How many molecules are there in 24 grams of HSO₃?

Which of this is not an effect of humans using a natural recourse?

katen-ka-za [31]

Answer:

it can allow more room for additional living things in the habitat

Explanation:

Use water for an example.

- Taking water can destroy a fish habitat.

- Using excess water can cause water to run out.

- Taking/using water leaves less amounts for others/organisms.

Taking water does not allow additional room for organisms in a habitat.

5 0

2 years ago

Which biogeochemical cycle has the least activity because the essential element is mostly stored in rock?

loris [4]

<span>The biogeochemical cycle that has the least activity because its essential element is mostly stored in rock is the phosphorus cycle. Phosphorus, unlike the other essential elements, is commonly found as a solid. This is why the atmosphere does not play a role in this cycle. Instead, phosphorus remains on land, and is mostly found in rocks and minerals.</span>

3 0

3 years ago

Read 2 more answers

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

Thallium has two stable isotopes, 203tl and 205tl. knowing that the atomic weight of thallium is 204.4, which isotope is the mor

Elza [17]

You have to figure out a way to write the two unknown abundances in terms of one variable.

The total abundance is 1 (or 100%). So if you say the abundance for the first one is X then the abundance for the second one has to be 1-X (where X is the decimal of the percentage so say 0.8 for 80%).

203(X) + 205(1-X) = 204.4

Then you just solve for X to get the percentage for TI-203.
And then solve for 1-X to get the percentage for TI-205.

After that the higher percentage would be the most abundant.

203x + 205 - 205x = 204.4
-2x + 205 = 204.4
-2x = -0.6

x = 0.3
1-x = 0.7

Then the TI-205 would have the highest percentage and would be the most abundant.

7 0

3 years ago

Please help !!!!!!!!

Salsk061 [2.6K]

1) CH2 (gas) + Br (solid) -> BrC (solid) + H2 (gas)
2) a) CH4 + Br2 -> CH3Br + HBr
2) b) methane + bromine is substitution because one hydrogen atom from methane is replaced by one bromine atom. addition reaction takes place when one molecule combines with another to form a larger molecule so therefore a molecule from X and bromine combine to form XBr.

6 0

2 years ago