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antiseptic1488 [7]
3 years ago
15

If 17.8 g grams of lithium (Li) are combined with 50.0 grams of water (H20) there is

Chemistry
1 answer:
Ad libitum [116K]3 years ago
3 0

Answer:

Explanation:

Li+H2O..................LiOH+H2

mass of Li=17.8 g

No of moles of Li=17.8/7=2.5

mass of H2O=50.0 g

No of moles of H2O=50.0/18=2.8

limiting reagent is Li so 1 mole of Li produce 1 mole of LiOH therefore 2.5 mole of Li produce 2.5 moles of LiOH now we have to convert it into grams

moles=given mass/molar mass

given mass=moles*molar mass

molar mass of LiOH=24

mass=2.5*24=60 gram

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1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

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3 years ago
Please help<br>Chemistry question
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Answer:

The SMRs can adjust their energy output using coolants and control rods.

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Veronika [31]

Answer:

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Explanation:

By the mass of the two elements in the solution, we determine the moles of each:

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Mole fraction solvent = Moles of solvent / Total moles

Total moles = Moles of solute + Moles of solvent

1.39 moles of solute + 0.543 moles of solvent = 1.933 moles → Total moles

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Answer:

Rate = 0.001615 Ms-1

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Rate = ?

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Rate = 1.58E-4 * 2.84 * 3.60

Rate = 0.001615 Ms-1

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3 years ago
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