All categories

3 years ago

Which of the following is true of solutes dissolving in water?

Chemistry

1 answer:

Talja [164]3 years ago

3 0

B is the answer to your question.

C2H4 is not capable of hydrogen bonding because the H's are attached to the Carbon, and the charge is 0.

Although HCl cannot hydrogen bond, that aspect does not hinder it's ability to dissolve. Because HCl is polar and so is water, the positive side of H2O will be attracted to the negative side of HCl, thus "tearing" the molecule apart. (Like dissolves like - polar dissolves polar)

Based on the Solubility rule, KBr is soluble because it contains a group 1 metal.

You might be interested in

Which type of chemical reaction does this equation exemplify?

garri49 [273]

The answer is combustion

4 0

3 years ago

PLEASE ANSWERRRRRRR

Neporo4naja [7]

Answer:

Explanation:

Because the silk took away the electrons to allow the rod to become positive. They cant be destroyed because of the law of conservation of charges. Also protons cannot move only electrons can.

8 0

3 years ago

Does the motion of the moon affects how we see it from Earth? *

irina [24]

Yes that’s why we see it in different shapes all the time

7 0

2 years ago

A sample of gas occupies a volume of 73.7 mL. As it expands, it does 133.7 J of work on its surroundings at a constant pressure

katen-ka-za [31]

In thermodynamics, work performed by a system is the energy transferred by the system to its surroundings. It can be calculated by the expression:

W = PdV

Integrating,

We will have,

W = P(V2 - V1)
133.7 (1 litre-atm / 101.325 Joule) ( 760 Torr / atm ) = 783 (V2 - .0737 )
V2 = 1.35 L

Hope this answers the question. Have a nice day.

7 0

3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago