chem lab

Ba(c2H3O2)2

Ba +2 + 2C2H3O2 -

average is calculated

x + x1 +x2 +x3 . . . / n

what is the mass percent of KCl if 4.96 g KCl is dissolved in 253 mL of water?

%KCl = 4.69 g KCl / 253 g * 100% = 1.854 KCl

concentrated HCl is 12.0 M how many mL of the concentration solution are required to prepare 83.0 Ml of 0.260 M solution

(0.260M)(83.0 mL)= (12.0 M) (V2) 21.58 = (12.0M)V2 V2= 1.80 mL

mass percent KClO3 in sample

(0.843 g KClO3 / 1.13g KClO3) * 100% = 74.6%

to make a cup of coffee, you heated 0.175 kg of water to change the water temperature form 23.0 C to 89.0 C calculated the quantity of heat required

(175kg)(4.184 J/g C(89.0 C -23.0C)= 48325.2 J

it takes 75 mL of a 2.5 M HCl solution to neutralize 55 mL of a base unknown concentration. Concentration

(2.5 M)(0.075 L) = (M2)(0.055 L) M2= 3.4 M

a 12.0 mL solution of 2.80 M HCl is mixed with 29.4 mL solution of 2.14 M HCl solution. assuming the HCl does not dissociate, what would be the concentration of HCl in the mixture?

(2.80)(0.012) + (2.114)(0.0294) = 0.0965 mol M= m/v 0.0965 / 0.0414 = 2.33 M

if the actual concentration of the HCl was 0.33 M what is your precent error

(3.3M -0.04 M) / 3.3 *100 = 3.0% error

percent error for theoretical and experimental mol of H2

0.00137 mol H2 - 0.00147 mol H2 * 100% = 6.80%

theoretical mole of H2

0.001467 mol H2 (1 mol H2 / 1 mol Mg) = 0.00146 mol H2

mass of KClO3

0.006875 mol KClO3 * 122.55 g KClO3 / 1 mol KClO3 = 0.843 g KClO3

moles of KClO3

0.010 mol O2 * 2 mol KClO3 / 3 mol O2 = 0.006875 mol KClO3

mole Mg used

0.0352 Mg (1 mol Mg / 1 mol Mg) = 0.001467

A gas has a density of 1.184 g/L what is the density is g/mL? what is the density is mg/mL? what is the density is mg/L?

1.184/1000 = 0.001184 g/mL 1.184 / 1 = 1.184 mg/mL 1.184 (1000) = 1184 mg/L

which reactant Is the limiting reagent

23.0 g NaCl / 58.44 g/mol * 1 mol AgCl / 1 mol NaCl * 143.32 agCl / 1 mol AgCl =

how many mol of H2 are in a flask with a pressure of 694 mmHg and volume of 3.50 L at 24.6 C

24.6 C + 273.15 = 297.75 K (0.91 atm)(3.50 L )= n (0.0821) (297.75 K) n=0.13 moles

how many grams of AgNO3 are needed to dissolve in water to prepare a 278 mL of 1.54 M solution?

278mL/ 1000L = 0.278 L M=n/v 1.54 M = n / 0.278 L = 0.42812 mol AgNO3 0.42812 mol AgNO3 * 169.87 g/mol / 1 mol AgNO3 = 72.73 g AgNO3

a gas has a pressure of 2.30 atm at 25.6 C if the volume remains constant, what would be the pressure when temperature reaches 63.9 C

298.75 k 63.9 C + 273.15 = 337.05 K 2.30 atm / 298.75 K = P2 / 337.05 K P2= 2.60 atm

reactions in water 2 AgNO3 + BaCl2 yields

2AgCl + Ba(NO3)2

(NH2)2SO4 dissolved in water

2NH4 + (aq) +SO4 -2

determine the molarity of HCl using the data called and the titration formula

molarity = mol/L solo (22mL)(1.0 M NaOH) / 10 = 2.2 M

how many moles of oxygen gas were evolved

mold O2 = 1.237 g O2 * 1 mol / 32g = 0.03866 mol O2

experimental mole of H2

n=PV/RT (0.9557 atm )(0.03461 L ) / (0.0821)(294.15K) = 0.00137 mol H2

NaBr

Na+ Br-

a 23.0 g sample of sodium chloride is mixed with 49.7 g sample of silver nitrate. enough water was added for the reaction to take place write a balanced chemical reaction, and state the physical states on each chemical

