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AlekseyPX
2 years ago
5

Which postulate of Dalton's theory is consistent with the following observation concerning the weights of reactants and products

? When 100 grams of solid calcium carbonate is heated, 44 grams of carbon dioxide and 56 grams of calcium oxide are produced.
Chemistry
1 answer:
hram777 [196]2 years ago
6 0

<u>Answer:</u> This illustrates law of conservation of mass.

<u>Explanation:</u>

Dalton's theory is based on mainly two laws which are law of conservation of mass and law of constant proportion.

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The chemical equation for the decomposition of calcium carbonate follows:

CaCO_3\rightarrow CaO+CO_2

We are given:

Mass of calcium carbonate = 100 grams

Mass of calcium oxide = 56 grams

Mass of carbon dioxide = 44 grams

Total mass on reactant side = 100  g

Total mass on product side = 56 + 44 = 100 g

As, the total mass on reactant side is equal to the total mass on product side.

Thus, this illustrates law of conservation of mass.

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Answer:

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7 0
1 year ago
Read 2 more answers
The pressure of a 609.64 gram sample of F2 in a 88.84 L container is measured to be 2770.96 torr. What is the temperature of thi
Oxana [17]

Answer : The temperature of the gas is, 245.9 K

Explanation :

To calculate the temperature of gas we are using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT

where,

P = pressure of gas = 2770.96 torr = 3.646 atm

Conversion used : (1 atm = 760 torr)

V = volume of gas = 88.84 L

T = temperature of gas = ?

R = gas constant = 0.0821 L.atm/mole.K

w = mass of gas = 609.64 g

M = molar mass of F_2 gas = 38 g/mole

Now put all the given values in the ideal gas equation, we get:

(3.646atm)\times (88.84L)=\frac{609.64g}{38g/mole}\times (0.0821L.atm/mole.K)\times (T)

T=245.9K

Therefore, the temperature of the gas is, 245.9 K

4 0
2 years ago
The list below includes some of the properties of butane, a common fuel. Identify the chemical properties in the list. Check all
Brilliant_brown [7]

<u>Full Question:</u>

The list below includes some of the properties of butane, a common fuel. Identify the chemical properties in the list. Check all of the boxes that apply.

denser than water

burns readily in air  

boiling point of –1.1°C  

odorless

does not react with water

burns readily in air

does not react with water

<h3><u>Explanation:</u></h3>

The type of alkane that is used in many products includes Butane. It is found as a natural gas in the environment. It is found on the deeper part of ground. It can be obtained by drilling process and gets used up in many of the products that is used for commercial purposes.

Molecular mass that is associated with butane is 58.12 g/mol. The boiling point of butane is -1 degree Celsius and -140 degree Celsius is its melting point. It is a liquefied gas and does not react with water. It will burn in air more readily.

6 0
2 years ago
Beer brewing begins with steeping grains in hot water, releasing the sugars inside. The sugar water is then heated to a boil and
user100 [1]

Answer:

The answers to the question are

a. 166.64 ° F

b. 217990.08 J/hour or 60.55 J/s = 60.55 watts

c. 13.C

Explanation:

a. To solve the question we list out the given variables thus

mass of grain = 16.5 lbs

Temperature of grain = 67 °F

Volume of hot water = 5 gals = ‪0.02273‬ m³

Equilibrium temperature of the mixture = 154 °F

Specific heat capacity of the grain = 0.44 times specific heat capacity  of water

Therefore we have

Heat supplied by hot water = heat gained by mixture

Density of the water = 997 kg/m³ which gives

Therefore the mass of the water = (Density of the water) × (Volume of the water) = (997 kg/m³) × ‪(0.02273‬ m³) = 22.66181 kg

Therefore the heat supplied by the water =22.66 kg×1000 g/kg ×4.2 J/g°C×(Tₓ -‪67.78 °C) = ‪7.48 kg×1000 g/kg×0.44×4.2 J/g°C×(67.78 -‪19.44)

= 95172 × (Tₓ -‪67.78 °C) =668205.7536 J

(Tₓ -‪67.78 °C) = 7.02 from where Tₓ = 74.80 °C = ‪166.64 ° F

The initial temperature (strike temperature) of the hot water = 74.80 °C = 166.64 ° F

b. Where the mixture lost two degrees we have

22.66 kg×1000 g/kg ×4.2 J/g°C×2 °C + ‪7.48 kg×1000 g/kg×0.44×4.2 J/g°C×2  °C = 217990.08 J therefore the average energy lost per unit time = 217990.08 J/hour or 60.55 J/s

c. To find out how much it cost we have

Heat energy required to raise 5 gallons of water from 110 °F to 166.64 °F we have

22.66 kg×1000 g/kg ×4.2 J/g°C×(74.8 °C-‪43.33 °C) = 2994745.92 J

Energy lost during the heating = 10% = 299474.59 J

Total energy supplied 2994745.92 J + 299474.59 J  = 3294220.5 J

Time for heating = 47 minutes, therefore rate of energy consumption = (3294220.5 J)/ (47×60) = 1168.163 Watt 1.168 kW

Cost of energy = 15.C per kilowatt-hour therefore 1.168 kW for 47 minutes will cost

1.168 kW ×47/60×15 = 13.C

therefore it cost 13.C to heat the 5 gallons of tap water initially at 110 ° F to the strike temperature 166.64 °F

6 0
2 years ago
a square boat made from iron has overall dimensions of 2.00 cm x 11.0 cm x 11.0 cm. it has a mass of 213 g. what is its density
Rus_ich [418]
Volume:

2.00 x 11.0 x 11.0 => 242 cm³

mass : 213 g

D = m / V

D = 213 / 242

D = 0.880 g/cm³

Answer B

hope this helps!
5 0
2 years ago
Read 2 more answers
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