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emmasim [6.3K]
2 years ago
11

A simple circuit with a battery connected to a buzzer by electrical wires is placed on a table. When the buzzer sounds, the tabl

e begins to ratte. Which form of energy makes the table rattle?
chemical
ОО
electrical
heat
mechanical
Chemistry
1 answer:
Ainat [17]2 years ago
7 0

Answer:

Mechanical

Explanation:

mechanical energy:

energy that an object has

because of its motion or

position

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Molarity is defined as the number of moles of solute in 1 L of solution 
number of NaOH moles present - 0.20 mol 
volume of solution - 150 mL 
since molarity is taken as number moles of solute in 1000 mL
if 150 mL contains - 0.20 mol 
then 1000 mL should contain - 0.20 / 150 x 1000 = 1.33 mol 
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The atomic number determines the identity of the atom. True or False?
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THE CORRDCT ANSWER IS TRUE
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Aluminum oxide reacts with hydrochloric acid to produce salt. What is the name of the salt produced?
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18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c
Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

   Percentage by mass of hydrogen: 0.77%

b)  Percentage by mass of chlorine: 80.37%

c) Molecular formula: C_{2} H Cl_{3}

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}

The same process is used to calculate the amount of hydrogen:

0.089g H_{2}O*\frac{2g H}{18g H_{2}O }  = 0.010g H

The percentage by mass of carbon and hydrogen are calculated as follows:

%C\frac{0.238g}{1.3g} *100%= 18.3%

%H\frac{0.010g}{1.3g} *100%=0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}= 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl=\frac{0.43g}{0.535g} * 100%= 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

18.3g C*\frac{1mol C}{12g C} = 1.52mol C

0.77g H*\frac{1mol H}{1g H} = 0.77mol H

80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl

Dividing each of the quantities above by the smallest (0.77mol), the  subscripts in a tentative formula would be

C=\frac{1.52}{0.77} = 1.97 ≈ 2

H = \frac{0.77}{0.77} = 1

Cl =\frac{2.27}{0.77}=2.94≈3

The empirical formula for the compound is:

C_{2} H Cl_{3}

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

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