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3 years ago

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As a chemical reaction occurs, the thermometer in the container records an increase in temperature. What is true of the reaction

?
The reaction is endothermic because heat was taken in by the reaction.

The reaction is exothermic because heat was released by the reaction.

The reaction is endothermic because heat was released by the reaction.

The reaction is exothermic because heat was taken in by the reaction.

2 answers:

iren2701 [21]3 years ago

7 0

The second one is correct

Serga [27]3 years ago

5 0

Answer:

The reaction is exothermic because heat was released by the reaction.

Explanation:

'exothermic' means a reaction released heat over the time of the experiment.

'endothermic' means that the reaction released cold over the experiment.

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What is the universe made of? Astronomers face an embarrassing conundrum: they don't know what 95% of the universe is made of.

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Answer:

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3 years ago

If a sample containing 18.1 g of NH3 is reacted with 90.4 g of

USPshnik [31]

Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃ = 18.1g

Mass of Cu₂O = 90.4g

Unknown:

Limiting reactant = ?

Mass of N₂ formed = ?

Solution:

The reaction equation is given as:

Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol

Molar mass of NH₃ = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = $\frac{18.1}{143.2}$ = 0.13moles

Number of moles of NH₃ = $\frac{90.4}{17}$ = 5.32moles

From this reaction;

1 mole of Cu₂O combines with 2 mole of NH₃

So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃

= 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

Mass = number of moles x molar mass

1 mole of Cu₂O will produce 1 mole of N₂

0.13 mole of Cu₂O will produce 0.13 mole of N₂

Mass = 0.13 x (2 x 14) = 3.64g

5 0

3 years ago

The amount of space tgat matter occupies is referred to as...

Marrrta [24]

It is either mass or volume

4 0

3 years ago

Read 2 more answers

For the decomposition of A to B and C, A(s)⇌B(g)+C(g) how will the reaction respond to each of the following changes at equilibr

lys-0071 [83]

Answer:

a. No change.

b. The equilibrium will shift to the right.

c. No change

d. No change

e. The equilibrium will shift to the left

f. The equilibrium will shift to the right

Explanation:

We are going to solve this question by making use of Le Chatelier´s principle which states that any change in a system at equilibrium will react in such a way as to attain qeuilibrium again by changing the equilibrium concentrations attaining Keq again.

The equilibrium constant for A(s)⇌B(g)+C(g)

Keq = Kp = pB x pC

where K is the equilibrium constant ( Kp in this case ) and pB and pC are the partial pressures of the gases. ( Note A is not in the expression since it is a solid )

We also use Q which has the same form as Kp but denotes the system is not at equilibrium:

Q = p´B x p´C where pB´ and pC´ are the pressures not at equilibrium.

a. double the concentrations of Q which has the same form as Kp but : products and then double the container volume

Effectively we have not change the equilibrium pressures since we know pressure is inversely proportional to volume.

Initially the system will decrease the partial pressures of B and C by a half:

Q = pB´x pC´ ( where pB´and pC´are the changed pressures )

Q = (2 pB ) x (2 pC) = 4 (pB x PC) = 4 Kp ⇒ Kp = Q/4

But then when we double the volume ,the sistem will react to double the pressures of A and B. Therefore there is no change.

b. double the container volume

From part a we know the system will double the pressures of B and C by shifting to the right ( product ) side since the change reduced the pressures by a half :

Q = pB´x pC´ = ( 1/2 pB ) x ( 1/2 pC ) = 1/4 pB x pC = 1/4 Kp

c. add more A

There is no change in the partial pressures of B and C since the solid A does not influence the value of kp

d. doubling the concentration of B and halve the concentration of C

Doubling the concentrantion doubles the pressure which we can deduce from pV = n RT = c RT ( c= n/V ), and likewise halving the concentration halves the pressure. Thus, since we are doubling the concentration of B and halving that of C, there is no net change in the new equilibrium:

Q = pB´x pC´ = ( 2 pB ) x ( 1/2 pC ) = K

e. double the concentrations of both products

We learned that doubling the concentration doubles the pressure so:

Q = pB´x pC´ = ( 2 pB ) x ( 2 pC ) = 4 Kp

Therefore, the system wil reduce by a half the pressures of B and C by producing more solid A to reach equilibrium again shifting it to the left.

f. double the concentrations of both products and then quadruple the container volume

We saw from part e that doubling the concentration doubles the pressures, but here afterward we are going to quadruple the container volume thus reducing the pressure by a fourth:

Q = pB´x pC´ = ( 2 pB/ 4 ) x (2 pC / 4) = 4/16 Kp = 1/4 Kp

So the system will increase the partial pressures of B and C by a factor of four, that is it will double the partial pressures of B and C shifting the equilibrium to the right.

If you do not see it think that double the concentration and then quadrupling the volume is the same net effect as halving the volume.

3 0

3 years ago

Carry out the following operations as if they were calculations of experimental results and express each answer in standard nota

Shalnov [3]

Answer:

1. $10.6\; \rm m$ (one decimal place.)

2. $0.79\; \rm g$ (two decimal places.)

3. $16.5\;\rm cm^2$ (three significant figures.)

Explanation:

<h3>1.</h3>

The first and second expressions are additions and subtractions. When adding two numbers, the accuracy of the result is given by the number of decimal places in it. The result should have as many decimal places as the input with the least number of decimal places.

For example, in the first expression:

$5.6792\;\rm m$ has four decimal places.
$0.6\; \rm m$ has only one decimal place.
$4.33\; \rm m$ has two decimal places.

Therefore, the result should be rounded to one decimal place. Note that these units are compatible for addition, since they are all the same. The result should have the same unit (that is: $\rm m$ .)

Therefore:

$\rm 5.6792\; m + 0.6\; m + 4.33\; m \approx 10.6\; \rm m$ . (Rounded to one decimal place.)

<h3>2.</h3>

Similarly:

$\rm 3.70\; \rm g$ has two decimal places.
$2.9133\; \rm g$ has four decimal places.

Therefore, the result should be rounded to two decimal places. Its unit should be $\rm g$ (same as the unit of the two inputs.)

$\rm 3.70\; g - 2.9133\; g \approx 0.79\; \rm g$ . (Rounded to two decimal places.)

<h3>3.</h3>

When multiplying two numbers, the accuracy of the result should be based on the number of significant figures in it. The result should have as many significant figures as the input with the least number of significant figures. In this expression:

$4.51\; \rm cm$ has three significant figures.
$3.6666\; \rm cm$ has five significant figures.

Therefore, the result should have only three significant figures.

The unit of the result is supposed to be the product of the units of the input. In this expression, that unit will be $\rm cm \cdot cm$ , which is occasionally written as $\rm cm^2$ .

$\rm 4.51 \; cm \times 3.6666 \; cm \approx 16.5\; \rm cm^2$ . (Rounded to three significant figures.)

7 0

3 years ago