Answer:
30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.
Explanation:
M1 = 3.00 M
M2 = 1.80 M
V2 = 50 .0 mL = 50 /1000 L = 0.05 L
V1 = unknown
In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.
So we use the equation:
M1 V1 = M2 V2
V1 = M2 V2 / M1
V2 = 1.80 * 0.05 / 3.0
V2 = 0.09 /3.0
V2 = 0.03 L or 30 mL
To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.
Answer:
d) cut the large sized Cu solid into smaller sized pieces
Explanation:
The aim of the question is to select the right condition for that would increases the rate of the reaction.
a) use a large sized piece of the solid Cu
This option is wrong. Reducing the surface area decreases the reaction rate.
b) lower the initial temperature below 25 °C for the liquid reactant, HNO3
Hugher temperatures leads to faster reactions hence this option is wrong.
c) use a 0.5 M HNO3 instead of 2.0 M HNO3
Higher concentration leads to increased rate of reaction. Hence this option is wrong.
d) cut the large sized Cu solid into smaller sized pieces
This leads to an increased surface area of the reactants, which leads to an increased rate of the reaction. This is the correct option.
The partial pressure of hydrogen is 0.31 atm
calculation
find the number of hydrogen moles the container, that is
25/100 x 6.4 =1.6 moles of hydrogen
find the partial pressure for hydrogen in 1.6 moles
that is 6.4 moles= 1.24 atm
1.6 moles= ?
by cross multiplication
1.6moles x1.24 atm/ 6.4 moles= 0.31 atm
Radioactivity another name for radioactive decay. Radioactivity refers to particles emitted from nuclei as a result of nuclear instability.
