Answer:

Explanation:
There are two main kinds of properties: chemical and physical.
Chemical properties, like the name suggests, have to be observed by changing the chemical composition.
That leaves <u>physical properties.</u> They can be measured without any chemical composition changes.
Some examples include: color, odor, mass, density, and volume. All can be measured with just the senses or measuring tools and no composition alterations are needed.
Answer:
C12H22O11(aq) + H2O(l) —> 4C2H5OH(aq) + 4CO2(g)
Explanation:
When aqueous sugar (sucrose) react with water in the presence of yeast, the following products are obtained as shown in the equation below:
C12H22O11(aq) + H2O(l) —> C2H5OH(aq) + CO2(g)
Now, we shall balance the equation as follow:
There are a total of 24 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 4 in front of C2H5OH as shown below:
C12H22O11(aq) + H2O(l) —> 4C2H5OH(aq) + CO2(g)
There are a total of 9 atoms of C on the right side and 12 atoms on the left side. It can be balance by putting 4 in front of CO2 as shown below:
C12H22O11(aq) + H2O(l) —> 4C2H5OH(aq) + 4CO2(g)
Now the equation is balanced.
Answer:

Explanation:
Chemical Equations are representations of chemical reactions in terms of the symbols and formulae of the elements and compounds involved. A chemical equation usually have the reactant at the left hand side while the product is on the right hand side.
A chemical Equation is of little or no value if is not in balanced equation. When an equation is balanced , the total number of atoms of any element on the left-hand side of it must be equal to the total number of atoms of that element on the right hand side.
in the given question; we are given a word problem of chemical symbol to compute and also to balance the chemical equation.
From below; the chemical equation can be written as:

From the above equation we will notice that it is not truly balanced ; so th balanced equation can be written as:

Answer:
0.1313 g.
Explanation:
- It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.
- Suppose that hydrogen behaves ideally and at STP conditions.
<u><em>Using cross multiplication:</em></u>
1.0 mol of hydrogen occupies → 22.4 L.
??? mol of hydrogen occupies → 1.47 L.
∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.
- Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:
<em>The no. of grams of hydrogen = no. of hydrogen moles x molar mass of hydrogen</em> = (6.563 x 10⁻² mol)(2.0 g/mol) = <em>0.1313 g.</em>