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asambeis [7]
3 years ago
11

A student performs three trials of a titration of an acid with an unknown concentration. She compares her measured concentration

to the actual value, and calculates a 25% error. Assuming she has performed all of the calculations correctly, name one aspect of her methods that may have led to such a high error.
Chemistry
1 answer:
Viktor [21]3 years ago
4 0

Answer:

There are many errors possible while titrating the acid of an unknown concentration with a base like NaOH.

Main error that leads to the error in results is misreading of the end point volume .

End point is when the reaction between the analyte and solution of known concentration has stopped .

Sometimes Burette is not straight enough to read the volume of the end point. One way to misread the volume of burette is by looking at the burette volume at an angle .

From above , volume seems to be higher. Indicators are used to indicate the color change of the reaction. In Acid-Base titrations , indicators first lighten up then changes its color.

So, error may have occurred in wrongly judging of the end point by color change of the indicator .

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fiasKO [112]

Answer:

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Explanation:

5 0
3 years ago
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The specific heat of copper is 0.093 cal/g0C. Calculate the temperature change that occurs if 28 g of copper at 25 0C absorbs 58
Umnica [9.8K]

Answer:

22.27 °C = ΔT

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m × c × ΔT

Given data:

mass = 28 g

heat absorbed = 58 cal

specific heat of copper =  0.093 cal/g .°C

temperature change =ΔT= ?

Solution:

Q = m × c × ΔT

58 cal = 28 g × 0.093 cal /g.°C × ΔT

58 cal = 2.604 cal.°C × ΔT

58 cal / 2.604 cal .°C = ΔT

22.27 °C = ΔT

5 0
3 years ago
Determine whether each statement is a description of a physical property or a chemical property. Please check the box that appli
sdas [7]
All except option two
5 0
3 years ago
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Write a ground state electron configuration for each neutral atom
Gre4nikov [31]

Answer:

Pb[lead] [Xe]4f^145d^106s^26p^2

U[uranium] 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

negative charge (−), and

(2) electron loss will result to a positive charge (+),

1s^2 2s^2 2p^6 3s^2 3p^6 = 18 electrons

Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6,

for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

Mercury has a regular electron configuration. It becomes:

[Xe]4f145d106s2

Explanation:

socratic.org helped me! I'm really sorry if this is wrong!

6 0
3 years ago
How many moles of BaSO4 are formed if 0.5 moles of Na2 SO4 react with 60 g of BaCl2? Na2 SO4 + BaCl2 → BaSO4 + 2NaCl
faust18 [17]
<span>the balanced equation for the reaction is as follows
Na</span>₂<span>SO</span>₄<span> + BaCl</span>₂<span> ----> 2NaCl + BaSO</span>₄ 
<span>stoichiometry of Na</span>₂<span>SO</span>₄<span> to BaCl</span>₂<span> is 1:1
first we need to find out which the limiting reactant is
limiting reactant is fully used up in the reaction.
                                                                           
number of Na2So4 moles - 0.5 mol number of BaCl2 moles - 60 g / 208 g/mol = 0.288 mol
since molar ratio is 1:1 equal number of moles of both reactants should react with each other
                                                                         
therefore BaCl2 is the limiting reactant and Na2SO4 is in excess. amount of product formed depends on number of limiting reactant present.
stoichiometry of BaCl</span>₂<span> to BaSO</span>₄<span> is 1:1.
                                                                           
therefore number of BaSO4 moles formed - 0.288 mol</span>
6 0
3 years ago
Read 2 more answers
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