There are five states of matter out of which we encounter three states of matter in our day today life
a) gas b) solid and c) liquid
the main difference between the three is of
a) Inter molecular forces of attraction
b) thermal energy
due to this
a) solid has high intermolecular forces and low thermal energy: thus they have fix shape and occupy fix volume
b) liquid has intermediate forces and medium themal energy. Thus they may have fixed volume and but no fix shape
c) gas has weak intermoelcular forces and high thermal energy. thus they have no fixed volume no fix shape
so in the given problem
the state of the substance D- Gas.
C-c-c-c-c
|
c
c-c-c-c-c
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c
c
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c-c-c-c
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c
c-c-c-c
| |
c c
Chemical change occur when two substances are combined and produces a new substance or decomposes into two or more substances which are entirely different from the original two substances.
There are three types of chemical changes. These are 1) Inorganic Changes, 2) Organic Changes, and 3) Biochemical Changes
Here are some examples of chemicsal changes.
If you combine Sodium and Water, chemical changes causes decomposition into Sodium Hydroxide and Hydrogen.
Sodium + Water ==> Sodium Hydroxide and Hydrogen
Na + H2O ====> NaOH and H
Another example of chemical change is:
Carbon Dioxide and Water will decompose into Sugar and Oxygen
Carbon Dioxide + Water ==> Sugar and Oxygen
CO2 + H2O ==> CnH2nOn (where n is between 3 and 7) and O
Answer:
Water pressure 0.5 atm
Total Pressure= 2.27 atm
Explanation:
To answer this problem, one has to realize that there are two processes that increase the temperature of the sealed vessel.
First, the dry air in the sealed vessel will be heated which will cause its pressure to increase and it can be determined by the equation:
P₁ x T₂ = P₂ x T₁ ∴ P₂ = P₁ x T₂ / T₁
For the second process, we have an amount of n moles of water which will be released when the copper sulfate is heated. In this case, to determine the value of the the water gas we will use the gas law:
PV = nRT ∴ P = nRT/V
n will we calculated from the quantity of sample.
2.50 g CuSo₄ 5H₂O x 1 mol/ 249.69 g = 0.01 mol CuSo₄ 5H₂O
the amount water of hydration is
= 0.01 mol CuSo₄ 5H₂O * 5 mol H₂O / 1 mol CuSo₄ 5H₂O
= 0.05 mo H₂O
pressure of dry air at the final temperature,
P₂ = 1 atm x 500 K/ 300 K = 1.67 atm
Pressure of water :
P (H₂O) 0.05 mol x 0.08206 Latm/kmol x 500 K/ 4 L = 0.5 atm
∴ Total Pressure = 1.67 atm
H2O Pressure = 0.5 atm
Answer:
0.032 L or 32 mL
Explanation:
Use the dilution equation M1V1 = M2V2
M1 = 9.0 M
V1 = This is what we're looking for.
M2 = 0.145 M
V2 = 2 L
Solve for V1 --> V1 = M2V2/M1
V1 = (0.145 M)(2 L) / (9.0 M) = 0.032 L
