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9966 [12]
3 years ago
5

Which of the following has the greatest mass? 1 mole of uranium (U) 1 mole of oxygen (O) 1 mole of sodium (Na) 1 mole of chlorin

e (Cl)
Chemistry
2 answers:
ELEN [110]3 years ago
4 0

Answer:

Uranium

Explanation:

Since all of the options have only 1 mole of each, simply compare their masses to find the greatest one. You'll find that Uranium is the greatest with about 238 amu.

Lelu [443]3 years ago
3 0
To determine which has the greatest mass, we need to find out the molecular mass of each element:

Uranium (U) = 238 g/mol
Oxygen (O) = 16 g/mol
Sodium (Na) = 23 g/mol
Chlorine (Cl) = 35.45 g/mol

Therefore, if we have one mole of each element, the element with the greatest mass is 238 grams or <u>Uranium.</u><span />
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A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.
sleet_krkn [62]

Explanation:

The given data is as follows.

      Pressure (P) = 760 torr = 1 atm

      Volume (V) = 720 cm^{3} = 0.720 L

     Temperature (T) = 25^{o}C = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

                                PV = nRT

   Total atoms present (n) = \frac{PV}{RT}

                                          = 1 \times \frac{0.720 L}{0.0821 \times 298}

                                           = 0.0294 mol

Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

               x + y = 0.0294

Also,      40x + 131y = 2.966

             x = 0.0097 mol

              y = (0.0294 - 0.0097)

                = 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = \frac{y}{(x+y)}

                               = \frac{0.0197}{0.0294}

                              = 0.67

Therefore, the mole fraction of Xe is 0.67.

6 0
3 years ago
The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO
Irina-Kira [14]

Answer:

Amount of excess Carbon (ii) oxide left over = 23.75 g

Explanation:

Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂

Molar mass of Fe₂O₃ = 160 g/mol;

Molar mass of Carbon (ii) oxide = 28 g/mol

From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g)  of carbon (ii) oxide

450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide

Therefore the excess reactant is carbon (ii) oxide.

Amount of excess Carbon (ii) oxide left over = 260 - 236.25

Amount of excess Carbon (ii) oxide left over = 23.75 g

5 0
3 years ago
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