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Firlakuza [10]
3 years ago
13

100 PNTS ANSWER ASAP PLZ The density of a particular small rubber ball is 1.1 g/cm3. It was dropped into a 100 mL beaker filled

with 50 mL each of ethanol and distilled water. Which is the most accurate description of what happened to the rubber ball in the beaker? A. It sank below the ethanol and distilled water. B. It floated on top of the ethanol and distilled water. C. It floated on top of the ethanol, but sank below the distilled water. D. It sank below the ethanol, but floated on top of the distilled water.
Chemistry
2 answers:
Klio2033 [76]3 years ago
6 0

Answer:

A. It sank below the ethanol and distilled water.

Explanation:

I just tok the quiz! :)

AlekseyPX3 years ago
3 0

Answer:

a

Explanation:

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VA 19.75-g sample was heated by 12.35 calories. The specific heat of the sample is 0.125 cal/g°C. What was the initial temperatu
SOVA2 [1]

Answer:

31.9 °C  

Explanation:

The formula for the heat q absorbed by an object is

q = mCΔT where ΔT = (T₂ - T₁)

Data:

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m = 19.75 g

C = 0.125 cal°C⁻¹g⁻¹

T₂ = 37.0 °C

Calculations

(a) Calculate ΔT

q = mCΔT

12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT

12.35 = 2.406ΔT °C⁻¹  

ΔT  = 12.35/(2.406 °C⁻¹) = 5.13 °C

(b) Calculate T₂

ΔT = T₂ - T₁

T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C

The original temperature was 31.9 °C.

 

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Answer:

Explanation:

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I don't really understand this worksheet question.​
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25/2 and 96/X
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