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Lostsunrise [7]
3 years ago
13

Which two of the following elements would you expect to be most similar: nitrogen, chlorine, barium, fluorine, and sulfur? 

Chemistry
2 answers:
Anon25 [30]3 years ago
8 0
Nitrogen clorine flourine and sulfur
myrzilka [38]3 years ago
6 0

Answer:

Chlorine and Flourine

Explanation:

It is important to identify the groups in which the elements being to. The elements in each group have the same number of electrons in the outer orbital. Those outer electrons are also called valence electrons.

Since elements in a group have the same number of valence electrons, they behave similarly in chemistry.

Nitrogen - Group 5

Chlorine - Group 7

Barium - Group 2

Flourine - Group 7

Sulphur - Group 6

This means the most similar elements would be flourine and Chlorine because they are both group 7 elements.

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2 years ago
etermine the molar concentration of magnesium ions in a solution formed by mixing 100.0 mL of 0.100 M MgCl, solution with 100.0
marysya [2.9K]

Answer:

Molar concentration of magnesium ions in the final solution is 0.2 M.

Explanation:

c=\frac{n}{V}

c = Concentration of the solution

n =   moles of the compound

V = volume of the solution in L

1) Molarity of MgCl_2 solution = 0.100 M

Volume of MgCl_2 solution = 100.0 mL = 0.1 L

Moles of MgCl_2 in 100 mL solution= n

0.100 M=\frac{n}{0.1L}

n=0.01 mol

MgCl_2(aq)\rightarrow Mg^{2+}(aq)+2Cl^-(aq)

1 mole of magnesium chloride gives 1 mol of magnesium ions and 2 moles of chloride ions.

Then 0.01 moles of magnesium chloride will give:

x=\frac{1}{1}\times 0.01 mol=0.01 mol magnesium ions.

2) Molarity of Mg_3(PO_4)_2 solution = 0.100 M

Volume of Mg_3(PO_4)_2 solution = 100.0 mL = 0.1 L

Moles of Mg_3(PO_4)_2 in 100 mL solution= n'

0.100 M=\frac{n'}{0.1L}

n'=0.01 mol

Mg_3(PO_4)_2(aq)\rightarrow 3Mg^{2+}(aq)+2PO_{4}^-(aq)

1 mole of magnesium phosphate gives 3 mol of magnesium ions and 2 moles of phosphate ions.

Then 0.01 moles of magnesium phosphate will give:

y=\frac{3}{1}\times 0.01 mol=0.03 mol magnesium ions.

After mixing both solutions:

Moles of magnesium ions = x + y = 0.01 mol + 0.03 mol = 0.04 mol

Total volume after mixing = 0.1 L + 0.1 L = 0.2 L

Molar concentration of magnesium ions in the final solution:[Mg^{2+}]

[Mg^{2+}]=\frac{0.04 mol}{0.2 L}=0.2 mol/L

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4 years ago
How many moles of a gas sample are in a 20.0 L container at 373 K and 203 kPa? The gas constant is 8.31 L−kPa/mol−K. A)0.33 mole
12345 [234]

Answer:

Option (C) 1.30 moles

Explanation:

The following data were obtained from the question:

Volume (V) = 20L

Temperature (T) = 373K

Pressure (P) = 203 kPa

Gas constant (R) = 8.31 L.kPa/mol.K.

Number of mole (n) =...?

The number of mole of the gas in the container can obtained by applying the ideal gas equation as illustrated below:

PV = nRT

Divide both side by RT

n = PV /RT

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Therefore, 1.30 mole of the gas is present in the container.

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Answer: The given statement is False.

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Answer: during a chemical reaction, matter cannot be created or destroyed

Explanation:

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