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anygoal [31]
3 years ago
10

A 29.2 mL sample of a 0.458 M aqueous hydrofluoric acid solution is titrated with a 0.337 M aqueous sodium hydroxide solution. W

hat is the pH at the start of the titration, before any sodium hydroxide has been added
Chemistry
1 answer:
Svetach [21]3 years ago
4 0

Answer:

3.51

Explanation:

Before any sodium hydroxide has been added, the pH is that of the aqueous hydrofluoric acid solution.

HF is a weak acid that dissociates according to the following equation.

HF(aq) ⇄ H⁺(aq) + F⁻(aq)      Ka = 6.76 × 10⁻⁴

We can find [H⁺] using the following expression.

[H⁺] = √(Ca × Ka)

where

Ca: concentration of the acid

Ka: acid dissociation constant

[H⁺] = √(Ca × Ka)

[H⁺] = √(0.458 × 6.76 × 10⁻⁴)

[H⁺] = 3.10 × 10⁻⁴ M

The pH is:

pH = -log [H⁺]

pH = -log 3.10 × 10⁻⁴ = 3.51

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A system is in equilibrium when the rate of the forward reaction is _____ the rate of the reverse reaction.
Fudgin [204]

Answer:

A system is in equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction.

Explanation:

In a system in equilibrium, the rupture and formation of new bonds from the formation and decomposition of substances must have the same reaction rate. Thanks to equal speeds we can say that a system in equilibrium is dynamic and in general every system tends to move spontaneously towards equilibrium. The environment never interferes.

For example.

N2 (g) + 3H2 (g) ---> 2NH3 (g) formation

2NH3 ----> N2 (g) + 3H2 (g) decomposition

N2 (g) + 3H2 (g)  <------>  2NH3 (g)     In equilibrium

8 0
3 years ago
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Write an expression that can be used to check the quotient of 646÷3
iVinArrow [24]

Answer:

215.33

Explanation:

A mathematical expression is an equation that connect one or more variables with constants.

The Quotient is the result of carrying out the division operation.let "y" be the quotient of dividing 646 by 3

therefore 646÷3 =y

\frac{646}{3}=y\\ 3y=646\\

y=\frac{646}{3}\\ y=215.33

5 0
3 years ago
Iodine-131 is a radioactive isotope. after 8.00 days, 50.2% of a sample of 131i remains. what is the half-life of 131i?
barxatty [35]
 The half life  of 131 i  is  8.06  days

   calculation

by  use  of concentration  time   equation  for  radioactive  decay

In Nt/No = -Kt

where  Nt/No  is the  fraction  of the sample   remaining  at  time T

convert 50.2%  in  decimal  = 50.2/100 = 0.502

therefore = In  0.502 = -  K 8.0 t
                     - 0.689 =-8.0 k
divide both side by 8.0

                       K =  0.860

t1/2  =  0.693/K

t1/2 = 0.693/0.860 =  8.06   days

3 0
3 years ago
Is this anwser correct
Hoochie [10]
Yes it is correct I think not for sure
3 0
3 years ago
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Which statement is true?
mihalych1998 [28]
B. Most rocks are composed of single mineral
3 0
3 years ago
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