Answer:
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Answer:
Explanation:
1 mol of ideal gas at STP occupies 22.4 (or 22.7 depending on the convention being used for STP) liters in volume. I will use 22.4 so 17.88*22.4 = 400.5 L
Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.
Given :
Energy , E = 330 J .
Initial temperature , 
.
Final temperature , 
.
Mass of benzene , m = 24.6 g .
To Find :
The molar hear capacity of benzene at constant pressure .
Solution :
Molecular mass of benzene , M = 78 g/mol .
Number of moles of benzene :
Energy required is given by :
Hence , this is the required solution .
