All categories

3 years ago

0.159 mol 2.25 M = A L of HCI

Chemistry

2 answers:

mylen [45]3 years ago

4 0

Answer will be
0.159 mol
2.25 M
= A L of HCI
.0707

Hope this helps.

Verizon [17]3 years ago

3 0

Answer:

hmmm?what?

Explanation:

ayusin mo flece:>

You might be interested in

Mercury, which is sometimes used in thermometers, has a density of 13.534 g/ml at room temperature. what volume of mercury conta

Sergeu [11.5K]

D - density: 13,534 g/ml
m - mass: 10g
V - volume: ??
_____________
d = m/V
V = m/d
V = 10/13,534
V = 0,7389 ml

:•)

6 0

3 years ago

Please answer quickly

Stolb23 [73]

Inherited it from his grandfather

8 0

3 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

Indicate which solution in each pair has the lower pH. Your response should be a four letter "word". The first letter should be

JulijaS [17]

Answer:

bcfh

Explanation:

HClO₄ reacts with water thus:

HClO₄ + H₂O → H₃O⁺ + ClO₄⁻

That means HClO₄ produce H₃O⁺ that decreases pH. That means the higher concentration of HClO₄ decreases pH. Thus, lower pH will be:

b) 0.2 M HClO4

The reaction of NaClO₄ is:

NaClO₄ + H₂O → OH⁻ + HClO₄ + Na⁺

The higher concentration of NaClO₄ the higher production of OH⁻ that increase pH, that means the lower concentration of NaClO₄ the lower pH, thus, the answer is:

c) 0.1 M NaClO or

HF reacts with water thus;

HF ⇄ H⁺ + F⁻

The equilibrium constant is:

k = [H⁺] [F⁻] / [HF] = 3,5x10⁻⁴

For HNO₂ equilibrium is:

HNO₂ ⇄ H⁺ + NO₂⁻

k = [H⁺] [NO₂⁻] / [HNO₂] = 4,5x10⁻⁴

As k value is higher for HNO₂, the concentration of H⁺ will be higher in this system doing the HNO₂ with the lower pH.

f) 0.1 M HNO2

NaOH is a strong base that produce OH⁻ that increase pH, pure water is neutral, thus, the lowe pH is:

h) pure water

I hope it helps!

7 0

3 years ago

Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39

nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ

2) 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39 kJ

3) Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3: Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

This equation we add to the second equation

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

ΔH° = 2*18 + (-39) = 36 - 39 = -3 kJ

This new equation we will divide by 3

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) → CO2(g) + 2FeO(s)

ΔH° =-3/3 = -1 kJ

Now we will substract this new equation from the first equation

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

The next equation we will divide by 2

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

CO(g) + FeO(s) → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

7 0

3 years ago