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worty [1.4K]
2 years ago
13

Larisa pumps up a soccer ball until it has a gauge pressure of 61 kilopascals. The volume of the ball is 5.2 liters. The air tem

perature is 32°C, and the outside air is at standard pressure. How many moles of air are in the ball?
Chemistry
1 answer:
kirill115 [55]2 years ago
7 0

The total ball pressure is sum of gauge pressure and atmospheric pressure

We know that atmospheric pressure = 101.325 kPa

so the total ball pressure = 61 kPa + 101.325 kPa = 162.325 kPa

Now we will use ideal gas equation as

PV = nRT

P = pressure = 162.325 kPa

V = 5.2 L

R = gas constant = 8.314 kPa  L / mol K

T = 32 C = 273.15 + 32 = 305.15 K

n = moles = ?

Moles = PV / RT = 162.325 kPa X 5.2 / 305.15 X 8.314 = 0.332 moles

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Eva buys a package of food, and the nutrition label says that
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Answer is 74,844 calories

8 0
2 years ago
A 1.0-gram sample of solid iodine is placed in a tube and the tube is sealed after all of the air is removed. The tube and the s
yKpoI14uk [10]

Answer:

27.0

Explanation:

Because Mass can neither be created nor be destroyed hence total mass of sample of iodine and tube remain equal as it is sealed.

7 0
2 years ago
How many grams of (nh4)3po4 are needed to make 0.250 l of 0.150 m (nh4)3po4? hints how many grams of (nh4)3po4 are needed to mak
NeX [460]
<span>There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. We calculate the mass of the solute by first determining the number of moles needed. And by using the molar mass, we can convert it to units of mass.

Moles </span>(nh4)3po4 = 0.250 L (0.150 M) = 0.0375 moles (nh4)3po4
Mass = 0.0375 mol (nh4)3po4 (149.0867 g / mol) = 5.59 g (nh4)3po4
5 0
2 years ago
What is the pH of a 0.28 M solution of ascorbic acid (Vitamin C)? (The values for Ka1 and Ka2 for ascorbic acid are 8.0×10−5 and
Tomtit [17]

Answer:

pH = 2.32

Explanation:

H2A + H2O -------> H3O+ + HA-    

Ka2 is very less so i am not considering that dissociation.

now Ka = 8.0×10−5

            = [H3O+] [HA-] / [H2A]

lets concentration of H3O+ = X then above equation will be

8.0×10−5 = [X] [X] / [0.28 -X]

8.0×10−5 = X2 /  [0.28 -X]

X2 + 8.0×10−5 X - 2.24 x 10−5

solve the quardratic equation

X =0.004693 M

pH = -log[H+}

    = -log [0.004693]

    = 2.3285

    ≅2.32

pH = 2.32

5 0
2 years ago
A pan containing 40 grams of water was allowed to cool from a temperature of 91.0C. If the amount of heat repressed is 1,300 jou
Vinvika [58]

Answer:

83°C

Explanation:

The following were obtained from the question:

M = 40g

C = 4.2J/g°C

T1 = 91°C

T2 =?

Q = 1300J

Q = MCΔT

ΔT = Q/CM

ΔT = 1300/(4.2x40)

ΔT = 8°C

But ΔT = T1 — T2 (since the reaction involves cooling)

ΔT = T1 — T2

8 = 91 — T2

Collect like terms

8 — 91 = —T2

— 83 = —T2

Multiply through by —1

T2 = 83°C

The final temperature is 83°C

3 0
2 years ago
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