Answer:
25 mM Tris HCl and 0.1% w/v SDS
Explanation:
A <em>10X solution</em> is ten times more concentrated than a <em>1X solution</em>. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:
(250 mM Tris HCl)/10 = 25 mM Tris HCl
(1.92 M glycine)/10 = 0.192 M glycine
(1% w/v SDS)/10 = 0.1% w/v SDS
Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.
<span>The balanced chemical equation for this reaction is:
2NaOH (aq)+H2SO4 (aq) → Na2SO4 (aq)+2H2O (l)
According to question, 60 ml of NaOH solution was used for neutralizing 40 ml of 0.50M H2SO4.
The no. of moles of H2SO4 is calculated using the equation:
mol solute = (molarity) (L soln)
mol H2SO4 = 0.50 M x 0.040 L = 0.02 moles of H2SO4
As per the equation, the number of moles of NaOH used is:
0.02 moles of H2SO4 (2 mol NaOH) (1 mol H2SO4) = 0.04 moles of NaOH
Therefore, using the given volume of NaOH, the concentration or molarity of NaOH can be calculated using the formula :
Molarity = mol solute/L soln = 0.04 mol NaOH/0.06 L = 0.67 M
Therefore, the concentration of NaOH is 0.67 M.</span>
Answer:
An observation
Explanation: observation can also involve the perception and recording of data via the use of scientific instruments. The term may also refer to any data collected during the scientific activity
Your bond Line notation is missing a carbon.
