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2 years ago

3. If the dartboard below is used to model an atom, which subatomic particles would be located at Z?

Chemistry

1 answer:

yarga [219]2 years ago

3 0

Answer : The protons and neutrons subatomic particles will be located at Z.

Explanation :

In the model of an atom, there are three subatomic particles. Protons, neutrons and electrons are the subatomic particles.

The protons and the neutrons subatomic particles are located inside the nucleus and the electrons subatomic particle are located around or outside the nucleus.

The protons are positively charged, electrons are negatively charged and neutrons are neutral that means it has no charge.

In the given dartboard, Z is the nucleus in which the protons and neutrons subatomic particles are present and x, w & y are the electrons because they are located around the nucleus.

Hence, the protons and neutrons subatomic particles will be located at Z.

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2 years ago

What is wrong with the following explanation of an experiment? When it was heated, the sample of carbon decomposed.

Drupady [299]

"Carbon" is an element. It is found in the fourth group of the periodic table, and it is a stable element. This means that it can not be decomposed via heating, because if an element were to break down, it would release its subatomic particles. The explanation was probably one used to describe the thermal decomposition of a compound into smaller compounds.

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3 years ago

Into which two smaller groups are plants divided.

vova2212 [387]

It’s B

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3 years ago

A sample of gas has a volume of 20.0 mL at STP. What will the volume be if the temperature is changed to 546 K and the pressure

Ostrovityanka [42]

The volume did not change, it remained at 20 ml

<h3>Further explanation</h3>

Given

20 ml a sample gas at STP(273 K, 1 atm)

T₂=546 K

P₂=2 atm

Required

The volume

Solution

Combined gas Law :

$\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}$

Input the value :

$\tt \dfrac{1\times 20}{273}=\dfrac{2\times V_2}{546}\\\\V_2=\dfrac{1\times 20\times 546}{273\times 2}\\\\V_2=20~ml$

The volume does not change because the pressure and temperature are increased by the same ratio as the initial conditions (to 2x)

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2 years ago

An alkyne with the molecular formula C5H8 was reduced with H2 and Lindlar's catalyst. Hydroboration-oxidation of the resulting a

irinina [24]

Since the addition of the H2O in the last step of hydroboration is anti-Markovnikov, the starting material is 1-pentyne.

The addition of H2 to C5H8 yields an alkene when a Lindlar catalyst is used. Recall that the Lindlar catalysts poisons the process so that the addition do not go on to produce an alkane.

When hydroboration is carried out on the alkene, we are told that a primary alcohol was obtained. We must note that in the last step of hydroboration, water is added in an anti- Markovnikov manner to yield the primary alcohol. Hence, the starting material must be 1-pentyne as shown in the image attached.

Learn more: brainly.com/question/2510654

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2 years ago