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likoan [24]
3 years ago
6

3. Heating by direct contact between particles is called

Chemistry
1 answer:
Yanka [14]3 years ago
8 0
I believe it is “Conduction” im not 100% sure tho, I haven’t done this since the 7th grade.
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Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces
Brilliant_brown [7]

Answer:

0.364

Explanation:

Let's do an equilibrium chart for the reaction of combustion of ammonia:

2NH₃(g) + (3/2)O₂(g) ⇄ N₂(g) + 3H₂O(g)

4.8atm 1.9atm 0 0 Initial

-2x -(3/2)x +x +3x Reacts (stoichiometry is 2:3/2:1:3)

4.8-2x 1.9-(3/2)x x 3x Equilibrium

At equilibrium the velocity of formation of the products is equal to the velocity of the formation of the reactants, thus the partial pressures remain constant.

If pN₂ = 0.63 atm, x = 0.63 atm, thus, at equilibrium

pNH₃ = 4.8 - 2*0.63 = 3.54 atm

pO₂ = 1.9 -(3/2)*0.63 = 0.955 atm

pH₂O = 3*0.63 = 1.89 atm

The pressure equilibrium constant (Kp) is calculated with the partial pressure of the gases substances:

Kp = [(pN₂)*(pH₂O)³]/[(pNH₃)²*(pO_2)^{3/2}]

Kp = [0.63*(1.89)³]/[(3.54)²*(0.955)^{3/2}]

Kp = 4.2533/11.6953

Kp = 0.364

7 0
3 years ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
4 years ago
Consider the dissolution equation of lead(II) chloride.
Vladimir79 [104]

Answer:

The solubility product of lead(II) chloride is 1.61\times 10^{-5}.

Explanation:

Concentration of lead (II) ions = [Pb^{2+}]=0.0159 M

Concentration of chloride ion = [Cl^-]=0.0318 M

PbCl_2(s)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

The expression of a solubility product will be given as:

K_{sp}=[Pb^{2+}][Cl^-]^2

=0.0159 M\times (0.0318 M)^2=1.61\times 10^{-5}

The solubility product of lead(II) chloride is 1.61\times 10^{-5}.

4 0
3 years ago
1.00 x 10^6 atoms of gold is equivalent to how many grams?
S_A_V [24]

Answer:

3.27 x 10⁻¹⁶ grams

Explanation:

moles Au = 1.00 x 10⁻⁶ Atoms / 6.02 x 10²³Atoms / mole = 1.66 x 10⁻¹⁸ mole Au

grams Au = 1.66 x 10⁻¹⁸ mole Au x 196.97 grams Au/mole Au

= 3.27 x 10⁻¹⁶ grams Au

8 0
4 years ago
What other types of evidence might chemists use to support claims that they have synthesized a particular substance?
Misha Larkins [42]
Help me plssssss.............
7 0
3 years ago
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