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suter [353]
3 years ago
8

Calculate the concentration of OH- in a solution that contains 3.9 x 10-4 M H3O+ at 25°C. Identify the solution as acidic, basic

or neutral. Calculate the concentration of OH- in a solution that contains 3.9 x 10-4 M H3O+ at 25°C. Identify the solution as acidic, basic or neutral. 3.9 × 10-4 M, neutral 2.7 × 10-2 M, basic 2.6 × 10-11 M, basic 2.7 × 10-2 M, acidic 2.6 × 10-11 M, acidic
Chemistry
1 answer:
aalyn [17]3 years ago
8 0

Answer:

[OH⁻] = 2,6x10⁻¹¹

Acidic

Explanation:

The kw in water is:

2 H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)

kw = [OH⁻] [H₃O⁺] = 1,00x10⁻¹⁴

If concentracion of H₃O⁺ is 3,9x10⁻⁴M:

[OH⁻] [3,9x10⁻⁴M] = 1,00x10⁻¹⁴

<em>[OH⁻] = 2,6x10⁻¹¹</em>

pH is defined as - log[H₃O⁺]. If pH>7,0 the solution is basic, if pH<7,0 solution is acidic, if pH=7,0 solution is neutral.

In this problem,

pH = - log [3,9x10⁻⁴M] = <em>3,4</em>

As pH is < 7.0, the solution is <em>acidic</em>

<em></em>

I hope it helps!

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5 0
3 years ago
Kendra and Jordan gather firewood. The wood is cool to the touch, but the campfire they make warms their hands. Which statement
Free_Kalibri [48]

Answer:

D

Explanation:

6 0
3 years ago
Directions: Each set of lettered choices below refers to the numbered statements immediately following it. Select the one letter
maksim [4K]

Answer : The only reaction (C) that shows that the same reactant undergoes both oxidation and reduction.

Explanation :

Disproportionation reaction : It is defined as the reaction in which the same reactant undergoes both oxidation and reduction reaction. It is a redox reaction.

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(A) The given balanced reaction is,

H_2SeO_4(aq)+2Cl^-(aq)+2H^+(aq)\rightarrow H_2SeO_3(aq)+Cl_2(g)+H_2O(l)

This reaction is a redox reaction but not disproportionation reaction because in this reaction there are two reactants in which chlorine shows oxidation and selenium shows reduction.

(B) The given balanced reaction is,

S_8(s)+8O_2(g)\rightarrow 8SO_2(g)

This reaction is a redox reaction but not disproportionation reaction because in this reaction there are two reactants in which sulfur shows oxidation and oxygen shows reduction.

(C) The given balanced reaction is,

3Br_2(aq)+6OH^-(aq)\rightarrow 5Br^-(aq)+BrO_3^-(aq)+3H_2O(l)

This reaction is a disproportionation reaction because in this reaction only one reactant bromine that shows both oxidation and reduction reaction.

(D) The given balanced reaction is,

Ca^{2+}(aq)+SO_4^{2-}(aq)\rightarrow CaSO_4(s)

This reaction is a combination reaction in which the two reactant react to give a single product. There is no changes in the oxidation state of calcium and sulfate.

(E) The given balanced reaction is,

PtCl_4(s)+2Cl^-(aq)\rightarrow PtCl_6^{2-}(aq)

This reaction is a combination reaction in which the two reactant react to give a single product. There is no changes in the oxidation state of platinum and chlorine.

Hence, the only reaction (C) that shows that the same reactant undergoes both oxidation and reduction.

6 0
3 years ago
If we mix 25 grans if sodium oxide with a large amount of potassium chloride, how many grams of sodium chloride should be produc
amid [387]

Answer:

47.2 g

Explanation:

Data given:

mass of Sodium oxide (Na₂O) = 25 grams

mass of potassium chloride (KCl) = excess

amount of sodium chloride (NaCl) = ?

Solution:

First we look for reaction of sodium oxide with potassium chloride.

                    Na₂O + 2 KCl -----------> 2 NaCl + K₂O

As we know that potassium chloride is in excess amount and sodium oxide is 25 g so it means sodium oxide is a limiting reactant and amount of sodium chloride depends on the amount of sodium oxide.

So, now we will look for mole mole ration of Na₂O to NaCl.

                    Na₂O + 2 KCl -----------> 2 NaCl + K₂O

                    1 mol                              2 mol

from above equation we come to know that 1 mole of Na₂O gives 2 moles of NaCl.

Now convert moles to mass

As we Know

Molar mass of Na₂O = 2(23) + 16 = 62 g/mol

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

So

                  Na₂O              +   2 KCl    ----------->      2 NaCl          +      K₂O

             1 mol (62 g/mol)                                   2 mol (58.5 g/mol)

                  62 g                                                          117 g

So, it means that 62 g of Na₂O produce 117 g of NaCl then how many grams of NaCl will be produce by 25 g of Na₂O

Apply unity formula

                           62 g of Na₂O ≅ 117 g of NaCl

                           25 g of Na₂O ≅ X g of NaCl

Do cross multiplication

                          X g of NaCl = 117 g x 25 g / 62 g

                          X g of NaCl = 47.2 g

So,

25 g of Na₂O gives 47.2 grams of NaCl.

4 0
3 years ago
The specific heat of gold is 0.031 calories/gram°C. If 10.0 grams of gold were heated and the temperature of the sample
IgorLugansk [536]

Answer:

6.2 calories

Explanation:

Data Given:

change in temperature = 20 °C

specific heat of gold = 0.031 calories/gram °C

mass of gold = 10.0 grams

Amount of Heat = ?

Solution:

Formula used

             Q = Cs.m.ΔT

Where:

Q = amount of heat

Cs = specific heat of gold = 0.031 calories/gram °C

m = mass

ΔT = Change in temperature

Put values in above equation

                Q = 0.031 calories/gram °C x 10.0 g x 20 °C

                Q = 6.2 calories

So option A is correct = 6.2 calories

6 0
3 years ago
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