Answer:
Here are a few more examples:
Smoke and fog (Smog)
Dirt and water (Mud)
Sand, water and gravel (Cement)
Water and salt (Sea water)
Potassium nitrate, sulfur, and carbon (Gunpowder)
Oxygen and water (Sea foam)
Petroleum, hydrocarbons, and fuel additives (Gasoline)
Heterogeneous mixtures possess different properties and compositions in various parts i.e. the properties are not uniform throughout the mixture.
Examples of Heterogeneous mixtures – air, oil, and water, etc.
Examples of Homogeneous mixtures – alloys, salt, and water, alcohol in water, etc.
Explanation:
Answer:
HCO₂
Explanation:
From the information given:
The mass of the elements are:
Carbon C = 26.7 g; Hydrogen H = 2.24 g Oxygen O = 71.1 g
To determine the empirical formula;
First thing is to find the numbers of moles of each atom.
For Carbon:

For Hydrogen:

For Oxygen:

Now; we use the smallest no of moles to divide the respective moles from above.
For carbon:

For Hydrogen:

For Oxygen:

Thus, the empirical formula is HCO₂
Answer:
solute and solvent
Explanation:
Because solutes dissolves in solvent to form a solution
Take 15/100 X 75 = The answer
Answer:
HCl
Explanation:
Given data:
Mass of Zn = 50 g
Mass of HCl = 50 g
Limiting reactant = ?
Solution:
Chemical equation:
Zn + 2HCl → ZnCl₂ + H₂
Number of moles of Zn:
Number of moles = mass / molar mass
Number of moles = 50 g/ 65.38 g/mol
Number of moles = 0.76 mol
Number of moles of HCl:
Number of moles = mass / molar mass
Number of moles = 50 g/ 36.5 g/mol
Number of moles = 1.4 mol
Now we will compare the moles of Reactant with product.
Zn : ZnCl₂
1 : 1
0.76 : 0.76
Zn : H₂
1 : 1
0.76 : 0.76
HCl : ZnCl₂
2 : 1
1.4 : 1/2×1.4 = 0.7
HCl : H₂
2 : 1
1.4 : 1/2×1.4 = 0.7
Less number of moles of product are formed by HCl it will act limiting reactant.