Answer:
3.91 moles of Neon
Explanation:
According to Avogadro's Law, same volume of any gas at standard temperature (273.15 K or O °C) and pressure (1 atm) will occupy same volume. And one mole of any Ideal gas occupies 22.4 dm³ (1 dm³ = 1 L).
Data Given:
n = moles = <u>???</u>
V = Volume = 87.6 L
Solution:
As 22.4 L volume is occupied by one mole of gas then the 16.8 L of this gas will contain....
= ( 1 mole × 87.6 L) ÷ 22.4 L
= 3.91 moles
<h3>2nd Method:</h3>
Assuming that the gas is acting ideally, hence, applying ideal gas equation.
P V = n R T ∴ R = 0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹
Solving for n,
n = P V / R T
Putting values,
n = (1 atm × 87.6 L)/(0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹ × 273.15K)
n = 3.91 moles
Result:
87.6 L of Neon gas will contain 3.91 moles at standard temperature and pressure.
<u>Given information:</u>
Concentration of HCl = 0.035 M
<u>To determine:</u>
pH of the solution
<u>Explanation:</u>
Hydrochloric acid, HCl is a strong acid. It will completely dissociate to give H+ and Cl- ions
HCl → H+ + Cl-
Hence the concentration of H+ = Cl- = 0.035M
Now, pH measures the strength of H+ ions in a given solution. It is expressed as:
pH = -log[H+]
pH (HCl) = -log(0.035) = 1.46
Ans: pH of 0.035M HCl is 1.46
Precipitation occurs when the product of the ion concentration exceeds the Ksp.
Answer:
C = 18.29 g
Explanation:
Given data:
Mass of beryllium needed = ?
Mass of nitrogen = 18.9 g
Solution:
Chemical equation:
3Be + N₂ → Be₃N₂
now we will calculate the number of moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 18.9 g/ 28 g/mol
Number of moles = 0.675 mol
Now we will compare the moles of nitrogen and Be from balance chemical equation.
N₂ : Be
1 : 3
0.675 : 3/1×0.675 = 2.03 mol
Number of moles of Be needed are 2.03 mol.
Mass of Beryllium:
Mass = number of moles × molar mass
Mass = 2.03 mol × 9.01 g/mol
Mass = 18.29 g
Given:
0.060 mol of NiC2O4
Ksp = 4 x 10⁻¹⁰
1.0 L of solution
Kf of Ni(NH3)6 2⁺ = 1.2 x 10⁹
<span>NiC2O4 + 6NH3 ⇋ Ni(NH3)6 2+ + 2O4 2- </span>
<span>NiC2O4 ⇋ Ni 2+ + C2O4 2- ...Ksp </span>
<span>Ni2+ + 6NH3 ⇋ Ni(NH3)6 2+...Kf </span>
Ksp * Kf = (4 x 10⁻¹⁰) * (1.2 x 10⁹) = 0.48
K = 0.48 = [Ni(NH3)6 2+][C2O4 2-] / [NH3]⁶<span>
</span>0.48 = (0.060)² / [NH3]⁶<span> ... (dissolved C2O4 2- = 0.060M)
</span><span>[NH3]</span>⁶<span> = (0.060)</span>²<span> / 0.48 = </span>0.0036 / 0.48 = 0.0075
NH3 = ⁶√0.0075
NH3 = 0.44 M
