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3 years ago

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

1 answer:

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

8 0

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

belka [17]

<span>In the field of science, usually, the product of an experiment is computed ahead to understand if it reached a specific objective. It would reach greater than 100% of percent yield if the factors include faster reaction rates; proper handling of the reactants, no outside contaminants, and the procedure of the experiment is followed smoothly. It would reach lesser than 100% percent yield if the experiment is not followed, external factors such as contamination from the environment (wind, moisture, etc). </span>

8 0

3 years ago

Will give branliest

fomenos

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.

5 0

2 years ago

If the density of water is 1g/cm3 what is its specific gravity

MakcuM [25]

Answer:

Specific gravity is the density of asubstance divided by the density of water. Since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density(but without any units).

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