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3 years ago

APP

Chemistry

1 answer:

RideAnS [48]3 years ago

4 0

Answer:

Explanation:

This question is about nuclear reactions. (a) (i) Define the term unified atomic mass unit. [1]. 1/12th mass of an atom of carbon-12.

The nuclear reaction can be written as: Mg2512+He42→H11+XAZ. where. A is the mass number and; Z is the atomic number of the new nuclide, X.

The mass defect for an isotope was found to be 0.410 amu/atom. Calculate the binding energy in kJ/mol of atoms. (1 J = 1 kg m2/s2): (a) 3.69 x 1010

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4 years ago

Would you expect trna or mrna to be more extensively hydrogen bonded? Why?

AveGali [126]

Answer:

tRNA

Explanation:

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4 years ago

Describe a procedure to separate a mixture of sugar black pepper and pebbles

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3 years ago

A gas at a pressure of 2.0 atm is in a closed container. Indicate the changes (if any) in its volume when the pressure undergoes

garik1379 [7]

Answer:

a) The volume increases

b) The volume decreases

c) The volume does not change

Explanation:

A gas at a pressure of 2.0 atm is in a closed container. Indicate the changes (if any) in its volume when the pressure undergoes the following changes at constant temperature and constant amount of gas.

When there is a change in the pressure (P) at constant temperature (T) and amount of gas (n), we can find the change in the volume (V) using Boyle's law.

P₁.V₁ = P₂.V₂

where,

1 refer to the initial state

2 refer to the final state

a) The pressure drops to 0.40 atm.

P₁.V₁ = P₂.V₂

(2.0 atm) . V₁ = (0.40 atm) . V₂

V₂ = 5 . V₁

The volume increases.

b) The pressure increases to 6.0 atm.

P₁.V₁ = P₂.V₂

(2.0 atm) . V₁ = (6.0 atm) . V₂

V₂ = 0.33 . V₁

The volume decreases.

c) The pressure remains at 2.0 atm.

P₁.V₁ = P₂.V₂

(2.0 atm) . V₁ = (2.0 atm) . V₂

V₂ = V₁

The volume does not change.

4 0

3 years ago

0.0500 mol of gas occupies a cylinder which is sealed on top by a moveable piston. The piston is circular, with a mass of 30.0 k

Viefleur [7K]

Answer:

The workdone by both N₂ and neon gas is 49.3 J

The change in internal energy of N₂ and neon gas is 125.6 J and 73.54 J respectively

The heat for N₂ and neon gas is 171.9 J and 122.84 J respectively.

Explanation:

Given that:

number of moles = 0.05 mole

mass of the piston = 30 kg

diameter = 5.00 cm = 0.05 m

Area (A) = πr²

$Area (A) = \pi*(\frac{0.05}{2})^2 \\ \\ Area (A) = 0.0019635 \\ \\ Area (A) = 19.635*10^{-4} \ m^2$

The piston is said to move from 30 cm - 40 cm

So, the change in volume ΔV is calculated as:

$=(40-30)*10^{-2} *19.635*10^{-4}$

$= 1.9635*10^{-4} \ m^3$

Outside the cylinder; the pressure $P_{air}= 1 \ atm$ = 101325 Pa

Thus, workdone $w_1$ = PΔV

= $101325*1.9635*10^{-4}$

= 19.90 J

The gravitational work $w_2 = mgh$

Given that the height (h) = 10 cm = 0.1 m

Then; $w_2 = 30*9.8*0.1$

$w_2 = 29.4 \ J$

The total workdone $w_{total}$ for both cases is:

$w_{total } =w_1 + w_2$

$w = (19.90 + 29.4) \ J$

$w =49.3 \ J$

The pressure of gas inside the cylinder is determined as:

$P_{in}.A = P_{out}.A +mg$

$(P_{in}-P_{out}) = \frac{mg}{A} \\ \\ P_{in} -10^5 = \frac{30*9.8}{19.635*10^{-4}} \\ \\ P_{in} = 149732.6203+10^5 \\ \\ P_{in} = 2.497*10^5 \ Pa$

a). assuming that the gas is N₂.

$C_v =\frac{5}{2}R$

Thus, the change in internal energy ΔU is given as:

$\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{5}{2}R \delta T \\ \\ \delta U = \frac{5}{2}nR \delta T$

Since $P_{in} \delta V = nR \delta T ; \ Then$ ;

$\delta \ U = \frac{5}{2} P_{in} \delta V \\ \\ \delta \ U = \frac{5}{2}*2.497*10^5 *1.9635*10^{-4} \\ \\ \delta \ U = 122.57 \ J$

ΔU ≅ 125.6 J

The heat Q = ΔU + W

Q = (122.6 + 49.3) J

Q = 171.9 J

b) In Neon gas:

$C_v = \frac{3}{2}R$

∴

change in internal energy is;

$\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{3}{2}R \delta T \\ \\ \delta U = \frac{3}{2}P_{in}.V$

$\delta U = \frac{3}{2}*2.497*10^5*1.9635*10^{-4}$

ΔU = 73.54 J

The heat Q = ΔU + W

Q = (73.54 + 49.3) J

Q = 122.84 J

5 0

3 years ago