Question

Please answer parts A-K. Thank you!

**Part A**

How many mm^3 are in 2.20 ft^3

**Part B**

A lead ball has a mass of 25 pounds and a density of 11.4 g/cm^3. What is the volume of the ball in ml?

**Part C**

How many low dose 81 mg aspirin tablets can be made from 25.0 pounds of aspirin?

**Part D**

How much heat in kJ is needed to raise the temperature of 60.0 gal of water from 25.0°C to 60.0°C? q = mcPT

**Part E**

Calculate the molar mass of C_2952H4664O_832N812S_8 Fe_4

**Part F**

What is the mass percent of each element in C_10H_15N?

**Part G**

How many moles of Cu are in 1.48 x 1025 Cu2 SO4 molecules?

**Part H**

If the actual amount of Li3PO4 produced in the above reaction is 4.53 grams, what is the percent yield?

**Part I**

What is the [H] and [OH-] of a solution with a pH of 7.4?

**Part J**

If an unknown molarity NaOH solution is titrated to a faint pink color with 0.407 g of KHP, what is the concentration of the NaOH?

**Part K**

Determine the molarity of an HCl solution if a 25.0 ml aliquot of the HCl required 28.9 ml of the NaOH from the previous question to get to a faint pink color.

Answer #1

A) 1 ft^3 = 2.832*10^7 mm^3

So, 2.20 ft^3 = 2.20 * 2.832 * 10^7 = 6.23*10^7 mm^3 ....Answer

B) 1 pounds = 453.952 grams

So, 25 pounds = 25*453.952 = 11339.8 grams

Volume = mass/density = 11339.8/11.4 = 994.72 cm^3

1 cm^3 = 1 ml

SO, 994.72 cm^3 = 994.72 ml ....Answer

C) 25 pounds = 11339.8*10^3 mg

So, number of tablets = 11339.8*10^3 / 81 = 139997 ....Answer

D) 1 gallon = 3785.412 ml

Density of water = 1 g/ml

So,

Mass of water = 3785.412 grams * 60 = 227125 g

Q = 227125 * 4.184 * (60 - 25) = 33260185 J = 33260.2 KJ ...Answer

E) Molar mass of C2952H4664O832N812S8Fe = 65321.46 g/mol

F) Molar mass C10H15N = 149.233 g/mol

% of C = 80.48%

% of H = 10.13%

% of N = 9.39 %

G) Moles of Cu = 2*1.48*10^25 / 6.023*10^23 = 0.49145*10^2 = 49.145 moles

Please answer all, thank you!
Part A
How many grams of hydrogen are collected in a reaction where
1.90 L of hydrogen gas is collected over water at a temperature of
40 ∘C and a total pressure of 750 torr?
mH2=_____g
Part B
Describe how you would make 100.0 mL of a 0.900 M NaOH solution
from a 13.0 M stock NaOH solution.
Dilute Value/Units
of the 13.0 M stock solution to a final volume of 100.0 mL.
Part C...

Q1: You have 21.7 mL of a 0.9 M solution of A. How
many moles of A do you have?
Q2: In a titration experiment, 9.3 mL of an aqueous HCl solution
was titrated with 0.4 M NaOH solution. The equivalence point in the
titration was reached when 9.5 mL of the NaOH solution was added.
What is the molarity of the HCl solution?
Q3: In the titration of HCl with NaOH, the equivalence point is
determined
a. from the point...

1b. Calculate the exact molarity of a solution of NaOH if 55.00
mL of it is needed to titrate an amount of KHP that equals your
average mass (that you stated above). Average is 50.0 (4 pts)
2a. What is the average molarity for your standardized NaOH
solution?_______________ (1 pt) Average: .0075 M
2b. If 1.20 grams of impure solid KHP sample required 2.53 mL of
your standardized NaOH to reach the end point, what was the percent
KHP in...

How many mL of 7.73 M NaOH are needed to prepare 324 mL of 0.146
M NaOH?
Your Answer:
Question 5 options:
Answer
units
What is the molarity of a NaOH solution if 27.1 mL are needed to
titrate a 0.6119 g sample of KHP?
Your Answer:
Question 6 options:
Answer
units

You will need the answer tio question 2 during the
experiment.
1) At the endpoint of a titration, the color of the solution
should be (a)bright pink, (b) pink, (c) faint pink, (d) colorless
?
2)We will calculate the amount of acid to use in each titration.
Assume that you are using 0.0512 M NaOH (aq). A good violume of
NaOH(aq) to use per titration is 15ml, From this molarity and
volume, the moles of NaOH can be calculated. Since...

After standardizing your NaOH solution, you have determined that
the concentration is actually 0.1125 M. What is the weight percent
concentration of KHP (MW=204.23) in your unknown sample of mass
1.625 g if it takes 38.40 mL of titrant to reach the equivalence
point?
A student weighs out 1.7500 grams of dried potassium hydrogen
phthalate standard (MW=204.23) and finds that it takes 39.05 mL to
reach the endpoint of the titration with the solution of NaOH. What
is the molarity...

Part A
What volume of 0.105 M HClO4 solution is needed to
neutralize 55.00 mL of 9.00×10−2M NaOH?
Part B
What volume of 0.120 M HCl is needed to neutralize 2.70
g of Mg(OH)2?
Part C
If 25.6 mL of AgNO3 is needed to precipitate all the Cl− ions in
a 0.770-mg sample of KCl (forming AgCl), what is the molarity of
the AgNO3 solution?
Part D
If 45.7 mL of 0.102 M HCl solution is needed to
neutralize a...

(a) How many milliliters of 0.165 M HCl are needed to
neutralize completely 35.0 mL of 0.101 M
Ba(OH)2 solution?
________ ml
(b) How many milliliters of 2.50 M
H2SO4 are needed to neutralize 50.0 g of
NaOH?
_________ mL
(c) If 56.8 mL of BaCl2 solution is needed to
precipitate all the sulfate in a 544 mg sample of
Na2SO4 (forming BaSO4), what is
the molarity of the solution?
_________M
(d) If 47.5 mL of 0.250 M HCl solution...

(a) How many milliliters of 0.165 M HCl are needed to neutralize
completely 35.0 mL of 0.101 M Ba(OH)2 solution? ______ml
(b) How many milliliters of 3.50 M H2SO4 are needed to
neutralize 75.0 g of NaOH?
_______mL
(c) If 55.8 mL of BaCl2 solution is needed to precipitate all
the sulfate in a 544 mg sample of Na2SO4 (forming BaSO4), what is
the molarity of the solution?
________M
(d) If 47.5 mL of 0.375 M HCl solution is needed...

In Part 1 of Lab 2 you will make and standardize a solution of
NaOH(aq). Suppose in the lab you measure the solid NaOH and
dissolve it into 100.0 mL of water. You then measure 0.2013 g of
KHP (KC8H5O4, 204.22 g/mol) and place it in a clean, dry 100-mL
beaker, and then dissolve the KHP in about 25 mL of water and add a
couple of drops of phenolphthalein indicator. You titrate this with
your NaOH(aq) solution and find...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 36 minutes ago

asked 49 minutes ago

asked 50 minutes ago

asked 54 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 3 hours ago

asked 3 hours ago

asked 4 hours ago

asked 4 hours ago