Ask question

Ask question

All categories

IgorLugansk [536]

3 years ago

10

Một hỗn hợp X gồm CH3OH; C2H5OH; phenol có khối lượng 28,9 gam phản ứng vừa hết với 100ml dung dịch NaOH 2M. Phần trăm theo khối

lượng phenol là ? Biết C = 12; O = 16; H = 1; Na = 23. *

1 answer:

slavikrds [6]3 years ago

3 0

Answer:

nNaOH = 0,2 mol

Trong hỗn hợp các chất đề bài cho chỉ có phenol tác dụng với NaOH

C6H5OH + NaOH → C6H5ONa + H2O

0,2 ← 0,2 (mol)

⟹ mphenol = 0,2.94 = 18,8 gam

⟹

%mC6H5OH=18,8/28,9 x100%= 65,05%

Explanation:

You might be interested in

Pure nitrobenzene freezes at 5.67 C. When 1.0g of ethanol (C2H6O) is mixed with 20.0g nitrobenzene, the freeze point drops to –0

sukhopar [10]

Answer: The freezing-point depression constant (Kf) of nitrobenzene is $5.7^0C/m$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(5.67-(-0.53)^0C=6.2^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte nitrobenzene)

$K_f$ = freezing point constant = ?

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent = 20 g = 0.02 kg

mass of solute (ethanol) = 1.0 g

Molar mass of ethanol = 46 g/mol

$6.2=1\times K_f\times \frac{1.0g}{46g/mol\times 0.02kg}$

$K_f=5.7^0C/m$

Thus freezing-point depression constant (Kf) of nitrobenzene is $5.7^0C/m$

5 0

3 years ago

A syringe contains 610 mL of CO at 310 K and 1.5 atm pressure. A second syringe contains 520 mL of N2 at 325 K and 3.5 atm. What

scoray [572]

Answer:

P = 1.21atm

Explanation:

Using PV = nRT, moles of both syringes is:

Moles CO:

n = PV / RT

n = 1.5atm*0.610L / 0.082atmL/molK*310K

n = 0.0360 moles

Moles N₂:

n = PV / RT

n = 3.5atm*0.520L / 0.082atmL/molK*325K

n = 0.0683 moles.

As in the container you mix both gases, moles in the container are:

n = 0.0360 + 0.0683 = <em>0.1043 moles</em>

Conditions of the container are:

V = 2.00L; T = 273.15K + 10°C = 283.15K; n = 0.1043 moles.

Thus, pressure is:

P = nRT / V

P = 0.1043mol*0.082atmL/molK*283.15K / 2.00L

<h3>P = 1.21atm</h3>

5 0

4 years ago

How does the altitude of the Sun change throughout the year in Reno, NV?

vampirchik [111]

Given what we know, we can confirm that in Reno, NV, the angle of the Sun is lower during the winter months and <u>higher </u>in the <u>sky </u>during the summer.

<h3>How can we know this?</h3>

We can know this about Reno, NV, given that it is irrelevant where we placed this question. the answer will be the same all across the world. It has to do with the way the Earth orbits the Sun as well as the tilt of the Earth. This causes the Sun to be higher in the sky during summer and lower in the winter months.

Therefore, we can confirm that the angle of the Sun is lower during the winter months and higher in the sky during the summer, both in Reno, NV and everywhere else.

To learn more about The Earth's orbit visit:

brainly.com/question/15022652?referrer=searchResults

5 0

3 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago

Which statement describes what is gained or lost during radioactive decay?

Karolina [17]

Answer:

The answer is

Atoms gain other atoms

Hope this helps you

3 0

3 years ago