Answer:
3.25×10²⁴ molecules
Explanation:
From the question given above, the following data were obtained:
Mass of H₂O = 97.2 g
Number of molecule of H₂O =?
From Avogadro's hypothesis, we understood that:
1 mole of H₂O = 6.02×10²³ molecules
Next, we shall determine the mass of 1 mole of H₂O. This can be obtained as follow:
1 mole of H₂O = (2×1) + 16
= 2 + 16
= 18 g
Thus,
18 g of H₂O = 6.02×10²³ molecules
Finally, we shall determine the number of molecules in 97.2 g of H₂O. This can be obtained as follow:
18 g of H₂O = 6.02×10²³ molecules
Therefore,
97.2 g of H₂O = 97.2 × 6.02×10²³ / 18
97.2 g of H₂O = 3.25×10²⁴ molecules
Thus, 97.2 g of H₂O contains 3.25×10²⁴ molecules.
<span>polar bonds due to high difference . In electronegativity of oxygen and hydrogen.</span>
<span>1 meter = 100 cm
so 1cm is 1/100 *100 = 1% error.
1 kilogram = 1000g
so 1g is 1/1000 * 100 = 0.1% error.
as 1%>0.1%, the 1cm in a meter is the greater error</span><span>
</span>
Answer:
Option (A) : Positive
Explanation:
A combustion reaction will always be associated with a change in entropy that is Positive due to the gaseous products released. Hence, there is a large positive entropy change.
This problem is being solved using Ideal Gas Equation.
PV = nRT
Data Given:
Initial Temperature = T₁ = 27 °C = 300 K
Initial Pressure = P₁ = constant
Initial Volume = V₁ = 8 L
Final Temperature = T₂ = 78 °C = 351 K
Final Pressure = P₂ = constant
Final Volume = V₂ = ?
As,
Gas constant R and Pressures are constant, so, Ideal gas equation can be written as,
V₁ / T₁ = V₂ / T₂
Solving for V₂,
V₂ = (V₁ × T₂) ÷ T₁
Putting Values,
V₂ = (8 L × 351 K) ÷ 300 K
V₂ = 9.38 L
