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Chemistry Hm 122

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Subject
Chemistry
School
Community and Technical College at West Virginia University Institute of Technology
Type
Homework
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CHEM 112 Activity #1
Due: September 9
th
, 2021
Name _______________________
(Units must be used on all measurements. Factor-label Mathematics must be used for credit!)
(1) - Examine the following chemical equation and answer the questions
C
4
H
6
O
6 (aq)
+ 2NaOH
(aq)
Na
2
C
4
H
4
O
6 (aq)
+ 2H
2
O
(l)
a) Calculate the molar mass of tartaric acid (C
4
H
6
O
6
).
M
C4H6O6
= ( 4xM
c
) + (6xM
H
) +(6x M
O
) = 4x 12 + 6x1 + 6 x 16 = 150 g/mol
b) How many grams of NaOH would be needed to make 15.0 L of 2.50 M NaOH solution?
The molar mass of NaOH
M
NaOH
= M
Na
+ M
O
+ M
H
= 14 +16+1 = 31 g/mol
The molar number of NaOH
C=n/V => n= C x V = 2.5 x 15 = 37,5 mol of NaOH
The grams of NaOH used to make the solution
n = m/M => m = n x M = 37,5 x 31 = 1162,5 g = 1,1625 Kg of NaOH
c) How many ml of 2.50 M NaOH are needed to react with 35.7 g of tartaric acid (C
4
H
6
O
6
)
n
tartaric acid
= m/M = 35.7/150 = 0.238 mol of tartaric acid
n
C4H6O6
= n
NaOH
/2 => n
NaOH
=2 x n
C4H6O6
= 2x 0.238 = 0.476 moles of NaOH
The concentration C = n/ V => the volume of NaOH = n/C = 0.476 /2.5 = 0.1904 L = 190.4 ml of
NaOH

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(2) Gold (Au) sells for $60,097.67/kg (2Sept21). If you needed $44,800 for a new car, how much
gold ore (which contains 80.45 % gold by mass) would be needed to contain enough pure gold
to purchase the vehicle. (only FACTOR LABEL will earn credit)
60097.67 dollars 1kg
44800 dollars m
1
= 44800/60097.67 = 0.74545 Kg = 745.45 g (which contains 80.45 % gold by
mass)
745.45 g 80.45 %
m
2
100% => m
2
= (745.45 x 100) / 80.45 = 926.6 g of gold ore
(3) If you are evaluating two different sets of measurements 0.952(7) atm and 0.977(5) atm. Are
they statistically equivalent?
No, they aren’t statistically equivalent.
(4) - The average person contains about 5.00 liters of blood. If the Henry’s law constant for blood (at
20
0
C) for N
2
and O
2
are 6.1x10
-4
mol/Latm and 1.3x10
-3
mol/Latm, respectively, answer the
following solubility questions.
a) Are these gases considered “soluble” or “insoluble” at sea level? EXPLAIN for credit.
At sea level the pressure decreases. The solubility of these gases depends on the pressure.
The solubility increases when the pressure increase. That’s why at sea level, these gases are
considered insoluble.
b) How many grams of N
2
and O
2
can be dissolved in a person’s blood stream at sea level?
(Remember the atmosphere consists of roughly 80 % N
2
and 20 % O
2
)
a)Grams dissolved of diazote (N
2
)
6.4 x 10
-4
moles 1L
n
N2
5 L => n
N2
= 5 x 6.4 x 10
-4
= 3.2 x 10
-3
moles of N
2

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CHEM 112 Activity #1 Due: September 9th, 2021 Name _______________________ (Units must be used on all measurements. Factor-label Mathematics must be used for credit!) (1) - Examine the following chemical equation and answer the questions C4H6O6 (aq) + 2NaOH (aq)→ Na2C4H4O6 (aq)+ 2H2O(l) a) Calculate the molar mass of tartaric acid (C4H6O6). MC4H6O6 = ( 4xMc) + (6xMH) +(6x MO) = 4x 12 + 6x1 + 6 x 16 = 150 g/mol b) How many grams of NaOH would be needed to make 15.0 L of 2.50 M NaOH solution? The molar mass of NaOH MNaOH= MNa + MO + MH = 14 +16+1 = 31 g/mol The molar number of NaOH C=n/V => n= C x V = 2.5 x 15 = 37,5 mol of NaOH The grams of NaOH used to make the solution n = m/M => m = n x M = 37,5 x 31 = 1162,5 g = 1,1625 Kg of NaOH c) How many ml of 2.50 M NaOH are needed to react with 35.7 g of tartaric acid (C4H6O6) ntartaric acid = m/M = 35.7/150 = 0.238 mol of tartaric acid n C4H6O6 = nNaOH /2 => nNaOH =2 x nC4H6O6 = 2x 0.238 = 0.476 moles of NaOH The concentration C = n/ V => the volume of NaOH = n/C = 0.476 /2.5 = 0.1904 L = 190.4 ml of NaOH (2) Gold (Au) sells for $60,097.67/kg (2Sept21). If you needed $44,800 for a new car, how much gold ore (which conta ...
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