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swat32
2 years ago
7

Winston predicts that certain people will be vaporized and that certain people will never be vaporized. Who? Why?

Chemistry
1 answer:
Margarita [4]2 years ago
6 0

Answer:

Hi

One of the most famous works George Orwell shows us in 1984 a society that is controlled by the Superstate, in such a way that the inhabitants cannot do or say anything without The Big Brother finding out. If anyone tries to oppose this system, it is vaporized and in less than three days society forgets that there was a person who conspired or worse, that existed.

The detentions happened at night. He woke up startled because one hand shook one shoulder, a flashlight focused on him and a circle of grim faces appeared around the bed. In most cases there was no process. People disappeared and always during the night. The name of the individual in question disappeared from all the records, any reference to what he had done was erased from everywhere and his passage through life was completely annulled as if he had never existed.

Explanation:

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How many minutes are in 3.5 years
-Dominant- [34]

Answer: 1,839,600 minutes.

Explanation: 3.5 years to minutes has been calculated by multiplying 3.5 years by 525,600

7 0
3 years ago
How many grams of XeF6 are required to react with 0.579 L of hydrogen gas at 6.46 atm and 45°C in the reaction shown below?
ankoles [38]
The chemical reaction equation for this is 

XeF6 + 3H2 ---> Xe + 6HF

Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.

we get n = 0.1433 moles H2

to get the mass of XeF6, 

we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.

0.1433 moles H2 x \frac{1 mole XeF6}{3 moles H2} x \frac{245.28 g XeF6}{1 mole XeF6} = 11.71 g XeF6

Therefore, 11.71 g of XeF6 is needed to completely react with 0.579 L of Hydrogen gas at 45 degrees Celcius and 6.46 atm.
3 0
3 years ago
A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from
sweet-ann [11.9K]

Answer:

T_i~=163.1 ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT

Q_A_u=m_A_u*Cp_A_u*deltaT

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT

Now we can <u>put the values into the equation</u>:

60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)

Now we can <u>solve for the initial temperature of gold</u>, so:

T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20

T_i~=163.1 ºC

I hope it helps!

5 0
2 years ago
These two questions please
zhenek [66]

Answer:

1.) 3

2.) 60 CM

Explanation:

1. Density=\frac{MASS}{VOLUME}= \frac{75}{25}

2. Length*Width*Height=3*10*2

6 0
3 years ago
Read 2 more answers
How many moles of NH3 can you make from 6.20 moles of N2?
german
4 I think


$ hope it welp
4 0
3 years ago
Read 2 more answers
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