60 ml of a 0.40 m solution of h2so4 is used to neutralize 0.3 m magnesium hydroxide solution, what is the ml of sodium hydroxide?
Answers
60 ml of a 0.40 m solution of h2so4 is used to neutralize 0.3 m magnesium hydroxide solution, the volume of NaOH required to neutralize the given amount of Mg(OH)2 is 160 ml.
To answer this question, we need to use the concept of stoichiometry. Stoichiometry is the calculation of the quantities of reactants and products in a chemical reaction. The balanced chemical equation for the neutralization reaction between sulfuricacid (H2SO4) and magnesium hydroxide (Mg(OH)2) is:
H2SO4 + Mg(OH)2 → MgSO4 + 2H2O
From the equation, we can see that one mole of H2SO4 reacts with one mole of Mg(OH)2. Therefore, the number of moles of Mg(OH)2 in 60 ml of 0.3 m solution is:
moles of Mg(OH)2 = concentration x volume = 0.3 x (60/1000) = 0.018 moles
Since one mole of Mg(OH)2 requires one mole of NaOH to neutralize it, we need 0.018 moles of NaOH. The concentration of the NaOH solution is not given, so we cannot directly calculate the volume of NaOH required. However, we can use the concentration and volume of the H2SO4 solution to find the number of moles of H2SO4 used in the neutralization reaction:
moles of H2SO4 = concentration x volume = 0.40 x (60/1000) = 0.024 moles
From the balanced equation, we know that one mole of H2SO4 reacts with two moles of H2O. Therefore, the number of moles of H2O produced in the reaction is:
moles of H2O = 2 x moles of H2SO4 = 0.048 moles
Since the reaction is neutralization, the same number of moles of H2O and H+ ions are produced. Therefore, the number of moles of H+ ions produced is also 0.048 moles. Each mole of NaOH can neutralize one mole of H+ ions. Therefore, the number of moles of NaOH required is:
moles of NaOH = 0.048 moles
If we assume that the concentration of the NaOH solution is also 0.3 m, then the volume of NaOH required is:
volume of NaOH = moles of NaOH / concentration = 0.048 / 0.3 = 0.16 L = 160 ml
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Answer:
C is it!
Explanation:
B. saponification
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D. neutralization
Answers
Answer: D. neutralization
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Answer: K only has 1 valence electron. It will leave with only a little effort, leaving behind a positively charged K^+1 atom.
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Answers
The standard enthalpy of formation occurs under the standard conditions of 298k and 1bar.
Answers
Answer:
18.46 L
Explanation:
From the question given above, the following data were obtained:
Initial volume (V1) = 12.0 L
Initial pressure (P1) = 760 torr
Final pressure (P2) = 0.650 atm
Final volume (V2) =..?
Next, we shall express the pressure in the same unit of measurement.
Thus, 760 torr is equivalent to 1 atm.
Finally, we shall the new volume of the balloon as follow:
Initial volume (V1) = 12.0 L
Initial pressure (P1) = 1 atm
Final pressure (P2) = 0.650 atm
Final volume (V2) =..?
P1V1 = P2V2
1 × 12 = 0.650 × V2
12 = 0.650 × V2
Divide both side by 0.65
V2 = 12/0.65
V2 = 18.46 L
Therefore, the new volume of the balloon is 18.46 L.
Answers
1. Double-Replacement reactions
2. Decomposition
3. Combustion
4. Syntesis
5. Single replacement
Further explanation
Given
Chemical equations
Required
Type of reaction
Solution
1. 2AgNO₃ + MgCl₂ ⇒ 2AgCl + Mg(NO₃)₂
Double-Replacement reactions. Happens if there is an ion exchange between two ion compounds in the reactant
2. 2KBr⇒2K +Br₂
Decomposition
Single compound breaks down in to 2 or more products
3. C₃H₈ + 5O₂ ⇒ 4H₂O + 3CO₂
Combustion
Hydrocarbon and Oxygen reaction and form water and carbon dioxide
4. NaO + H₂O ⇒ NaOH
Syntesis
2 or more reactants combine to form a single product
5. Zn + CuCl₂ ⇒ZnCl₂ + Cu
Single replacement
One element replaces another element from a compound