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3 years ago

What happens to the electron configuration when you move down a family?

Chemistry

1 answer:

stiv31 [10]3 years ago

4 0

Answer:

The number of energy levels increases as you move down a group as the number of electrons increases.

Explanation:

Each subsequent energy level is further from the nucleus than the last. Therefore, the atomic radius increases as the group and energy levels increase. 2) As you move across a period, atomic radius decreases.

You might be interested in

How many elements have scientists discovered

Luda [366]

Answer:

118

Explanation:

The discovery of the 118 chemical element known to exist us of 2021

5 0

3 years ago

Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

4 years ago

Which of the following is not a reason why actual yield is less than theoretical yield?

Gnom [1K]

Your answer is B, conservation of mass

Recall that percent yield is given by: %Yeild = actual yeild/theoretical yeild x100

During experiments, there are errors made:

• uncertainty in measurements

• losses of reactants and products

• impurity in reactants

• losses during separation (e.g. filtration or purification)

• Some side reactions might also happen.

Among the given options, only conservation of mass does not contribute to a lower actual yield compared to the theoretical yield.

5 0

3 years ago

CaCO3(s)+2H*(aq) →Ca2+ (aq)+H2001+CO262)

Effectus [21]

Answer:CO2(g) will be formed at a faster rate in experiment 2 because more H+ particles can react per unit time

Explanation:

8 0

3 years ago

Yet a third pair of compounds of manganese and oxygen is 50.48% and 36.81% oxygen respectively. In what small whole number ratio

Mariulka [41]

Answer:

The number ratio is 4:7

Explanation:

Step 1: Data given

Compound 1 has 50.48 % oxygen

Compound 2 has 36.81 % oxygen

Molar mass oxygen = 16 g/mol

Molar mass manganese = 54.94 g/mol

Step 2: Calculate % manganes

Compound 1: 100 - 50.48 = 49.52 %

Compound 2: 100 - 36.81 = 63.19 %

Step 3: Calculate mass

Suppose mass of compounds = 100 grams

Compound 1:

50.48 % O = 50.48 grams

49.52 % Mn = 49.52 grams

Compound 2:

36.81 % O = 36.81 grams

63.19 % Mn = 63.19 grams

Step 4: Calculate moles

Compound 1

Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles

Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles

Compound 2

Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles

Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles

Step 5: calculate mol ratio

We will divide by the smallest amount of moles

Compound 1

O: 3.155/0.9013 = 3.5

Mn: 0.9013 / 0.9013 = 1

Mn2O7

Compound 2

O: 2.301 / 1.150 = 2

Mn: 1.150 / 1.150 = 1

MnO2

The number ratio is 2:3.5 or 4:7

7 0

3 years ago