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Dominik [7]
2 years ago
11

To gravimetrically analyze the silver content of a piece of jewelry made from an alloy of Ag and Cu, a student dissolves a small

, pre-weighed sample in HNO3(aq). Ag+(aq) and Cu2+(aq) ions form in the solution.Which of the following should be the next step in analytical process?Centrifuging the solution to isolate the heavier ionsAdding enough base solution to bring the pH up to 7.0Adding a solution containing an anion that forms an insoluble salt with only one of the metal ionsEvaporating the solution to recover the dissolved nitrates
Chemistry
1 answer:
Ipatiy [6.2K]2 years ago
8 0

Answer:

Adding a solution containing an anion that forms an insoluble salt with only one of the metal ions.

Explanation:

The student have in solution Ag⁺ and Cu²⁺ ions but he just want to analyze the silver, that means he need to separate ions.

Centrifuging the solution to isolate the heavier ions <em>FALSE </em>Centrifugation allows the separation of a suspension but Ag⁺ and Cu²⁺ are both soluble in water.

Adding enough base solution to bring the pH up to 7.0 <em>FALSE </em>At pH = 7,0 these ions are soluble in water and its separation will not be possible.

Adding a solution containing an anion that forms an insoluble salt with only one of the metal ions <em>TRUE </em>For example, the addition of Cl⁻ will precipitate the Ag⁺ as AgCl(s) allowing its separation.

Evaporating the solution to recover the dissolved nitrates. <em>FALSE</em> . Thus, you will obtain the nitrates of these ions but will be mixed doing impossible its separation.

I hope it helps!

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You try to measure out exactly 5.0 milliliters of water by eye into each of five test tubes. When you go back to check the volum
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Answer:

The volumes are both, accurate and precise.

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In the measurement of a set, precision refers to how much coincidence exists in the measurements of an specific value, as the measurements are close, we can say the volumes are precise.

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6 0
1 year ago
Victoria has a crate of vegetables that weighs 100 newtons. She exerts a force of 100 newtons to lift the crate with a pulley. W
Dennis_Churaev [7]

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7 0
2 years ago
Read 2 more answers
The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3
Oksanka [162]

Answer : The mass of NaN_3 required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of N_2 gas = 1.00 atm

V = Volume of N_2 gas = 113 L

n = number of moles N_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 85^oC=273+85=358K

Putting values in above equation, we get:

1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K

n=3.84mol

Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced reaction we conclude that

As, 3 mole of N_2 produced from 2 mole of NaN_3

So, 3.84 moles of N_2 produced from \frac{2}{3}\times 3.84=2.56 moles of NaN_3

Now we have to calculate the mass of NaN_3

\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3

Molar mass of NaN_3 = 65 g/mole

\text{ Mass of }NaN_3=(2.56moles)\times (65g/mole)=166.4g

Therefore, the mass of NaN_3 required is, 166.4 grams.

3 0
2 years ago
One of the emission spectral lines for Be31 has a wavelength of 253.4 nm for an electronic transition that begins in the state w
max2010maxim [7]

Explanation:

The given data is as follows.

   \lambda = 253.4 nm = 253.4 \times 10^{-9}m      (as 1 nm = 10^{-9})

            n_{1} = 5,        n_{2} = ?

Relation between energy and wavelength is as follows.

                    E = \frac{hc}{\lambda}

                       = \frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{253.4 \times 10^{-9}}

                       = 0.0784 \times 10^{-17} J

                       = 7.84 \times 10^{-19} J

Hence, energy released is 7.84 \times 10^{-19} J.

Also, we known that change in energy will be as follows.

     \Delta E = -2.178 \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}

where, Z = atomic number of the given element

 7.84 \times 10^{-19} J = -2.178 \times (4)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}

    \frac{7.84 \times 10^{-19} J}{34.848} = \frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}

      0.02 + 0.04 = \frac{1}{n^{2}_{1}}

                      n_{1} = \sqrt{\frac{1}{0.06}}

                          = 4

Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.

4 0
1 year ago
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