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inysia [295]
3 years ago
6

Fill in the chart to describe and give examples of physical changes.

Chemistry
2 answers:
frosja888 [35]3 years ago
8 0

Answer:

here are some examples of physical change!!!

Explanation:

-An ice cube melting into water in your drink.

-Freezing water to make ice cubes.

-Boiling water evaporating.

-Hot shower water turning to steam.

-Steam from the shower condensing on a mirror.

stellarik [79]3 years ago
7 0

Answer:

Examples:

Melting an ice cube.

Boiling water.

Mixing sand and water.

Breaking a glass.

Dissolving sugar and water

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I *think* through weathering and erosion. Hope this helped
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It is B nevermind I thought it was c but it wasn't.

5 0
3 years ago
What is the name of this hydrocarbon? 2–methylpentyne 2–dimethylpentene 2,3–dimethylpentane 3,2–dimethylpentane
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3 years ago
you have a solution of water and salt. if the total solution weighs 50 grams and the salt weighs 4 grams. what is the percent co
Natali5045456 [20]

Answer:

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Explanation:

I believe. Hope this was helpful.

8 0
3 years ago
Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25
Veseljchak [2.6K]

Answer: The molecular formula will be C_6H_6O_6

Explanation:

Mass of CO_2 = 17.95 g

Mass of H_2O= 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, =\frac{12}{44}\times 17.95=4.89g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, =\frac{2}{18}\times 4.87=0.541g of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H =  0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.407}{0.407}=1

For H =\frac{0.541}{0.407}=1

For O = \frac{0.410}{0.407}=1

The ratio of C : H : O = 1: 1  : 1

Hence the empirical formula is CHO.

Hence the empirical formula is CHO

The empirical weight of CHO = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = \frac{44.0}{0.25}\times 1=176g

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6

The molecular formula will be=6\times CHO=C_6H_6O_6

5 0
3 years ago
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