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3 years ago

7

A graduated cylinder is filled to 10.0 mL with water and a piece of granite is placed in the cylinder displacing the level to 23

.7mL . what is the volume of the granite piece in cubic centimeters?

1 answer:

Rus_ich [418]3 years ago

8 0

Answer:

13.7 cubic centimeters

Explanation:

23.7-10=13.7mL

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A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,

Dafna1 [17]

Answer : The concentration of $A,B\text{ and }C$ at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.

Explanation :

The given chemical reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

$\Delta G^o=-RT \ln k$

where,

$\Delta G^o$ = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = $25^oC=273+25=298K$

Now put all the given values in the above relation, we get:

$-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k$

$k=5.45$

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

Initially 0.30 0.40 0

At equilibrium (0.30-x) (0.40-x) x

The expression of equilibrium constant will be,

$k=\frac{[C]}{[A][B]}$

$5.45=\frac{x}{(0.30-x)\times (0.40-x)}$

By solving the term x, we get

$x=0.168\text{ and }0.716$

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of $A$ at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of $B$ at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of $C$ at equilibrium = x = 0.168 M

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3 years ago

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saul85 [17]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

What kind of bond holds together atoms within a molecule?

sesenic [268]

I think it would be these three answers ionic , covalent , and polar covalent

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3 years ago

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Can someone help Plz

Marrrta [24]

Answer: A. 1.60 liters

Explanation:

$2Na_2O_2+2CO_2\rightarrow 2Na_2CO_3+O_2$

As can be seen from the balanced chemical equation, 2 moles of $CO_2$ produce 1 mole of $O_2$ .

According to Avogadro's law, every 1 mole of the gas occupies 22.4 liters at STP.

Thus $2$ moles of $CO_2$ occupies $22.4\times 2=44.8L$ at STP and produce 1 mole of $O_2$ i.e. 22.4 L

44.8 L of $CO_2$ will produce 22.4 L of $O_2$

Thus $3.20L$ of $CO_2$ will produce $\frac{22.4}{44.8}\times 3.20=1.60L$

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2 years ago

1. A student sees tiny bubbles clinging to the inside of an unopened plastic bottle full of carbonated soft drinks. The student

Lina20 [59]

Answer:

1) The bubbles will grow, and more may appear.

2)Can A will make a louder and stronger fizz than can B.

Explanation:

When you squeeze the sides of the bottle you increase the pressure pushing on the bubble, making it compress into a smaller space. This decrease in volume causes the bubble to increase in density. When the bubble increases in density, the bubble will grow and more bubbles will appear. Therefore, Changing the pressure (by squeezing the bottle) changes the volume of the bubbles. The number of bubbles doesn't change, just their size increases.

Carbonated drinks tend to lose their fizz at higher temperatures because the loss of carbon dioxide in liquids is increased as temperature is raised. This can be explained by the fact that when carbonated liquids are exposed to high temperatures, the solubility of gases in them is decreased. Hence the solubility of CO2 gas in can A at 32°C is less than the solubility of CO2 in can B at 8°C. Thus can A will tend to make a louder fizz more than can B.

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