Using standard heats of formation, calculate the standard enthalpy change for the following reaction.

malfutka [58]

What question do you need help with

5 0

3 years ago

two perfumes A and B were released at the same time and you are standing 7.5metres away from both of them.molarmass of perfume A

kifflom [539]

17.40 sec is the time will take to smell second perfume after diffusion takes place

Acc. to Graham's law of Diffusion

Diffusion of Gas inversely proportional to square root of its Molecular mass.

<u>rb (Perfume B)</u>

ra(perfume A)

= $\sqrt{\frac{Ma}{Mb} }$ its equation (1)

Give molar mass of Perfume A = 275 g/mol

molar mass of Perfume B= 205g/mol putting value in (1)

» $\frac{rb}{ra}=\sqrt{\frac{275}{205} }$

$\frac{rb}{ra} =1.16$ <em> its eq (2)</em>

» Perfume B will defuse 1:16 times faster than perfume A.

Hence, perfume B will be first smelled by Person.

Sf Equal volume V of two goes diffuse in t1 and t2 sec. respectively ton

$\frac{ra}{rb}=\frac{tb}{ta}$

Now,, from eq(2) toto

1/1:15 = $\frac{tb}{ta}$

ta

Given tb =the smell of Perfume B as it diffuse faster

ta= 1.15 x 15 see

=ta2=17.40 sec

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4 0

1 year ago

What is the molarity of a solution that contains 2.634 mol KCI in 25.2 L solution?O a. 0.105O b. 0.00141O c. 7.73O d. 105

Vikentia [17]

Explanation:

To solve this question, we need to use the following formula:

M = n/V

So:

M = ??

n = 2.634 mol

V = 25.2 L

M = 2.634/25.2

M = 0.105 mol/L

Answer: a. 0.105

4 0

1 year ago

A vessel with a volume of 32. 5 l contains 2. 80 g of n2 gas, 0. 403 g of h2 gas, and 79. 9 g of ar gas. at 25°c, what is the to

d1i1m1o1n [39]

The total pressure of the gases in the vessel is found out by using the partial pressures of individual gases and adding them up. The total pressure of the gases in the vessel is 1.72 atm.

The Partial pressure of the gases is the individual pressure of the gases present in a vessel containing a mixture of gases. The total pressure can be found out by adding partial pressure of all gases.

Given:

Volume of vessel, V= 32.5L

Mass of Nitrogen gas = 2.8g

Mass of Hydrogen gas = 0.403g

Mass of Argon gas = 79.9g

We know that:

Molar Mass of Nitrogen gas = 28g

Molar Mass of Hydrogen gas = 2g

Molar Mass of Argon gas = 40g

∴ Moles of gas = Given mass / Molar mass

∴ Using the Ideal Gas Equation:

PV = nRT

where, P is the Pressure of gas

V is the volume of gas

n is the moles of gas

R is the Universal Gas constant

T is the temperature

Applying the above equation of all three gases indivudually,

∴ For Nitrogen,

P₁V = n₁RT

⇒ P₁ × 32.5 = 2.8/28 × 0.082 × 298

⇒ P₁ = 0.07atm

∴ For Hydrogen,

P₂V = n₂RT

⇒ P₂ × 32.5 = 0.4/2 × 0.082 × 298

⇒ P₂ = 0.15 atm

∴ For Argon,

P₃V = n₃RT

⇒ P₃ × 32.5 = 79.9/40 × 0.082 × 298

⇒ P₃ = 1.5 atm

∴ Total pressure of vessel, P = P₁ + P₂ + P₃

P = 1.72 atm

The total pressure of the gases in the vessel is 1.72 atm.

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6 0

1 year ago

Which are isotopes? an atom that has an atomic number of 20 and a mass number of 42 is an isotope of an atom that has which are

olga55 [171]

Isotopes have the same atomic number or same number Protons, but different mass number or neutrons.

You can say atomic number or protons.

22 neutrons 20 protons
atomic number 20 and a mass number 40
Atomic number 20 and mass # 42

6 0

3 years ago

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