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Zolol [24]
3 years ago
15

A chemistry student in lab needs to fill a temperature-control tank with water. The tank measures 24.0 cm long by 21.0 cm wide b

y 13.0 cm deep.
In addition, as shown in the sketch below, the student needs to allow 2.0 cm between the top of the tank and the top of the water, and a round-bottom flask

with a diameter of 10.5 cm will be just barely submerged in the water.

Calculate the volume of water in liters which the student needs. Round your answer to the nearest 0.1 L. ol.2 cm

M

flask

OL

6

water
Chemistry
1 answer:
Natalija [7]3 years ago
6 0

Answer:

The required volume of water the student needs is 4.9 litres of water

Explanation:

From the diagram related to the question, we have;

The dimensions of the tank are;

Length of tank = 24.0 cm = 0.24 m

Width of tank = 21.0 cm = 0.21 m

Depth of tank = 13.0 cm. = 0.13 m

Allowance provided between the top of the tank and the top of the water = 2.0 cm

Diameter of the round bottom flask, D = 10.5 cm = 0.105 m

Therefore, the radius of the round bottom flask, r = 0.105/2 = 0.0525 m

Therefore we have;

Depth of water in the tank = Depth of tank - Allowance provided between the top of the tank and the top of the water

∴ Depth of water in the tank = 13.0 - 2.0 = 11.0 cm = 0.11 m

Given that the flask is immersed in the water contained in the tank to raise the tank water level, we have;

Volume of water + Volume of flask in the tank = Length of tank × Width of tank × Depth of water in the tank

Volume of water + Volume of flask in the tank =  0.24 × 0.21 × 0.11 = 0.005544 m³ = 0.005544 m³× 1000 l/m³ = 5.544 l

The volume of the spherical flask = 4/3·π·r³ = 4/3·π·0.0525³ = 6.06×10⁻⁴ m³ = 6.06×10⁻⁴ m³ × 1000 l/m³ = 0.606 l

The required volume of water the student needs , V = Volume of water + Volume of flask in the tank - The volume of the spherical flask = 5.544 l - 0.606 l = 4.9 l.

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Find [H+] of a 0.056 M hydrofluoric acid solution. Ka = 1.45 x 10-7
brilliants [131]

Answer:  [H^+] of 0.056 M HF solution is 8.96\times 10^{-5}

Explanation:

HF\rightarrow H^+F^-

 cM              0             0

c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 0.056 M and \alpha = ?

K_a=1.45\times 10^{-7}

Putting in the values we get:

1.45\times 10^{-7}=\frac{(0.056\times \alpha)^2}{(0.056-0.056\times \alpha)}

(\alpha)=0.0016

[H^+]=c\times \alpha

[H^+]=0.056\times 0.0016=8.96\times 10^{-5}  

Thus [H^+] of 0.056 M HF solution is 8.96\times 10^{-5}

8 0
3 years ago
A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of
Bad White [126]

Answer:

a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}

The number of moles of the ideal gas is:

n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}

n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}

n = 5.541\,mol

The final temperature is:

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}

T_{2} = 360.955\,K

The final pressure is:

P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}

T_{2} =  347.348\,K

The final volume is:

V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}

V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

V_{2} = 0.14\,m^{3}

4 0
4 years ago
What is the most soluble salt of the following set? What is the most soluble salt of the following set?
DiKsa [7]

Answer:

(c) AgCN\ (K_{sp} =6.0\times 10^{-17})

Explanation:

The solubility product of a solid is the amount of solid dissociates into its respective ions in the solution. Thus more the value of the Ksp, the more is the salt soluble in the solvent.

So,  Given that:-

Sn(OH)_2\ (K_{sp} =1.6\times 10^{-19})

Al(OH)_3\ (K_{sp} =1.9\times 10^{-33})

AgCN\ (K_{sp} =6.0\times 10^{-17})

Fe(OH)_3\ (K_{sp} =2.6\times 10^{-39})

The salt having highest value of Ksp is AgCN. So, it is most soluble.

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3 years ago
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l l
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8 0
3 years ago
A mixture of sodium bicarbonate and ammonium bicarbonate is 75.9% bicarbonate by mass. what is the mass percent of sodium bicarb
Lorico [155]
Answer: 24.1%,  under below assumptions.

Justification:

The question is quite ambiguous, because one of the data is not clearly stated. It says that the mixture consists on two compounds:

- sodium bicarbonate, and
- ammonium bicarbonate

.After, it says that it is 75.9 % bicarbonate, but it does not specify which bicarbonate, it might be the sodium bicarbonate or the ammonium bicarbonate. It is apparent that you omitted that information by error.

Given that later, the question is <span>what the mass percent of sodium bicarbonate is in the mixture, it is supposed that the 75.9% content is of ammonium bicarbonate.

With that said, you can calculate the mass percent of sodium bicarbonate, because there are only two compounds and so you know that both add up the 100% of the mixture.

In formulas:

100% = %m/m sodium bicarbonate + %m/m ammonium bicarbonate = 100%

=> % m/m sodium bicarbonate = 100% - % m/s ammonium bicarbonate

=> % sodium bicarbonate = 100% - 75.9% = 24.1%

Answer: 24.1%
</span>
8 0
3 years ago
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