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aleksklad [387]
1 year ago
14

For a typical titration, 0. 010 m naoh is the titrant (in the buret). if the initial buret reading is 2. 45 ml, and the final bu

ret reading is 18. 70 ml, how much naoh was used for the titration?
Chemistry
1 answer:
pickupchik [31]1 year ago
5 0

16.25 ml NaOH was used for the titration.

<h3>Titration:</h3>

"The process of calculating the quantity of a material A by adding measured increments of substance B, the titrant, with which it reacts until exact chemical equivalency is obtained (the equivalence point)" is the definition of titration.

A titration is a method for figuring out the concentration of an unknown solution by using a solution with a known concentration. Until the reaction is finished, the titrant (the known solution) is typically added from a buret to a known volume of the analyte (the unknown solution).

0. 010 m NaOH is the titrant (in the buret).

The initial buret reading is 2. 45 ml

The final buret reading is 18. 70 ml

The volume of NaOH = 18.70ml - 2.45 ml

                                    = 16.25 ml

Therefore, 16.25 ml NaOH was used for the titration.

Learn more about titration here:

brainly.com/question/27107265

#SPJ4

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Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f
Korvikt [17]

Answer:

55.2kgNa_{3}AlF_{6}

Explanation:

1. First balance the equation for the synthesis of cryolite:

Al_{2}O_{3}_{(s)}+6NaOH_{(l)}+12HF_{(g)}=2Na_{3}AlF_{6}+9H_{2}O_{(g)}

2. Find the limiting reagent between the Al_{2}O_{3},NaOH and HF

- First calculate the number of moles of each compound using its molar mass and the mass that reacted completely:

13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{101.96gAl_{2}O_{3}}*\frac{1000g}{1kg}=131molesAl_{2}O_{3}

55.4kgNaOH*\frac{1molNaOH}{40kgNaOH}*\frac{1000g}{1kg}=1385molesNaOH55.4kgHF*\frac{1molHF}{20kgHF}*\frac{1000g}{1kg}=2770molesHF

- Divide the number of moles obtained between the stoichiometric coefficient of each compound in the chemical reaction:

Al_{2}O_{3}:\frac{131}{1}=131

NaOH:\frac{1385}{6}=231

HF:\frac{2770}{12}=231

The Al_{2}O_{3} is the limiting reagent because it has the smallest number.

3. Find the mass of cryolite produced:

13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{0.10196kgAl_{2}O_{3}}*\frac{2molesNa_{3}AlF_{6}}{1molAl_{2}O_{3}}*\frac{0.20994kgNa_{3}AlF_{6}}{1molNa_{3}AlF_{6}}=55.2kgNa_{3}AlF_{6}

3 0
4 years ago
Can someone help me I need help please!
zavuch27 [327]

Answer:

Explanation:

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example 1 -2= 1 so there is the atomic number of 1 and atomic mass of 2

let me know if this helps

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a pure silver ring contains 2.60 *10^22 silver atoms. How many moles of silver atoms does it contain?
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Moles of silver = 2.6 x 10^22/6.02 x 10^23 = 0.043 moles. 

Hope this helps, have a great day ahead!
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The answer to your question is Gallium, or GA. The trick is to always check if they are along the same vertical column. This indicates that they have the same Valence electrons, which is involved in all the bonding and define the electronegative status of the atom.

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