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Ivenika [448]
4 years ago
14

What is the concentration of 10.00 mL of HBr if it takes 16.73 mL of a 0.253 M LiOH solution to neutralize it?

Chemistry
1 answer:
Leni [432]4 years ago
3 0
First. let's write the reaction formula: HBr +LiOH ----> LiBr + H₂O

let's get the moles of LiOH first

moles= Molarity x Liters

moles= 0.253 M x 0.01673 Liter= 0.00423 moles LiOH

using the balanced equation, you can see that 1 mol LiOH is equal to 1 mol HBr. so:

0.00423 mol LiOH = 0.00423 mol HBr

now let's find the concentration

molarity= mol/ Liters

0.00423 mol/ 0.01000 Liters= 0.423 M
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In case of low-mass stars,the outer layers of the low mass stars are expelled  as the core collapses such that the outer layers form a planetary nebula. The core remains as a white dwarf and finally become a black dwarf as it cools down. A low mass star consumes its core hydrogen and turns it into helium over its lifetime.

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Which of the following of elements would have all four valence electrons
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Water could be made to boil lower than its normal boiling point of 100 degrees Celsius at 92 degrees Celsius by?
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Water could be made to boil lower than its normal boiling point of 100 degrees Celsius at 92 degrees Celsius by lowering the atmosphere or external  pressure.

The boiling point is the temperature at which the vapor pressure of liquid equals the external or atmosphere pressure.

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3 0
4 years ago
Be sure to answer all parts. find the most acidic hydrogen in tert−butyl methyl ketone and write a chemical equation for the pro
sleet_krkn [62]
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In <span>tert−butyl methyl ketone there are two carbons at each alpha position. Among these two carbons only methyl carbon contains hydrogen atoms while the second one is bonded to further three carbons making it Quaternary carbon. The base abstracts proton from methyl group and enolate is formed. 

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7 0
3 years ago
A chemistry graduate student is given 125.mL of a 0.20M acetic acid HCH3CO2 solution. Acetic acid is a weak acid with =Ka×1.810−
Over [174]

Answer : The mass of sodium acetate is, 1.097 grams.

Explanation : Given,

The dissociation constant for acetic acid = K_a=1.8\times 10^{-5}

Concentration of acetic acid (weak acid)= 0.20 M

volume of solution = 125. mL

pH = 4.47

First we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (1.8\times 10^{-5})

pK_a=5-\log (1.8)

pK_a=4.74

Now we have to calculate the concentration of sodium acetate (conjugate base or salt).

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

Now put all the given values in this expression, we get:

4.47=4.74+\log (\frac{[Salt]}{0.20})

[Salt]=0.107M

Now we have to calculate the mass of sodium acetate.

\text{Concentration}=\frac{\text{Mass of }NaCH_3CO_2\times 1000}{\text{Molar mass of }NaCH_3CO_2\times \text{Volume of solution (in mL)}}

0.107M=\frac{\text{Mass of }NaCH_3CO_2\times 1000}{82g/mol\times 125mL}

\text{Mass of }NaCH_3CO_2=1.097g

Therefore, the mass of sodium acetate is, 1.097 grams.

6 0
4 years ago
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