The mass of an object is constant, but its weight may change depending on:

Answer:

C15H24O

Explanation:

TO GET THE EMPIRICAL FORMULA, WE NEED TO KNOW THE MASSES AND CONSEQUENTLY THE NUMBER OF MOLES OF EACH OF THE INDIVIDUAL CONSTITUENT ELEMENTS.

FIRSTLY, WE CAN GET THE MASS OF THE CARBON FROM THAT OF THE CARBON IV OXIDE. WE NEED TO KNOW THE NUMBER OF MOLES OF CARBON IV OXIDE GIVEN OFF. THIS CAN BE CALCULATED BY DIVIVDING THE MASS BY THE MOLAR MASS OF CARBON IV OXIDE. THE MOLAR MASS OF CARBON IV OXIDE IS 44G/MOL

HENCE, THE NUMBER OF MOLES OF CARBON IV OXIDE IS 4.122/44 WHICH EQUALS 0.094. SINCE THERE IS ONLY ONE ATOM OF CARBON IN CO2 THEN THEY HAVE EQUAL NUMBER OF MOLES AND THUS THE NUMBER OF MOLES OF CARBON IS 0.094. WE CAN THEN PROCEED TO CALCULATE THE MASS OF CO2 PRESENT. THIS CAN BE CALCULATED BY MULTIPLYING THE NUMBER OF MOLES BY THE ATOMIC MASS UNIT. THE ATOMIC MASS UNIT OF CARBON IS 12. HENCE, THE MASS OF CO2 PRESENT IS 12 * 0.094 = 1.128g

WE CAN NOW GET THE MASS OF THE HYDROGEN BY MULTIPLYING THE NUMBER OF MOLES OF WATER BY 2 AND ALSO ITS ATOMIC MASS UNIT

TO GET THE NUMBER OF MOLES OF WATER, WE SIMPLY DIVIDE THE MASS BY THE MOLAR MASS. THE MOLAR MASS OF WATER IS 18g/mol.  The NUMBER OF MOLES IS THUS 1.350/18 = 0.075

THE NUMBER OF MOLES OF HYDROGEN IS TWICE THAT OF WATER SINCE IT CONTAINS 2 ATOMS PER MOLECULE OF WATER. ITS NUMBER OF MOLES IS THUS 0.075*2 = 0.15 MOLE

THE MASS OF HYDROGEN IS THUS 0.075 * 2 * 1 = 0.15g

WE CAN NOW FIND THE MASS OF OXYGEN BY SUBTRACTING THE MASSES OF HYDROGEN AND CARBON FROM THE TOTAL MASS.

MASS OF OXYGEN = 1.376-0.15-1.128 = 0.098g

THE NUMBER OF MOLES OF OXYGEN IS THUS 0.098/16 = 0.006125

WE CAN NOW USE THE NUMBER OF MOLES TO OBTAIN THE EMPIRICAL FORMULA.

WE DO THIS BY DIVIDING EACH BY THE SMALLEST NUMBER OF MOLES WHICH IS THAT OF THE OXYGEN.

C = 0.094/0.006125 = 15

H = 0.15/0.006125 = 24

O = 1

THE EMPIRICAL FORMULA IS THUS C15H24O

The combustion of 1.376 g of butylated hydroxytoluene (BHT) produced 4.122 g CO2 and 1.350 g H2O. Calculations yield an empirical formula of CH2O, indicating one carbon, two hydrogen, and one oxygen atom.

To determine the empirical formula of butylated hydroxytoluene (BHT), we can follow these steps:

1. **Find moles of CO2 and H2O produced:**

  \[ \text{moles of } CO_2 = \frac{\text{mass of } CO_2}{\text{molar mass of } CO_2} \]

  \[ \text{moles of } H_2O = \frac{\text{mass of } H_2O}{\text{molar mass of } H_2O} \]

2. **Find the mole ratio:**

  Divide the moles of each element (C, H, and O) in CO2 and H2O by the smallest number of moles.

3. **Write the empirical formula:**

  Use the mole ratios to write the empirical formula.

Let's perform the calculations:

\[ \text{Molar mass of } CO_2 = 12.01 \, \text{(C)} + 2 \times 16.00 \, \text{(O)} = 44.01 \, \text{g/mol} \]

\[ \text{Molar mass of } H_2O = 2 \times 1.01 \, \text{(H)} + 16.00 \, \text{(O)} = 18.02 \, \text{g/mol} \]

\[ \text{moles of } CO_2 = \frac{4.122 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.0938 \, \text{mol} \]

\[ \text{moles of } H_2O = \frac{1.350 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.0749 \, \text{mol} \]

Divide by the smallest number of moles (0.0749) to get a ratio close to 1:1:

\[ \text{C} : \text{H} : \text{O} \approx 1.25 : 1 : 1 \]

The ratio is approximately 1:1:1, so the empirical formula is CH2O.

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Answer:

Two  

Step-by-step explanation:

There are two ways to determine the number sets of equivalent protons in a compound.  

(a) By substitution

Replace each H with a different group and see if you get a different compound

For example, the six CH₃ H atoms are an equivalent set, because, if I replace any one of them, I will always get the same compound: 1-chloromethyl-4-methylbenzene (Fig. 1).

Similarly, the four aromatic H atoms are an equivalent set. If I replace any one of them, I will always get the same compound: 2-chloro-1,4-dimethylbenzene (Fig. 2).

Thus, there are two sets of equivalent protons in p-xylene.

(b) By symmetry

Two atoms are equivalent if one can be converted into the other by a symmetry operation on the molecule.

The methyl hydrogens can be interconverted by a combination of reflections about the AB and CD mirror planes and byrotations about the C-C bonds to the ring (Fig. 3).

Similarly, the four aromatic H atoms can be interconverted by a combination of reflections about the AB and CD mirror planes.

Again, we find two sets of equivalent protons.

Magnetizing steel is a physical change, not a chemical change because it does not change the chemical composition of the steel.

A chemical change is a change in which new substances are formed. This is because the atoms in the original substances are rearranged to form new atoms or molecules. There is no chemical change when magnetizing steel. For example, when iron rusts, it reacts with oxygen to form a new substance called iron oxide.

A physical change is a change in which the form of a substance is changed, but the substanceitself is not changed. For example, when you melt ice, you are changing the form of water from a solid to a liquid, but the water itself is not changed.

When you magnetize steel, you are simply aligning the magnetic domains in the steel. Magnetic domains are regions of a material that have a net magnetic field. When the magnetic domains are aligned, the steel becomes magnetized.

Magnetizing steel does not change the chemical composition of the steel. The steel is still made up of the same atoms, and the atoms are still arranged in the same way. The only difference is that the magnetic domains are aligned, which gives the steel its magnetic properties.

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