write it is that kJ is in the product - or that its delta H is a negative

equivalent (Eq)

- Amount of ion that provides one mole of positive or negative charge - if a ion has positive or negative charge of 2, then it has a 2Eq/ 1 mole - Eq can also be in 2mEq/1 L

Oxidation reduction reaction can occur in

- Combination - Decomposition - Single replacement > It cannot be in double replacement

Soluble and insoluble ionic compounds

- If they don't satisfy this chart and they will form a solid or a precipitate - If they do satisfy the chart then they will disassociate and form ions (aq) - you have to cross your two compounds together to from the new compounds for the the product side

solute concentration

- In the initial and diluted solution > moles of solute are the same

limiting reactant

- Is the substance that is used up first - Limits the amount of product that can form - Theoretical yield is always the amount of product produced by the limiting reactant

in a balancing a chemical equation

- No atoms are lost or gained - The number of atoms on the reactant side is equal to the number of atoms on the product side of each element - modify only the coefficients to balance an equation

R for the ideal gas law

- PV/nT = R - R (given P in atm) = 0.0821 L x atm/ mole x K

given P, n, T, V (Pressure, number of moles, temperature, and volume)

- PV= nRT - remember T is in K so °C + 273 = K - ideal gas law

percent yield

- Ratio of actual yield to theoretical yield multiplied by 100 - The answer is in %

Mole of Elements in a chemical compound

- Subscriptions are used to tell how much moles are in an element and are used write conversion factors for mole of each element in 1 mole of a compound - 1 mole of aspirin C9H8O4 > 9 moles of C/1 mole C9H8O4 > 8 moles of H/1 mole C9H8O4 > 4 moles of O/1 mole C9H8O4

strong electrolyte

- dissociates completely when dissolved in water - ionic compounds (has metals in it "left side of the table")

weak electrolyte

- dissociates partially in water, Mostly molecules with a few ions - has non-metals with hydrogen

Remember

- if we are given P and V then then we do P1V1 = P2V2

nonelectrolyte

- no dissociation and only produces molecules - non-metals that contain carbon

Soubility

- the maximum amount of solute that dissolves in a specific amount of solvent - Depends on temperature

Avogadro's number

1 mole = 6.02 x 10^23 particles > 1 mole/6.02 x 10^23 particles or vice versa - can be used for molecules, atoms or particles or anything

balancing a chemical equation

1, Al(s) + S(s) -> Al2S3 (s) - not balanced > reactant side - Al = 1 - S = 1 > product side - Al = 2 - S = 3 2, put a coefficient of 2 in front of Al and 3 in front of S > 2Al(s) + 3S(s) -> Al2S3 (s) - balanced > reactant side - Al = 2 - S = 3 > product side - Al = 2 - S = 3

How many equivalents are there in 2.5 moles of SO4^-2?

1, How many equivalent is in SO4^-2? > Since there is -2 charge, it has: > 2 Eq/ 1 mole of SO4^-2 2, plug the given with the Eq conversion factor > 2.5 moles SO4^-2 x 2 Eq/ 1 mole of SO4^-2 = 5.0 Eq SO4^-2

Balance the following chemical equation

1, Na3(PO4) (aq) + MgCl2 (aq) -> Mg3(PO4)2 (s) + NaCl (aq) 2, file what we have - reactant side > Na= 3 > PO4= 1 > Mg = 1 > Cl = 2 - product side > Na = 1 > PO4 = 2 > Mg = 3 > Cl = 1 - not balanced 2, balance metals first - add 3 to MgCl2 > 1 x 3 = 3 Mg > 2 x 3 = 6 Cl 3, put a 6 infront of NaCl > 1 x 6 = 6 Na > 1 x 6 = 6 Cl 4, put a 2 infront of Na3PO4 > 2 x 3 = 6 Na > 2 x 1 = PO4 5, our new equation > 2Na3(PO4) (aq) + 3MgCl2 (aq) -> Mg3(PO4)2 (s) + 6NaCl (aq) 6, list all the atoms of 2Na3(PO4) (aq) + 3MgCl2 (aq) -> Mg3(PO4)2 (s) + 6NaCl (aq) - reactant side > 2 x 3 = 6 Na > 1 x 1 = 1 PO4 > 3 x 1 = 3 Mg > 3 x 2 = 6 Cl - product side > 6 x 1 = 6 Na > 2x 1 = 2 PO4 > 3 x 1 = 3 Mg > 6 x 1 = 6 Cl - balanced

given P, V and T (Pressure, volume and temperature)

1, P1V1/T1 = P2V2/T2 2, it cross multiples and so we have > T2P1V1 = T1P2V2 3, then solve for something by dividing - Combined gas law - remember T is in K so °C + 273 = K

