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2 years ago

If the universe is expanding but the expansion is slowing down, what will the eventual

1 answer:

6 0

Mark Brainliest please

Answer:

The big freeze or heat death
But the expansion would stretch everything out until everything is evenly spread through the universe. When that happens, objects can gain or lose energy. The gas clouds that form stars would dissipate, black holes would evaporate and eventually even light particles would fizzle out.

Check for more info regarding death of universe

Doesn’t allowing me to put the link here. Check your comment

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Please help quick

Rama09 [41]

Answer:

c = 0.898 J/g.°C

Explanation:

1) Given data:

Mass of water = 23.0 g

Initial temperature = 25.4°C

Final temperature = 42.8° C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Specific heat capacity of water is 4.18 J/g°C

ΔT = 42.8°C - 25.4°C

ΔT = 17.4°C

Q = 23.0 g × × 4.18 J/g°C × 17.4°C

Q = 1672.84 j

2) Given data:

Mass of metal = 120.7 g

Initial temperature = 90.5°C

Final temperature = 25.7 ° C

Heat released = 7020 J

Specific heat capacity of metal = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 25.7°C - 90.5°C

ΔT = -64.8°C

7020 J = 120.7 g × c × -64.8°C

7020 J = -7821.36 g.°C × c

c = 7020 J / -7821.36 g.°C

c = 0.898 J/g.°C

Negative sign shows heat is released.

7 0

3 years ago

The mass of an object is described in what??

inessss [21]

<span>The mass of an object is measured in either grams or kilograms. Mass is best described as the amount of matter, or "stuff," in a solid, and is different from weight (which is the force of gravity on an object). Since mass is used with solids, it will be measured in grams or kilograms (rather than in something like liters, which would be used with the volume of a liquid). To measure mass, you can use a balance, for example a triple balance beam.</span>

6 0

3 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago

Help meeeeehahahshshs

natka813 [3]

What grade are you in I’m in 9th ;-;

7 0

3 years ago

In which of the following situstions would convection currents occur

andriy [413]

I believe your answer should be C because the heat would move from the soup to the spoon.

5 0

3 years ago

Read 2 more answers