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3 years ago

Aqueous solution of copper sulphate is acidic . Why?

Chemistry

1 answer:

andreev551 [17]3 years ago

7 0

Answer:

Explanation:

Aqueous solution of copper sulphate

is acidic in nature, as when copper sulphate on dissolving with water, following chemical reaction takes place :

CuSO4 + 2 H2O ====> Cu(OH)2 + H2 SO4

As we can see, copper hydroxide and sulphuric acid is generated in this chemical reaction.Copper hydroxide is a weak base whereas sulphuric acid is a very strong acid which results into increase in acidity of the solution.

Thus, the aqueous solution of copper sulphate is acidic in nature.

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Which element does not have the same number of electrons in its outermost shell as the other elements in its group?(1 point)

Natalija [7]

Answer:

Helium

Explanation:

7 0

2 years ago

GIVING BRAINLIEST One mole of hydrogen gas (H2), reacts with one mole of bromine Br2(g) to produce 2 moles of hydrogen bromide g

JulsSmile [24]

Answer:

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Given, that 1 mole of $H_2$ gas and 1 mole of $Br_2$ liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

$H_2(g) + Br_2(l)\rightarrow 2HBr(g) ,\Delta H_{f}^o= -72.58kJ$

Divide the equation by 2.

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

4 0

3 years ago

In the titration of 50. 0 mL of 0. 400 M HCOOH with 0. 150 M LiOH, how many mL of LiOH are required to reach the equivalence poi

mart [117]

The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL

<h3>Balanced equation </h3>

HCOOH + LiOH —> HCOOLi + H₂O

From the balanced equation above,

The mole ratio of the acid, HCOOH (nA) = 1

The mole ratio of the base, LiOH (nB) = 1

<h3>How to determine the volume of LiOH </h3>

Molarity of acid, HCOOH (Ma) = 0.4 M
Volume of acid, HCOOH (Va) = 50 mL
Molarity of base, LiOH (Mb) = 0.15 M

Volume of base, LiOH (Vb) =?

MaVa / MbVb = nA / nB

(0.4 × 50) / (0.15 × Vb) = 1

20 / (0.15 × Vb) = 1

Cross multiply

0.15 × Vb = 20

Divide both side by 0.15

Vb = 20 / 0.15

Vb = 133.3 mL

Thus, the volume of the LiOH solution needed is 133.3 mL

Learn more about titration:

brainly.com/question/14356286

8 0

1 year ago

The AP Biology teacher is measuring out 638.0 g of dextrose (C6H12O6) for a lab. How many moles of dextrose is this equivalent t

Katena32 [7]

The AP Biology teacher is measuring out 638.0 g of dextrose (C6H12O6) for a lab the moles of dextrose is this equivalent to is 3.6888 moles.

<h3>What are moles?</h3>

A mole is described as 6.02214076 × 1023 of a few chemical unit, be it atoms, molecules, ions, or others. The mole is a handy unit to apply due to the tremendous variety of atoms, molecules, or others in any substance.

To calculate molar equivalents for every reagent, divide the moles of that reagent through the moles of the restricting reagent. The calculation is follows:

655/12 x 6 + 12+ 16 x 6
= 655/ 180 = 3.6888 moles.