Answer:
124.56 moles of Hydrogen atoms.
Explanation:
We'll begin by calculating the number of moles of ethane that contains 1.25×10²⁵ molecules. This can be obtained as follow:
From Avogadro's hypothesis, 1 mole of any substance contains 6.02x10²³ molecules. This implies that 1 mole of ethane also contains 6.02x10²³ molecules.
Thus, 6.02x10²³ molecules are present in 1 mole of ethane.
Therefore, 1.25×10²⁵ molecules are present in = 1.25×10²⁵/6.02x10²³ = 20.76
Therefore, 20.76 moles of ethane contains 1.25×10²⁵ molecules.
Finally, we shall determine the number of mole of Hydrogen in 20.76 moles of ethane. This can be obtained as follow:
Ethane has formula as C2H6.
From the formula, 1 mole of ethane, C2H6 contains 6 moles of Hydrogen atoms.
Therefore, 20.76 moles of ethane will contain = 20.76 × 6 = 124.56 moles of Hydrogen atoms.
Therefore, 1.25×10²⁵ molecules of ethane contains 124.56 moles of Hydrogen atoms.
The most possible answer is <span>electronic configuration i had the same question in </span>Page 23, Chapter 3: Ionic and Covalent Compounds. hope my answer was helpful;)
Answer:
48.5 kJ
Explanation:
Let's consider the sublimation of carbon dioxide at its sublimation temperature, that is, its change from the solid to the gaseous state.
CO₂(s) → CO₂(g)
The molar mass of carbon dioxide is 44.01 g/mol. The moles corresponding to 66.0 g are:
66.0 g × (1 mol/44.01 g) = 1.50 mol
The heat of sublimation for carbon dioxide is 32.3 kJ/mol. The heat required to sublimate 1.50 moles of carbon dioxide is:
1.50 mol × (32.3 kJ/mol) = 48.5 kJ
Answer:
T = 42.08 °C
Explanation:
Using the expression,
Wherem
is the activation energy
R is Gas constant having value = 8.314 J / K mol
Thus, given that, = 45.6 kJ/mol = 45600 J/mol (As 1 kJ = 1000 J)
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (30 + 273.15) K = 303.15 K
So,
Conversion to °C as:
T(K) = T( °C) + 273.15
So,
315.23 = T( °C) + 273.15
<u>T = 42.08 °C</u>
Answer: Molarity of a solution : 0.416 M
In our question LiI(Lithium Iodide) is solute and we need to find the moles of LiI.
Given grams of LiI = 27.8g , Volume of solution = 500 mL
Molar mass of LiI = 133.85 g/mol
Moles of LiI = 0.208 mol
To find molarity we need volume of solution in 'L' , so we need to convert 500 mL to L.
1 L = 1000 mL
Volume = 0.5 L
Molarity = 
or 0.416 M