Nitrogen gas reacts with hydrogen gas to produce ammonium (NH3) gas. How many liters of NH3 can be produced at 0.93 atm and 24°C from a 16.0 g sample of nitrogen gas (N2) and am exess gas

1, given > 0.93 atm > 24°C + 273 = 297K > 16.0 g of N2 and we need to find how much liters of NH3 can be produced 2, So first we need to find moles of NH3 from g of N2 2a, plan > g of N2 -> molar mass of N2 -> mole-mole factor of NH3 and N2 -> moles of NH3 2b, find molar mass of N2 > 1 mole N2/ 28.02 g of N2 2c, find mole-mole factor for NH3 and N2 (use the coefficients as the number of moles) > 2 moles of NH3/ 1 mole of N2 2d, now find the moles of NH3 with the given g of N2 > 16.0 g N2 x 1 mole N2/ 28.02 g of N2 x 2 moles of NH3/ 1 mole of N2 = 1.14 moles of NH3 3, now we have > 0.93 atm > 24°C + 273 = 297K > 1.14 moles of NH3 - and we need to find how much liters of NH3 can be produced 4, use the ideal has law which is PV= nRT > we want V so solve for V by dividing P to both sides > V = nRT/P 5, now find our R, from PV= nRT > rearrange R , PV = RnT > now solve for R, by dividing both sides by nT > now we have PV/nT = R 6, now since we are given P in atm, our L will equal 0.0821 L 7, now fill PV/nT = R in, > atm x 0.0821 L/ moles x K 8, now fill in our given values for V = nRT/P > V = 1.14 moles NH3 x atm x 0.0821 L/ moles x K x 297 K/ 0.93 atm = 30. L NH3 (rounded to least sigfigs)

Given the following reaction, N2(g) + 3H2 -> 2NH3(g), Calculate the amount of ammonia, NH3, That can be formed when 2.50 g of nitrogen gas, N2, Reacts with 2.00 g of hydrogen gas, H2,

1, given > 2.50 g N2 > 2.00 g H2 and we need to calculate the amount of NH3 that can be produced 2, we need to do two conversion for this one > conversion 1, g of N2 -> molar mass of N2 -> mole-mole factor of N2 and NH3 -> molar mass of NH3 > conversion 2, g of H2 -> molar mass of H2 -> mole-mole factor of H2 and NH3 -> molar mass of NH3 3, do N2 first 3a, find the molar mass of N2, > 28.02 g N2/ 1 mole of N2 3b, find the mole-mole factor for N2 and NH3, (remember that mole-mole factor uses the coefficients as the number of moles from each substance in the balanced equation ) > 1 mole of N2/2 mole of NH3 3c, find the molar mass of NH3 (remember that with molar mass we don't worry about the coefficients, just the subscript > 1 mole of N/ 14.01 g N > 1 mole of H/ 1.008 g H - now use the subscript to find how much moles are in each element > N, has no subscript so it has 1 mole of N > H, has a subscript of 3 so it has 3 moles of H - now multiply the moles to each element's molar mass > 1 mole of N x 14.01 g N/1 mole of N = 14.01 g N > 3 moles of H3 x 1.008 g H/1 mole of H3 = 3.024 g of H3 - now add them up > 14.01 g N + 3.024 g of H3 = 17.03 g of NH3 (rounded to least dp) - molar mass conversion factor for NH3 > 1 mole NH3/ 17.03 g of NH3 4, now calculate the amount of NH3 with N2 > 2.50 g N2 x 1 mole of N2/28.02 g N2 x 2 mole of NH3/1 mole of N2 x 17.03 g of NH3/1 mole NH3 = 3.04 g NH3 (round the least sigfig) 5, now do H2 5a, find the molar mass of H2 > 1 mole H2/ 2.016 g of H2 5b, find the mole-mole factor for H2 and NH3, (remember that mole-mole factor uses the coefficients as the number of moles from each substance in the balanced equation ) > 3 mole of H2/2 mole of NH3 5c, find the molar mass of NH3 (remember that with molar mass we don't worry about the coefficients, just the subscript > 1 mole of N/ 14.01 g N > 1 mole of H/ 1.008 g H - now use the subscript to find how much moles are in each element > N, has no subscript so it has 1 mole of N > H, has a subscript of 3 so it has 3 moles of H - now multiply the moles to each element's molar mass > 1 mole of N x 14.01 g N/1 mole of N = 14.01 g N > 3 moles of H3 x 1.008 g H/1 mole of H3 = 3.024 g of H3 - now add them up > 14.01 g N + 3.024 g of H3 = 17.03 g of NH3 (rounded to least dp) - molar mass conversion factor for NH3 > 1 mole NH3/ 17.03 g of NH3 6, now calculate the amount of NH3 with H2 > 2.00 g H2 x 1 mole H2/ 2.016 g of H2 x 2 mole of NH3/ 3 mole of H2 x 17.03 g of NH3/1 mole NH3 = 11.3 g of NH3 (rounded to least sigfigs) 7, now evaluate what we have > H2 = 11.3 g of NH3 > N2 = 3.04 g of NH3 8, the Limiting reactant and theoretical yield is > N2 since its the least amount so therefore the max amount of NH3 that can be formed is 3.04 g of NH2

