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MaRussiya [10]
3 years ago
6

Elements in the same blank on the periodic table have the same number of valence electrons

Chemistry
2 answers:
aleksandr82 [10.1K]3 years ago
7 0

Answer:

  • Yes, it is true that elements in the same blank on the periodic table have the same number of valence electrons.
  • As, the periodic table is the systematic arrangement of the different elements based upon the different elements properties (both physical and chemical). As the groups are vertical blocks in which those elements are arranged which posses or has the same number of electrons inside there electronic configuration.

nikitadnepr [17]3 years ago
6 0
Elements in the same group on the periodic table have the same number of valence electrons. The "groups" are the column (or rows). groups are vertically and periods are horizontally. 
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How many grams of NaF form when .5 mol of HF reacts with excess Na2SiO3?
IgorLugansk [536]

Answer:

5.25g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is shown below:

Na2SiO3 + 8HF → H2SiF6 + 2NaF + 3H2O

From the balanced equation above,

8 moles of HF reacted to produce 2 moles of NaF.

Therefore, 0.5 moles of HF will react to produce = (0.5 x 2)/8 = 0.125 mole of NaF.

Next, we shall convert 0.125 mole of NaF to grams.

This is illustrated below:

Mole of NaF = 0.125 mole

Molar mass of NaF = 23 + 19 = 42g/mol

Mass of NaF =..?

Mass = mole x molar mass

Mass of NaF = 0.125 x 42

Mass of NaF = 5.25g

Therefore, 5.25g of NaF is produced from the reaction.

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melamori03 [73]

<u>Answer:</u> The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For nickel:</u>

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol

For the given chemical reaction:

3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)

  • <u>For nickel (II) oxide:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from \frac{3}{3}\times 0.252=0.252mol of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g

  • <u>For aluminium:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from \frac{2}{3}\times 0.252=0.168mol of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

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3 years ago
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