The missing labels are:
- CuCO₃(s), H₂SO₄(aq): reactants.
- +: plus sign.
- CuSO₄(aq), H₂O(l), CO₂(g): products.
- (s): solid.
- (aq): aqueous.
- (l): liquid.
- (g): gaseous.
<h3>What is a chemical equation?</h3>
It is a way to represent a chemical reaction.
Let's consider the following chemical equation.
CuCO₃(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l) + CO₂(g)
The missing labels are:
- CuCO₃(s), H₂SO₄(aq): reactants. They are on the left side of the equation.
- +: plus sign. It separates substances.
- CuSO₄(aq), H₂O(l), CO₂(g): products. They are on the right side of the equation.
- (s): solid.
- (aq): aqueous.
- (l): liquid.
- (g): gaseous.
Learn more about chemical equations here: brainly.com/question/26227625
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Answer:
[C₆H₅NH₃⁺] = 0.0399 M
Explanation:
This excersise can be easily solved by the Henderson Hasselbach equation
C₆H₅NH₃Cl → C₆H₅NH₃⁺ + Cl⁻
pOH = pKb + log (salt/base)
As we have value of pH, we need to determine the pOH
14 - pH = pOH
pOH = 8.43 (14 - 5.57)
Now we replace data:
pOH = pKb + log ( C₆H₅NH₃⁺/ C₆H₅NH₂ )
8.43 = 9.13 + log ( C₆H₅NH₃⁺ / 0.2 )
-0.7 = log ( C₆H₅NH₃⁺ / 0.2 )
10⁻⁰'⁷ = C₆H₅NH₃⁺ / 0.2
0.19952 = C₆H₅NH₃⁺ / 0.2
C₆H₅NH₃⁺ = 0.19952 . 0.2 = 0.0399 M
Since the mole ratio is 1 to 1 for this reaction, you can just use M1V1 = M2V2 to solve, given that 1 refers to HBr and 2 refers to LiOH.
M1 = ?
V1 = 0.40 L
M2 = 0.50 M
V2 = 0.20 L
M1(0.40) = (0.50)(0.20)
Solve for M1 —> M1 = (0.50)(0.20)/0.40 = 0.25 M HBr
Explanation:
Which is a pure substance?
1. soda
2. gasoline
3. salt water
4. carbon dioxide
carbon dioxide
Bromine, a liquid at room temperature, has a boiling point of 58°C and a melting point of -7.2°C. Bromine can be classified as a
1. compound.
2. impure substance.
3. mixture.
4. pure substance.
pure substance.
They have seven electrons in their valence shell, so halogens are very reactive.
Hope this helps! :)
