A sample of matter must be copper if(1) each atom in the sample has 29 protons(2) atoms in the sample react with oxygen
(3) the sample melts at 1768 K
(4) the sample can conduct electricity

87.75 grams of sodium chloride (NaCl) are required to prepare 500. mL of 3.00 M solution.

What is Mole?

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter.

A mole is defined as the amount of substance containing the same number of atoms, molecules, ions, etc. as the number of atoms in a sample of pure 12C weighing exactly 12 g.

Given,

Volume of NaCl = 500ml

Concentration = 3M

Moles = Concentration × volume

= 3 × 0.5

= 1.5 moles

Mass = moles × molar mass

= 1.5 × 58.5

= 87.75 g

Therefore, 87.75 grams of sodium chloride (NaCl) are required to prepare 500. mL of 3.00 M solution.

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The mass mass number is 22 and the charge of the atom is +2. The correct answer is option A

How to determine the mass number and charge?

1. The mass number of the atom can be obtained as follow:

• Number of protons = 12
• Number of neutrons = 12
• Mass number =?

Mass number = number of proton + number of neutron

= 12 + 12

= 24

2. The charge of the atom can be obtained as follow:

• Number of protons = 12
• Number of electrons = 10
• Charge =?

Charge = number of protons - number of electrons

= 12 - 10

= +2

Thus, the mass number is 24 and the charge is +2. The correct answer to the question is option A

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Answer: Option is the correct answer which is 4) Empirical, structural, molecular

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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