The molarity is count by dividing the mole of the solute within 1 liter of solvent. In this case, the KNO3 is 16.8g with 101.11 g/mol molar mass. Then we need to find the mol first. The calculation would be: 16.8g / (101.11g/mol)= 0.0166 mol.
Then the molarity would be: 0.0166mol/ 0.3l= 0.0498= 0.0553 M
Answer:
34.3 g
Explanation:
Step 1: Write the balanced equation
2 CH₃CH₂OH ⇒ CH₃CH₂OCH₂CH₃ + H₂O
Step 2: Calculate the moles corresponding to 50.0 g of CH₃CH₂OH
The molar mass of CH₃CH₂OH is 46.07 g/mol.
50.0 g × 1 mol/46.07 g = 1.09 mol
Step 3: Calculate the theoretical moles of CH₃CH₂OCH₂CH₃ produced
The molar ratio of CH₃CH₂OH to CH₃CH₂OCH₂CH₃ is 2:1. The moles of CH₃CH₂OCH₂CH₃ theoretically produced are 1/2 × 1.09 mol = 0.545 mol.
Step 4: Calculate the real moles of CH₃CH₂OCH₂CH₃ produced
The percent yield of the reaction is 85%.
0.545 mol × 85% = 0.463 mol
Step 5: Calculate the mass corresponding to 0.463 moles of CH₃CH₂OCH₂CH₃
The molar mass of CH₃CH₂OCH₂CH₃ is 74.12 g/mol.
0.463 mol × 74.12 g/mol = 34.3 g
Answer: In metallic bonds, the mobile electrons surrounding the positive ions are called <u><em>dipole</em></u>.
Answer:
3.5 × 10⁵ g of salt
Explanation:
<em>What is the mass (grams) of salt in 10.0 m³ of ocean water?</em>
We have this data:
- 1.000 mol salt is equal to 58.44 g salt
- 1.0 L of ocean water contains 0.60 mol of salt
We will need the following relations:
We can use proportions:
Answer:
All the numbers in a chemical formula are significant, It is because the numbers in a chemical formula denote the number of different atoms present in the particular compound.
For example in H2SO4 there are 2 H 1 S and 4 O. This means 1 H2SO4 has 2 hydrogen 1 sulphur and 4 oxygen.
