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ella [17]
2 years ago
15

the student performed a second trial and accidentally added more sodium hydroxide to the flask than was needed to reach the end

point but recorded the final volume. would this error increase, decrease or have no effect on the calculated acid concentration for the second trial?
Chemistry
1 answer:
Harrizon [31]2 years ago
4 0

A greater volume of acid will be needed to neutralize the sodium hydroxide thereby making the calculated concentration of acid for the second trial to decrease.

Titration is usually a way of determining the concentration of an unknown solution by accurately measuring the volume of a known solution that reacts completely with the unknown solution.

The point when reaction is completed is indicated by the end point of the reaction. If more base (sodium hydroxide) is added to the flask, then more volume of acid will be needed to neutralize it thereby making the calculated concentration of acid for the second trial to decrease.

Learn more: brainly.com/question/1527403

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An equilibrium mixture of O2, SO2 and SO3 contains equal concentrations of SO2 and SO3.
AleksAgata [21]

Answer:

[O₂(g)] = 0.0037M

Explanation:

            2SO₂(g) + O₂(g) => 2SO₃(g)

Conc:   [SO₂(g)]   [O₂(g)]     [SO₃(g)]  and  [SO₂(g)] = [SO₃(g)]

Kc =  [SO₃(g)]²/[O₂(g)][SO₂(g)]² => Kc = 1/[O₂(g)] = 270 if [SO₂(g)] = [SO₃(g)]

∴ [O₂(g)] = (1/270)M = 0.0037M  

7 0
3 years ago
3. A compound was analyzed and found to contain 9.8 g of nitrogen, 0.70 g of
Nesterboy [21]

Answer:

HNO₃

Explanation:

Data given

Nitrogen = 9.8 g

Hydrogen =  0.70 g

Oxygen = 33.6 g

Empirical formula = ?

Solution:

Convert the masses to moles

For Nitrogen

Molar mass of N = 14 g/mol

                              no. of mole = mass in g / molar mass

Put value in above formula

                          no. of mole = 9.8 g/ 14 g/mol

                          no. of mole = 0.7

                           mole of N = 0.7 mol

For Hydrogen

Molar mass of H = 1 g/mol

                     no. of mole = mass in g / molar mass

Put value in above formula

                     no. of mole = 0.70 g/ 1 g/mol

                      no. of mole = 0.7

mole of H = 0.7 mol

For Oxygen

Molar mass of O = 16 g/mol

                       no. of mole = mass in g / molar mass

Put value in above formula

                       no. of mole = 33.6 g / 16 g/mol

                       no. of mole = 2.1

mole of O = 2.1 mol

Now we have values in moles as below

N = 0.7

H = 0.7

O = 2.1

Divide the all values on the smallest values to get whole number ratio

N = 0.7 / 0.7 = 1

H = 0.7 / 0.7 = 1

O = 2.1 / 0.7 = 3

So all have following values

N = 1

H = 1

O = 3

So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.

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A. the number dots shown on the lewis dot diagrams for the element in the period
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Anuta_ua [19.1K]
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The balanced chemical reaction is written as:

PbS + 3/2O2 = PbO + SO2
4 0
3 years ago
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