4) The volume of the HCl used is 9.500 mL while the volume of the NaOH used is 3.800 mL.
5) Molarity of sodium hydroxide is obtained from; Molarity of HCl * 1/2
What is titration?By reacting an unknown component with a known quantity of a different chemical known as a titrant, titration is a laboratory procedure used to measure the concentration of an unknown substance, often a solute dissolved in a liquid.
The endpoint of a titration can be detected in a number of ways, depending on the specific titration being performed.
4)
Volume of the Acid used = Initial reading - Final reading = 25.00 - 15.50 = 9.500 mL
Volume of the base used = 8.80 - 5.00 = 3.800 mL
5)
We know that the mole ratio is 1:2 and the implication of this is that the set up to obtain the molarity of the sodium hydroxide solution is Molarity of HCl * 1/2
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Choose the equation below that is balanced correctly.
S8 +24 028 SO3
S8+ 12 0₂8 SO3
6 S8+8 026 SO3
2 S8 +3 022 SO3
The balanced equation for the reaction between sulfur (S₈) and oxygen (O₂) to form sulfur trioxide (SO₃) is 2S₈ + 16O₂ → 16SO₃.
What is the balanced chemical equation?Balancing chemical equations involves the addition of stoichiometric coefficients to the reactants and products.
The balanced equation for the reaction between sulfur (S₈) and oxygen (O₂) to form sulfur trioxide (SO₃) is determined as;
2S₈ + 16O₂ → 16SO₃
From the reactants side we can see that sulfur is 16 and also 16 in the product side. The number of oxygen in the reactant side is 32 and also 32 in the product side.
Thus, the balanced equation for the reaction between sulfur (S₈) and oxygen (O₂) to form sulfur trioxide (SO₃) is 2S₈ + 16O₂ → 16SO₃.
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Does anyone know the answer to this question
Answer:
A
Explanation:
If Hydrogen is H₂ There will be two silver
and is Carbon is C There will only be one gray
and if Oxygen is O₃ There will be three red
2 NaN3 → 2 Na + 3 N
Given 9.98 grams of N2, how many moles of NaN3 are produced?
0.238 moles of NaN₃ are produced from 9.98 grams of N₂.
What is the moles of NaN₃ produced?The moles of he mass of NaN₃ produced
The balanced equation for the reaction is:
2 NaN₃ → 2 Na + 3 N₂
The molar ratio between NaN₃ and N₂ is 2:3, which means that for every 2 moles of NaN₃, 3 moles of N₂ are produced.
The mole ratio is used to determine how many moles of NaN₃ are produced from 9.98 grams of N₂.
First, we need to convert the mass of N₂ to moles:
moles of N₂ = mass of N2 / molar mass of N₂
moles of N₂ = 9.98 g / 28.02 g/mol
moles of N₂ = 0.356 mol
moles of NaN₃ = (2/3) * moles of N₂
moles of NaN₃ = (2/3) * 0.356 mol
moles of NaN₃ = 0.238 mol
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