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2 years ago

Which intermolecular force is the strongest?

Chemistry

2 answers:

il63 [147K]2 years ago

7 0

Let us understand the types of given intermolecular forces

a) dipole-dipole : this is present between two polar molecules having permanent dipole

eXample : dipole dipole interaction between HCl-HCl molecule

b) dipole induced dipole : here a polar molecule having a dipole induces dipole in another non polar molecule due to attraction /repulsion for electrons and attraction / repulsion for nucleus.

c) hydrogen bond : the interaction between a hydrogen atom bonded with a highly electr-onegative element (F,O or N) and another highly electro-negative element (F,O or N)

Example Water-water

d) london dispersion : these are weekest of all the intermcolecular forces and are between the electrons of one molecule / atom and nucleus of another molecules.

The strongest intermolecular forces is hydrogen bond

Reptile [31]2 years ago

4 0

The answer is C. Hydrogen Bond

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Answer the following questions based on the reaction below: NaOH(aq) + KHP(s) --> NaKP(aq)+H2O(I)

Natali [406]

Answer:

a) 1.44g

b) 77.3%

Explanation:

1. Chemical balanced equation (given)

$NaOH(aq)+ KHP(s)\rightarrow NaKP(aq)+H_2O(l)$

2. Mole ratio

$1molNaOH(aq):1molKHP(s)$

This is, 1 mol of NaOH will reacts with 1 mol of KHP.

3. Find the number of moles in 72.14 mL of the base

$Molarity=\text{number of moles of solute}/\text{volume of solution in liters}$

$\text{Volume of solution}=72.14mL=0.07214liters$

$\text{Number of moles of NaOH}=0.0978M\times 0.07214liter=0.007055mol$

4. Find the number of grams of KHP that reacted

The number of moles of KHP that reacted is equal to the number of moles of NaOH, 0.007055 mol

Convert moles to grams:

mass = number moles × molar mass = 0.007055mol × 204.23g/mol

mass = 1.4408 g.

You have to round to 3 significant figures: 1.44 g (because the molarity is given with 3 significant figures).

5. Find the percentage of KHP in the sample

The percentage is how much of the substance is in 100 parts of the sample.

The formula is:

% = (mass of substance / mass of sample) × 100

% = (1.4408g/ 1.864g) × 100 = 77.3%

7 0

2 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

2 years ago

Please Help Me!

avanturin [10]

Answer:

B) Cu(s) is metal that reacts spontaneously with NiCl2(aq).

4 0

2 years ago

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Calculate the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature (645oc). assume

baherus [9]

Answer : 7.87 X $10^{-6}$ .

Explanation : To find the fraction of schottky defects in the given lattice of CsCl,

we use the formula, $\frac{N_{s} }{N}} = exp ( \frac{-Q_{s}}{2KT} )$

on solving with the given values , $Q_{s}= 1.86 eV$ and T as 645 + 273 K and rest are the constants.

$\frac{N_{s} }{N}} = exp ( \frac{-1.86 eV}{2 X (8.62 X 10^{-5}) X (645 +273)} )$

we get the answer as 7.87 X $10^{-6}$ .

6 0

2 years ago

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For an atom’s electrons, how many energy sublevels are present in the principal energy level n = 4?

defon

"n" represents principal energy level or principal quantum number. Principal energy levels are then subdivided into sublevels. The level of principal energy level is always equal to number of sublevels. Therefore if principal energy level = 4, then there are also 4 sublevels present, namely: 4s, 4p, 4d and 4f sublevels.

3 0

2 years ago