Answer:
![\boxed {\boxed {\sf 2.76 \ mol}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%202.76%20%5C%20mol%7D%7D)
Explanation:
Molarity is found by dividing the moles of solute by liters of solution.
![molarity = \frac {moles}{liters}](https://tex.z-dn.net/?f=molarity%20%3D%20%5Cfrac%20%7Bmoles%7D%7Bliters%7D)
We know the molarity is 1.2 M (mol\liter) and there are 2.3 liters of solution. Substitute the known values into the formula.
![1.2 \ mol/liter= \frac {x}{2.3 \ liters}](https://tex.z-dn.net/?f=1.2%20%5C%20mol%2Fliter%3D%20%5Cfrac%20%7Bx%7D%7B2.3%20%5C%20liters%7D)
Since we are solving for x, we must isolate the variable. It is being divided by 2.3 and the inverse of division is multiplication. Multiply both sides by 2.3 liters.
![2.3 \ liters *1.2 \ mol/liter= \frac {x}{2.3 \ liters}* 2.3 \ liters\\2.3 *1.2 \ mol= x\\2.76 \ mol =x](https://tex.z-dn.net/?f=2.3%20%5C%20liters%20%2A1.2%20%5C%20mol%2Fliter%3D%20%5Cfrac%20%7Bx%7D%7B2.3%20%5C%20liters%7D%2A%202.3%20%5C%20liters%5C%5C2.3%20%2A1.2%20%5C%20mol%3D%20x%5C%5C2.76%20%5C%20mol%20%3Dx)
In a solution with a molarity of 1.2 and 2.3 liters of solution, there are 2.76 moles.
Difference in density between the two liquids
Answer:
13.94moles of Na₂O
Explanation:
The balanced reaction expression is given as:
4Na + O₂ → 2Na₂O
Given parameters:
Number of moles of O₂ = 6.97moles
Unknown:
Number of moles of Na₂O
Solution:
To solve this problem;
1 mole of O₂ will produce 2 moles of Na₂O ;
6.97 moles of O₂ will produce 6.97 x 2 = 13.94moles of Na₂O
Answer:
The molar mass of a substance is defined as the mass in grams of 1 mole of that substance. One mole of isotopically pure carbon-12 has a mass of 12 g. ... That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 1023 atoms, molecules, or formula units of that substance.
Explanation:
Answer:
63.05% of MgCO3.3H2O by mass
Explanation:
<em>of MgCO3.3H2O in the mixture?</em>
The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:
<em>Mass water:</em>
3.883g - 2.927g = 0.956g water
<em>Moles water -18.01g/mol-</em>
0.956g water * (1mol/18.01g) = 0.05308 moles H2O.
<em>Moles MgCO3.3H2O:</em>
0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =
0.01769 moles MgCO3.3H2O
<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>
0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O
<em>Mass percent:</em>
2.448g MgCO3.3H2O / 3.883g Mixture * 100 =
<h3>63.05% of MgCO3.3H2O by mass</h3>