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nika2105 [10]
3 years ago
15

What do oxygen, silicon, and selenium have in common? How might this relate to their organization on the periodic table?

Chemistry
1 answer:
Sonbull [250]3 years ago
8 0

Answer:

They all belong to the p block in the periodic table

Explanation:

Let us examine the electronic configuration of each element;

Oxygen - [He] 2s2 2p4

Silicon - [Ne] 3s2 3p2

Selenium - [Ar] 4s2 3d10 4p4

A common thread that joins all the elements listed above is that they all belong to the p-block in the periodic table. They could be collectively referred to as p-block elements.

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How many moles of SnF₂ will be produced along with 48 grams of H₂? *
KIM [24]

Answer: 24 moles of SnF_2 are produced.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

{\text{Moles of}  H_2}=\frac{48g}{2g/mol}=24moles

Sn+2HF\rightarrow SnF_2+H_2

According to stoichiometry :

1 mole of H_2 is accompanied with = 1 mole of SnF_2

Thus 24 moles of H_2  is accompanied with =\frac{1}{1}\times 24=24moles  of SnF_2

Thus 24 moles of SnF_2 are produced.

8 0
3 years ago
4.3 moles of a gas are at a temperature of 28 degrees * C with a pressure of 1.631 atm. What volume does the gas occupy?
Shkiper50 [21]

Answer:

65.2L

Explanation:

Using the general gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (Litres)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (Kelvin)

According to the information provided in this question,

P = 1.631 atm

V = ?

n = 4.3 moles

T = 28°C = 28 + 273 = 301K

Using PV = nRT

V = nRT/P

V = 4.3 × 0.0821 × 301 ÷ 1.631

V = 106.26 ÷ 1.631

V = 65.15

Volume of the gas = 65.2L

7 0
3 years ago
What is a daughter cell
Leno4ka [110]
The cells that result from the reproductive division of one cell during mitosis or meiosis
6 0
3 years ago
Read 2 more answers
In an electron-dot structure of an element, the dots are used to represent ________.
Vlad [161]
Dots are used to represent electrons.
3 0
3 years ago
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is ________ M sodium ion and ________ M sulfate i
omeli [17]

Answer:

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

Explanation:

Step 1: Data given

Volume = 500 mL = 0.500 L

The concentration sodium sulfate = 2.104 M

Step 2: The equation

Na2SO4 → 2Na+ + SO4^2-

For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)

Step 3: Calculate the concentration of the ions

[Na+] = 2*2.104 M = 4.208 M

[SO4^2-] = 1*2.104 M = 2.104 M

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

8 0
3 years ago
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