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2 years ago

What subatomic particles make up the nucleus of an atom (select ALL that are in the nucleus)

Chemistry

1 answer:

Ilia_Sergeevich [38]2 years ago

3 0

3 and 4 i would think

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GIVING BRAINLIEST!! PLEASE HELP

LenaWriter [7]

Answer:

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Explanation:

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3 years ago

Is ok2 gas a element compound or homogeneous or heterogeneous mixture?

ivolga24 [154]

I'm guessing you mean O2 gas? This is an elemental compound. It can't be a mixture because it is composed of only 1 element. Hope I helped!

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2 years ago

Which part of dalton's theory did thomson's findings dispute? atoms of different elements have different masses. atoms of the sa

skad [1K]

Answer is: the atom is the smallest known particle of matter.
John Dalton claimed that atom is indestructible and all atoms of a given element are identical in mass and properties.
Thomson discovered electron and found the first evidence for isotopes of a stable element.

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3 years ago

Read 2 more answers

L20_UENAxleFsh-VHfg9rLfZsitva3L3UKGjmF1B299/vie Settle a dispute. X Anya claims that this challenge can be solved with a single

Fynjy0 [20]

Answer:

Neither arre correct

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Neither Anya nor Braden are correct. This is because if you use 90 degrees, 180 degrees, or even 270 degrees you will not get the exact image, which means that the image will not be found by just a rotation because there will be a curve in the image. You can solve it if you can do 90 degree rotation and translation.

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2 years ago

The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42

denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

$k = Ae^{\frac{-Ea}{RT}}$

Where:

k: is the rate constant

A: is the frequency factor

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

$k_{1} = k_{2}$

$Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}$ (1)

By solving equation (1) for T₁ we have:

$T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C$

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!

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3 years ago