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denis-greek [22]

3 years ago

5

Can someone help me with the question in the image? if it is correct i will mark as brainliest

1 answer:

Debora [2.8K]3 years ago

5 0

Answer:

I would think the correct answer would be B and D.

Explanation:

The information the institutions gathered doesn't tell whether the temperature increased or decreased so it couldn't be A. And the data that is gathered by each institute is similar to one another, so it couldn't be C. Therefore, you're left with B and D.

I hope this helps!

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A system of gas at low density has an initial pressure of 1.90 × 10 5 1.90×105 Pa and occupies a volume of 0.18 m³. The slow add

Bad White [126]

Answer:

ΔU = -6.2 × 10⁴ J

Explanation:

The system absorbs 965 J of heat, that is, q = 965 J.

The work (w) can be calculated using the following expression.

w = -P . ΔV

where,

P is the external pressure

ΔV is the change in the volume

w = - (1.90 × 10⁵ N/m²) × (0.51 m³ - 0.18 m³) = -6.3 × 10⁴ J

The change in the internal energy (ΔU) is:

ΔU = q + w = 965 J + (-6.3 × 10⁴ J) = -6.2 × 10⁴ J

8 0

3 years ago

Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.84×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.14×10−4, what is the

Nimfa-mama [501]

Answer: The value of equilibrium constant for reaction is, $1.42\times 10^{-2}$

Explanation:

The given chemical equations are:

(1) $PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$ ; $K_3=1.84\times 10^{-10}$

(2) $AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq)$ ; $K_4=1.14\times 10^{-4}$

Now we have to calculate the equilibrium constant for chemical equation as:

$PbCl_2(aq)+2Ag^{+}(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq)$ ; $K=?$

We are reversing reaction 2 and multiplying reaction 2 by 2 and then adding both reaction, we get the final reaction.

The equilibrium constant for the reverse reaction will be the reciprocal of that reaction.

If the equation is multiplied by a factor of '2', the equilibrium constant of that reaction will be the square of the equilibrium constant.

If we are adding equations then the equilibrium constants will be multiplied.

The value of equilibrium constant for reaction is:

$K=(\frac{1}{K_4})^2\times K_3$

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

$K=(\frac{1}{1.14\times 10^{-4}})^2\times (1.84\times 10^{-10})$

$K=1.42\times 10^{-2}$

Hence, the value of equilibrium constant for reaction is, $1.42\times 10^{-2}$

7 0

3 years ago

How many nucleons are in an Iodine element? Okay

Lady_Fox [76]

Answer:

there are 74

Explanation:

3 0

2 years ago

What is the major product in this reaction

Butoxors [25]

Answer:

I think option A is right answer

4 0

2 years ago

Formula of sodium bicarbonate<br>please help me with this

Stels [109]

Answer:

NaHCO₃

Explanation:

Sodium bicarbonate (baking soda) is a chemical compound with the formula NaHCO₃.

5 0

2 years ago