Answer:
1.32 mole
Explanation:
The following data were obtained from the question:
Volume of solution = 2.2L
Molarity of solution = 0.60M
Mole of Li3PO4 =..?
Molarity is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is represented as:
Molarity = mole /Volume
With the above formula we can easily calculate the number of mole of Li3PO4 as shown below:
Molarity =mole /Volume
0.6 = mole of Li3PO4 /2.2
Cross multiply
Mole of Li3PO4 = 0.6 x 2.2
Mole of Li3PO4 = 1.32 mole
Therefore, 1.32 mole of Li3PO4 is contained in the solution.
Answer:
To do this question. we first have to find the mass% of nitrogen in N2O and then, using that percentage, we can simply find the number of moles of N from the number of moles of N2O
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<u>Mass % of Nitrogen:</u>
Mass% of nitrogen = (Molar mass of N2 / Molar mass of N2O)*100
Mass% nitrogen = (28 / 44)*100
Mass% of Nitrogen = 0.63 * 100
Mass% nitrogen = 63%
<u>Mass of Nitrogen:</u>
So, now we can say that in any given mass of N2O. 63% of the total mass is the mass of Nitrogen
Hence, total mass * 63/100 = Mass of Nitrogen
Replacing the variables
40 * 0.63 = Mass of Nitrogen
Mass of Nitrogen = 25.2 grams
Answer: 41.2 atm
Explanation
To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,

where,
are the initial pressure and temperature of the gas.
are the final pressure and temperature of the gas.
We are given:

Putting values in above equation, we get:

The maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas is 41.2
Answer:
(3) 34.5grammes of C2H6O
(4) 5.56 x 10^22 Formula units
(5) 0.5 moles of CaCO3
Explanation:
(3) molar mass of C2H6O = 24 + 6 + 16 = 46 g/mol
mass of C2H6O =0.75 mole x 46 g/mole = 34.5 grammes
(4) Formula units of NaCl = (5.4/ 58.5) x 6.022x10^23 = 5.56 x 10^22
(4) moles = mass / molar mass
molar mass of CaCO3 = 40 + 12 + 48 = 100 g/mol
moles of CaCO3 = 50/100 = 0.5 moles
The answer is [OH⁻] = 1 x 10⁻⁸.
To find OH⁻, divide the ionic product of water by [H₃O⁺] as :
<u>OH⁻ + H₃O⁺ = H₂O</u>
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- [OH⁻] = 1 x 10⁻¹⁴ / 1 x 10⁻⁶
- [OH⁻] = 1 x 10⁻⁸