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yarga [219]
3 years ago
10

(a) Which elements are most likely to form ionic compounds? (b) Which metallic elements are most likely to form cations with dif

ferent charges?
Chemistry
1 answer:
elena-s [515]3 years ago
3 0

Answer:

A. Generally ionic compounds are formed metal and non metal elements since they forms opposite charged ions due to difference in their Electronegativity values.

NaCl, MgCl2, Ca(OH)2 etc are ionic compounds.

B. Generally transition elements (D elements) forms cations with different charges or simply they exist in different oxidation states due to availability of empty d orbitals.

Examples are Iron exsits as Fe+2 and Fe+3

Mn exists as Mn+2 , Mn+4, Mn+6 ,Mn+7 etc.

You might be interested in
Stoichiometry Problems!
lisov135 [29]

Hey there!

C₆H₁₂O₆(s) + 6O₂(g) => 6CO₂(g) + 6H₂0(l)

a.)

First we need to find how many molecules of oxygen gas we need.

Every one molecule of C₆H₁₂O₆ will react with six molecules of O₂. So, if we have 3.011 x 10²³ molecules of C₆H₁₂O₆, we need six times that of oxygen.

3.011 x 10²³ x 6 = 18.066 x 10²³ = 1.8066 x 10²⁴

So we need 1.8066 x 10²⁴ molecules of O₂. We need to find the volume of this in liters.

At STP, one mole of a gas occupies 22.4 liters. Let's find the number of moles we have of O₂.

(1.8066 x 10²⁴) ÷ (6.022 x 10²³) = 3 moles

3 x 22.4 = 67.2

67.2 liters of O₂ is needed.

b.)

Okay, so to find the percent yield, we need to find the theoretical yield and the actual yield. We are given the actual yield, so what we need is the theoretical yield.

For every one mole of C₆H₁₂O₆, theoretically 6 moles of H₂O will be produced.

Let's convert grams to moles for C₆H₁₂O₆:

1 gram / 180 grams = 0.0055556 moles C₆H₁₂O₆

Theoretically, 6 times that is the moles of H₂O produced:

0.0055556 x 6 = 0.033333 moles H₂O

Molar mass of H₂O is 18.015, so let's find grams:

0.033333 x 18.015 = 0.600 grams H₂O

So we have our theoretical yield, 0.600, and our actual yield, 0.303.

0.303 ÷ 0.600 = 0.505

Convert to a percent: 0.505 x 100 = 50.5%

The percent yield is 50.5%.

Hope this helps!

4 0
3 years ago
Round off 00907506 to four significant figures.
KIM [24]
The answer of the question is 9075
6 0
3 years ago
Identify the atom below
natulia [17]

easy there are three levels so you look at the periodic table and go to the third row now the last level on this atom is 1 electron so your answer is <u>Na</u>

4 0
3 years ago
Both c5h12 and c5h11oh can be used as fuels. predict which compound would release a greater amount of heat per gram when it unde
ruslelena [56]
Are you doing a research on C5 burning?I am Zy from China.
4 0
3 years ago
The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.751 g sample of e
fredd [130]

<u>Answer:</u> The empirical formula for the given compound is C_{4}H_{10}O

<u>Explanation:</u>

The chemical equation for the combustion of ether follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=4.159g

Mass of H_2O=2.128g

Mass of sample = 1.751 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 4.159 g of carbon dioxide, \frac{12}{44}\times 4.159=1.134g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.128 g of water, \frac{2}{18}\times 2.128=0.236g of hydrogen will be contained.

Mass of oxygen in the compound = (1.751) - (1.134 + 0.236) = 0.381 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.134g}{12g/mole}=0.0945moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.236g}{1g/mole}=0.236moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.381g}{16g/mole}=0.0238moles

  • <u>Step 2:</u> 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 0.0238 moles.

For Carbon = \frac{0.0945}{0.0238}=3.97\approx 4

For Hydrogen = \frac{0.236}{0.0238}=9.91\approx 10

For Oxygen = \frac{0.0238}{0.0238}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 10 : 1

Hence, the empirical formula for the given compound is C_{4}H_{10}O

7 0
3 years ago
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