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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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Octane is a liquid component of gasoline. Given the following vapor pressures of octane at various temperatures, estimate the bo
Hitman42 [59]

Answer:

110.8 ºC

Explanation:

To solve this problem we will make use of the Clausius-Clayperon equation:

lnP = - ΔHºvap/RT + C

where P is the pressure, ΔHºvap is the enthalpy of vaporization, R is the gas constant, T is the temperature, and C is a constant of integration.

Now this equation has a form y = mx + b where

y = lnP

x = 1/T

m = -ΔHºvap/R

Now we have to assume that ΔHºvap remains constant which is a good asumption given the narrow range of temperatures in the data ( 104-125) ºC

Thus what we have to do is find the equation of the best fit for this data using a  software as excel or your calculator.

T ( K)               1/T                  ln P

377               0.002653       5.9915

384              0.002604       6.2115

390              0.002564       6.3969

395              0.002532       6.5511

398              0.002513        6.6333

The best line has a fit:

y = -4609.5 x  + 18.218

with R² = 0.9998

Now that we have the equation of the line, we simply will substitute for a pressure of 496 mm in Leadville.

ln(496) = -4609.5(1/Tb) + 18.218

6.2066 = -4609.5(1/Tb) +18.218

⇒ 1/Tb = (18.218 - 6.2066)/4609.5 = 0.00261

Tb = 383.76 K  = (383.76 -273)K = 110.8 ºC

Notice we have touse up to 4 decimal places since rounding could lead to an erroneous answer ( i.e boiling temperature greater than 111, an impossibility given the data in the question). This is as a result of the value 496 mmHg so close to 500 mm Hg.

Perhaps that is the reason the question was flagged.

7 0
2 years ago
The dissociation of calcium carbonate has an equilibrium constant of Kp = 1.16 at 1073 K . CaCO3(s) ⇄ CaO(s) + CO2(g) If you pla
tino4ka555 [31]

Answer:

<h3>Pressure of CO_2 in the container=1.6 atm</h3>

Explanation:

First balance the chemical equation:

CaCO_3(s) ⇄  CaO(s) + CO_2(g)

two components are solid so these two will not exert any kind of pressure in the container so at equilibrium only CO2 will apply pressure on the container

Therefore only partial pressure of CO2 will be taken for the calculation of equilibrium pressure constant i.e. Kp

K_p=[CO_2]

[CO_2]=p

K_p=p

p=K_p = 1.16atm

Pressure of CO_2 in the container=1.6 atm

8 0
2 years ago
Using the following thermochemical equation, determine the amount of heat produced from the combustion of 24.3 g benzene (C6H6).
Anna007 [38]

Answer:

ΔH = -976.5 kJ

Explanation:

For the reaction given, there are 2 moles of benzene (C6H6). The heat of this reaction is -6278 kJ, which means that the combustion of 2 moles of benzene will lose 6278 kJ of heat. It is an exothermic reaction.

The value of ΔH, the enthalpy, is a way of measurement of the heat, and it depends on the quantity of the matter (number of moles).

So, 24.3 g of benzene has :

n = mass/ molar mass

n = 24.3/78.11

n = 0.311 moles

2 moles ------------ -6278 kJ

0.311 moles ----------- x

By a simple direct three rule:

2x = -1953.08

x = -976.5 kJ

3 0
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Which of the following concerning second ionization energies is true?
Dennis_Churaev [7]

Answer:

D

Explanation:

The fact is that the both elements belong to different groups in the periodic table. Mg is in group 12 while Al is in group 13. The outer most electron configuration for Mg2+=[Ne]3s0

Al+= [Ne]3s13p0

It is evident that in Al+, the second electron is to be removed from a filled 3s subshell which is energetically unfavourable. Unlike In Mg2+ where the two electrons are removed. There is always a tendency towards a quick loss of all the valence electrons in the valence shell. It will be more difficult to remove an electron from a filled shell.

5 0
1 year ago
Calculate the first and second velocities of the car with one washer attached to the pulley, using the formulas v1 = 0.25 m / t1
labwork [276]

Answer: .11  .28

Explanation:

I was on this lesson

8 0
2 years ago
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