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zlopas [31]
3 years ago
10

When aqueous solutions of ________ are mixed, a precipitate forms.

Chemistry
1 answer:
xeze [42]3 years ago
7 0

Answer:

A) NiBr₂ and AgNO₃  

Explanation:

We can use the solubility rules to decide which reaction gives the precipitate.

The important ones for this problem are:

  1. Salts of Group I elements and of ammonia are soluble.  
  2. Nitrates are  soluble.
  3. Halides are soluble. Important exceptions the silver halides.
  4. Most sulfates are soluble.
  5. Most hydroxides are slightly soluble.

The possible products from each reaction are

A) Ni(NO₃)₂ and AgBr; B) NaBr and KI; C) KCl and Cr₂(SO₄)₃

D) KNO₃ and Ba(OH)₂; E) LiI and Cs₂CO₃.

A) Ni(NO₃)₂ and AgBr

Ni(NO₃)₂ — soluble (Rule 2)

     AgBr — insoluble (an exception to Rule 3)

When aqueous solutions of NiBr₂ and AgNO₃ are mixed, a precipitate forms.

B) NaBr and KI

NaBr — soluble (Rule 1)

    KI — soluble (Rule 1)

C) KCl and Cr₂(SO₄)₃

        KCl — soluble (Rule 1)

Cr₂(SO₄)₃ — soluble (Rule 4)

D) KNO₃ and Ba(OH)₂

   KNO₃ — soluble (Rule 1)

Ba(OH)₂ — soluble (Rule 5)

E) LiI and Cs₂CO₃

       LiI — soluble (Rule 1)

Cs₂CO₃ — soluble (Rule 1)

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Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el
Dovator [93]

Answer: 406 hours

Explanation:

Q=I\times t

where Q= quantity of electricity in coloumbs

I = current in amperes = 39.5 A

t= time in seconds = ?

The deposition of copper at cathode is represented by:

Cu^{2+}+2e^-\rightarrow Cu

96500\times 2=193000 Coloumb of electricity deposits 1 mole of copper

i.e. 63.5 g of copper is deposited by = 193000 Coloumb

Thus 19.0 kg or 19000 g of copper is deposited by = \frac{193000}{63.5}\times 19000=57748032 Coloumb

57748032=39.5\times t

t=1461975sec=406hours    (1hour=3600s)

Thus it will take 406 hours to plate 19.0 kg of copper onto the cathode if the current passed through the cell is held constant at 39.5 A

3 0
3 years ago
Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

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How do you convert between the volume of a gas at STP and the number of moles of the gas?
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Standard temperature and pressure (STP) means a temperature of 0°c and a pressure of 1 atmosphere (atm). The molar gas volume is used to convert between the number of moles of a gas and the volume of the gas at STP. One mole of a gas occupies a volume of 22400 cm³ or 22.4 liters at STP according to the molar gas volume.
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gtnhenbr [62]

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Answer:

Explanation:

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