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valentina_108 [34]
3 years ago
15

A student wants to examine a substance without changing its identity. Which of the following properties of the substance should

the student examine?
A. Ductility, because it can be observed by altering the shape of the substance
B. Flammability, because it can be observed by altering the shape of the substance
C. Ductility, because it can be observed by changing the atoms in a molecule of the substance
D. Flammability, because it can be observed by changing the atoms in a molecule of the substance
Chemistry
2 answers:
finlep [7]3 years ago
6 0

Answer:

A

Explanation:

trust me

Vinvika [58]3 years ago
3 0

Answer:

A

Explanation:

Ductility, because it can be observed by altering the shape of the substance

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The heaviest element to be created by exothermic nuclear fusion is:
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Outline the steps needed to determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.
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Answer : The limiting reactant is O_2

Explanation : Given,

Mass of C_3H_8 = 30.0 g

Mass of O_2 = 75.0 g

Molar mass of C_3H_8 = 44 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_3H_8 and O_2.

\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{30.0g}{44g/mole}=0.682moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{75.0g}{32g/mole}=2.34moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

From the balanced reaction we conclude that

As, 5 mole of O_2 react with 1 mole of C_3H_8

So, 2.34 moles of O_2 react with \frac{2.34}{5}\times 1=0.468 moles of C_3H_8

From this we conclude that, C_3H_8 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Therefore, the limiting reactant is O_2

5 0
3 years ago
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