Answer:
Graduated cylinders are designed for accurate measurements of liquids with a much smaller error than beakers. They are thinner than a beaker, have many more graduation marks, and are designed to be within 0.5-1% error. ... Therefore, this more precise relative of the beaker is just as critical to almost every laboratory.
Explanation:
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Answer:
The amount of HC₂H3₃2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.
Explanation:
Equation of the reaction between acetic acid, HC₂H₃O₂(aq) and sodium hydroxide, NaOH(aq) is given below:
CH₃COOH (aq) + NaOH (aq) ----> CH₃COONa (aq) + H₂O
The equation of the reaction shows that acetic acid andsodium hydroxide will react in a 1:1 ratio
Since the concentration of NaOH was not given, we can assume that the concentration is 0.01 M
Moles of NaOH in 5.0 mL of 0.01 M NaOH = 0.01 × 5/1000 = 0.00005 moles
Moles of NaOH in 1.0 mL of 0.01 M NaOH = 0.01 ×1/1000 = 0.0001 moles
Ratio of moles of NaOH in 5.0 mL to 1.0 mL = 0.00005/0.00001 = 5
There are five times more moles of NaOH in 5.0 mL than in 1.0 mL and this means that 5 times more the quantity of HC₂H₃O2(aq) required to react with 1.0 mL NaoH is needed to react with 5.0 mL NaOH.
Therefore, the amount of HC₂H₃O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.
Answer:
V= 37.0 mL
Explanation:
First find the moles of the known substance (CuCl2)
n= cv
where
n is moles
c is concentration
v is volume ( in litres)
n= 0.108 × 0.0946
n=0.0102168
Using the mole ratio in the balanced reaction, we can find the moles of Na3PO4
n (Na3PO4)= n (CuCl2) × 2/3
=0.0102168 × 2/3
=0.0068112
Now we have all the necessary values to calculate the volume
v=n/c
v= 0.0068112/0.184
v= 0.0370173913 L
v= 37.0 mL
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