To find a number from a percentage

Divide by 100

oxidation-reduction reaction

Electrons are transferred from one substance to another

reverse osmosis

Flow of water from high particle concentration to low power concentration

Osmosis

Flow of water from low particle concentration to high particle concentration

In an oxidation and reduction reaction

If the element is by itself it has a charge of zero

To find percentage

Multiply by 100

Need to find mmHg for R

P for mmHg = 760 > 760. mmHg x 22.4 L/1 mole x 273 K > 760. x 22.4/ 273k = 62.4 L (rounded to the least sigfig) > R = 62.4 L x mmHg/ mole x K

remember that with that laws that require division

P or V will also be at the top

given P and T only (Pressure and temperature)

P1/T1 = P2/T2 - gay lussacs law - remember T is in K so °C + 273 = K

given P and V only (Pressure and volume)

P1V1=P2V2 - Boyles law

Dalton's Partial Pressure

Pt=P1+P2+P3...

RIG

Reduction is gain of electrons

Isotonic solutions

Solution which exerts the same osmotic pressure as body fluids

actual yield

The amount of product actually obtained > Usually given

theoretical yield

The maximum amount of product which is calculated using the balanced equation (also is the limiting reactant)

excess reactant

The reactant that does not completely react and is left over at the end of the reaction

given V and T only (Volume and temperature)

V1/T1 = V2/T2 - charles law - remember T is in K so °C + 273 = K

Given V and n only (Volume and number of moles)

V1/n1 = V2/n2 - avogadros law

Water solubility equation

Water solubility = g of solute/ 100 g water

Cu(s) + 2AgNO3(aq) -> 2Ag(s) + Cu(NO3)2(aq)

single replacement - new element formed

mole-mole factor

the relationship between the number of moles of two substances in an equation and we use the coefficient of those Substances as their moles

Solubility of solid

when temp increases the solubility of solids increase

Hypertonic solution

when the solution is hyper that means it has more solute concentration then the RBC, which causes The red blood cells to shrink as Water rushes out to equalize, This is called crenation

Hypotonic solution

when the solution is hypo which means that the solution has less solute concentration compared to the red blood cells, Causing all the water to rush into the red blood cells to even it out, Causing the red blood cells to experience hemolysis as it become hypO

Zinc reacts with HCI to produce hydrogen gas (H2) and ZnCl2, Zn(s) + 2HCl (aq) -> H2(g) + ZnCl(aq), How many liters of a 1.50 M HCl solution Completely react with 5.32 g of zinc (Zn)

1, given > Molarity = 1.50 M HCl > 5.32 g of Zn > find how many liters HCl has 2, plan > g of Zn -> molar mass of Zn -> mole-mole factor for Zn and HCl -> molarity of HCl -> L of HCl 3, find the molar mass of Zn, (molar mass in always in 1 mole) > 1 mole Zn/65.41 g of Zn 4, find the mole-mole factor for Zn and HCl using the coefficients in the balanced equation > 1 mole of Zn/ 2 moles of HCl 5, find the Molarity conversion factor for HCl (mole/1 L) > given 1.50 M HCl > 1.50 moles HCl/ 1 L of solution 7, now that we have all our conversion factors, solve. > 5.32 g Zn x 1 mole Zn/65.41 g of Zn x 2 moles of HCl/1 mole of Zn x 1 L of solution/ 1.50 moles HCl = 0.108 L of solution ( round to the least sigfig)

2Cu(s) + O2(g) -> 2CuO(s)

1, if the element is by itself then it has a charge of zero > 2Cu^0(s) + O2^0 (g) -> 2CuO(s) 2, now in order for CuO to be together the charge needs to equal to zero, > O has a charge of -2 > Cu has a charge of 2 3, now re write with all the charges > 2Cu^0(s) + O2^0 (g) -> 2Cu^+2O^-2(s) 4, now oxidation is loss of electrons > we see Cu had a charge of zero and it lost two electrons to become positive 2, so Cu is oxidation 4, for reduction, reduction is gain of electrons > O started with a zero charge and gained two electrons to have -2, so its a reduction 5, Oxidation = Cu reduction = O

