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2 years ago

True or false: Atoms and mass are conserved in nature True False

Chemistry

1 answer:

katovenus [111]2 years ago

5 0

Answer:

It’s true

Explanation:

If we account for all reactants and products in a chemical reaction, the total mass will be the same at any point in time in any closed system. ... The Law of Conservation of Mass holds true because naturally occurring elements are very stable at the conditions found on the surface of the Earth.

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What would happen to the pressure if the volume were reduced to 0.5 l and the temperature increased to 260 ∘c?

Vsevolod [243]

<span>The ideal gas law. PV=nRT pressure x volume = moles x Faraday's constant x Temp Kelvin (C+273) Original data Pressure 1 atmosphere Volume 1 liter Temp 25C = 298K New data Volume 0.5 liter pressure X Temp 260C = 533K P1v1T1 = P2v2T2 plug and chug. (1)(1)(293) = (x)(0.5)(533) Solve for X, which is the new pressure. </span>

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2 years ago

Your brain and spinal cord are joined to the other parts of your body through __________.

Lostsunrise [7]

Answer is A

You never run down adjacent to you spinal cord and bones. That’s where they are the most protected. There are also holes in the pelvis bone for nerves to pass through

5 0

2 years ago

Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h

bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

We know that,

The relation between the $C_p\text{ and }C_v$ for an ideal gas are :

$C_p-C_v=R$

As we are given :

$C_p=28.253J/K.mole$

$28.253J/K.mole-C_v=8.314J/K.mole$

$C_v=19.939J/K.mole$

Now we have to calculate the entropy change of the gas.

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

$\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J$

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. $(w=-pdV)$

(C) Heat during the process will be,

$q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J$

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0

3 years ago

A sample of propane (C3H8) has a mass of 0. 47 g. The sample is burned in a bomb calorimeter that has a mass of 1. 350 kg and a

Nady [450]

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The specific heat has been expressed as:

$q=mc\Delta T$

<h3 /><h3>Computation for the heat absorbed</h3>

The iron and calorimeter are in side the closed system. Thus, the energy released by the sample, has been equivalent to the energy absorbed by the calorimeter.

$q_{released}=q_{absorbed}\\
q_{released}=m_{calorimeter}\;c_{calorimeter}\;\Delta T$

The given mass of calorimeter has been, $m_{calorimeter}=1350\;\rm g$

The specific heat of the calorimeter has been, $c_{calorimeter}=5.82\;\rm J/g^\circ C$

The change in temperature of the calorimeter has been, $\Delta T=2.87^\circ \rm C$

Substituting the values for heat released:

$q_{released}= 1350\;\text g\;\times\;5.82\;\text J/\text g^\circ \text C\;\times\;2.87^\circ \text C\\
q_{released}=22,549.5\;\text J\\
q_{released}}=22.54\;\rm kJ$

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

Learn more about specific heat, here:

brainly.com/question/2094845

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2 years ago

HELP PLEASE I WILL GIVE 100 PONTS

pochemuha

Answer:

boiling point

Explanation:

5 0

3 years ago

Read 2 more answers