NaCl (aq) + AgNO3 (aq) yields NaNO3 (aq) + AgCl (s)

a 56.3 sample of water was heated to 51.2 C and added to 45.8 g of 20.3 C water in a calorimeter. assume that no heat was lost to the surroundings including the calorimeter calculate the final temperature of the resulting solution

56.3 g * 4.184 J/g C (T-51.2 C)= -(45.8 g *4.184 J/g C)(T - 20.3) T= 37.3 C

a dry beaker weighing 45.48 g is completely filled with water and reweighed filled flask was found to have a mass of 78.648 g at is 25 C. the mass of water required to fill the flask is

78.648 g - 45.48 g = 33.17

at what temperature can a mixture of sodium nitrate and potassium sulfate be separated most efficiently

80 C

49.7 g AgNO3 / 169.87 g/mol * 1 mol AgCl / 1 mol AgNO3 * 143.32 AgCl / 1 mol AgCl 41..9

AgNO3 is limiting reagent

HNO3

H+ NO3+ OH-

a 1000.0 mL sample of lake water is titrated using 0.100 mL of a 0.100 M base solution. wha is the molarity of the acid in the lake water

M= (0.1)(0.1mL)/ 1000 mL = 0.00001 M

accuracy

how close to the correct value the measurement is

precision

how closely the measured number agree with each other in a given measurement

combined gas law

joined tighter become the combined gas law. if more than two properties are changing, this law is used. any properties that remain constant will drop out of the equation prose is in atm and volume in liters for gas laws calculations

mass of KClO3

mass test tube= 38.7g mass of test tube, MnO2= 38.75g mass of test tube, MnO2 a dnKClO3 = 39.2 g mass of test tube MnO2 and product after the second heating = 38.9g what is the mass of KClO3? mass of first tube - mass of test MnO2 0.45 mass of oxygen produced KClO3 mass of oxygen produced at end of experiment 38.9 -39.2 = 0.30 mass of oxygen what was the mols of oxygen produced mol= mass/mol mass 0.30. 32 g mol mass 0.30/32 g = 9.375 *10^3 0.009375 Boyles law which quantity remains constant - temperature Charles-pressure gay Lussac's - volume in the gas laws experiment, the plot of volume vs. pressure was linear - false it was a curve a gas with a pressure of 4.67 atm at 23.0 degrees celsius was heated to a temperature 38.0 degrees Celsius what is the pressure atm of the gas at this temperature ? PV=nRT P1=4.67atm T= convert to kelvin 296K T2= 311K P2=? P1/t1 =P2/t2 4.67/ 296 = p2/311 a 0.165 g sample of Mg reacted with HCl in a 500 mL sealed flask at 22.0 degrees celsius. the pressure of H2 was measured to be 715 mmHg. what is the theoretical mold of H2? balance equation Mg+2HCl msss of mag/ mol of mg when using pv=nrt the pressure is expressed in -- what is the mass of KCl if 4.96 g KCl is dissolved in 253mL of water %mass KCl= 4.96g/ (4.96 +253)g *100 = 1.92 the quality of the solute dissolved in a solvent can be expressed in molarity (M) the molarity is defined as moles of solute divided by volume of the solution. we use breakers to imply the concentration for instance if the cocnentrantionof NaCl is 4.91 M we can show it as molarity= mass of solute / volume of soulte c= n/v c=m/MV if 2.00g of NaOH is dissolved in 248mL of a solution, the concentration of NaOH can be calculated by finding the moles of NaOH mass= 2.00g molar mass= 40 g/mol volume = 248 mL c= 2.00/40 * 248*10^-3. in liters 0.202 M prepare 2.50 L of a 0.240 M NaOH from a 3.00 M NaOH solution (3.00M)(Vi)=(0.240M)(2.50L) Vi= 200mL what is the mass percent of KCl if 4.96 g KCl is dissolved in 253 mL of water %mass KCl = 4.96g / (4.96 +253)g * 100 = 1.92 if 75 ml of a 2.5 M HCL is titrated with NaOH solution to neutralize 55mL of NaOH calculate the concentration an unknown basic solution volume= final volume - initial volume volume = 55mL MaVa=MbVb (75 mL)(2.5 M)=(5.5m)(Mb) = 3.4 M

use Boyles law for the following calculation. a sample of gas which occupies a volume of 22.4 mL at a pressure of 1.21 atm is subjected to a decrease in pressure by 82.0 mmHg

p1= 1.21 atm p2= 82.0 mmHg 82.0 mmHg * 1 atm = 0.108 atm 1.21 atm. - 0.108 atm = 1.102 atm (1.21 atm )(22.4 mL) = 1.102 atm V2= 24.6 mL

