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emmainna [20.7K]
2 years ago
10

Which of these did your answer include? High boiling and melting points: Hydrogen bonds increase the amount of energy required f

or phase changes to occur, thereby raising the boiling and melting points. High specific heat: Hydrogen bonds increase the amount of energy required for molecules to increase in speed, thereby raising the specific heat. Lower density as a solid than as a liquid: Hydrogen bonds increase the volume of the solid by holding molecules apart, thereby decreasing the density. High surface tension: Hydrogen bonds produce strong intermolecular attractions, which increase surface tension.
Chemistry
2 answers:
scZoUnD [109]2 years ago
7 0

Here we have to get the right answers which include the given phrase.

The correct answers are as following:

High boiling and melting points: Hydrogen bond increase the amount of energy required for phase changes to occur, thereby raising the boiling and melting points.

High specific heat: Hydrogen bond increase the amount of energy required for molecules to increase the speed, thereby raising the specific heat.

High surface tension: Hydrogen bonds produce strong inter molecular attractions, which increase surface tension.

The incorrect answer:

Lower density as a solid than as a liquid: actually, density of solid is more than density of liquid as hydrogen bonds in solid produce strong inter molecular attractions among molecules, which aggregates molecules together, hence volume of associated molecules reduces. Therefore, density of solid is more than that of liquid.

marishachu [46]2 years ago
5 0

Answer:

A,B, D

Explanation:

just took the test

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Calculate the radius ratio for NaBr if the ionic radii of Na + and Br − are 102 pm and 196 pm , respectively. radius ratio: Base
Fudgin [204]

Answer : The expected coordination number of NaBr is, 6.

Explanation :

Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.

This is represented by,

\frac{r_{cation}}{r_{anion}}

When the radius ratio is greater than 0.155, then the compound will be stable.

Now we have to determine the radius ration for NaBr.

Given:

Radius of cation, Na^+ = 102 pm

Radius of cation, Br^- = 196 pm

\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.

The relation between radius ratio and coordination number are shown below.

Therefore, the expected coordination number of NaBr is, 6.

8 0
2 years ago
A solution is prepared by adding 100 mL of 1.0 M HC2H3O2 (aq) to 100 mL of 1.0 M NaC2H3O2 (aq). The solution is stirred and its
DIA [1.3K]

Answer:

(C) H3O+(aq) + C2H3O2−(aq) -> HC2H3O2(aq) + H2O(l)

Explanation:

A buffer is a solution of a weak acid and its salt. It mitigates against changes in acidity or alkalinity of a system. A buffer maintains the pH at a constant value by switching the equilibrium concentration of the conjugate acid or conjugate base respectively.

Addition if an acid shifts the equilibrium position towards the conjugate acid side while addition of a base shifts the equilibrium position towards the conjugate base side.

5 0
2 years ago
0.50 mol A, 0.60 mol B, and 0.90 mol C are reacted according to the following reaction
algol [13]

Reactant C is the limiting reactant in this scenario.

Explanation:

The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.

Balanced chemical reaction is:

A + 2B + 3C → 2D + E

number of moles

A = 0.50 mole

B = 0.60 moles

C = 0.90 moles

Taking A as the reactant

1 mole of A reacted to form 2 moles of D

0.50 moles of A will produce \frac{2}{1} = \frac{x}{0.50}

thus 0.50 moles of A will produce 1 mole of D

Taking B as the reactant

2 moles of B reacted to form 2 moles of D

0.60 moles of B reacted to form x moles of D

\frac{2}{2} = \frac{x}{0.6}

x = 2 moles of D is produced.

Taking C as the reactant:

3 moles of C reacted to form 2 moles of D

O.9 moles of C reacted to form x moles of D

\frac{2}{3} = \frac{x}{0.9}

= 0.60 moles of D is formed.

Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.

5 0
2 years ago
If a pharmacist dissolves 1.2 grams of a medicinal agent in 60 ml of a cough syrup having a specific gravity of 1.20, what is th
guapka [62]

Mass of medicinal agent taken = 1.2 g

the volume is 60 mL

Specific gravity = 1.20

So the mass of solution = specific gravity X volume = 1.20 * 60 = 72g

Now if we have increased the volume by 0.2 so the new volume = 60.2

New mass = 72 + 1.2 = 73.2  

Specific gravity = mass /  volume = 73.2 / 60.2 = 1.22 g/mL

7 0
2 years ago
Given that at 25.0 ∘C Ka for HCN is 4.9×10−10 and Kb for NH3 is 1.8×10−5, calculate Kb for CN− and Ka for NH4+. Enter the Kb val
neonofarm [45]

Explanation:

Using the expression :

K_a\times K_b=K_w

Where, K_w is the dissociation constant of water.

At 25\ ^0C, K_w=10^{-14}

Thus, for HCN , K_a=4.9\times 10^{-10}

<u>K_b for CN⁻ can be calculated as:</u>

K_a\times K_b=K_w

4.9\times 10^{-10}\times K_b=10^{-14}

K_b=2.0\times 10^{-5}

Thus, for NH₃ , K_b=1.8\times 10^{-5}

<u>K_a for NH_4^+ can be calculated as:</u>

K_a\times K_b=K_w

K_a\times 1.8\times 10^{-5}=10^{-14}

K_a=5.6\times 10^{-10}

5 0
2 years ago
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