This will be classified as light on the API scale due to the large percentage of lighter fractions such as paraffins and naphthenes.
Answer:
164.3g of NaCl
Explanation:
Based on the chemical equation:
CaCl2 + 2NaOH → 2NaCl + Ca(OH)2
<em>where 1 mole of CaCl2 reacts with 2 moles of NaOH</em>
To solve this question we must convert the mass of CaCl2 to moles. Using the chemical equation we can find the moles of NaCl and its mass:
<em>Moles CaCl2 -Molar mass: 110.98g/mol-</em>
156.0g CaCl₂ * (1mol / 110.98g) = 1.4057 moles CaCl2
<em>Moles NaCl:</em>
1.4057 moles CaCl2 * (2mol NaCl / 1mol CaCl2) = 2.811 moles NaCl
<em>Mass NaCl -Molar mass: 58.44g/mol-</em>
2.811 moles NaCl * (58.44g / mol) = 164.3g of NaCl
55.9 kPa; Variables given = volume (V), moles (n), temperature (T)
We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:
<em>pV = nRT</em>
Solve for <em>p</em>: <em>p = nRT/V</em>
R = 8.314 kPa.L.K^(-1).mol^(-1)
<em>T</em> = (265 + 273.15) K = 538.15 K
<em>V</em> = 500.0 mL = 0.5000 L
∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa
Answer:
The answer is "
"
Explanation:
Given:
Molarity= number of moles
because it is 1 Liter
therefore,
it takes 20 mL of Tris.
So, take 
