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3 years ago

PLS HELPPP why is CH3OH more polar than CH3CH2OH

Chemistry

1 answer:

dusya [7]3 years ago

8 0

Answer:

Explanation:

Polarity is about differencens in electronegativity. CH bonds have around the same electronegativity value so a CH bond is nonpolar. The more CH bonds there are in a molecule, the more nonpolar it is. Since CH3CH2OH has more carbon-hydrogen bonds than CH3OH, it is more nonpolar. With the same reasoning, since CH3OH has less CH bonds, it's more polar.

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strontium-90 has a half life of 29 years. if a site held 4000 kg of this isotope, approximately what mass of strontium-90 would

Ludmilka [50]

Answer:

125g

Explanation:

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3 years ago

The mass of 0.10 mole of methane

saveliy_v [14]

Answer:

A methane molecule is made from one carbon atom and four hydrogen atoms. Carbon has a mass of 12.011 u and hydrogen has a mass of 1.008 u. This means that the mass of one methane molecule is 12.011 u + (4 × 1.008u), or 16.043 u. This means that one mole of methane has a mass of 16.043 grams.

メタン分子は、1つの炭素原子と4つの水素原子から作られています。炭素の質量は12.011uで、水素の質量は1.008uです。これは、1つのメタン分子の質量が12.011 u +（4×1.008u）、つまり16.043uであることを意味します。これは、1モルのメタンの質量が16.043グラムであることを意味します。^>^

5 0

3 years ago

Consider the titration of 1L of 0.36 M NH3 (Kb=1.8x10−5) with 0.74 M HCl. What is the pH at the equivalence point of the titrati

worty [1.4K]

Answer:

Explanation:

The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid

Firstly, we write the equation of reaction between ammonia and hydrochloric acid.

NH3(aq)+HCl(aq)→NH4Cl(aq)

Ionically:

HCl + NH3 ---> NH4 + Cl-

Firstly, we calculate the number of moles of the ammonia as follows:

from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M

An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.

NH4+ H2O ⇄ NH3 H3O+

I 0.242 0 0

C -X +x +X

E 0.242-X X X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka

To find this, we use the mathematical equation below

K a ⋅ K b = K w

, where K w- the self-ionization constant of water, equal to

10 ^-14 at room temperature

This means that you have

K a = K w.K b = 10 ^− 14 /1.8 * 10^-5 = 5.56 * 10^-10

Ka = [NH3][H3O+]/[NH4+]

= x * x/(0.242-x)

Since the value of Ka is small, we can say that 0.242-x ≈ 0.242

Hence, K a = x^2/0.242 = 5.56 * 10^-10

x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10

x = 0.00001161895

[H3O+] = 0.00001161895

pH = -log[H3O+]

pH = -log[0.00001161895 ] = 4.94

7 0

3 years ago

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.8

Lubov Fominskaja [6]

Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is $pK_a =7.82$

Explanation:

From the question we are told

The volume of base is $V_B = 26.mL = 0.0260L$

The pH of solution is $pH = 7.82$

The concentration of the acid is $C_A = 0.1M$

From the pH we can see that the titration is between a strong base and a weak acid

Let assume that the the volume of acid is $V_A = 18mL= 0.018L$

Generally the concentration of base

$C_B = \frac{C_AV_A}{C_B}$

Substituting value

$C_B = \frac{0.1 * 0.01800}{0.0260}$

$C_B= 0.0692M$

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

$HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}$

Now before the reaction the number of mole of base is

$No \ of \ moles[N_B] = C_B * V_B$

Substituting value

$N_B = 0.01300 * 0.0692$

$= 0.0009 \ moles$

Now before the reaction the number of mole of acid is

$No \ of \ moles = C_B * V_B$

Substituting value

$N_A = 0.01800 *0.1$

$= 0.001800 \ moles$

Now after the reaction the number of moles of base is zero i.e has been used up

this mathematically represented as

$N_B ' = N_B - N_B = 0$

The number of moles of acid is

$N_A ' = N_A - N_B$

$= 0.0009\ moles$

The pH of this reaction can be mathematically represented as

$pH = pK_a + log \frac{[base]}{[acid]}$

Substituting values

$7.82 = pK_a +log \frac{0.0009}{0.0009}$

$pK_a =7.82$

8 0

3 years ago

The electrons formed from the aerobic oxidation of glucose are

SVEN [57.7K]

Answer:

i and ii

Explanation:

In the aerobic oxidation of glucose, the electrons formed are transferred to O2 after several others transfer reactions like passing through coenzymes NAD+ and FAD

4 0

3 years ago