How to balance a chemical equation

1, make sure the reactants and products have the correct chemical formulas 2, if an element occurs in only one compound on both sides of the equation, balance it first. If there is more than one such element, balance metals before non metals. 3, if an element occurs as a free element on either side of the chemical equation, balance it last. Always balance free elements by adding the coefficient of the free element. 4, balance polyatomic ions as single chemical species 5, check to make sure the equation is balanced by summing the total number of each type of atom on both sides of the equation 6, remember to change only the coefficient's to balancing a chemical equations, Never change the subscripts of the chemical formulas 7, Define a coefficient must be the lowest whole number

Calculate the molar mass of Li2CO3

1, molar mass is over 1 mole, find the mass in g of each element using the atomic mass > 6.941 g Li/1 mole Li > 12.01 g of C/1 mole C > 16.00 g O/ 1 mole of O 2, now use the subscript to see how much moles each element has in Li2CO3 (molar mass uses only subscripts) > Li, has a subscript of 2 so it has 2 mole of Li > C, has no subscript so there is just 1 mole of C > O, has a subscript of 3 so it has 3 moles of O 3, now multiply the moles to the molar mass of each respective element conversion factor > 2 mole of Li x 6.941 g Li/1 mole Li = 13.88 g of Li (rounded to least sigfig) > 1 mole of C x 12.01 g of C/1 mole C = 12.01 g of C (rounded to least sigfig) > 3 moles of O x 16.00 g O/ 1 mole of O = 48.00 g of O (rounded to least sigfig) 4, now add up the grams from Li, C, and O > 13.88 g of Li + 12.01 g of C + 48.00 g of O = 73.89 g of Li2CO3 (rounded by least dp) 5, now put it into a molar mass conversion factor for Li2CO3 > 1 mole of Li2CO3/ 73.89 g of Li2CO3 or vice versa

Calculate heat changes, 2H2O(l) -> 2H2(g) + O2(g) delat H = 572kJ

1, this is an endothermic reaction due to a positive delta H value 2, if this chemical equation is balanced then we can make our heat as a conversion factor (use coefficients for number of moles) > 2 moles H2O/572 kJ > 2 moles of H2/572 kJ > 1 mole O2/ 572kJ

1L

1000mL

molar volume at STP

22.4 L/ 1 mole of gas

1 atm

760 mmHg or 760 torr

Need to find atm for R

> 1 atm x 22.4 L/ 1 mole x 273 K > 22.4 L/ 273 K = 0.0821 L (round to least sig fig) R = 0.0821 L x atm/ mole x K

identifying a balanced equation

> Al(s) + S(s) -> Al2S3 (s) - not balanced > reactant side - Al = 1 - S = 1 > product side - Al = 2 - S = 3

percent yield equation

> Percent yield (%) = actual yield/ theoretical yield x 100 > Just remember "Actual over"

Concentration and volume for the Initial and diluted solution

C1V1 = C2V2 > C1V1 = initial > C2V2 = diluted - Concentration may be in terms of: > Percent concentration > Molarity

Solubility of gases

When temperature increases the solubility of gases decrease

Ca(s) + Cl2(g) -> CaCl2(s)

combination reaction - 2 reactants and 1 product

CH4(g) + 2O2(g) --(delta)--> CO2(g) + 2H2O (g) + energy

combustion reaction - heat is released

Fe2S3(s) -> 2Fe(s) + 3S(s)

decomposition reaction - 1 reactant and two products

BaCl2(aq) + K2SO4(aq) -> BaSO4(s) + 2KCl (aq)

double replacement - solid form

If you're not asked to calculate the amount of product, what is the quickest way to find the limiting reactant

ex, What is the limiting reactant of the reaction 2Na3PO4(aq) + 3MgCl2(aq) -> Mg3(PO4)2 + 6NaCl (aq), if we are given 4 moles of NaPO4(aq) mixed with 5 moles of MgCl2(aq) 1, pick a product you want to compare to, NaCl 2, the mole-mole factor for NaCl and Na3PO4 is (use coefficients), > 6 moles NaCl/ 2 moles Na3PO4 3, the mole-mole factor for NaCl and MgCl2(aq) is (use coefficients), > 6 moles NaCl/ 3 moles MgCl2 4, now find NaCl using the given Na3PO4 > 4 moles of Na3PO4 x 6 moles NaCl/ 2 moles Na3PO4 = 12 moles NaCl 5, now find NaCl using the given MgCl2(aq) > 5 moles MgCl2(aq) x 6 moles NaCl/ 3 moles MgCl2 = 10 moles of NaCl 6, Determine the limiting reactant > MgCl2 = 10 moles of NaCl > Na3PO4 = 12 moles NaCl - so MgCl2 is the limiting recant since it is less

If we have multiple products in a reaction, does it matter which product we choose to determine the limiting/excess reactant

no it doesn't if you use product 1 then stick with it

oIl

oxidation is loss of electrons