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chubhunter [2.5K]
2 years ago
8

In the compound potassium nitrate (KNO3), the atoms within the nitrate ion are held together with bonding, and the potassium ion

and nitrate ion are held together by bonding.

Chemistry
1 answer:
sveta [45]2 years ago
7 0

In the compound potassium nitrate (KNO3), the atoms within the nitrate ion are held together with COVALENT bonding, and the potassium ion and nitrate ion are held together by IONIC bonding.

A covalent bond, also called a molecular bond, is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs. Covalent bond is formed between two non-metals.

Ionic bonds form when one atom gives up one or more electrons to another atom. It is the complete transfer of valence electron(s) between oppositely charged atoms.  Ionic bond is formed between metal (electropositive element) and non-metal(electronegative element)

In nitrate ions the Nitrogen (N) and Oxygen (O) both are non-metals and it involves the sharing of electron pairs between N and O atoms, so the bonding in Nitrate (NO_{3}^{-}) ion is covalent bonding.

In potassium nitrate , Potassium (K) is a metal and Nitrate (NO_{3}^{-}) ion is non-metal and it involves the complete transfer of valence electron between oppositely charged atoms (K+) and (NO_{3}^{-}). So the bonding between Potassium and Nitrate is Ionic bonding.

NOTE : Bonding between Non-metals is Covalent bonding.

Bonding between Metal and Non-metals is Ionic bonding.



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Barium sulfate, BaSO4, is a white crystalline solid that is insoluble in water. It is used by doctors to diagnose problems with
pshichka [43]

Answer:2

Explanation:

Ba(OH)2 contains two oxygen atoms

BaSO4 contains four oxygen atoms.

This means that barium sulphate contains two more oxygen atoms than barium hydroxide in its formula. This is clearly seen from the two formulae shown above.

5 0
1 year ago
A 0.300 g piece of copper is heated and fashioned into a bracelet. The amount of energy transferred by heat to the copper is 66,
Alja [10]

Answer:

X=600 degrees celsius

Explanation:

We can evaluate by using the equation ==> Q=m(c)(delta T)

-We know the mass is 0.3 g

-Q= 66,300 J

-C= 390 j/g 0C

-Therefore, we need to solve for delta T (which will be represented by x)

 Q= mct

66,300 J= 0.3 g (390 J/g 0C) (t)

The unit of measurement (g-grams) can be crossed out on the left side of our expression.

66,300 J=0.3(390 J 0C) (t)

-If we convert into a proportional relationship:

66,300 J                117 J/0C x

------------       =        -----------

117 J/0C                 117 J/0C

-Joules on both sides can be crossed off

*117 J/0C came from the product in terms of 0.3(390 J 0C)

*If we find the quotient of 66,300 J and 117 J/0C, we get 566.6 repeating

-You may keep the answer as it is, although for simplicity reasons, we can round to the nearest hundred.

-Therefore, x=600.

-----Remember, x represented the delta t in our equation. (I made notice of that when evaluating).

Hope this helps! :)

3 0
1 year ago
A 0.271 g sample of an unknown vapor occupies 294 ml at 140.°c and 847 mmhg. the empirical formula of the compound is ch2. what
sdas [7]
When the molar mass M = mass (g)/ no.of moles (Mol)

∴ moles= 0.271 g / M

By using the gas equation:

PV = n RT

when P is the pressure = 847 mmHg / 760 = 1.11 atm

V is the volume = 0.294 L

n = 0.271 / M

R is constant = 0.0821 

T= 140+273 = 413 K

so by substitution:

when n = PV/RT

∴ 0.271/ M = 1.11 atm *0.294 L/ 0.0821 *413

∴ M = 28 


when the empirical formula of CH2 = 12+2 = 14 

∴ the exact no.of moles = 28/14 = 2

∴the molecular formula = 2(CH2) = C2H4

6 0
2 years ago
(4) strontium (sr) has an fcc crystal structure, an atomic radius of 0.215 nm and an atomic weight of 87.62 g/mol. calculate the
OleMash [197]
<span>2.59 g/cm^3 For a face centered cubic crystal, there is 1 atom at each corner that's shared between 8 unit cells. And since there's 8 corners, that gives 8*1/8 = 1 atom per unit cell. Additionally, there are 6 faces, each with 1 atom, that's shared between 2 cells. So 6*1/2 = 3. So each unit cell has a mass of 1 + 3 = 4 atoms. The size of the unit cell will be equal to either the diameter of one atom along the edge, or the diameter of 2 atoms as the diagonal across one face of the cube, whichever results in the larger unit cell. Taking that into consideration, I will use the value of 2 for the diagonal of a face of the unit cell, resulting in the length of an edge of the unit cell being sqrt(2^2/2) = sqrt(2) = 1.414213562 times the atomic diameter. So 1.414213562 * 2 * 0.215 nm = 0.608 nm So the volume of a single unit cell is (0.608 nm)^3 Avogadro's number of atoms will require 6.0221409x10^23 / 4 = 1.50554x10^23 unit cells and will have a mass of 87.62 grams. The volume will be 1.50554x10^23 * (0.608x10^-7 cm)^3 = 1.50554x10^23 * 0.224755712x10^-22 cm^3 = 33.83776414 cm^3 So the density is approximately 87.62 g/33.83776414 cm^3 = 2.589414585 g/cm^3, when rounded to 3 significant figures is 2.59 g/cm^3.</span>
7 0
1 year ago
Dimethylhydrazine is a carbon-hydrogen-nitrogen compound used in rocket fuels. When burned in an excess of oxygen, a 0.312 gg sa
Stolb23 [73]

Answer:

CH₄N

Explanation:

Given that;

mass of the sample =  0.312 g

mass of CO2 = 0.458 g

mass of H2O =  0.374 g

nitrogen content of a 0.486 gg sample is converted to 0.226 gg N2N2.

Let start with calculating the respective numbers of moles of Carbon Hydrogen and Nitrogen from the given data.

numbers of moles of Carbon from CO2 = \frac{mass of CO_2}{molarmass}*\frac{1 mole of C}{1 mole of CO_2}

= \frac{0.458}{44}*\frac{1mole of C}{1 mole of CO_2}

= 0.0104 mole

numbers of moles of hydrogen from H2O = \frac{mass of H_2O}{molarmass}*\frac{2 mole of H}{1 mole of H_2O}

= \frac{0.374}{18.02}*\frac{2 mole of H}{1 mole of H_2O}

= 0.02077 × 2

= 0.0415 mole

The nitrogen content of a 0.486 g sample is converted to 0.226 g N2

Now, in 1 g of the sample; The nitrogen content = \frac{0.226}{0.486}*1

in 0.312 g of the sample, the nitrogen content will be; \frac{0.226}{0.486}*0.312

= 0.1450 g of N2

number of moles of N2 = \frac{mass}{molar mass}* \frac{2 mole}{1 mole}

= \frac{0.1450}{28.0134} *\frac{2 mole}{1 mole}

= 0.0103 mole

Finally to determine the empirical formula of Carbon  Hydrogen and Nitrogen; we have:

                                Carbon         Hydrogen           Nitrogen

number of moles      0.0104         0.0415                 0.0103

divided by the

smallest number      \frac{0.0104}{0.0103}             \frac{0.0415}{0.0103}                    \frac{0.0103}{0.0103}

of moles

                                   1        :           4           :               1

∴ The empirical formula = CH₄N

8 0
1 year ago
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