Ask question

Ask question

All categories

3 years ago

7

If anyone can answer this that will be amazing !!

1 answer:

Sergeu [11.5K]3 years ago

4 0

Right, left, same, left (will put the rest in comments)

You might be interested in

Find the grams in 1.26 x 10^-4 mole of HC2 H3O2

ivolga24 [154]

Molar mass :

HC₂H₃O₂ = 1 + 12*2 + 1 * 3 + 16 * 2 = 60 g/mol

1 mole <span>HC₂H₃O₂ -------------- 60 g
</span>1.26x10-⁴ mole ----------------- mass

mass = 1.26x10-⁴ * 60

mass = 0.00756 g of <span>HC₂H₃O₂</span>

hope this helps!

7 0

3 years ago

Cameron is visiting a new country while traveling around the world. He is extremely fascinated by the cuisine of this country. H

lina2011 [118]

Not exactly chemistry but ok.

Phó is one of the famous dish in Vietnamese Culture

Hence, the answer is Vietnamese Cuisine

8 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

What is the function of the nervous system? What is the basic unit of the nervous system?

s2008m [1.1K]

Answer:

basic unit of le nervous system is ur nerves brother nah im jp ur brain bcs if u didnt have a brain u wouldnt feel anything :D

Explanation:

3 0

2 years ago

Read 2 more answers

3. compounds in which elements are held together by covalent bonds...keeps their own properties

stich3 [128]

The anserr is c that what i think

3 0

3 years ago