The symbol is U and its atomic number is 92
<h3>
Answer:</h3>
= 5.79 × 10^19 molecules
<h3>
Explanation:</h3>
The molar mass of the compound is 312 g/mol
Mass of the compound is 30.0 mg equivalent to 0.030 g (1 g = 1000 mg)
We are required to calculate the number of molecules present
We will use the following steps;
<h3>Step 1: Calculate the number of moles of the compound </h3>

Therefore;
Moles of the compound will be;

= 9.615 × 10⁻5 mole
<h3>Step 2: Calculate the number of molecules present </h3>
Using the Avogadro's constant, 6.022 × 10^23
1 mole of a compound contains 6.022 × 10^23 molecules
Therefore;
9.615 × 10⁻5 moles of the compound will have ;
= 9.615 × 10⁻5 moles × 6.022 × 10^23 molecules
= 5.79 × 10^19 molecules
Therefore the compound contains 5.79 × 10^19 molecules
Answer:
44.8 L
Explanation:
Ideal Gas Equation -
i.e.,
PV = nRT
where,
P = pressure
V = volume
n = moles
R = universal gas constant
T = temperature
Using the information given in the question, Volume of the gas can be calculated -
P = 101.3 kPa
V = ?
n = 2.00 moles
R = 8.31
T = 0 degree C = 273.15 K
Using the above data, and putting the data in the respective formula -
PV = nRT
101.3 kPa * V = 2.00 moles * 8.31 * 273.15 K
V = 44.8 L
Hence, the volume of the given gas = 44.8 L
Mass of Oxygen required : 24 grams
<h3>Further explanation</h3>
Given
3 moles of H
1.5 moles of O
3 moles of H₂O
Required
Mass of O
Solution
Reaction
2H₂ + O₂ ⇒ 2H₂O
Mass of Oxygen for 1.5 moles of O :
= mol x Ar O
= 1.5 moles x 16 g/mol
= 24 grams
False; an ionic bond generally occurs between a nonmetal and a metal; for example, NaCl.