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borishaifa [10]
2 years ago
14

2.50 g of an unknown base is dissolved in 289 mL of water and has a concentration of 0.216 M. What is the identification of the

unknown base?A. NaOHB. LiOHC. KOH
Chemistry
1 answer:
AnnyKZ [126]2 years ago
3 0

Answer:

The answer to your question is: A. NaOH

Explanation:

Data

mass = 2.5g of unknown base

volume = 289 ml = 0.289 l

concentration = 0.216 M

What is the base= ?

Formula

 Molarity (M) = # moles / Volumen (V)

 # moles = M x V

Process

# moles = 0.216 x 0.289 = <u>0.062     in the solution</u>

MW NaOH = 23 + 16 + 1 = 40 g

MW LiOH = 7 + 16 + 1 = 24 g

MW KOH = 39 + 16 + 1 = 56 g

Rule of three

                        40 g of NaOH ------------------- 1 mol

                        2.5 g               --------------------  x

                 x = 2.5 x 1 /40 = <u>0.062 moles</u>  is the same amount of moles in the

                                                                    solution is the answer

                         24 g of LiOH ----------------- 1 mol

                         2.5 g         ---------------------   x

                x = 2.5 x 1 / 24 = 0.104 moles

                      56 g of KOH ------------------ 1 mol

                      2.5 g             --------------------  x

                x = 2.5 x 1 / 56 = 0.045

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Lithium has an atomic mass of 6.941 amu. Lithium has two common isotopes. The one isotope has a mass of 6.015 amu and a relative
koban [17]

Answer:

The atomic mass of second isotope is 7.016

Explanation:

Given data:

Average Atomic mass of lithium = 6.941 amu

Atomic mass of first isotope = 6.015 amu

Relative abundance of first isotope = 7.49%

Abundance of second isotope = ?

Atomic mass of other isotope = ?

Solution:

Total abundance = 100%

100 - 7.49 = 92.51%

percentage abundance of second isotope = 92.51%

Now we will calculate the mass if second isotope.

Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

6.941 = (6.015×7.49)+(x×92.51) /100

6.941 =  45.05235 + (x92.51) / 100

6.941×100 = 45.05235 + (x92.51)

694.1 - 45.05235   = (x92.51)

649.04765 = x 92.51

x = 485.583 /92.51

x = 7.016

The atomic mass of second isotope is 7.016

3 0
2 years ago
The diagram shows the scales used for recording temperatures. The labels for the scales are missing. 3 thermometers are oriented
QveST [7]

Answer d

Explanation:

8 0
2 years ago
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Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l
Zepler [3.9K]

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

We are given:

\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol

Putting values in above equation, we get:

-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]

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4 0
2 years ago
Which is the stronger acid in each of the following pairs? Explain your reasoning.
Makovka662 [10]

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Explanation:

The PKa tool relative to Ph are used to contrast the pairs.

The pKa of phenol is 10. The pKa of p-hydroxybenzaldehyde is 9.24

The pKa for meta-cyanophenol is 8.61 and the pKa for para-cyanophenol is 7.95. 

The pKa value of o-fluorophenol is 8.7, while that of the p-fluorophenol is 9.9. It's obvious that the inductive effect is more dominant at ortho-position, which results in a more acidic nature

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