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Thepotemich [5.8K]
3 years ago
7

The density of corn syrup is 1.38 g/cm3.. . Which substance will float in corn syrup? A. graphite (2.16 g/cm3). B. glycerin (1.2

6 g/cm3). C. copper (8.9 g/cm3). D. silver (10.5 g/cm3)
Chemistry
2 answers:
PolarNik [594]3 years ago
7 0
"Glycerin" is the one substance among the following choices given in the question that will <span>float in corn syrup. The correct option among all the options that are given in the question is the second option or option "B". I hope that this is the answer that has actually come to your desired help.</span>
9966 [12]3 years ago
7 0

Answer: B. glycerin (1.26 g/cm3).

Explanation: The substances with higher value of density than the density of corn syrup will sink and the substances with lower value of density than the density of corn syrup will float.

A.  graphite : has a density of 2.16g/cm^3 which is higher than the density of corn syrup which is 1.38g/cm^3, Thus it will sink.

B. glycerin: has a density of 1.26g/cm^3 which is lower than the density of corn syrup which is 1.38g/cm^3, Thus it will float.

C. copper :has a density of 8.9g/cm^3 which is higher than the density of corn syrup which is 1.38g/cm^3, Thus it will sink.

D. silver: has a density of 10.5g/cm^3 which is higher than the density of corn syrup which is 1.38g/cm^3, Thus it will sink.

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CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of
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Answer:

129.73 g of CaBr₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr

Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of CaBr₂ = 40 + (80×2)

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Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g

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= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

SUMMARY :

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e

Xg of CaBr₂ = (200 × 48)/74

Xg of CaBr₂ = 129.73 g

Thus, 129.73 g of CaBr₂ were consumed.

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