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tangare [24]
3 years ago
11

Ayudenme porfa doy corona xd

Chemistry
1 answer:
vova2212 [387]3 years ago
5 0

Answer:

e) 5

Explanation:

Because it's H

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How much energy is required to heat a frozen can of juice (360 grams- mostly water) from 0 degrees Celsius ( the temperature of
Usimov [2.4K]

Answer:

1,100,160J or 262.94 kcal

Explanation:

The juice is frozen at 0 degrees Celsius and I assume that it will become gas at 100 degrees Celsius. So we change the form of the water from solid to liquid, then to gas. That means we have to find out how much heat needed to change water form too, not only the heat needed to increase its temperature.

The latent heat of water is 4.2J/g °C while the heat of fusion is 334 J/g and the heat of vaporization is 2260 J/g. The energy needed will be:

360g * 4.2J/g °C * (110-0°C ) + 360g * 334 J/g + 360g * 2260 /g = 1,100,160J or 262.94 kcal.

8 0
3 years ago
The vaporization of 1 mole of liquid water (system) at 100.9 C, 1.00 atm, is endothermic.
yan [13]

Answer:

a) The work done is 10.0777 kJ

b) The water's change in internal energy is -122.1973 kJ

Explanation:

Given data:

1 mol of liquid water

T₁ = temperature = 100.9°C

P = pressure = 1 atm

Endothermic reaction

T₂ = temperature = 100°C

1 mol of water vapor

VL = volume of liquid water = 18.8 mL = 0.0188 L

VG = volume of water vapor = 30.62 L

3.25 moles of liquid water vaporizes

Q = heat added to the system = -40.7 kJ

Questions: a) Calculate the work done on or by the system, W = ?

b) Calculate the water's change in internal energy, ΔU = ?

Heat for 3.25 moles:

Q_{1} =3.25*(-40.7)=-132.275kJ

The work done:

W=-nP*delta(V)=-3.25*101.33*(0.0188-30.62)=10077.6637J=10.0777kJ

The change in internal energy:

delt(U)=W+Q_{1} =10.0777-132.275=-122.1973kJ

7 0
3 years ago
2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is.
Andrej [43]

2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.  

The reaction of 2-bromo-3,4-dimethylpentane is combined with t-butoxide forms 2 alkene in the elimination reaction due to steric hindrance.  The least stable alkene 3,4 dimethyl - 1- pentene is easy to make. the  t-butoxide is (CH₃)₃CO⁻. The reaction involves in this reaction is E2 elimination reaction. This reaction involves the one step reaction. The product will also form that is 3,4 dimethyl - 2 - pentene.  so the reaction involve Elimination reaction and the product due to steric hindrance is 3,4 dimethyl - 1- pentene

Thus, 2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.  

To learn more about  t-butoxide here

brainly.com/question/12303978

#SPJ4

4 0
1 year ago
A gas at STP occupies 22.4 L if the temperature is changed to 260 K and the pressures changed it to 0.50 ATM what will the new v
asambeis [7]

Answer:

The new volume will be 42, 7 L.

Explanation:

We use the gas formula, which results from the combination of the Boyle, Charles and Gay-Lussac laws. According to which at a constant mass, temperature, pressure and volume vary, keeping constant PV / T. The conditions STP are: 1 atm of pressure and 273 K of temperature.

P1xV1/T1 =P2xV2/T2

1 atmx 22,4 L/273K = 0,5atmx V2/260K

V2=((1 atmx 22,4 L/273K )x 260K)/0,5 atm= 42, 67L

3 0
3 years ago
Calculate the wavelength for the transition from n = 4 to n = 2, and state the name given to the spectroscopic series to which t
finlep [7]

Answer:

The wavelength for the transition from n = 4 to n = 2 is<u> 486nm</u> and the name  name given to the spectroscopic series belongs to <u>The Balmer series.</u>

Explanation

lets calculate -

Rydberg equation-   \frac{1}{\pi } =R(\frac{1}{n_1^2} -\frac{1}{n_2^2})

where ,\pi is wavelength , R is Rydberg constant ( 1.097\times10^7), n_1 and n_2are the quantum numbers of the energy levels. (where n_1=2 , n_2=4)

Now putting the given values in the equation,

                \frac{1}{\pi }=1.097\times10^7\times(\frac{1}{2^2} -\frac{1}{4^2} )=2056875m^-^1

    Wavelength \pi =\frac{1}{2056875}

             =4.86\times10^-^7 = 486nm

<u>    Therefore , the wavelength is 486nm and it belongs to The Balmer series.</u>

8 0
3 years ago
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