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Mariana [72]
3 years ago
5

Does observing a physical property of a substance change the identity of the substance? Explain.

Chemistry
1 answer:
nikklg [1K]3 years ago
4 0
No because observing a physical property of a substance is just observing like color or texture. They don't change. Physical properties do not change the composition of a substance and neither do physical changes.
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Convert 32.56 km/hr into ft/hr
Gekata [30.6K]
Note that
1 m = 3.2808 ft

Therefore
1 km = 3280.8 ft
and
32.56 \,  \frac{km}{h} = (32.56 \,  \frac{km}{h})*(3280.8 \,  \frac{ft}{km}) =  1.0682 \, \times 10^{5} \, \frac{ft}{h}

Answer: 1.0682 x 10⁵ ft/hr

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2 years ago
What are some factors that might cause our percent yield to be greater than 100%? What are some factors that might cause it to b
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3 years ago
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60 Grams are produced due to the fact that it may not produce more due to the certain capacity that varies on the size of the cube.

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2 years ago
If the density of water is 1g/cm3 what is its specific gravity
MakcuM [25]

Answer:

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8 0
3 years ago
Read 2 more answers
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is
Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

7 0
3 years ago
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