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3 years ago

35 Which element reacts with oxygen to form ionic bonds?(1) calcium (3) chlorine

Chemistry

2 answers:

QveST [7]3 years ago

8 0

Covalent bonds exist between two nonmetal elements while ionic bonds exist between a metal and a nonmetal element. In this case, oxygen is a nonmetal and in order to create an ionic bond, the other substance should be a metal. The answer therefore to this problem is calcium.

kow [346]3 years ago

3 0

$\boxed{\left( 1 \right){\text{ calcium}}}$ reacts with oxygen to form ionic bonds.

Further Explanation:

Covalent bonds are the types of bonds that involve the sharing of electrons between the bonded atoms. This bond is usually formed between two or more non-metals.

Ionic bonds are the chemical bonds that involve the transference of electrons from one atom to another. These bonds are generally formed between metals and non-metals. In these types of bonds, ions are formed by the atoms. Cations are formed by the loss of electrons and anions result from the gain of electrons.

Oxygen is a non-metal. In order to form an ionic bond with oxygen, the other element has to be a metal.

Hydrogen, chlorine, and nitrogen are non-metals so these can form covalent bonds with oxygen. But calcium is a metal and it forms ionic bonds with oxygen. Calcium loses two electrons to form ${\text{C}}{{\text{a}}^{2 + }}$ and these electrons are gained by oxygen, resulting in the formation of ${{\text{O}}^{2 - }}$ . The attraction between ${\text{C}}{{\text{a}}^{2 + }}$ and ${{\text{O}}^{2 - }}$ is an ionic bond and the resulting compound is CaO (For structure, refer to the attached image).

Learn more:

Identification of ionic bonding: brainly.com/question/1603987
What type of bond exists between phosphorus and chlorine? brainly.com/question/81715

Answer details:

Grade: High School

Chapter: Ionic and covalent compounds

Subject: Chemistry

Keywords: ionic bond, covalent bond, nitrogen, hydrogen, chlorine, calcium, oxygen, CaO, metal, non-metal.

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4 years ago

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Acrylonitrile () is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. If 1

pychu [463]

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

First off.. not a chem board.. but n e way.

This is a limiting reagent problem.

set it up as a DA problem.(Dimension Analysis)

Start with what you want.

you want Grams of acrylonitrile (C3H3N)

so start with that (Using ACL in place of Acrylonitrile.. just for ease of typing)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

solve that you wiill get grams of Acrylonitrile created by 15 grams oc C3H6 = 18.9g

Same setup for the two other reactants.

so i did it and for

oxygen I got 11.04 grams

and for Ammonia i got 15.29 grams

So the most you can make is 11.04 grams because if you have ot make any more .. you will have to get more O2 .. but since you have only 10 grams of it .. that is the most u can make in this reaction.

Both the other reactants are in excess.

rate brainliest pls

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3 years ago

If a compound has a composition of 82% nitrogen and 18% hydrogen, what is the empirical formula for this compound

Damm [24]

Answer: The empirical formula for the given compound is $NH_3$

Explanation : Given,

Percentage of H = 18 %

Percentage of N = 82 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 18 g

Mass of N = 82 g

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

Step 1: Converting the given masses into moles.

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{18g}{1g/mole}=18moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{82g}{14g/mole}=5.8moles$

Step 2: 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 5.8 moles.

For Hydrogen = $\frac{18}{5.8}=3.10\approx 3$

For Nitrogen = $\frac{5.8}{5.8}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of H : N = 3 : 1

Hence, the empirical formula for the given compound is $NH_3$

3 0

3 years ago

What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115 g of

Citrus2011 [14]

Answer:

freezing point (°C) of the solution = - 3.34° C

Explanation:

From the given information:

The freezing point (°C) of a solution can be prepared by using the formula:

$\Delta T = iK_fm$

where;

i = vant Hoff factor

the vant Hoff factor is the totality of the number of ions in the solution

Since there are 1 calcium ion and 2 nitrate ions present in Ca(NO3)2, the vant Hoff factor = 3

$K_f$ = 1.86 °C/m

m = molality of the solution and it can be determined by using the formula

$molality = \dfrac{mole \ of \ solute }{kg \ of \ solvent }$

which can now be re-written as :

$molality = \dfrac{mole \ of \ Ca(NO_3)_2}{kg \ of \ water}$

$molality = \dfrac{\dfrac{mass \ of \ \ Ca(NO_3)_2}{molar \ mass of \ Ca(NO_3)_2} }{kg \ of \ water}$

$molality = \dfrac{\dfrac{11.3 \ g }{164 \ g/mol} }{0.115 \ kg }$

molality = 0.599 m

∴

The freezing point (°C) of a solution can be prepared by using the formula:

$\Delta T = iK_fm$

$\Delta T =3 \times (1.86 \ ^0C/m) \times (0.599 \ m)$

$\Delta T =3.34^0 \ C$

$\Delta T =$ the freezing point of water - freezing point of the solution

3.34° C = 0° C - freezing point of the solution

freezing point (°C) of the solution = 0° C - 3.34° C

freezing point (°C) of the solution = - 3.34° C

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3 years ago

How many moles of ammonium nitrate are there in 32.5 mL of a0.125 M NH4NO3 solution?

Nady [450]

Answer: The number of moles of ammonium nitrate is 0.004 moles.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

We are given:

Molarity of $NH_4NO_3$ solution = 0.125 M

Volume of solution = 32.5 mL

Putting values in above equation, we get:

$0.125M=\frac{\text{Moles of }NH_4NO_3\times 1000}{32.5mL}\\\\\text{Moles of }NH_4NO_3=0.004mol$

Hence, the number of moles of ammonium nitrate is 0.004 moles.

8 0

3 years ago