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2 years ago

Acids which contain only hydrogen and a single other element are referred to as.

Chemistry

1 answer:

Pachacha [2.7K]2 years ago

8 0

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A binary acid is an acid that consists of hydrogen and one other element.

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The following substances dissolve when added to water. Classify the substances according to the strongest solute-solvent interac

Norma-Jean [14]

Answer:

CuCl2-Ion-dipole forces

CuSO4-Ion-dipole forces

NH3-Dipole-dipole forces

CH3OH-Dipole-dipole forces

Explanation:

Water consists of a dipole. The water molecule contains a positive end and a negative end. The positive ion attracts the negative dipole of water while the positive dipole in water interacts with the negative ion of an ionic substance. This explains the dissolution of ionic substances in water.

Copper II chloride and copper sulphate are ionic substances hence they dissolve by the mechanism described above.

Molecules consisting of dipoles dissolves by interaction of the molecule's dipoles with the dipoles in water. For example, methanol interacts with water through hydrogen bonding which is involves molecular dipoles

3 0

3 years ago

A soccer player applies a force of 56.6 N to a soccer ball while kicking it. If the ball has a mass of 0.46 kg, what is the acce

lubasha [3.4K]

Answer:

$a=123.04\ m/s^2$

Explanation:

Given that,

Force applied to a soccer player, F = 56.6 N

The mass of the ball, m = 0.46 kg

We need to find the acceleration of the soccer ball. The force acting on the ball is given by :

F = ma

Where

a is the acceleration

$a=\dfrac{F}{m}\\\\a=\dfrac{56.6 }{0.46 }\\\\a=123.04\ m/s^2$

So, the required solution is $123.04\ m/s^2$ .

3 0

3 years ago

Why Al is a member of group 13 rather than group 3?

Levart [38]

Answer:

Boron and Aluminium

Explanation:

Boron and Aluminium are present in Group 13 of the modern periodic table. Group 13 (IUPAC System) can also be referred to as Group III-A. Logically, Boron and Aluminum can't be placed alongwith elements such as Yttrium as they don't exhibit properties of a transition metal.

8 0

3 years ago

. Using improved chemistry equipment in the late 1700s, chemists

Alex_Xolod [135]

Answer:vC

Explanation:

8 0

3 years ago

An acetate buffer solution is prepared by combining 50. mL of a 0.20 M acetic acid, HC2H3O2 (aq), and 50. mL of 0.20 M sodium ac

Gala2k [10]

Answer:

B. CH3COOH pH > 4.7 (4.8)

Explanation:

CH3COOH + NaOH ↔ CH3COONa + H2O
CH3COONa + NaOH ↔ CH3COONa

∴ mol NaOH = (5 E-3 L)*(0.10 mol/L) = 5 E-4 mol

⇒ mol CH3COOH = (0.05 L)*(0.20 mol/L) = 0.01 mol

⇒ C CH3COOH = (0.01 mol - 5 E-4 mol) / (0.105 L)

⇒ C CH3COOH = 0.0905 M

∴ mol CH3COONa = (0.05 L )*(0.20 mol/L) = 0.01 mol

⇒ C CH3COONa = (0.01 mol + 5 E-4 mol) / (0.105 L )

⇒ C CH3COONa = 0.1 M

∴ Ka = ([H3O+]*(0.1 + [H3O+])) / (0.0905 - [H3O+]) = 1.75 E-5

⇒ 0.1[H3O+] + [H3O+]² = (1.75 E-5)*(0.0905 - [H3O+])

⇒ [H3O+]² 0.1[H3O+] = 1.584 E-6 - 1.75 E-5[H3O+]

⇒ [H3O+]² + 0.1000175[H3O+] - 1.584 E-6 = 0

⇒ [H3O+] = 1.5835 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = - Log (1.5835 E-5)

⇒ pH = 4.8004 > 4.7

7 0

3 years ago