Answer:
- <u><em>Ratio of the mass carbon that combines with 1.00 g of oxygen in compound 2 to the mass of carbon that combines with 1.00 g of oxygen in compound 1 = 2</em></u>
Explanation:
First, detemine the mass of oxygen in the two samples by difference:
- mass of oxygen = mass of sample - mass of carbon
Item Compound 1 Compound 2
Sample 80.0 g 80.0 g
Carbon 21.8 g 34.3 g
Oxygen: 80.0 g - 21.8g = 58.2 g 80.0 g - 34.3 g = 45.7 g
Second, determine the ratios of the masses of carbon that combine with 1.00 g of oxygen:
- For each sample, divide the mass of carbon by the mass of oxygen determined above:
Sample Mass of carbon that combines with 1.00 g of oxygen
Compound 1 21.8 g / 58.2 g = 0.375
Compound 2 34.3 g / 45.7 g = 0.751
Third, determine the ratio of the masses of carbon between the two compounds.
- Divide the greater number by the smaller number:
- Ratio = 0.751 / 0.375 = 2.00 which in whole numbers is 2
Answer:
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Explanation:
Answer: A
Explanation: Beta particles have a charge of -1
CBr4 is a symmetric tetrahedral molecule so it will be non-polar.
Answer:
The specific heat of the metal is 2.09899 J/g℃.
Explanation:
Given,
For Metal sample,
mass = 13 grams
T = 73°C
For Water sample,
mass = 60 grams
T = 22°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
, water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.
Since, Specific heat of water = 4.184 J/g°C
Let Cp be the specific heat of the metal.
Substituting values,
(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)
By solving, we get Cp =
Therefore, specific heat of the metal sample is 2.09899 J/g℃.
