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ArbitrLikvidat [17]
3 years ago
13

How could you use these solutions to determine the identities of each metal powder?

Chemistry
1 answer:
Anastaziya [24]3 years ago
5 0

Answer:

1.) The nitric acid solution will oxidize and thus dissolve _*(Zn and Pb)*_. This will allow to identify _**Pt**_.

2) To distinguish between _*(Zn and Pb)*_, we can use the nickel nitrate.

3) The nickel nitrate solution will oxidize and thus dissolve _**Zn**_ and will not oxidize or dissolve _**Pb**_.

Explanation:

1) Unlike Zinc and Lead, Platinum does not react with Nitric acid. So, it will be the only metal from step 1 that doesn't react. Pt is identified in this manner.

2) Nickel is higher than Lead in the activity series, but Zinc is higher than both of them in the activity series. This selectivity can be used to distinguish between Zinc and Lead metal powders.

3) Because Zinc is higher than Nickel in the activity series, it means that Zinc metal can and will displace Nickel from Nickel Nitrate solution. Therefore the Nickel Nitrate solution will oxidize and dissolve the Zinc metal.

But, there will be no reaction with the Lead metal powders sample as Pb is lower than Ni in the activity series, so, Nickel Nitrate solution will not oxidize or dissolve the Lead metal powders.

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(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction
ale4655 [162]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2}   \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu}   \\\\m_{Cu}=0.380gCu

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3}   \\\\m_{AgNO_3}=2.03gAgNO_3

Best regards!

5 0
3 years ago
From Young's experiment, how does the frequency of light affect the number of lines observed?
Marina CMI [18]

Answer : (C) "Higher frequencies have larger spaces between lines".

Explanation:

In Young's experiment, the condition for constructive interference is given by :

dsin\theta=n\lambda.........(1)

n is order or number of lines observed

d is distance between slits

\theta is the angle between the path and the line from screen to the slits.

We also know that, c=\nu \lambda

or

\lambda=\dfrac{c}{\nu}

where,

c is the speed of light

\nu is frequency

\lambda is wavelength

So, equation (1) turns into

dsin\theta=n\dfrac{c}{\nu}

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So,

\nu\propto n

or

Higher frequencies have larger spaces between line.

So, correct option is (C).

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Fog forms when a water vapor changes to the___ state of matter?
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Answer:

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