Ask question

Ask question

All categories

3 years ago

15

Chose one that best fits the question

2 answers:

kramer3 years ago

6 0

Answer:

Explanation:

The fourth one.

ch4aika [34]3 years ago

6 0

Answer:

C

Explanation:

Liquid gas plasma solid! Hope's this helps!

You might be interested in

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

How many grams are in 1.946 moles of nacl

Ilia_Sergeevich [38]

Answer:

113.8g

Explanation:

Statement of problem: mass of 1.946mole of NaCl

Given parameters:

Number of moles of NaCl = 1.946mole

Unknown: mass of NaCl

Solution

To find the mass of NaCl, we apply the concept of moles which expresses the relationship between number of moles and mass according to the equation below:

Number of moles = $\frac{mass}{molar mass}$

To find the molar mass of NaCl:

the atomic mass of Na = 23g

atomic mass of Cl = 35.5g

Molar mass of NaCl = (23 + 35.5) = 58.5gmol⁻¹

Mass of NaCl = Number of moles x molar mass of NaCl

Mass of NaCl = 1.946 x 58.5 = 113.8g

7 0

2 years ago

Suppose you are a food chemist working for a company that makes and manufactures soda. Your job is to create a new soft drink wi

Mekhanik [1.2K]

Answer:

The answer to your question is given after the questions so I just explain how to get it.

Explanation:

a)

Get the molecular weight of Phosphoric acid

H₃PO₄ = (3 x 1) + (31 x 1) + (16 x 4)

= 3 + 31 + 64

= 98 g

98 g ----------------- 1 mol

0.045 g --------------- x

x = (0.045 x 1) / 98

x = 0.045 / 98

x = 0.00046 moles or 4.6 x 10 ⁻⁴

b)

Molarity = $\frac{moles}{volume}$

Molarity = $\frac{0.00046}{0.35}$

Molarity = 0.0013 or 1.31 x 10⁻³

c)

Formula C₁V₁ = C₂V₂

V₁ = C₂V₂ / C₁

Substitution

V₁ = (0.0013)(1) / 0.01

Simplification and result

V₁ = 0.0013 / 0.1

V₁ = 0.13 l = 130 ml

7 0

3 years ago

On a graph, which type of line shows a direct proportion?

leonid [27]

Direct Proportion and The Straight Line Graph.Straight<span> line graphs that go through the origin, like the one immediately below, show that the quantities on the graph are in direct proportion.
thanks
cbuck763

</span>

8 0

3 years ago

How are elements arranged in the Periodic Table?

Furkat [3]

By their properties.

8 0

3 years ago

Read 2 more answers