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2 years ago

11. The electron configuration for phosphorous is [Ar]3s23p4. TRUE FALSE

Chemistry

1 answer:

user100 [1]2 years ago

5 0

Answer:

False

Explanation:

Phosphorus is number 15 on the periodic table, so its electronic configuration is:

${1s}^{2} {2s}^{2} {2p}^{6} {3s}^{2} {3p}^{3}$

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20.1 g of aluminum and 219 g of chlorine gas react until all of the aluminum metal has been converted to AlCl3. The balanced equ

Romashka-Z-Leto [24]

Answer:

The amount of Cl2 gas left , after the reaction goes to completion is : 139.655 grams

Explanation:

Molar mass : It is the mass in grams present in one mole of the substance.

Moles of the substance is calculated by:

$Moles=\frac{Mass}{Molar\ mass}$

$2Al(s)+3Cl_{2}(g)\leftarrow 2AlCl_{3}(g)$

According to this equation:

2 mole of Al = 3 mole of Cl2 = 2 mole of AlCl3

Molar mass of Al = 27.0 g/mol

Mass of Al = 20.1 gram

Moles of Al present in the reaction :

$Moles=\frac{Mass}{Molar\ mass}$

$Moles=\frac{20.1}{26.98}$

Moles of Al = 0.744

Similarly calculate the moles of Cl2

Molar mass of Cl2 = 71.0 g/mol

Mass = 219 gram

$Moles=\frac{Mass}{Molar\ mass}$

$Moles=\frac{219}{70.98}$

Moles of Cl2 = 3.08 moles

According to equation,

2 mole of Al reacts with = 3 mole of Cl2

1 moles of Al reacts with = 3/2 mole of Cl2

0.744 moles of Al reacts with = 3/2(0.744) moles of Cl2

= 1.116 moles of Cl2

But actually present Cl2 = 3.08 moles

Hence Al is the limiting reagent , and Cl2 is the excess reagent.

The whole Aluminium Al get consumed during the reaction.

The amount of Cl2 in excess = Total Cl2 - Cl2 consumed

Cl2 in excess = 3.08 - 1.116 = 1.964 moles

Cl2 in grams = 1.964 x 70.90 = 139.655 grams

6 0

4 years ago

3.06 x 10? Gg =Blank g

kicyunya [14]

Answer:

30.6

Explanation:

7 0

3 years ago

Read 2 more answers

In one of your classes, you were asked to start a program that would help the government in environmental conservation,

kirill [66]

The answer is C

Because using crocodile skin won’t benefit the environment

5 0

3 years ago

Quinone, an oxygenated hydrocarbon, is a chemical used in the photography industry. When a 0.144-g sample is analyzed by combust

earnstyle [38]

Answer:

C3H2O1

Explanation:

The following data were obtained from the question:

Mass of compound = 0.144 g

Mass of CO2 = 0.352 g

Mass of H2O = 0.0480 g

Empirical formula of compound =?

Next, we shall determine the mass of carbon (C), hydrogen (H) and oxygen (O) present in the compound. This can be obtained as follow:

For carbon (C):

Molar mass of CO2 = 12 + (2×16)

Molar mass of CO2 = 44 g/mol

Mass of CO2 = 0.352 g

Mass of C in the compound

= 12/44 × 0.352

= 0.096 g

For Hydrogen (H):

Molar mass of H2O = (2×1) + 16

Molar mass of H2O = 18 g/mol

Mass of H2O = 0.0480 g

Mass of H in the compound

= 2/18 × 0.0480

= 0.0053 g

For oxygen (O):

Mass of O = mass of compound – (mass of C + mass of H)

Mass of compound = 0.144 g

Mass of C = 0.096 g

Mass of H = 0.0053 g

Mass of O = 0.144 – (0.096 + 0.0053)

Mass of O = 0.144 – 0.1013

Mass of O = 0.0427 g

Finally, we shall determine the empirical formula of the compound as follow:

C = 0.096 g

H = 0.0053 g

O = 0.0427 g

Divide both side by their molar mass

C = 0.096 / 12 = 0.008

H = 0.0053 / 1 = 0.0053

O = 0.0427 / 16 = 0.0027

Divide by the smallest

C = 0.008 / 0.0027 = 3

H = 0.0053 / 0.0027 = 2

O = 0.0027 / 0.0027 = 1

Therefore, the empirical formula of the compound is C3H2O1

4 0

4 years ago

Draw the Lewis structure for SO2. How

juin [17]

Answer:

It should be 9

Explanation:

That’s how I remember drawing them I took chemistry last year. Tell me if it’s good.

5 0

3 years ago