You should never heat chemicals in a closed system (closed bottle or flask) because it might explode.
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Answer: Tides are the regular, alternating rise and fall of sea level caused by the gravitational pull of the moon and sun. The changing of the tide is often rapid and dramatic. On a smaller scale, similar motions occur on large lakes, in the atmosphere, and even within the solid earth.
Explanation: can I please have brainliest?
Answer:
Explanation:
The gas ideal law is
PV= nRT (equation 1)
Where:
P = pressure
R = gas constant
T = temperature
n= moles of substance
V = volume
Working with equation 1 we can get
The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.
or
(equation 2)
The cylindrical container has a constant pressure p
The volume is the volume of a cylinder this is
Where:
r = radius
h = height
(pi) = number pi (3.1415)
This cylinder has a radius, r and height, h so the volume is
Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:
Replacing these values in the equation 2 we get:
(equation 2)
The reaction between NaOH and Cu(NO₃)₂ is as follows
2NaOH + Cu(NO₃)₂ ---> 2NaNO₃ + Cu(OH)₂
Q1)
stoichiometry of NaOH to Cu(NO₃)₂ is 2:1
this means that 2 mol of NaOH reacts with 1 mol of Cu(NO₃)₂
the mass of Cu(NO₃)₂ reacted - 0.8024 g
molar mass of Cu(NO₃)₂ is 187.56 g/mol
therefore the number of Cu(NO₃)₂ moles that have reacted
- 0.8024 g/ 187.56 g/mol = 0.00427 mol
according to the stoichiometry , number of NaOH moles - 0.00427 mol x 2
then number of NaOH moles that have reacted - 0.00855 mol
In a 3.0 M NaOH solution, 3 moles are in 1000 mL of solution
Then volume required for 0.00855 mol - 1000 x 0.00855 /3 = 2.85 mL
2.85 mL of 3.0 M NaOH is required for this reaction
Q2)
Assuming that there's 100 % yield of Cu(OH)₂ , we can directly calculate the mass of Cu(OH)₂ formed from the number of moles of reactants that were used up.
Stoichiometry of Cu(NO₃)₂ to Cu(OH)₂ is 1:1
this means that 1 mol of Cu(NO₃)₂ gives a yield of 1 mol of Cu(OH)₂
the number of Cu(NO₃)₂ moles that reacted - 0.00427 mol
Therefore an equal amount of moles of Cu(OH)₂ were formed
Then amount of Cu(OH)₂ moles produced - 0.00427 mol
Mass of Cu(OH)₂ formed - 0.00427 mol x 97.56 g/mol = 0.42 g
A mass of 0.42 g of Cu(OH)₂ was formed in this reaction
