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umka2103 [35]
3 years ago
10

A fixed number of moles of an ideal gas are kept in a container of volume V and an absolute temperature T. If T and V are both d

oubled, the new r.m.s. velocity of the molecules in the gas will be _________ times the original r.m.s. molecular velocity. A: 1/4 B: 1/2 C: 1/sqrt(2) D: equal to E: sqrt(2) F: 2 G: 4 If T doubles while V is held constant, the new net internal energy of the gas will be ________ times the original internal energy of the gas.
Chemistry
1 answer:
liubo4ka [24]3 years ago
8 0

Answer:

If the temperature and volume ot a gas increases, the r.m.s. velocity of the molecules in the gas will be 2  times the original r.m.s. molecular velocity.

If T doubles while V is held constant, the new net internal energy of the gas will be 2 times the original internal energy of the gas.

Explanation:

Temperature and root mean square velocity are directly proportional to one anoth. If the temperature increases, root mean square velocity also increases and vice versa, while temperature is also directly proportional to the internal energy of the gas molecules, higher the temperature, higher will be the internal energy and lower the temperature so internal energy will be decreased.

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How would you classify weather?
STALIN [3.7K]

Answer:

Weather can have many classification, how hot how windy how cold how humid. There are however different classifications for different types of weather. Hope this helps :)

4 0
3 years ago
A sucrose solution is prepared to a final concentration of 0.250 M . Convert this value into terms of g/L, molality, and mass %.
MArishka [77]

Answer:

A. 85.6 g

= 0.0856 kg.

B. 0.00027 mol/g

= 0.27 mol/kg.

C. 8.39 %

Explanation:

Given:

Molar concentration = 0.25 M

Molar weight of sucrose = 342.296 g/mol

Density of solution = 1.02 g/mL

Mass of water = 934.4 g.

Density in g/l = 1.020 g/ml * 1000ml/1 l

= 1020 g/l

Mass of solution in 1 l of solution = 1020 g

Mass of solution = mass of solvent + mass of solute

Mass of sucrose = 1020 - 934.4

= 85.6 g of sucrose in 1 l of solution.

A.

Density of sucrose = mass/volume

= molar mass/molar concentration

= 342.296 * 0.25

= 85.6 g/l

Number of moles = mass/molar mass

= 85.6/342.296

= 0.25 mol

B.

Molality = number of moles of solute/mass of solvent

= 0.25/934.4

= 0.00027 mol/g

C.

% mass of sucrose = mass of sucrose/total mass of solution * 100

= 85.6/1020 * 100

= 8.39 %

6 0
3 years ago
A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperatur
Ymorist [56]

<u>Answer:</u> The specific heat of metal is 0.821 J/g°C

<u>Explanation:</u>

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of metal = 30 g

m_2 = mass of water = 100 g

T_{final} = final temperature = 25°C

T_1 = initial temperature of metal = 110°C

T_2 = initial temperature of water = 20.0°C

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]

c_1=0.821J/g^oC

Hence, the specific heat of metal is 0.821 J/g°C

8 0
3 years ago
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5 0
2 years ago
Some type of path is necessary to join both half-cells in order for electron flow to occur.
klemol [59]

The given statement, some type of path is necessary to join both half-cells in order for electron flow to occur, is true.

Explanation:

Flow of electrons is possible with the help of a conducting medium like metal wire.

A laboratory device which helps in completion of oxidation and reduction-half reactions of a galvanic or voltaic cell is known as salt bridge. Basically, this salt bridge helps in the flow of electrons from anode to cathode and vice-versa.

If salt bridge is not present in an electrochemical cell, the electron neutrality will not be maintained and hence, flow of electrons will not take place.

Thus, we can conclude that the statement some type of path is necessary to join both half-cells in order for electron flow to occur, is true.

4 0
3 years ago
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