All categories

2 years ago

Identify the precipitate (if any) that forms when KOH and Cu(NO3)2 are mixed

1 answer:

3 0

When KOH and Cu(NO3)2 are mixed it yields copper(II) hydroxide and potassium nitrate. The balanced reaction is:

Cu(No3)2(aq) + 2 KOH(aq) = Cu(OH)2(s) + 2 KNo3(aq)

As we can see from the equation a solid is formed therefore a precipitate is formed which is the copper (II) hydroxide which has a blue to purple appearance. This can be observed since copper (II) hydroxide has a low solubility in aqueous solution.

You might be interested in

A water molecule is composed of a large oxygen atom bonded to two small hydrogen atoms. Which of the following would make the be

sleet_krkn [62]

The best and most correct answer among the choices provided by your question is the third choice or letter C.

The best model of a water molecule would be: Two small, plastic balls attached to a larger plastic ball by toothpicks

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

5 0

2 years ago

Read 2 more answers

Compound w , c6h13cl, undergoes base-promoted e2 elimination to give a single c6h12 alkene, y. compound x, c6h13br, undergoes a

CaHeK987 [17]

Answer: W must be 5-chloro-2-methylpentane. It can give only 4-methy-1-pentene (Y) upon dehydrohalogenation: X must be 4-chloro-2-methylpentane. Dehydrohalogenation yields both Y and 4-methyl-2-pentene. (Z)

5 0

2 years ago

A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The t

TEA [102]

Explanation:

The reaction equation will be as follows.

$Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)$

So, according to this equation, 1 mole $Ca(OH)_{2}$ = 2 mol HBr = 1 mol $CaBr_{2}$

Therefore, calculate the number of moles of calcium hydroxide as follows.

No. of moles of $Ca(OH)_{2}$ = $V \times Molarity$

= $50 \times 0.6$

= 30 mmol

Similarly, calculate the number of moles of HBr as follows.

No. of moles of HBr = $M \times V$

= $50 \times 0.6$

= 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of $CaBr_{2}$ = $30 \times \frac{1}{2}$

= 15 mmol

Hence, calculate the amount of heat released as follows.

Heat released in the reaction(q) = $m \times s \times \Delta T$

as, m = mass of solution

and, Density = $\frac{mass}{volume}$

or, mass = Density × Volume

= 1.08 g/ml \times (50 + 50) ml

= 108 g

where, s = specific heat of solution = 4.18 j/g.k

and, change in temperature $\Delta T$ = $(26 - 23)^{o}C$

= $3
^{o}C$

Hence, the heat released will be as follows.

q = $m \times s \times \Delta T$

q = $108 \times 4.18 \times 3^{o}C$

= 1354.32 joule

or, = 1.354 kJ (as 1 kJ = 1000 J)

Also, $\Delta H_{rxn}$ = $\frac{-q}{n}$

= $\frac{-1.354}{15 \times 10^{-3}}$

= -90.267 kJ/mol

Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.

6 0

2 years ago

What are the coefficients that will balance this skeleton equation? AlCl3 + NaOH → Al(OH)3 + NaCl

almond37 [142]

Answer:

The coefficients are 1, 3, 1, 3

Explanation:

The equation;

AlCl3 + NaOH → Al(OH)3 + NaCl

can be balanced by using the coefficients 1, 3, 1, 3, such that the balanced equation will be;

AlCl3 + 3NaOH → Al(OH)3 + 3NaCl

Chemical equations needs to be balanced so as to follow the law of conservation of mass. This occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

3 0

2 years ago

Be sure to answer all parts. Consider the following reaction: Cd(s) + Fe2+(aq) → Cd2+(aq) + Fe(s) E o (Fe2+ / Fe) = −0.4400 V, E

Bezzdna [24]

Answer:

EMF = -0.17 V

Explanation:

For the reaction :

Cd(s) + Fe2+(aq) → Cd2+(aq) + Fe(s)

we will be using the Nersnt equation to calculate Ecell:

Ecell = Eº -0.0592 V/2 ln Q

where Q = ( Fe⁺² )/ (Cd²⁺)

and (Fe²⁺) and (Cd²) are the molar concentration of Fe²⁺ and Cd²⁺

Reduction

Fe²⁺ + 2e⁻ ⇒ Fe(s) Eº red = -0.4400 V

Oxidation

Cd (s) ⇒ Cd²⁺ + 2e⁻ Eº ox = +0.4000 V

Eºcell = Eox + Ered = 0.4000 V + (-0.4400 V) = -0. 0400 V

Ecell = Eº -0.0692 V/2 ln Q = -0.0400 V - 0.0592/2 ln ( 0.80/0.010 ) = -0.17V

No because Eºcell is negative and ΔºG will be positive since

ΔºG = -nFEºcell

5 0

2 years ago