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Reika [66]
3 years ago
6

I NEED HELP ON EARTH SCIENCE !!!!!

Chemistry
2 answers:
Cloud [144]3 years ago
8 0
Where is the assignment?? OR worksheet
il63 [147K]3 years ago
6 0
What kind of problem

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Which of the following is NOT true regarding Rutherford's Gold Foil experiment?
erastova [34]

Answer:

The area around the nucleus must be of low mass.

Explanation:

Rutherford`s experiment showed that there are some positive charges in the center of the atoms, and because they are all together, they will give a great mass to the atom.

It was quite different from Thomson`s experiment, in which it was thought that the negative charges were mixed with the positive charges, around the atom (like a Pudding Model). In Rutherford`s experiment, because the direction of beta particles, it was the prediction of the positive nucleus.

Hope this info is useful.

8 0
3 years ago
List four examples of diffusion seen in daily life.
babunello [35]

Answer:

Spraying perfume in one corner of the room and the smell travels to the other side of the room

Explanation:

6 0
3 years ago
PLZ HURRY
Aliun [14]

Hello.

The answer is: False

The current theory of the formation of our solar system is that the planets formed more or less in their present orbits and they were made balls of gas but they were never comets captured by the sun.

Have a nice day

5 0
3 years ago
Read 2 more answers
What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su
anastassius [24]

Answer:

V=27992L=28.00m^3

Explanation:

Hello,

In this case, the combustion of methane is shown below:

CH_4+2O_2\rightarrow CO_2+2H_2O

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3

Best regards.

8 0
3 years ago
What is the temperature of 1.2 moles of helium gas at 2.57 ATM if it occupies 15.5L of volume
agasfer [191]
We will assume helium to behave as an ideal gas and apply the ideal gas law:
PV = nRT
For pressure measured in atmospheres and volume measured in liters, the value of the molar gas constant is 0.082. Therefore:
T = PV / nR
T = (2.57 x 15.5) / (1.2 x 0.082)
T = 404.8 Kelvin
8 0
3 years ago
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