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joja [24]
2 years ago
14

A valve to a new chamber in the bottle is opened, and the gas expands to 2 x 10-3 m3. (The gas does no work in this process beca

use the gas molecules don't have anything moving to push on.) After a while, the parts of the gas re-equilibrate, without exchanging heat with the outside. What is the new temperature, T? T =
Chemistry
1 answer:
Fed [463]2 years ago
8 0

Answer:

The temperature doesn't change

Explanation:

Given that:

dU=dQ - dW

Where

  • U: internal energy
  • W: work
  • Q: heat

If the gas does no work and doesn't exchange heat with the outside there insn't a variation of the internal energy.

Internal energy is realted to the energy in the molecules, they vibration and theregore the temperature.

So, if there isn't a change in the internal energy you may say that there isn't a change in the temperature.

The increase in the volume is balanced by a decrease of the pression.

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(g) On the graph in part (d) , carefully draw a curve that shows the results of the second titration, in which the student titra
atroni [7]

Answer:

Here's what I get  

Explanation:

(g) Titration curves

I can't draw two curves on the same graph, but I can draw two separate curves for you.  

The graph in part (d) had an equivalence point at 20 mL.

In the second titration, the NaOH was twice as concentrated, so the volume to equivalence point would be half as much — 10 mL.

The two titration curves are below.

(h) Evidence of reaction

HCl and NaOH are both colourless.

They don't  evolve a gas or form a precipitate when they react.

The student probably noticed that the Erlenmeyer flask warmed up — a sign of a chemical change.

4 0
2 years ago
A rigid, 28-L steam cooker is arranged with a pressure relief valve set to release vapor and maintain the pressure once the pres
MaRussiya [10]

Answer:

\Delta S_{source}>-1.204\frac{kJ}{K}

Explanation:

Hello!

In this case, given the initial conditions, we first use the 10-% quality to compute the initial entropy:

s_1=s_{f,175kPa}+q*s_{fg,175kPa}\\\\s_1=1.4850\frac{kJ}{kg*K} +0.1*5.6865\frac{kJ}{kg*K}=2.0537\frac{kJ}{kg*K}

Now the entropy at the final state given the new 40-% quality:

s_2=s_{f,150kPa}+q*s_{fg,150kPa}\\\\s_2=1.4337\frac{kJ}{kg*K} +0.4*5.7894\frac{kJ}{kg*K}=3.7495\frac{kJ}{kg*K}

Next step is to compute the mass of steam given the specific volume of steam at 175 kPa and the 10% quality:

m_1=\frac{0.028m^3}{(0.001057+0.1*1.002643)\frac{m^3}{kg} } =0.274kg\\\\m_2=\frac{0.028m^3}{(0.001053+0.4*1.158347)\frac{m^3}{kg} } =0.0603kg

Then, we can write the entropy balance:

\Delta S_{source}+\frac{Q}{T_1} -\frac{Q}{T_2} +s_2m_2-s_1m_1-s_{fg}(m_2-m_1)>0

Whereas sfg stands for the entropy of the leaving steam to hold the pressure at 150 kPa and must be greater than 0; thus we plug in:

Which is such minimum entropy change of the heat-supplying source.

Best regards!

3 0
2 years ago
How many hydrogen atoms are in the following molecule of ammonium sulfide? (NH4)2S
Lelechka [254]
The are eight hydrogen atoms in ammonium sulfide because there are 2 molecules of ammonium.
3 0
2 years ago
Read 2 more answers
A gas of 190 mL at a pressure of 74 atm can be expected to change its pressure when its volume changes to 30.0 mL. Express its n
kozerog [31]

Answer : The new pressure of the gas will be, 468.66 atm

Explanation :

Boyle's Law : This law states that pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}     (At constant temperature and number of moles)

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure of the gas = 74 atm

P_2 = final pressure of the gas = ?

V_1 = initial volume of the gas = 190 ml

V_2 = final volume of the gas = 30 ml

Now we put all the given values in the above formula, we get the final or new pressure of the gas.

74atm\times 190ml=P_2\times 30ml

P_2=468.66atm

Therefore, the new pressure of the gas will be, 468.66 atm

4 0
2 years ago
Read 2 more answers
The average density of a carbon-fiber-epoxy composite is 1.615 g/cm3. the density of the epoxy resin is 1.21 g/cm3 and that of t
ra1l [238]
The composite material is composed of carbon fiber and epoxy resins. Now, density is an intensive unit. So, to approach this problem, let's assume there is 1 gram of composite material. Thus, mass carbon + mass epoxy = 1 g.

Volume of composite material = 1 g / 1.615 g/cm³ = 0.619 cm³
Volume of carbon fibers = x g / 1.74 g/cm³ 
Volume of epoxy resin = (1 - x) g / 1.21 g/cm³ 

a.) V of composite = V of carbon fibers + V of epoxy resin
0.619 = x/1.74 + (1-x)/1.21
Solve for x,
x = 0.824 g carbon fibers
1-x = 0.176 g epoxy resins

Vol % of carbon fibers = [(0.824/1.74) ÷ 0.619]*100 =<em> 76.5%</em>

b.) Weight % of epoxy = 0.176 g epoxy/1 g composite * 100 = <em>17.6%</em>
     Weight % of carbon fibers = 0.824 g carbon/1 g composite * 100 = <em>82.4%</em>
4 0
2 years ago
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