100.0 g of liquid copper (molar mass 63.546 g/mol; melting point 1358 K; density 8.02 g/mL) is placed in a rigid container of vo
1 answer:
Answer:
8.912x10^-18
Explanation:
-dn/dt = pANa/2piMRT
100 g = initial copper
Number of moles = 100/63.546
= 1.5736
Mass of copper left = 100-10.0168
= 89.9832
Moles = 89.9832/63.546
= 1.4160
dn = 1.4160-1.5736
= -0.1576
dt = 2 hrs
A = 3.23mm² = 3.23x10^-6
M = 63.546
T = 0.0821
T = 1508k
Na = 6.023x10²³
When we insert all these into the formula above
We get
P = 8.912x10^-18atm
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Answer:- 2.39 mL are required.
Solution:- It's a dilution problem and to solve this type of problems we use the dilution equation:
Where, and
are molarities of concentrated and diluted solutions and
and
are their respective volumes.
= 1.10M
= 5.00mM = 0.005M (since, mM stands for milli molar and M stands for molar. 1M = 1000mM)
= ?
= 525 mL
Let's plug in the given values in the formula:
So, 2.39 mL of 1.10M are needed to make 525 mL of 5.00mM solution.