When was you first body count

The type of compound that is most likely to contain a covalent bond is ________.

max2010maxim [7]

The type of substance that is most likely to contain a covalent bond is ONE THAT IS COMPOSED OF ONLY NON METALS.
Covalent bond is a type of chemical bond in which electron pairs are shared among the participating elements in order to achieve the octet form. Covalent bond is usually found among non metals.

6 0

3 years ago

Please show all work:

Tom [10]

If 40.0 grams of magnesium is reacted with an excess of nitric acid. 3.3 g of hydrogen gas will be produced.

<h3>What is Stoichiometry ?</h3>

Stoichiometry helps us use the balanced chemical equation to measure quantitative relationships and it is to calculate the amounts of products and reactants that are given in a reaction.

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

Now we have to write the balanced equation

Mg + 2HNO₃ → Mg(NO₃)₂ + H₂

According to Stoichiometry

$40.9\ \text{g Mg} \times \frac{1\ \text{mol Mg}}{24.31\ \text{g Mg}} \times \frac{1\ \text{mol}\ H_2}{1\ \text{mol Mg}} \times \frac{2.02\ g\ H_2}{1\ \text{mol}\ H_2}$

= 3.3 g H₂

Thus from the above conclusion we can say that If 40.0 grams of magnesium is reacted with an excess of nitric acid. 3.3 g of hydrogen gas will be produced.

Learn more about the Stoichiometry here: brainly.com/question/16060223

#SPJ1

8 0

2 years ago

CaCO3(s) ∆→CaO(s) + CO2(g).

sveta [45]

Answer:

74.9%.

Explanation:

Relative atomic mass data from a modern periodic table:

Ca: 40.078;
C: 12.011;
O: 15.999.

What's the <em>theoretical</em> yield of this reaction?

In other words, what's the mass of the CO₂ that should come out of heating 40.1 grams of CaCO₃?

Molar mass of CaCO₃:

$M(\text{CaCO}_3) = 40.078 + 12.011 + 3 \times 15.999 = 100.086\;\text{g}\cdot\text{mol}^{-1}$ .

Number of moles of CaCO₃ available:

$\displaystyle n(\text{CaCO}_3) =\frac{m}{M} = \frac{40.1}{100.086} = 0.400655\;\text{mol}$ .

Look at the chemical equation. The coefficient in front of both CaCO₃ and CO₂ is one. Decomposing every mole of CaCO₃ should produce one mole of CO₂.

$n(\text{CO}2) = n(\text{CaCO}_3)= 0.400655\;\text{mol}$ .

Molar mass of CO₂:

$M(\text{CO}_2) = 12.011 + 2\times 15.999 = 44.009\;\text{g}\cdot\text{mol}^{-1}$ .

Mass of the 0.400655 moles of $\text{CO}_2$ expected for the 40.1 grams of CaCO₃:

$m(\text{CO}_2) = n\cdot M = 0.400655 \times 44.009 = 17.632\;\text{g}$ .

What's the <em>percentage</em> yield of this reaction?

$\displaystyle \textbf{Percentage}\text{ Yield} = \frac{\textbf{Actual}\text{ Yield}}{\textbf{Theoretical}\text{ Yield}}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} = \frac{13.2}{17.632}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} =74.9\%$ .

7 0

3 years ago

PLEASE HELP ME !!! :(

Akimi4 [234]

Answer:

You need two potassium atom to form an ionic bond with one sulfur atom. Pottassium ions are formed by donating losing their electrons in the outermost shell, thr lost electrons are gained by sulphur atom to form 2K+ and S2-, respectively. The attraction between the positive ions and negative ions gives rise to the ionic bond and the molecular formula is K2S

Explanation:

4 0

2 years ago

Who created the "plum pudding model" of the atom?

Zepler [3.9K]

It was first proposed by J. J. Thomson in 1904.

8 0

3 years ago