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2 years ago

What are the elements in an ionic bond?

Chemistry

2 answers:

miss Akunina [59]2 years ago

5 0

Answer:

Ionic bonds require an electron donor, often a metal, and an electron acceptor, a nonmetal. Ionic bonding is observed because metals have few electrons in their outer-most orbitals. By losing those electrons, these metals can achieve noble gas configuration and satisfy the octet rule.

Explanation:

KIM [24]2 years ago

4 0

Answer:

Metal and non-metal

Explanation:

For exp, sodium (metal) and chloride (non-metal) form an ionic bond to make NaCl.

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at stp which of following would have the same number of molecules a 1 l of c2h4 gas? a. 0.5 of H2 b. 1L of Ne c.2L of H2O d.3L o

sashaice [31]

Answer:d

Explanation:

4 0

3 years ago

0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati

erastova [34]

Answer:

$[Ag^{+}]=4.2\times 10^{-2}M$

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

$BaCrO_4 (s)\leftrightharpoons Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ba^{2+}][CrO_{4}^{2-}]$

$1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]$

$[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}$

Now,

$Ag_{2}CrO_4(s) \leftrightharpoons 2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]$

$1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})$

$[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}$

$[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M$

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

7 0

3 years ago

What is the molarity of 50.84 g of Na2CO3 dissolved in 0.400 L solution?

borishaifa [10]

Answer:

1.20 M

Explanation:

Convert grams of Na₂CO₃ to moles. (50.84 g)/(105.99 g/mol) = 0.4797 mol

Molarity is (moles of solute)/(liters of solvent) = (0.4797 mol)/(0.400 L) = 1.20 M

4 0

3 years ago

A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.

Lubov Fominskaja [6]

Answer:

$Ag_2SO_4$

Explanation:

Formula for the calculation of no. of Mol is as follows:

$mol=\frac{mass\ (g)}{molecular\ mass}$

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

$mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol$

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

$mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol$

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

$mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol$

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

$Ag, \frac{0.05305}{0.02657} \approx 2$

$S, \frac{0.02657}{0.02657} \approx 1$

$O, \frac{0.10594}{0.02657} \approx 4$

Therefore, empirical formula of the compound = $Ag_2SO_4$

7 0

3 years ago

Read 2 more answers

hydrolysis of decapeptide P with the enzyme trypsin affords the following fragments: Glu-Gly-Lys, Gln-Val-Ile, Ala-Ser-Phe-Lys.

ehidna [41]

Answer:

The sequence of an amino acid P is:

Glu-Gly-Lys-Ala-Ser-Phe-Lys-Gln-Val-Ile

Explanation:

Fragments obtained on hydrolysis of decapeptide P by the action of an enzyme named trypsin:

Glu-Gly-Lys,
Gln-Val-Ile
Ala-Ser-Phe-Lys

Fragments obtained on hydrolysis of decapeptide P by the action of an enzyme named chymotrypsin:

Lys-Gln-Val-Ile,
Glu-Gly-Lys-Ala-Ser-Phe

In order to determine the sequence of protein P , we will arrange fragments in such a way so that common fragments or the common parts of fragments should come under each other.

On arranging these fragments :

Glu-Gly-Lys-Ala-Ser-Phe

Glu-Gly-Lys

Ala-Ser-Phe-Lys

Lys-Gln-Val-Ile

Gln-Val-Ile

The sequence of an amino acid P is:

Glu-Gly-Lys-Ala-Ser-Phe-Lys-Gln-Val-Ile

3 0

3 years ago