It is called the Fertilization
Given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.
<h3>How to calculate mass of substances?</h3>
The mass of a substance can be calculated using the following steps:
Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
1 mole of Cu react with 2 moles of AgNO3
- Molar mass of AgNO3 = 169.87 g/mol
- Molar mass of Cu = 63.5g/mol
moles of AgNO3 = 262g/169.87g/mol = 1.54mol
1.54 moles of AgNO3 will react with 0.77 moles of Cu.
mass of Cu = 0.77 × 63.5 = 48.97g
Therefore, given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.
Learn more about mass at: brainly.com/question/6876669
Answer:
Explanation:
The density of a gas can be obtained using the gas ideal equation and the molar mass of the gas.
This is the decution of the final formula:
Now, you just need to substitute values:
- R = 0.08206 atm-liter / k-mol
- d = 32.0 g/mol × 0.9869 atm / [0.08206 atm-liter/k-mol × 273.15K]
- d = 1.4 g/liter (using two significant figures)
As you see, I have not used the 4.8 grams datum. That is because the density of the gases may be calculated from the temperature, pressure and molar mass of the gas, using the ideal gas equation.
Since, you have the mass of gas, you might use this other procedure:
- Volume of 1 mol of gas at STP: about 22.4 liter/mol
- Mass of 1 mol of oxygen gas: 32.0 g/mol (the molar mass)
- number of moles in 4.8 g of oxygen = 4.8 g / 32.0 g/mol = 0.15 mol
- Volume of 0.15 mol of oxygen: 0.15 mol × 22.4 liter/mol = 3.36 liter
- Density = mass / volume = 4.8 g / 3.36 liter = 1.4 g/liter (same result)
<u>Answer:</u> The mass of iron (III) nitrate is 11.16 g/mol
<u>Explanation:</u>
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
We are given:
Molarity of solution = 0.3556 M
Molar mass of Iron (III) nitrate = 241.86 g/mol
Volume of solution = 129.8 mL
Putting values in above equation, we get:
Hence, the mass of iron (III) nitrate is 11.16 g/mol