Answer:
184.113 g/mol
Explanation: The atomic mass of Mg is 24.3 amu. The atomic mass of bromine is 79.9. Therefore, the formula weight of MgBr2 equals 24.3 amu + (2 × 79.9 amu), or 184.1 amu. Because a substance's molar mass has the same numerical value as its formula weight, the molar mass of MgBr2 equals 184.1 g/mol.
Answer:
4190.22 L = 4.19 m³.
Explanation:
- For the balanced reaction:
<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>
It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>
- Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:
no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.
- Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:
<u><em>Using cross multiplication:</em></u>
5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.
??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.
∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.
- Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 60.95 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).
∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.
Answer:
no, and next time take it right
Explanation:
Periodic Trend:
The Atomic radius of atoms generally decreases from left to right across a period
Group Trend:
The atomic radius of atoms generally increases from top to bottom within a group. As atomic number increases down a group, there is a increase in the positive nuclear charge, however the co-occurring increase in the number of orbitals wins out, increasing the atomic radius down a group in the periodic table
Answer :
The Atom with the greatest atomic radius is chlorine. Fluorine can be ruled out because it is in the same period as oxygen and further to the right down the period. Chlorine has the largest atomic size because it is farthest down the group of any of the above elements listed.
