How many grams are in 26 moles of<br> magnesium hydroxide?

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

liubo4ka [24]

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.

8 0

3 years ago

When can the park be reopen to people and animals. Explain calculations and answers. Explain the concept.

julsineya [31]

Answer:

in 23 years

Explanation:

5 0

3 years ago

Read 2 more answers

A chemist prepares a sample of helium gas at a certain pressure, temperature and volume and then removes all but a fourth of the

cupoosta [38]

Answer:

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

Explanation:

Pressure in the container is P and volume is V.

Temperature of the helium gas molecules = $T_1$

Molecules helium gas = x

Moles of helium has = $n_1= \frac{x}{N_A}$

PV = nRT (Ideal gas equation)

$PV=n_1RT_1$ ...[1]

After removal of helium gas only a fourth of the gas molecules remains and pressure in the container and volume should remain same.

Molecules of helium left after removal = $\frac{x}{4}$

Moles of helium has left after removal = $n_2= \frac{x}{4\times N_A}$

$PV=n_2RT_2$ ...[2]

$n_1RT_1=n_2RT_2$

$\frac{x}{N_A}\times T_1=\frac{x}{4\times N_A}\times T_2$

$T_1=\frac{T_2}{4}$

$T_2=4T_1$

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

6 0

3 years ago

Consider the reaction. 2 Al ( s ) + Fe 2 O 3 ( s ) heat −−→ Al 2 O 3 ( s ) + 2 Fe ( l ) If 17.3 kg Al reacts with an excess of F

IrinaVladis [17]

Answer:

32.7 kilograms of aluminium oxide will be produced.

Explanation:

$2 Al ( s ) + Fe_2O_3 ( s ) +heat\rightarrow Al_2O_3 ( s ) + 2 Fe ( l )$

Mass of aluminum = 17.3 kg = 17300 g (1 kg = 1000 g )

Moles of aluminium = $\frac{17300 g}{27 g/mol}=640.74 mol$

According to reaction, 2 moles of aluminum gives 1 mole of aluminum oxide,then 640.74 moles of aluminum will give:

$\frac{1}{2}\times 640.74 mol= 320.37 mol$ of aluminum oxide

Mass of 320.37 moles of aluminum oxides:

320.37 mol × 102 g/mol = 32,677.74 g = 32.67774 kg ≈ 32.7 kg

32.7 kilograms of aluminium oxide will be produced.

6 0

3 years ago

When does a star reach equilibrium?

Snezhnost [94]

Actually, that does not happen until the protostar becomes a star when nuclear ignition starts and is maintained. It takes awhile for new star to go through its T-Tauri stage and settle down on the main sequence.

<span>A STAR does not reach hydrostatic equilibrium until it on the main sequence. Otherwise, it would remain a brown dwarf with not enough mass to to maintain nuclear fusion for more than 3,000 to 10,00 years. </span>

5 0

3 years ago

Read 2 more answers

How many grams are in 26 moles of magnesium hydroxide?