All categories

3 years ago

I NEED HELP REALLY BAD AND THIS IS DUE IN 5 MINSSS

Chemistry

1 answer:

fgiga [73]3 years ago

7 0

Answer:

Explanation:

The number of valence electrons in the atom is 5 as can be seen from the ionization energy.

Ionization energy is the measure of the readiness of an atom to lose an electron. The lower the value the easier it is for an atom to lose an electron and vice versa.

After the removal of the first electron it becomes increasingly more difficult to remove succeeding electrons because of the imbalance between the positive nuclear charge and the remaining electrons.

Therefore, since we have 5 ionization energies given, the number of valence electrons in this atom is 5.

You might be interested in

How can the structure of atoms be connected to a colorful firework display?

olasank [31]

These metallic substances are actually metal salts, which produce luminescent light of different colors when they are dispersed in the air. This light is produced by electrons inside the metal atoms

Hope this helped!

3 0

3 years ago

Which quantity could best be measured with a meter stick?

Ilya [14]

Answer:

Explanation:

A meter stick measures distances ( like centimeters) so....B

it does NOT measure mass , volume or VERY large distances (Like earth's diameter)

3 0

2 years ago

Read 2 more answers

Research help?

prisoha [69]

Answer:

They tenderize the product by coating and weakening the gluten bonds within the structure.

Even though they contain little or no moisture, they provide the illusion of wetness. ...

They enable browning.

They help move heat through the product, perpetuating the baking process.

Explanation:

4 0

3 years ago

A sample of O2(g) is placed in an otherwise empty, rigid container at 4224 K at an initial pressure of 4.97 atm, where it decomp

frutty [35]

Answer:

The value of $K_p$ at 4224 K is 314.23.

Explanation:

$O_2(g)\rightleftharpoons 2O(g)$

Initially

4.97 atm 0

At equilibrium

4.97 - p 2p

At initial stage, the partial pressure of oxygen gas = =4.97 atm

At equilibrium, the partial pressure of oxygen gas = $p_{O_2}=0.28 atm$

So, 4.97 - p = 0.28 atm

p = 4.69 atm

At equilibrium, the partial pressure of O gas = $p_{O}=2p=2\times 4.69 atm=9.38 atm$

The expression of $K_p$ is given as :

$K_p=\frac{(p_{O})^2}{(p_{O_2})}$

$K_p=\frac{(9.38 atm)^2}{0.28 atm}=314.23$

The value of $K_p$ at 4224 K is 314.23.

5 0

3 years ago

A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

ycow [4]

Answer: The molecular formula for the compound is $C_6H_6$

Explanation:

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.68 moles.

For Carbon = $\frac{7.68}{7.68}=1$

For Hydrogen = $\frac{7.74}{7.68}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is $CH$

Calculating the molar mass of the compound:

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = $62.3637\text{ L. torr }mol^{-1}K^{-1}$

T = temperature of the gas = $100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol$