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3 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]3 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. How many grams of zinc will be formed if 32.0 g of copper reacts with zinc nitrate? Copper(I)nitrate is the other product.

Yakvenalex [24]

Answer:

16.46 g.

Explanation:

It is a stichiometry problem.

We should write the balance equation of the mentioned chemical reaction:

2Cu + Zn(NO₃)₂ → Zn + 2Cu(NO₃).

It is clear that 2.0 moles of Cu reacts with 1.0 mole of Zn(NO₃)₂ to produce 1.0 mole of Zn and 2.0 moles of Cu(NO₃).

We need to calculate the number of moles of the reacted Cu (32.0 g) using the relation:

n = mass / molar mass

The no. of moles of Cu = mass / atomic mass = (32.0 g) / (63.546 g/mol) = 0.503 mol.

Using cross multiplication:

2.0 moles of Cu produces → 1.0 mole of Zn, from the stichiometry.

0.503 mole of Cu produces → ??? mole of Zn.

The no. of moles of Zn produced = (1.0 mol)(0.503 mol) / (2.0 mol) = 0.2517 mol.

∴ The grams of Zn produced = no. of moles x atomic mass of Zn = (0.2517 mol)(65.38 g/mol) = 16.46 g.

3 0

3 years ago

A solid mixture consists of 47.6g of KNO3 (potassium nitrate) and 8.4g of K2SO4 (potassium sulfate). The mixture is added to 130

IgorLugansk [536]

Answer: No crystals of potassium sulfate will be seen at 0°C for the given amount.

Explanation:

We are given:

Mass of potassium nitrate = 47.6 g

Mass of potassium sulfate = 8.4 g

Mass of water = 130. g

Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g

This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water

Applying unitary method:

In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams

So, in 130 grams of water, the amount of potassium sulfate dissolved will be $\frac{7.4}{100}\times 130=9.62g$

As, the soluble amount is greater than the given amount of potassium sulfate

This means that, all of potassium sulfate will be dissolved.

Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.

7 0

3 years ago

Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)

Reptile [31]

Answer: The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]$

For the given chemical reaction:

$C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]$

We are given:

$\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J$

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

4 0

3 years ago

A student performs three trials of a titration of an acid with an unknown concentration. She compares her measured concentration

Viktor [21]

Answer:

There are many errors possible while titrating the acid of an unknown concentration with a base like NaOH.

Main error that leads to the error in results is misreading of the end point volume .

End point is when the reaction between the analyte and solution of known concentration has stopped .

Sometimes Burette is not straight enough to read the volume of the end point. One way to misread the volume of burette is by looking at the burette volume at an angle .

From above , volume seems to be higher. Indicators are used to indicate the color change of the reaction. In Acid-Base titrations , indicators first lighten up then changes its color.

So, error may have occurred in wrongly judging of the end point by color change of the indicator .

4 0

3 years ago

What is an example of entropy from everyday life?

Deffense [45]

A Campfire is an example, the solid wood becomes ash.

8 0

3 years ago