Answer:
E - Be and O
A - Mg and N
E - Li and Br
F - Ba and Cl
B - Rb and O
Explanation:
Be and O
Be is a metal that loses 2 e⁻ to form Be²⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form BeO (E-MX).
Mg and N
Mg is a metal that loses 2 e⁻ to form Mg²⁺ and N is a nonmetal that gains 3 e⁻ to form O³⁻. For the ionic compound to be neutral, it must have the form Mg₃N₂ (A-M₃X₂).
Li and Br
Li is a metal that loses 1 e⁻ to form Li⁺ and Br is a nonmetal that gains 1 e⁻ to form Br⁻. For the ionic compound to be neutral, it must have the form LiBr (E-MX).
Ba and Cl
Ba is a metal that loses 2 e⁻ to form Ba²⁺ and Cl is a nonmetal that gains 1 e⁻ to form Cl⁻. For the ionic compound to be neutral, it must have the form BaCl₂ (F-MX₂).
Rb and O
Rb is a metal that loses 1 e⁻ to form Rb⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form Rb₂O (B-M₂X).
Answer:
0.02 moles.
Explanation:
volume of H₂ gas at R.T.P = 480 cm³
Where
R.T.P = room temperature and pressure
molar volume of gas at = 24000 cm³
no. of moles of hydrogen = ?
Solution:
formula Used
no. of moles = volume of gas / molar volume
put values in above equation
no. of moles = 480 cm³ / 24000 cm³/mol
no. of moles = 0.02 mol
So,
no. of moles of hydrogen in 480 cm³ is 0.02 moles.
Answer:
The correct answer is 10.939 mol ≅ 10.94 mol
Explanation:
According to Avogadro's gases law, the number of moles of an ideal gas (n) at constant pressure and temperature, is directly proportional to the volume (V).
For the initial gas (1), we have:
n₁= 1.59 mol
V₁= 641 mL= 0.641 L
For the final gas (2), we have:
V₂: 4.41 L
The relation between 1 and 2 is given by:
n₁/V₁ = n₂/V₂
We calculate n₂ as follows:
n₂= (n₁/V₁) x V₂ = (1.59 mol/0.641 L) x 4.41 L = 10.939 mol ≅ 10.94 mol
Density, Volume and Mass
3. A metal weighing 7.101 g is placed in a graduated cylinder containing 33.0 mL of water. The water
level rose to the 37.4 mL mark.
a) Calculate the density of the metal (in g/mL).
b) If you were to do this with an equal mass of aluminum (d = 2.7 g/mL), how high would the water rise?
