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3 years ago

5

Explain why atoms would gain or lose electrons in order to resemble the elements in the particular group you named above. Consid

er what you learned about elements in that group.

1 answer:

Leokris [45]3 years ago

4 0

Answer:

The number of electrons in the outermost shell of an element is represented in the periodic table as the group number that element is situated in the number of electrons in all shells of an element is represented in the periodic table as the element's atomic number.

Explanation:

I hope that helps

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6. Who stated that matter is not composed of particles

Elena L [17]

After careful consideration your answer is...

Leucippus and Democritus

*Hope I helped*

~Alanna~

8 0

3 years ago

Read 2 more answers

If 125g of KClO3 is heated, what is the total mass of the products?

Andrej [43]

Given parameters:

Mass of KClO₃ = 125g

Unknown:

Total mass of the products = ?

When KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;

2KClO₃ → 2KCl + 3O₂

All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.

The total mass of the products must give us 125g according to this law of conservation of matter.

Now to find the masses of each product,

Find the number of moles of the given reactant:

Number of moles = $\frac{mass}{molar mass}$

molar mass of KClO₃ = 39 + 35.5 + 3(16) = 122.5g/mol

So number of moles of KClO₃ = $\frac{125}{122.5}$ = 1.02moles

2. Now, using this number of moles, find the number of moles of the products using this value;

2 moles of KClO₃ produced 2 moles of KCl

1.02 moles of KClO₃ will also produce 1.02moles of KCl

2 moles of KClO₃ produced 3 moles of O₂

1.02 moles of KClO₃ will produce $\frac{1.02 x 3} {2}$ mole = 1.53 moles of O₂

3. Now find the masses of each product;

Mass = number of moles x molar mass

molar mass of KCl = 39 + 35.5 = 74.5g/mol

molar mass of O₂ = 16 x 2 = 32g/mol

Mass of KCl = 74.5 x 1.02 = 75.99g

Mass of O₂ = 32 x 1.53 = 48.96g

Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g

= 124.95g

This value is approximately the same as that of mass of KClO₃

7 0

3 years ago

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago

PLEASE HELP<br> two ways chemistry affects you in your daily life?

Fofino [41]

Answer:

when u do the dishes and when you take a shower i thank if you put thos in a full sentence they will be good answer

Explanation:

8 0

3 years ago

Which subatomic particle gives off visible light when it drops back down to a lower energy state?

icang [17]

Answer:

Option C = electron

Explanation:

Electrons are responsible for the production of colored light.

Electron:

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

Symbol= e-

Mass= 9.10938356×10⁻³¹ Kg

It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

How electrons produce the colored light:

Excitation:

When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.

De-excitation:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.

Other process may involve,

Fluorescence:

In fluorescence the energy is absorbed by the electron having shorter wavelength and high energy usually of U.V region. The process of absorbing the light occur in a very short period of time i.e. 10 ∧-15 sec. During the fluorescence the spin of electron not changed.

The electron is then de-excited by emitting the light in visible and IR region. This process of de-excitation occur in a time period of 10∧-9 sec.

Phosphorescence:

In phosphorescence the electron also goes to the excitation to the higher level by absorbing the U.V radiations. In case of Phosphorescence the transition back to the lower energy level occur very slowly and the spin pf electron also change.

5 0

2 years ago