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Which acid/base pair will give an equivalence point that cannot be predicted solely by a general knowledge of acid and base stre

sleet_krkn [62]

The acid/base pair that give equivalence point that Cannot be predicted by general knowledge is NaOH and HCI ONH.

An Acid is a substances that is corrosive in nature and turn blue lithmus paper to red which it react with base to produce salt and water.

Acid dissolve metals.

Base is a substance that turn red lihthmus paper to blue and react with acid to produce salt and water.

Therefore, The acid/base pair that give equivalence point that Cannot be predicted by general knowledge is NaOH and HCI ONH.

The question is incomplete as the options were not given. The options were gotten from another website.

Select the correct answer below:

ONaOH and HCI ONH,

HC ONH, and CH, COOH

NaOH and Christmas, COOH

Learn more about acid and base below.

brainly.com/question/2506771

8 0

3 years ago

Please can I have help filling in this table<br> I’ll be so grateful!!!

Nookie1986 [14]

Proton relative mass is +1
Neutron relative mass is 0
Electron relative mass is -1

6 0

3 years ago

Can someone answer theses problems? Only do the Even number problems,Im in a rush so please help me.

8090 [49]

The answer to the questions. All 15 of them.

7 0

4 years ago

A student carries out the same titration but uses an indicator instead of a pH meter. If the indicator changes color slightly pa

creativ13 [48]

Answer:

Too high a value

Explanation:

HA + NaOH ⟶ NaA +H₂O

If the student has gone slightly past the equivalence point, they have added too much base.

The moles of HA are directly proportional to the moles of NaOH, so the moles of acid that the student calculates will be too high.

The calculated concentration of acid will also be too high.

4 0

4 years ago

The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper

LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

$t_{1/2}=\frac{1}{k\times a_0}$

k = rate constant =?

$a_0$ = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

a) Initial concentration of XY = $a_o=0.100 M$

Rate constant of the reaction = k = $6.96\times 10^{-3} M^{-1} s^{-1}$

Half life of the reaction is:

$t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}$

$=1,436.78 s$

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

$\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}$

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

$\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = $6.20\times 10^{-2} M$

$\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become $6.20\times 10^{-2} M$ .

e) Initial concentration of $SO_2Cl_2= 0.050 M$

Final concentration after time t = x

t = 55.0 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

7 0

3 years ago