A solid object has a density of 1.3 g/mL in which liquids will it float explain ?

One mole of an ideal gas, for which CV,m = 3/2R, initially at 298 K and 1.00 × 105 Pa undergoes a reversible adiabatic compressi

oksian1 [2.3K]

Answer:

final temperature (T2) = 748.66 K
ΔU = w = 5620.26 J
ΔH = 9367.047 J
q = 0

Explanation:

ideal gas:

PV = RTn

reversible adiabatic compression:

δU = δq + δw = CvδT

∴ q = 0

∴ w = - PδV

⇒ δU = δw

⇒ CvδT = - PδV

ideal gas:

⇒ PδV + VδP = RδT

⇒ PδV = RδT - VδP = - CvδT

⇒ RδT - RTn/PδP = - CvδT

⇒ (R + Cv,m)∫δT/T = R∫δP/P

⇒ [(R + Cv,m)/R] Ln (T2/T1) = Ln (P2/P1) = Ln (1 E6/1 E5) = 2.303

∴ (R + Cv,m)/R = (R + (3/2)R)/R = 5/2R/R = 2.5

⇒ Ln(T2/T1) = 2.303 / 2.5 = 0.9212

⇒ T2/T1 = 2.512

∴ T1 = 298 K

⇒ T2 = (298 K)×(2.512)

⇒ T2 = 748.66 K

⇒ ΔU = Cv,mΔT

⇒ ΔU = (3/2)R(748.66 - 298)

∴ R = 8.314 J/K.mol

⇒ ΔU = 5620.26 J

⇒ w = 5620.26 J

H = U + nRT

⇒ ΔH = ΔU + nRΔT

⇒ ΔH = 5620.26 J + (1 mol)(8.314 J/K.mol)(450.66 K)

⇒ ΔH = 5620.26 J + 3746.787 J

⇒ ΔH = 9367.047 J

8 0

3 years ago

Read the picture :) please and thank you!

lana [24]

It's "C" a sample of dust particles at 0 Pascals

5 0

3 years ago

In the presence of hydrogen bond in most molecules is responsible for the following except

BlackZzzverrR [31]

There are no options so I'll just give my answer. Intermolecular hydrogen bonding is responsible for the high boiling point of water. The presence of hydrogen bonds can cause an anomaly in the normal succession of states of matter for certain mixtures of chemical compounds as temperature increases or decreases.

8 0

3 years ago

Describe the appearance of the material. Include its color and state (solid or liquid). Pour ¼ cup of water into a small, clear

Alex777 [14]

Answer:

Here the given material is taken and mixed with water.

Explanation:

The amount of material and water taken are same. Hence if it is not soluble in water it should make a dense and flowy paste like material and if it is soluble in water it should this and thicker density of water should remain.

If the amount of water that we are taking is more than the material will float in water if it is not soluble and lighter than water or would sink if it is heavier than water.

5 0

3 years ago

Read 2 more answers

If you feed 100 kg of N2 gas and 100 kg of H2 gas into a

torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of $N_2$ = 100 kg = 100000 g

Mass of $H_2$ = 100 kg = 100000 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $H_2$ = 2 g/mole

Molar mass of $NH_3$ = 17 g/mole

First we have to calculate the moles of $N_2$ and $H_2$ .

$\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles$

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $N_2$ react with 3 mole of $H_2$

So, 3571.43 moles of $N_2$ react with $3571.43\times 3=10714.29$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $N_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 1 mole of $N_2$ react to give 2 mole of $NH_3$

So, 3571.43 moles of $N_2$ react to give $3571.43\times 2=7142.86$ moles of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg$

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0

3 years ago