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hoa [83]
3 years ago
5

ANSWER ASAP (20 POINTS)

Chemistry
1 answer:
SIZIF [17.4K]3 years ago
5 0

Answer:

a chemical formula

Explanation:

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How much heat does a 100. g sample of copper absorb when its temperature increases by 30.0°C? The specific heat of copper is 0.3
erica [24]

Answer:

\boxed {\boxed {\sf B. \ 1170 \ Joules }}

Explanation:

We are asked to find how much heat a sample of copper absorbs when the temperature is increased.

Since we know the mass, temperature increase, and specific heat capacity, we can use the following formula to calculate heat.

q= mc \Delta T

The mass of the copper sample is 100 grams, the temperature is changed or increased by 30.0 degrees Celsius, and the specific heat of copper is 0.39 Joules per gram degrees Celsius.

  • m= 100 g
  • c= 0.39 J/g °C
  • ΔT= 30.0 °C

Substitute the values into the formula.

q= (100 \ g )(0.39 \ J/g \textdegree C ) (30.0 \textdegree C )

Multiply the first two values. Note that the units of grams cancel.

q= 39 \ J/ \textdegree C (30.0 \textdegree C )

Multiply again, this time the units of degrees Celsius cancel.

q= 1170 \ J

The copper sample absorbs <u>1170 Joules</u> of heat and <u>Choice B </u>is correct.

7 0
3 years ago
Work is equal to the what used and multiplyed by the what over which the force was applied
Allushta [10]
W=Fd
Work is equal to the force multiplied by the distance travelled.
4 0
3 years ago
a gas that exerts a pressure of 215 torr in a container with a volume of 51.0 mL will exert a pressure of ? torr when transferre
zhannawk [14.2K]
To calculate the new pressure, we can use Boyle’s law to relate these two scenarios (Boyle’s law is used because the temperature is assumed to remain constant). Boyle’s law is:

P1V1 = P2V2,

Where “P” is pressure and “V” is volume. The pressure and volume of the first scenario is 215 torr and 51 mL, respectively, and the second scenario has a volume of 18.5 L (18,500 mL) and the unknown pressure - let’s call that “x”. Plugging these into the equation:

(215 torr)(51 mL) =(“x” torr)(18,500 mL)
x = 0.593 torr

The final pressure exerted by the gas would be 0.593 torr.

Hope this helps!
3 0
3 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
2 years ago
Why pie bond are not perticipate in hybridaization​
DedPeter [7]
The first bond between two atoms is always a sigma bond and the other bonds are always pi bonds and a hybridized orbital cannot be involved in a pi bond. Thus we need to leave one electron (in case of Carbon double bond) to let the Carbon have the second bond as a pi bond.
3 0
2 years ago
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