gay-lussac's law

pressure and temperature are directly related, assuming constant volume. as the temperature increases in a system with fixed volume, the molecules move faster and have more collisions with the container, leading to increased pressure

precision of a set numbers may be expressed as

standard deviation calculated by taking th square root of the sum divided by (n-1)

mass of kennels

subtract mass of empty dish from mass of kennels 40g -20g = 20g mass of one kennel 20 g / 10 =2 grams each kennel weighs 2 grams mass of empty dish + 10 popcorn = 50 grams find mass of popcorn empty dishes + 10g - 25-20 = 5 g 5g / 10 = 0.5 g each popcorn weights 0.5 g all other it is 5 grams mostiur in dish d=m/v find mass mass percent 35 g KCl dissolved in 100 g water. mass percent of KCl? %= m/ toal mass * 100 35 g / 135 g * 100 % could see 35 g dissolved in water density of water is 1 mass of water will be 100 mL conversions of units calculate density of mass or volume when given two of the find mass percent AgNO3 + BaCl yields AgCl + Ba(NO3) BaCl2 + H2SO4 yields BaSO4 + HCl

How did the temperature of solution change upon addition of first 10 g KCl ?

the 250 mL beaker was 25 degrees Cesius. When the KCl solution was added, the temperature decreased to 21 degrees Celsius. It indicated the solution is endothermic

how many mL of water are mixed with 13.0 mL 1.75 M NaCl solution to prepare a solution of 0.510 M NaCl solution?

1.75(v1)=13.0 mL (0.510)M V1= 3.79 mL

if the original mass of the sample was 4.39 g, calculate the preset KClO3 in the original sample

3.159 g / 4.39 g *100% = 71.9%

a test tube contenting a mixture of KCl, KClO3, and MnO2 having a total weight of 23.584 g was heated to decompose the KClO3. After heating, the mass was found to be 22.347 g

302 g

calculate mass precipitate formed

49.7 g AgNO3 / 169.87 g/mol * 1 mol AgCl / 1 mol AgNO3 * 143.32 AgCl / 1 mol AgCl = 41.9

Describe on the solubility of each compound given below glucose in 100 mL water at 10C sodium acetate in 50mL water at 20 C Potassium sulfate in 100 mL water at 62 C

60 g/mL 24g 20 g/mL

calculate the number of moles and the mass of KClO3 originally present in the test tube

mass of KClO3 : 0.02577 mol KClO3 * 122.55 g KClO3 = 3.159 g KClO3

which two compounds have the same solubility values at 48 C

sodum acetate KBr

HI

H + I -

balanced chemical reaction between Mg and HCl

Mg(s) + 2HCl (aq) yields MgCl2 (aq) + H2(g)

ideal gas law

P1V1 / n1T1 = P2V2 /nT2. = r PV=nRT R = 0.0821

volume of H2 at STP condition

P1V1 /T1 = P2V2/ T2 (0.9557 atm )(0.0346 L) / (294.15K ) = (1.0atm)(v2) / (273.15) 0.001124162 = (1.0)(v2)/(273.15) = 0.030706485 0.030706485 = (1.0)(V2) v2= 0.030706485 L

what is the mass of a substance if the density and volume values are 4.15 g/mL and 12.6 mL

d= 4.15 g/mL V=12.6 mL 4.15 g/mL (12.6 mL) = 52.3 g

base on the molarity of the acid calculated above, what is th PH of the lake water?

pH = -log[H+] pH = -log10[0.00001] = 5

how did the temperature of solution change upon addition 10 g Ce2(SO4)3

the temperature of the 250 mL beaker was 25 degrees celsius. when the ce2(S)4)3 solution added, the temperature increased to 28.2 degree clesisus. it indicated the solution is exothermic.

when barium chloride and ammonium carbonate are mixed, a precipitate of barium carbonate is formed. write the total ionic and the net ionic equations for this reaction

total 2NH4 - (aq) + CO3 -2 (aq) + Ba2+ (aq) + 2Cl - (aq) yields BaCO3 (s) + 2NH4 - (aq) + 2Cl - (aq) net CO3 -2 (aq) + Ba2+ (aq) yields BaCO3 (s)

Avogadro's law

volume and moles are directly related, assuming constant temperature and pressure. more moles simply means more gas particles, and more gas particles will occupy more volume

Charles law

volume is directly related to temperature of a gas, assuming constant pressure. as the temperature increases, the gas particles go faster, so they collide more often with the sides of the container. in order to keep the pressure constant as the temperature rises, the volume must expand to keep the number of collisions the same