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3 years ago

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Chemistry

2 answers:

Anastasy [175]3 years ago

8 0

Hey have a great day!

Arada [10]3 years ago

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Answer:

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Explanation:

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A container is at a pressure of 3 atm and a temperature of 280K. What is the new temperature when the pressure is reduced to 1.5

GenaCL600 [577]

Answer:

140 K

Explanation:

Step 1: Given data

Initial pressure of the gas (P₁): 3 atm

Initial temperature of the gas (T₁): 280 K

Final pressure of the gas (P₂): 1.5 atm

Final temperature of the gas (T₂): ?

Step 2: Calculate the final temperature of the gas

We have a gas whose pressure is reduced. If we assume an ideal behavior, we can calculate the final temperature of the gas using Gay-Lussac's law.

T₁/P₁ = T₂/P₂

T₂ = T₁ × P₂/P₁

T₂ = 280 K × 1.5 atm/3 atm = 140 K

6 0

3 years ago

Two moles of an ideal gas are placed in a container whose volume is 2.3 x 10^-3 m3. The absolute pressure of the gas is 6.9 x 10

PtichkaEL [24]

Answer:

$K.E.=1.97\times 10^{-21}\ J$

Explanation:

Given that:-

Pressure = $6.9\times 10^5\ Pa$

The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = $\frac {1}{101325}$ P (atm)

Given the value of pressure = 43,836 Pa

So,

$6.9\times 10^5\ Pa$ = $\frac{6.9\times 10^5}{101325}$ atm

Pressure = 6.80977 atm

Volume = $2.3\times 10^{-3}\ m^3$ = 2.3 L ( 1 m³ = 1000 L)

n = 2 mol

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

6.80977 atm × 2.3 L = 2 mol × 0.0821 L.atm/K.mol × T

⇒T = 95.39 K

The expression for the kinetic energy is:-

$K.E.=\frac{3}{2}\times K\times T$

k is Boltzmann's constant = $1.38\times 10^{-23}\ J/K$

T is the temperature

So, $K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 95.39\ J$

$K.E.=1.97\times 10^{-21}\ J$

3 0

3 years ago

The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sampl

AlekseyPX

Answer: Concentration of $NH_3$ in the equilibrium mixture is 0.31 M

Explanation:

Equilibrium concentration of $H_2$ = 0.729 M

The given balanced equilibrium reaction is,

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

Initial conc. x 0 0

At eqm. conc. (x-2y) M (y) M (3y) M

The expression for equilibrium constant for this reaction will be:

3y = 0.729 M

y = 0.243 M

$K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}$

Now put all the given values in this expression, we get :

$K_c=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}$

$0.967=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}$

$x=0.80$

concentration of $NH_3$ in the equilibrium mixture = $0.80-2\times 0.243=0.31$

Thus concentration of $NH_3$ in the equilibrium mixture is 0.31 M

3 0

3 years ago

What two molecules were condensed in an aldol condensation to produce (ch3)3cch=chcoch3?

inessss [21]

The given compound is being synthesized by condensing Acetone and Pivaldehyde (Trimethylacetaldehyde).

First Acetone is treated with Base, the base abstracts the mildly acidic proton present at alpha position to carbonyl group. The resulting specie called enolate act as a nucleophile and attacks on highly reactive aldehyde which upon dehydration yields the Aldol Product. The Reaction is as follow,

5 0

3 years ago

Bases turn litmus paper what color? a) red b) blue c) yellow d) pink

yKpoI14uk [10]

Answer:
blue
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.......

5 0

3 years ago

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