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SCORPION-xisa [38]
3 years ago
5

For an alloy that consists of 93.1 g copper, 111.7 g zinc, and 4.0 g lead, what are the concentrations of (a) Cu, (b) Zn, and (c

) Pb in weight percent? The atomic weights of Cu, Zn, and Pb are 63.54, 65.39, and 207.2 g/mol, respectively.
Chemistry
1 answer:
Andrew [12]3 years ago
5 0

Answer:

(a) weight percent of Cu = 44.59%

(b) weight percent of Zn = 53.49%

(c) weight percent of Pb = 1.91%

Explanation:

Given: Alloy contains- Cu=93.1 g, Zn=111.7 g, Pb=4.0 g

Therefore, the total mass of the alloy (m) = mass of Cu (m₁) + mass of Zn (m₂)+ mass of Pb (m₃)

m= 93.1 g + 111.7 g + 4.0 g = 208.8 g

(a) weight percent of Cu = (m₁ ÷ m)× 100% =  (93.1 g ÷ 208.8 g)× 100% =44.59%

(b) weight percent of Zn = (m₂ ÷ m)× 100% =  (111.7 g ÷ 208.8 g)× 100% =53.49%

(c) weight percent of Pb = (m₃ ÷ m)× 100% =  (4.0 g ÷ 208.8 g)× 100% =1.91%

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Ahat [919]

Answer : The balanced chemical equation is,

2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)

Explanation :

Rules for the balanced chemical equation in basic solution are :

  • First we have to write into the two half-reactions.
  • Now balance the main atoms in the reaction.
  • Now balance the hydrogen and oxygen atoms on both the sides of the reaction.
  • If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.
  • If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion (OH^-) at that side where the less number of hydrogen are present.
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The half reactions in the basic solution are :

Reduction : MnO_4^-(aq)+2H_2O(l)+3e^-\rightarrow MnO_2(s)+4OH^-(aq) ......(1)

Oxidation : Br^-(aq)+6OH^-(aq)\rightarrow BrO_3^-(aq)+3H_2O(l)+6e^-  .......(2)

Now multiply the equation (1) by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)

8 0
3 years ago
Which of the following substances (with specific heat capacity provided) would show the greatest temperature change upon absorbi
JulijaS [17]

Answer:

Pb is the substance that experiments the greatest temperature change.

Explanation:

The specific heat capacity refers to the amount of heat energy required to raise in 1 degree the temperature of 1 gram of substance. The highest the heat capacity, the more energy it would be required. These variables are related through the equation:

Q = c . m . ΔT

where,

Q is the amount of heat energy provided (J)

c is the specific heat capacity (J/g.°C)

m is the mass of the substance

ΔT is the change in temperature

Since the question is about the change in temperature, we can rearrange the equation like this:

\Delta T = \frac{Q}{c.m}

All the substances in the options have the same mass (m=10.0g) and absorb the same amount of heat (Q=100.0J), so the change in temperature depends only on the specific heat capacity. We can see in the last equation that they are inversely proportional; the lower c, the greater ΔT. Since we are looking for the greatest temperature change, It must be the one with the lowest c, namely, Pb with c = 0.128 J/g°C. This makes sense because Pb is a metal and therefore a good conductor of heat.

Its change in temperature is:

\Delta T = \frac{q}{c.m} = \frac{100.0 J}{0.128 J/g.C . 10.0g } = 78.1

5 0
3 years ago
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Answer:

5.28 L

Explanation:

Step 1:

Data obtained from the question.

Volume of acid (Va) = 5.12L

Molarity of acid (Ma) = 2.75M

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Volume of base (Vb) =.?

Step 2:

The balanced equation for the reaction

2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

From the balanced equation above,

Mole ratio of the acid (nA) = 2

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Step 3:

Determination of the volume of the base.

This is illustrated below:

MaVa/MbVb = nA/nB

2.75 x 5.12 / 4 x Vb = 2/3

Cross multiply

4 x 2 x Vb = 2.75 x 5.12 x 3

Divide both side by 4 x 2

Vb = (2.75 x 5.12 x 3)/(4 x 2)

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topjm [15]

Answer:

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Then, taking into account the yield of the reaction (82 % = 0.82) and the percent of aluminun in the ore (71% = 0.71), you can write the following equation:

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(mass of ore)    (% of Al in the ore)        (yield)        ( Al metal to obtain)

You must just simplify, solve and compute:

  • 0.71 × 0.82 × X = 1,000
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Round to two significant figures; 1,700 kg = 1.7 × 10³ kg of ore ← answer.

6 0
3 years ago
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