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erma4kov [3.2K]
2 years ago
12

A mixture of gases A2 and B2 are introduced to a slender metal cylinder that has one end closed and the other fitted with a pist

on that makes a gas-tight seal so that the gases are a closed system. The cylinder is submerged in a large beaker of water whose temperature is 25∘C, and a spark is used to trigger a reaction in the cylinder. At the completion of the reaction, the piston has moved downward, and the temperature of the water bath has increased to 28∘C. If we define the system as the gases inside the cylinder, which of the following best describes the signs of q, w, and ΔE for this reaction?
(a) q < 0, w < 0, ΔE < 0
(b) q < 0, w > 0, ΔE < 0
(c) q < 0, w > 0, the sign of ΔE cannot be determined from the information given
(d) q > 0, w > 0, ΔE > 0
(e) q > 0, w < 0, the sign of ΔE cannot be determined from the information given
Chemistry
1 answer:
REY [17]2 years ago
3 0

Answer:

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

Explanation:

Determination of sign of q

Temperature of the water bath before the reaction = 25 °C

Temperature of the water bath after the completion of the reaction = 28 °C

After the completion of the reaction, temperature of the water bath is increased that means heat is released during the reaction and flows out of the system.

If heat is absorbed by the system, then q is indicated by positive sign and if heat is released by the system, then q is indicated by negative sign.

As in the given case, heat is released by the system, so sign of q is negative, or q < 0

Determination of sign of w

After the completion of the reaction, piston moved downward, that means volume of the system decreases or compression occur. During the compression, work is done on the system.

if work is done on the system, sign of w is positive.

If work is done by the system, sign of w is negative.

In the given case, work is done on the system, therefore sign of w is positive, or w > 0

Determination of sign of ΔE

Relationship between ΔE, q and w is given by first law of thermodynamics:

ΔE = q + w

In this case, q is positive and w is negative, so the sign of ΔE depends of magnitude of q and w. As magnitude of w and q cannot be determined in this case, thus, the given information is insufficient for the determination of sign of ΔE.

So, among the given option, option c is correct.

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

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4 0
2 years ago
How many grams of KClO3 are needed to produce of 4.26 moles of O2? 2 KClO3 2 KCl + 3 O2 a. 348 g b. 136 g c. 174 g d. 522 g e. 7
JulsSmile [24]
Ok so this is what we know :

2KClO3 -> 2KCl + 3O2         (Always check if equation is balanced - in this                                               case it is)
                              4.26moles
So we know that we have 4.26 moles of oxygen (O2). Now lets look at the ratio between KClO3 and O2.
We see that the ratio is 2:3 meaning that we need 2KClO3 in order to produce 3O2.
Therefore divide 4.26 by 3 and then multiply by 2.
4.26/3 = 1.42
1.42 * 2 = 2.84
Now we know that the molarity of KClO3 is 2.84 moles.
Multiply by R.M.M to find how many grams of KClO3 we have.

R.M.M of KClO3
K- 39
Cl- 35.5
3O- 3 * 16 -> 48
---------------------------
                      <span>122.5
</span>2.84 * 122.5 = 347.9 grams therefore the answer is (a)
                       348 grams needed of KClO3 to produce 4.26 moles of O2.
Hope this helps :).

8 0
2 years ago
Read 2 more answers
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2 years ago
A student carries out the precipitation reaction shown below, starting with 0.030 moles of calcium nitrate. The final mass of th
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Answer:

a. Ca₃(PO₄)₂.

b. 0.010 moles of Ca₃(PO₄)₂ can we expect to be produced

c. 3.1g of Ca₃(PO₄)₂

d. Percent yield = 93.5%

Explanation:

a. Based on the reaction:

3Ca(NO₃)₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6NaNO₃(aq)

<em>3 moles of calcium nitrate reacts with 2 moles of sodium phosphate producieng 1 mole of calcium phosphate.</em>

<em />

As you can see, Ca₃(PO₄)₂ is a solid product -(s)-, that means when the reaction occurs the precipitate produced is the solid,

<h3>Ca₃(PO₄)₂</h3><h3 />

b. As 3 moles of calcium nitrate produce 1 mole of calcium phosphate and there are 0.030 moles of calcium nitrate

0.030 moles Ca(NO₃)₂ × (1 mol Ca₃(PO₄)₂ / 3 moles Ca(NO₃)₂) =

<h3>0.010 moles of Ca₃(PO₄)₂ can we expect to be produced</h3><h3 />

c. As molar mass of Ca₃(PO₄)₂ is 310.18g/mol, the mass of 0.010 moles (The expected mass) is;

0.010 moles Ca₃(PO₄)₂ × (310.18g / mol) =

<h3>3.1g of Ca₃(PO₄)₂</h3><h3 />

d. The percent yield is defined as 100 times the ratio between the obtained yield (That is 2.9g of precipitate, Ca₃(PO₄)₂) and the expected yield, 3.1g of Ca₃(PO₄)₂:

\frac{2.9g}{3.1g} *100

<h3>Percent yield = 93.5%</h3>
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