All categories

3 years ago

The color and boiling point of a substance are both examples of

Chemistry

1 answer:

BartSMP [9]3 years ago

5 0

Answer:

Instensive physical properties

Explanation:

bc they don't depend the amount of matter in the substance

You might be interested in

10. Lead nitrate solution mixed with sodium sulfate solution forms lead sulfate as a

Leviafan [203]

Answer:

88%.

Explanation:

The percentage yield of lead sulfate in this experiment is 88% if 2.53 is divided by 2.85 and multiply by 100. The percentage yield can be calculated when the experimental yield is divided by theoretical yield and then multiply by 100. The percentage yield tells us about the actual yield that is gained in the end of experiment which is lower than theoretical yield.

8 0

3 years ago

A gas at 45 o C occupies 5.22 L. At what temperature will the volume be 6.87 L?

vfiekz [6]

Hhbbb. Bbhbbbbhhbbjjnjnmmmkkkkjn

5 0

3 years ago

Which layer of the atmosphere is between the mesosphere and the exosphere?

Genrish500 [490]

C. Thermosphere is the correct answer

6 0

2 years ago

Calculate the pH of a solution that contains 2.7 M HF and 2.7 M HOC6H5. Also, calculate the concentration of OC6H5- in this solu

borishaifa [10]

Answer:

$\large \boxed{\mathbf{1.36; 3.6 \times 10^{\mathbf{-9}}}\textbf{mol/L}}$

Explanation:

The HF is about five million times as strong as phenol, so it will be by far the major contributor of hydronium ions. We can ignore the contribution from the phenol.

1 .Calculate the hydronium ion concentration

We can use an ICE table to organize the calculations.

HF + H₂O ⇌ H₃O⁺ + F⁻

I/mol·L⁻¹: 2.7 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 2.7 - x x x

$K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}] \text{F}^{-}]} {\text{[HF]}} = 7.2 \times 10^{-4}\\\\\dfrac{x^{2}}{2.7 - x} = 7.2 \times 10^{-4}\\\\\text{Check for negligibility of }x\\\\\dfrac{2.7}{7.2 \times 10^{-4}} = 4000 > 400\\\\\therefore x \ll 2.7\\\dfrac{x^{2}}{2.7} = 7.2 \times 10^{-4}\\\\x^{2} = 2.7 \times 7.2 \times 10^{-4} = 1.94 \times 10^{-3}\\x = 0.0441\\\text{[H$_{3}$O$^{+}$]}= \text{x mol$\cdot$L$^{-1}$} = \text{0.0441 mol$\cdot$L$^{-1}$}$

2. Calculate the pH

$\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}$

3. Calculate [C₆H₅O⁻]

C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺

2.7 x 0.0441

$K_{\text{a}} = \dfrac{0.0441x} {2.7} = 1.6 \times 10^{-10}\\\\0.0441x = 1.6 \times 10^{-10}\\x = \dfrac{1.6 \times 10^{-10}}{0.0441} = \mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}\\\text{The concentration of phenoxide ion is $\large \boxed{\mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}}$}$

6 0

3 years ago

My last question its easy but i dont get it lol help plz

pashok25 [27]

There is no question on here lol

8 0

2 years ago

Read 2 more answers