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3 years ago

Which one of these is the accepted name for the compound n2o5?

1 answer:

6 0

C is the correct answer. For nonmetal-nonmetal compounds, use the Greek prefixes

Mono - 1
Di - 2
Tri - 3
Tetra - 4
Penta - 5
Hexa - 6
Hepta - 7

And so on...

The reason D is not right is because a and o cannot be next to each other in between prefix and element.

Hope this helps!

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Define "Mutarotation"

zubka84 [21]

Answer:

Mutarotation refers to the change in the optical rotation or optical activity of a solution due to the change in the equilibrium of the two anomers. It depends upon the optical activity and ratio of the anomeric forms in the solution.

To measure the optical rotation of a given solution, a polarimeter can be used and thus the ratio of the anomeric forms can be calculated.

3 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

Students are studying the effects of beach pollution by counting populations of seagulls at two different beach locations. One l

WARRIOR [948]

The weather/wind direction

6 0

3 years ago

What elements are being reduced and oxidized in this chemical formula?

dangina [55]

Answer: The element Na (Sodium) is getting oxidized and Hydrogen is getting reduced.

Explanation:

Oxidation reactions are the reactions in which addition of oxygen takes place.

Reduction reactions are the reactions in which loss of oxygen takes place.

For a given reaction:

$2Na(s)+2H_2O(l)\rightarrow H_2(g)+2NaOH(aq.)$

Sodium is getting oxidized because there is an addition of reaction with that element.

Hydrogen is getting reduced because there is a removal of oxygen with that element.

3 0

3 years ago

Which process produces the energy that is used by solar panels?

Montano1993 [528]

A. Combustion since it can’t be nuclear….?

8 0

2 years ago

Read 2 more answers