Answer:
<em> = 0.2 mL.</em>
Explanation:
Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:
where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.
Let C1 = 0.5M, V2 = ?
C2 = 0.010M; V2 = 10mL
⇒Volume of stock solution to be diluted, V2
= 
× 0.010
<em> = 0.2 mL.</em>
Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.
I hope this was a detailed explanation given the missing details of "Trial 1" in the question.
Answer: Out of the given options 
is expected to have the highest viscosity.
Explanation:
The resistance occurred in the flow of a liquid substance is called viscosity.
More stronger is the intermolecular forces present in a substance more will be its resistance in its flow. Hence, more will be its viscosity.
For example, has strong intermolecular hydrogen bonding than the one's present in 
and 
. This is because two-OH groups are present over here.
Thus, we can conclude that out of the given options is expected to have the highest viscosity.
Acetic acid activates the bromine and makes it a better electrophile.
<h3>What is bromination?</h3>
When a substance undergoes bromination, bromine is added to the compound as a result of the chemical reaction. After bromination, the result will have different properties from the initial reactant.
<h3>Why is 15M acetic acid used as a solvent for bromination?</h3>
DCM (dichloromethane) requires more time. Acetic acid has protons that can give one of the Br (bromine) a positive charge and activate it. There is a brief loss of aromaticity that calls for high energy activation.
Refer to the attached image for bromination reaction.
Learn more about bromination here:
brainly.com/question/26428023
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The final temperature : 345 K
<h3>
Further explanation
</h3>
Given
475 cm³ initial volume
600 cm³ final volume
Required
The final temperature
Solution
At standard temperature and pressure , T = 273 K and 1 atm
Charles's Law :
When the gas pressure is kept constant, the gas volume is proportional to the temperature
V₁/T₁=V₂/T₂
Input the value :
T₂=(V₂T₁)/V₁
T₂=(600 x 273)/475
T₂=345 K
%yield = 88.5%
<h3>Further explanation</h3>
Given
Reaction
Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)
Required
The percent yield
Solution
mol AgNO₃(MW=169,87 g/mol) :
= mass : MW
= 127 : 169.87
= 0.748
mol Ag from equation :
= 2/2 x mol AgNO₃
= 2/2 x 0.748
= 0.748
Mass Ag (theoretical) :
= mol x Ar Ag
= 0.748 x 108
= 80.784
% yield = (actual/theoretical) x 100%
%yield = 71.5/80.784 x 100%
<em>%yield = 88.5%</em>
