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lana66690 [7]
3 years ago
5

Explain how dalton’s atomic theory offered a convincing explanation of the observation that mass is conserved in chemical reacti

ons
Chemistry
1 answer:
podryga [215]3 years ago
6 0

Dalton's atomic theory was based on the law of conservation of mass which states that the matter can neither be created nor be destroyed but it can only transformed into one form or another. In a chemical reaction, total mass of the reactants will be equal to the total mass of the products.

Taking an example,

methane+oxygen\rightarrow \text{carbon dioxide}+water

CH_4+2O_2\rightarrow CO_2+2H_2O

Mass of CH_4 = 16g

Mass of 2O_2 = 2\times 32g

Total mass on reactant side = (16+2\times 32)

Total mass on reactant side = 80g

Mass of CO_2 = 44g

Mass of 2H_2O=2\times 18g

Total mass on product side = (44+2\times 18)

Total mass on product side = 80g

It is seen from the above example that the

total mass on reactant side = total mass on product side.


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1. How much heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C?
diamong [38]

Answer: An amount of 40980.48 J/g^{o}C heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.

Explanation:

Given: mass of lead = 4.64 kg

Convert kg into grams as follows.

1 kg = 1000 g\\4.64 kg = 4.64 kg \times \frac{1000 g}{1 kg}\\= 4640 g

T_{1} = 150^{o}C

T_{2} = 219^{o}C

The standard value of specific heat of lead is 0.128 J/g^{o}C.

Formula used to calculate heat is as follows.

q = m \times C \times \Delta T

where,

q = heat energy

m = mass of substance

C = specific heat of substance

\Delta T = change in temperature

Substitute the value into above formula as follows.

q = m \times C \times \Delta T\\= 4640 g \times 0.128 J/g^{o}C \times (219 - 150)^{o}C\\= 40980.48 J/g^{o}C

Thus, we can conclude that 40980.48 J/g^{o}C heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.

8 0
3 years ago
The common name for butanone, a readily available solvent, is​
azamat

Answer:

The common name for 2-butanone, a readily available solvent, is B-butanone

3 0
3 years ago
What reagent could be used to separate Br- from NO3- when added to an aqueous solution containing both?
joja [24]
<span>AgNO3(aq)
hope it helps
</span>
4 0
4 years ago
Read 2 more answers
What volume in litres will 38gr of F2 occupy at 0.999 bar and 273 K
Oliga [24]

Answer:

V=22.68L

Explanation:

Hello,

In this case, we use the ideal gas equation to compute the volume as shown below:

PV=nRT\\\\V=\frac{nRT}{P}

Nonetheless we are given mass, for that reason we must compute the moles of gaseous fluorine (molar mass: 38 g/mol) as shown below:

n=38 g *\frac{x}{y}  \frac{1mol}{38 g} =1mol

Thus, we compute the volume with the proper ideal gas constant, R:

V=\frac{1mol*0.083\frac{bar*L}{mol*K}*273K}{0.999bar} \\\\V=22.68L

Best regards.

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3 years ago
Is the sun typical of these nearby stars
marshall27 [118]
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