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frozen [14]
3 years ago
7

Suppose you were titrating approximately 0.1 m hcl solution with standardized naoh solution (0.0989m). in your lab manual, you w

ere instructed to add 15.00 ml hcl, but you accidentally added 25.00 ml instead. how would this mistake affect the ph at equivalence point?
Chemistry
2 answers:
Semenov [28]3 years ago
5 0
It alwasys chages becuse of magical bull in chem. ask the chem godzzz im aright  
MArishka [77]3 years ago
5 0

Answer is: will not change the pH of equivalence point.

Balanced chemical reaction: HCl + NaOH → NaCl + H₂O.

In this reaction pH of equivalence point will be always 7.

Equivalence point is the point which there is stoichiometrically equivalent amounts of acid and base.  

Chemist can draw pH curve (graph showing the change in pH of a solution, which is being titrated) for titration and determine equivalence point.

Near equivalence point indicator should change color, so we must pick indicator who change color near pH of equivalence point.

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The Lewis dot model of a metal atom has two dots. Which of the following is true about the atom?
a_sh-v [17]

Answer:

It is present in the second group of the periodic table

Explanation:

7 0
3 years ago
Read 2 more answers
Assume that temperature and number of moles of gas are constant in this problem.
IrinaVladis [17]

Question:

a. a direct linear relationship

b. an inverse linear relationship

c. a direct nonlinear relationship

d. an inverse nonlinear relationship

Answer:

The correct option is;

d. An inverse nonlinear relationship

Explanation:

From the universal gas equation, we have;

P·V = n·R·T

Where we have the temperature, T and the number of moles, n constant, therefore, we have

P×V = Constant, because, R, the universal gas constant is also constant, hence;

P×V = C

P = \frac{C}{V}

Since P varies with V then the graphical relationship will be an inverse nonlinear as we have

V P    C

1 5         5

2 2.5      5

3 1.67     5

4 1.25     5

5 1          5

6 0.83    5

7 0.7      5

8 0.63    5

9 0.56    5

10 0.5      5

Where:

V = Volume

P = Pressure

C = Constant = 5

P = C/V

The graph is attached.

4 0
3 years ago
What product(s) are formed during this chemical reaction? (Will be two answers)
lianna [129]

Answer:

BK. BK. BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK

3 0
3 years ago
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

7 0
3 years ago
in order to find the molar mass of an unknown compound, a research scientist prepared a solution of 0.930 g of an unknown in 125
PtichkaEL [24]

Answer:

Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

Explanation:

Let's apply the formula for freezing point depression:

ΔT = Kf . m

ΔT = 74.2°C - 73.4°C → 0.8°C

Difference between the freezing T° of pure solvent and freezing T° of solution

Kf = Cryoscopic constant → 5.5°C/m

So, if we replace in the formula

ΔT = Kf . m → ΔT / Kf = m

0.8°C / 5.5 m/°C = m → 0.0516 mol/kg

These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg

0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:

Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

3 0
3 years ago
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