Name the following molecule: W(H3AlO4)3

How many sigma and pi bonds in propionic bond

Ierofanga [76]

Propanoic acid formula is ch ch 2 so it has 8 bonds

7 0

2 years ago

If a hazardous bottle is labeled 20.2 percent by mass Hydrochloric Acid, HCl, with a density of 1.096 g/mL, calculate the molari

SIZIF [17.4K]

Answer:6M

Explanation:

From Co= 10pd/M

Where Co= molar concentration of raw acid

p= percentage by mass of raw acid=20%

d= density of acid=1.096g/cm3

M= molar mass of acid=36.5

Co= 10×20×1.096/36.5=6M

5 0

2 years ago

A hot gas flowing through a pipeline can be considered as a:________

k0ka [10]

Answer:

B) irreversible process

Explanation:

The process given here is irreversible.

7 0

2 years ago

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

Now put all the given values in above formula, we get:

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

Now put all the given values in above formula, we get:

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

1 year ago

A system delivers 225 j of heat to the surroundings while delivering 645 j if work calculate the change in the internal Chang

Charra [1.4K]

Heat given out to the surroundings by the system = 225 J

Work done by the system on the surroundings = 645 J

According to the energy conservation, the energy can neither be created nor it can be destroyed, it can transform from one form to another. Hence, the energy which is lost to the surrounding as a work done and heat came from the internal energy of the system.

Hence, the change in the internal energy = - 225 - 645 = - 870 Joules

Negative sign means that the internal energy of the system is decreased by 870 Joules

3 0

2 years ago

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