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Ulleksa [173]
3 years ago
8

Suppose now that you wanted to determine the density of a small crystal to confirm that it is boron. From the literature, you kn

ow that boron has a density of 2.34 g/cm^3. How would you prepare 20.0 mL of the liquid mixture having that density from pure samples of CHCI3 ( d= 1.492 g/mL) and CHBr3( d= 2.890 g/mL)? (Note: 1 mL = 1 cm^3.)
Chemistry
1 answer:
Phantasy [73]3 years ago
3 0

Answer:

the volume of  CHCI3 = 7.87 ml

the volume of  CHBr3 = 12.13 ml

Explanation:

From the given information:

We all know that 1 g/cm^3 = 1 g/ml

The density of boron = 2.34 g/ml

The Volume of the liquid mixture  = 20 ml

Recall that:

Density = mass/volume

Mass = Density × Volume

Mass = 2.34 g/ml × 20 ml

Mass = 46.8 g

Suppose the volume of  CHCI3 be Y and the Volume of CHBr3 be 20 - Y

Then :

Y (1.492) + 20-Y(2.890) = 46.8

1.492Y + 57.8 - 2.890Y = 46.8

- 1.398 Y = -11

Y = -11/ - 1.398

Y = 7.87 ml

Therefore, the volume of CHCI3  7.87 ml

the volume of  CHBr3 = 20 - Y

= 20 - 7.87

= 12.13 ml

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EastWind [94]

Answer:

158.5g Zn are produced

Explanation:

To solve this question we have to find the moles of Aluminium. With the moles of Aluminium and the balanced reaction we can find the moles of Zn and its mass as follows:

<em>Moles Al -Molar mass: 26.98g/mol</em>

43.6g Al* (1mol/26.98g) = 1.616 moles Al

<em>Moles Zn:</em>

1.616 moles Al * (3mol Zn / 2mol Al) =

2.424 moles Zn are produced

<em>Mass Zn -Molar mass: 65.38g/mol-</em>

2.424 moles Zn * (65.38g / mol) =

<h3>158.5g Zn are produced</h3>
6 0
3 years ago
Calculate the density of chlorine gas in g/l, at stp.
vichka [17]
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<span>At STP : p = 1 atm and T = 273.15 K 
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8 0
4 years ago
Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indi
AlexFokin [52]

Explanation:

The given cell reaction is as follows.

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Hence, reactions taking place at the cathode and anode are as follows.

At anode ; Oxidation-half reaction : In(s) \rightarrow In^{3+}(aq) + 3e^{-} ...... (1)

At cathode; Reduction-half reaction : Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s) ....... (2)

Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.

Therefore, net cell reaction is as follows.

      2In(s) \rightarrow 2In^{3+}(aq) + 6e^{-}

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Net reaction: 2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)

Thus, we can conclude that the overall cell reaction is as follows.

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3 years ago
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kow [346]

Answer:

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Hope this helps

6 0
3 years ago
Read 2 more answers
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5 0
3 years ago
Read 2 more answers
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