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sasho [114]
3 years ago
14

Which Salt is least soluble at 50 C

Physics
1 answer:
Anvisha [2.4K]3 years ago
4 0

The salt c e_{2}\left(s o_{4}\right)_{3} is the least soluble salt at 50^{\circ} \mathrm{C}.

Answer: Option 1

<u>Explanation: </u>

As it is clear that c e_{2}\left(s o_{4}\right)_{3}, a gas so the solubility of it decreases on increasing the temperature of water. Because gas molecule starts to evaporate on attaining heat with the increase of temperature thus leading to reduction in the solubility of them.

As most of the gas molecules get vaporised the solubility in water gradually decreases for this compound. It is also clear from the solubility graphs of the various salts.

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If an astronaut weighs 900 n on earth, what does he weigh on the planet venus?
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Use w=m*g value of g is 1.67m/s^2
5 0
3 years ago
a roller coaster is at the top of a hill and rolls to the top of a lower hill.If mechanical energy is conserved,on the top of wh
pychu [463]
Hello!!

Here we have a simple matter of conservation of energy. ME=PE+KE.

At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.

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7 0
3 years ago
What's the minimum Out PUT WORK<br> required to raise 14,0m3 of water 26.0m?
BartSMP [9]

Answer:

3.57 MJ

Explanation:

ASSUMING it's fresh water with density of 1000 kg/m³

W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J

Salt water would require more.

3 0
2 years ago
What’s the answer for this problem?
pickupchik [31]
The answer is always true a



4 0
3 years ago
A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres
Lemur [1.5K]

Answer:

V_2 = 0.125 m^3

Work done =  = 5 kJ

Explanation:

Given data:

volume of nitrogen v_1 = 0.08 m^3

P_1 = 150 kPa

T_1 = 200 degree celcius = 473 Kelvin

P_2 = 80 kPa

Polytropic exponent n = 1.4

\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}

putting all value

\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}

\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS

polytropic process is given as

P_1 V_1^n = P_2 V_2^n

150\times 0.08^{1.4} = 80 \times V_2^{1.4}

V_2 = 0.125 m^3

work done = \frac{P_1 V_1 -P_2 V_2}{n-1}

= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}

                  = 5 kJ

4 0
3 years ago
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