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Sonbull [250]
3 years ago
9

A 45.9 g sample of a metal is heated to 95.2°c and then placed in a calorimeter containing 120.0 g of water (c = 4.18 j/g°c) at

21.6°c. the final temperature of the water is 24.5°c. which metal was used?
Chemistry
1 answer:
VladimirAG [237]3 years ago
8 0

Answer:

Iron

Explanation:

Heat released by the metal sample will be equivalent to the heat absorbed by  water.

But heat = mass × specific heat capacity × temperature change

Thus;

Heat released by the metal;

= 45.9 g × c ×(95.2 -24.5) , where c is the specific heat capacity of the metal

= 3245.13c joules

Heat absorbed by water;

= 120 g × 4.18 J/g°C × (24.5-21.6)

= 1454.64 joules

Therefore;

3245.13c joules = 1454.64 joules

c = 1454.64/3245.13

  = 0.448 J/g°C

The specific heat capacity of the  metal sample is 0.448 J/g°C. The metal use is most likely, Iron.

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The type of reaction in a voltaic cell is best described as a
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If 34.5 g of Copper reacts with 70.2 g of silver nitrate, according to the following
Vlada [557]

Answer:

44,55 can be produced.

Explanation:

First, we balanced the equation

1Cu + 2AgNO3 → 1Cu(NO3)2 + Ag

Then, we find the moles of each reagent

mol Cu = \frac{34,5g}{63,55\frac{g}{mol} } = 0,543 mol\\

mol AgNO3 = \frac{70,2g}{169,87\frac{g}{mol} } = 0,413 mol\\

Now, we find the limiting reagent from the quantities of product that can be formed from each reagent

mol AgNO3= 0,543mol Cu . \frac{2 mol AgNo3}{1 mol Cu} = 1,086 mol

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1,086 moles of AgNO3 is necessary for each mole of Cu since we have 0.413 moles of Ag(NO3), the nitrate is the limiting reagent

the value of the limiting reagent determines the amount of product that is generated

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4 years ago
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What is the percent composition of carbon in heptane, C7H16?
JulsSmile [24]

Answer:

  • <em>Option d, 84%</em>

Explanation:

<u>1) Determine the molar mass of the heptane</u>

element   # of atoms    atomic mass        total mass      

  C              7                  12.011 g/mol         7×12.011 g/mol = 84.077 g/mol

  H             16                  1.008 g/mol         16×1.008 g/mol = 16.128 g/mol

                                                                                               -------------------

                                                                   Molar mass =    100.205 g/mol

<u>2) Percent of carbon, %</u>

  • % = (mass of element / molar mass )×100

           = (84.077 g/mol / 100.205 g/mol )×100 = 83.83 %

  • Round to two significant figures: 84% ← answer

4 0
3 years ago
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