Ask question

Ask question

All categories

3 years ago

5

16. Chemists can synthesize new materials using

1 answer:

Alex787 [66]3 years ago

8 0

Answer:

Yes Concurred,but where is the question

You might be interested in

A chemist has a small amount of compound ( b.p. = 65 oC) that must be fractionally distilled. Yet, the Chemist does not want to

koban [17]

Answer:

The chemist can either:

a. Use a small fractionation apparatus.

b. Add a compound with a much higher boiling point.

Explanation:

Using a smaller fractionation apparatus or Vigreux column will help to minimize loss of the distillate.

If a compound with a higher boiling point is added, the vapors of this liquid will displace the vapors of this small amount of compound with a lower boiling point. This compound with a higher boiling point is known as a Chaser.

6 0

3 years ago

What is the reason behind that the melting and boiling point of iron is more than that of sodium? Write any two differences betw

liraira [26]

Answer:

difference between metal and alloy is that the metal is a pure substance whereas the alloy is a mixture of two or more components.

Explanation:

Mostly the metalloids have the appearance just like the metallic appearance And also they are the brittle one's . Boron and silicon are the example. Note; By combining with other metals metalloid can form the alloy

5 0

2 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 is one way that scientist communicate the result of an experiment

arlik [135]

Answer:

wove oral visalize

Explanation:

7 0

3 years ago

ICMP scanning involves in checking for the live systems, which can be done by sending the following ping scan request to a host.

abruzzese [7]

Answer:

ICMP Echo Request

Explanation:

ICMP Echo Request is a form of probe or message sent by a user to a destination system.

5 0

3 years ago