All categories

3 years ago

Identify one type of noncovalent bond present in each solid.

Chemistry

1 answer:

NeX [460]3 years ago

7 0

Answer:

Explanation:

The question is not complete, the cmplete question is:

Identify one type of noncovalent bond present in each solid.

1) Table salt (NaCl) 2) Graphite (repeating)

a. hydrogen bonds

b. ionic interactions

c. van der Waals interactions

d. hydrophobic interactions

Answer:

1) Table salt

b. ionic interactions

Ionic bond are formed between atoms with incomplete outermost shell. Some atoms add electrons to their outermost shell to make the shell complete hence making it a negative ion while some atoms loses their electron to make the outermost shell complete becoming a positive ion. In NaCl, sodium (Na) has 1 electron in its outermost shell which it transfers to Cl which has 7 electrons in the outermost shell. Hence after the bonding the outermost shell of the atoms become complete.

2) Graphite

c. Van Der Waals interaction

Van der waal forces are weak interaction between molecules that exist between close atoms. Carbon atoms in graphite planes have covalent bond, these graphite planes are known as graphenes. Bonds between graphenes are very weak and are van der waals forces.

You might be interested in

30 treadmills to 36 elliptical machines

scoundrel [369]

Answer:

what? that's 66 total, 6 more elliptical machines, a 1 to 1.2 ratio

but I don't know what else you would mean

7 0

2 years ago

Read 2 more answers

What is the concentration of no3- ions in a solution prepared by dissolving 15.0 g of ca(no3)2 in enough water to produce 300. m

r-ruslan [8.4K]

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻

M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol

15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol

We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M

Concentration of NO3⁻ is 0.667 M.

4 0

3 years ago

Balance Fe3O4 + H2 = Fe + H2O

Snezhnost [94]

Fe3O4 + 4H2 = 3Fe + 4H2O

Fe3O4 + 4H2SO4 = Fe2(SO4)3 + FeSO4 + 4H2O

3 0

3 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago

Draw the structure(s) of the major organic product(s) of the following reaction. 1. lithium diisopropylamide / hexane 2. 1 eq. C

Jobisdone [24]

Answer:

See explanation below

Explanation:

You are not providing the starting material, however, I manage to find a similar question to this, so I'm gonna use it as a basis to help you answer yours.

Now let's analyze what is happening in the reaction so we can predict the final product.

We have a ketone here, reacting at first with LDA. This is a very strong base that is commonly used in reactions with ketones and aldehydes to promove a condensation. To do this, as LDA is a strong base it will occur firts an acid base reaction, substracting the most acidic hydrogen in the molecule (Which in this case, is the Beta hydrogen of the carbonile). This will cause an enolate formation.

Then, this enolate will react with the CH3I and form a new product. The final result would be a ketone with a methyl group now attached. In the picture 2, you have the mechanism and final product.

Hope this helps

6 0

3 years ago