Can you add more context to your question i’m confused
The question is incomplete, the complete question is;
Calculate the volume of concentrated HCL (12 M) needed to convert 1.5 g of sodium benzoate back to benzoic acid.
Answer:
8.33 * 10^-4 L
Explanation:
Equation of the reaction;
C6H5COONa + HCl -------->>>> C6H5COOH + NaCl
Number of moles of sodium benzoate = mass/molar mass
molar mass of sodium benzoate =144.11 g/mol
mass of sodium benzoate = 1.5 g
Number of moles of sodium benzoate = 1.5g/144.11 g/mol
Number of moles of sodium benzoate = 0.01 moles
Since the reaction is 1:1, 0.01 moles of HCl reacted.
but;
n = CV
n = number of moles
C = concentration
V = volume
V = n/C
V = 0.01 moles/12 M
V = 8.33 * 10^-4 L
Answer:
<h2>We get the following information from a chemical equation :--</h2>
1. We get the symbol and formulas of the reactants and products.
2. The physical state of the reactants and products.
3. From this reaction, we can know the molecular formula of ammonia, water and nitrogen.
4. Here all the species are in a gaseous state.
5. The reaction takes place at a high temperature and pressure.
6. Zinc metal reacts with dilute sulphuric acid to form zinc sulphate and hydrogen gas. This equation is written as:
Zinc metal reacts with dilute sulphuric acid to form zinc sulphate and hydrogen gas. This equation is written as: Zn+H2SO 4→ZnSO 4 +H2.
the gold would float because the gold is lighter then the silver. the silver is more haver then the gold. so the gold will float and silver will not float
Answer:
Cu (s) + 2H₂SO₄(aq) —> CuSO₄(s) + 2H₂O (l) + SO₂(s)
Explanation:
Copper metal => Cu
Sulphuric acid => H₂SO₄
copper (II) sulfate => CuSO₄
Water => H₂O
sulfur dioxide => SO₂
The equation for the reaction between copper metal and aqueous sulfuric acid can be written as follow:
Cu + H₂SO₄ —> CuSO₄ + H₂O + SO₂
The above equation can be balance as follow:
Cu + H₂SO₄ —> CuSO₄ + H₂O + SO₂
There are 2 atoms of S on the right side and 1 atom on the left side. It can be balance by writing 2 before H₂SO₄ as shown below:
Cu + 2H₂SO₄ —> CuSO₄ + H₂O + SO₂
There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by writing 2 before H₂O as shown below:
Cu + 2H₂SO₄ —> CuSO₄ + 2H₂O + SO₂
Thus, the equation is balanced.
Include the phase of each reactant and product, we have:
Cu (s) + 2H₂SO₄(aq) —> CuSO₄(s) + 2H₂O (l) + SO₂(s)