Why don’t water and oil mix when the two liquids are poured into the same container?

Great question!

Physical changes change the appearance of something or someone.

For example, let's say 2 cats get into a fight. One is left clean in the end of this, with no scratches or bites anywhere, so nothing has happened to it as a physical change.

However, the other cat is left scarred and with bite marks everywhere. All this has left the cat's once-beautiful coat now ragged and torn up. This is an example of a physical change.

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Answer:

affect the size, shape, and phase of a substance

Explanation:

Answer:

848.1 g/mol

Explanation:

Data given:

Standard of atomic mass of Ca = 50 amu

molar mass of Aluminium Acetate = ?

Solution:

Relative atomic mass represented by Ar. It is ratio of actual mass with respect to the 1/12th mass of C-12 but here Ca-50 is standard.

Formula of  Aluminium Acetate = Al(CH₃COO)₃

In order to calculate Ar

first we will calculate 1/50 of Ca

As we know

         mass of 1/12 of C-12 = 1.993 x 10⁻²⁶/12 = 1.661 x 10⁻²⁷ Kg

So, for Ca-50

        mass of 1/50 of Ca-50 = 1.993 x 10⁻²⁶/50 = 3.986 x 10⁻²⁸ Kg

Now

 Relative Atomic mass for element X = rest mass of "X"/ 3.986 x 10⁻²⁸ Kg . . . . . . (1)

First we have to know the relative atomic masses of Aluminium, carbon, hydrogen, oxygen atoms involve in Aluminium Acetate formula with respect to new standard Ca-50

By using equation-1 we can calculate Ar for which we have reported rest masses of atoms as below

Rest mass of Aluminium = 4.48 x 10⁻²⁶ Kg

Rest mass of carbon = 1.993 x 10⁻²⁶ Kg

Rest mass of hydrogen = 1.608 x 10⁻²⁷ Kg

Rest mass of oxygen = 2.657x10⁻²⁶ Kg

Now put values in equation 1 for each atom

• Ar for Aluminium= 4.48 x 10⁻²⁶Kg / 3.986 x 10⁻²⁸ Kg

                     Ar for Aluminium = 112.5  amu

• Ar for Carbon= 1.993 x 10⁻²⁶ Kg / 3.986 x 10⁻²⁸ Kg

                     Ar for Carbon= 50 amu

• Ar for Hydrogen = 1.608 x 10⁻²⁷ Kg / 3.986 x 10⁻²⁸ Kg

                     Ar for hydrogen = 4  amu

• Ar for Oxygen = 2.657x10⁻²⁶ Kg / 3.986 x 10⁻²⁸ Kg

                     Ar for Oxygen = 66.6  amu

Now find the molar mass ofAl(CH₃COO)₃

molar mass of Aluminium Acetate = Al(CH₃COO)₃

Al(CH₃COO)₃ = 112.5 + 3 (50 + 3(4) + 50 + 66.6 +66.6)

Al(CH₃COO)₃ = 112.5 + 3 (50 + 12 + 50 + 66.6 +66.6)

Al(CH₃COO)₃ = 112.5 + 3 (245.2)

Al(CH₃COO)₃ = 112.5 + 735.6

Al(CH₃COO)₃ = 848.1 g/mol

molar mass of Aluminium Acetate = 848.1 g/mol

71.8 g of iron (III) oxide (Fe₂O₃) were produced from 3 moles of magnesium oxide (MgO).

Explanation:

We have the following chemical reaction:

2 FeCl₃ + 3 MgO → Fe₂O₃ + 3 MgCl₂

We see from the chemical reaction that 3 moles of magnesium oxide (MgO) will produce 1 mole of iron (III) oxide (Fe₂O₃).

number of moles = mass / molar weight

mass = number of moles × molar weight

mass of Fe₂O₃ = 1 × 71.8 = 71.8 g

Learn more about:

number of moles

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Answer:

47 mL

Explanation:

By the image, the beaker graduation will from 45 from 50 mL, passing by 5 markings. So each mark has 1 mL. To read the volume, we have see the meniscus, which is the concave superficie of the liquid. The meniscus is above the second mark, so it passed 2 mL from the 45 line, the volume is 47 mL.