<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg
<em>Molar concentration
</em>
Assume you have 1 L solution.
Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)
= 1190 g solution
Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)
= 84.01 g NaHCO3
Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)
= 1.14 mol NaHCO3
<em>c</em> = 1.14 mol/1 L = 1.14 mol/L
<em>Molal concentration</em>
Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg
<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg
the largest is francium, Z=87 , or caesium, Z=55 ; and the smallest is definitely helium, Z=2 .
Explanation:
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Answer:
1.67 gradius Celcius
Explanation:
Please see the step-by-step solution in the picture attached below.
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Answer:
Its passive
Explanation:
Facilitated diffusion is a type of passive transport that allows substances to cross membranes with the assistance of special transport proteins.
The number of moles in 3.20 x 10² formula units of calcium iodide is 0.053 moles.
<h3>How to calculate number of moles?</h3>
The number of moles in the formula units of a substance is calculated by dividing the formula unit by Avogadro's number.
According to this question, 3.20 x 10² formula units are in calcium iodide. The number of moles is as follows:
no of moles = 3.20 x 10²² ÷ 6.02 × 10²³
no of moles = 0.53 × 10-¹
no of moles = 0.053 moles
Therefore, the number of moles in 3.20 x 10² formula units of calcium iodide is 0.053 moles.
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