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Fudgin [204]
3 years ago
13

When an acetic acid solution is titrated with sodium hydroxide, the slope of the titration curve (pH versus volume of NaOH added

) increases when sodium hydroxide is first added. This change shows that ________
Chemistry
1 answer:
Tanya [424]3 years ago
6 0

Answer:

When an acetic acid solution is titrated with sodium hydroxide, the slope of the titration curve (pH versus volume of NaOH added) increases when sodium hydroxide is first added. This change shows that <u>acetic acid is being converted to sodium acetate.</u>

Explanation:

<u>ABOUT TITRATION -</u> Titration is a chemical analysis method for determining the amount of a constituent in a sample by adding an exact known quantity of another substance to the measured sample with which the desired constituent reacts in a specific, known proportion. A burette, which is essentially a long, graduated measuring tube with a stopcock and a delivery tube at the bottom end, is used to gradually administer a standard solution of titrating reagent, or titrant.

It is feasible to identify a good visual color indicator for many titration processes that will signify the end point at, or very near to, the equivalence point.

Acid-base titrations, precipitation titrations, complex-formation titrations, and oxidation-reduction (redox) titrations are examples of such titrations, which are categorised according to the nature of the chemical reaction that occurs between the sample and the titrant.

Titrations of metal ions with the reagent disodium ethylenediaminetetraacetate are the most important titrations based on complex-formation processes (a salt of edetic acid, or EDTA).

<u>Hence , the answer is that acetic acid is converted into sodium acetate.</u>

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g How many moles of NaOH are present in a sample if it is titrated to its equivalence point with 44.02 mL of 0.0885 M H2SO4? 2Na
Lisa [10]

Answer:

n_{base}=3.90x10^{-3}molNaOH

Explanation:

Hello!

In this case, since the sulfuric acid and sodium hydroxide react in a 1:2 mole ratio, given the reaction, we realize they have the following mole ratio at the equivalence point:

2*n_{acid}=n_{base}

Which in terms of concentrations and volumes is:

2*M_{acid}V_{acid}=n_{base}

Thus, we can plug in the volume and concentration of acid to find the moles of base:

n_{base}=0.04402L*0.0885\frac{mol}{L} \\\\n_{base}=3.90x10^{-3}molNaOH

Best regards!

8 0
3 years ago
What is the density of helium gas at 17 degrees celcius and 773 torr?
valina [46]

Answer:

0.172g/L

Explanation:

Step 1:

Data obtained from the question:

Temperature (T) = 17°C

Pressure (P) = 773 torr

Step 2:

Conversion to appropriate unit:

For pressure :

760 torr = 1 atm

Therefore, 773 torr = 773/760 = 1.02 atm

For temperature:

Temperature (Kelvin) = temperature (celsius) + 273

temperature (celsius) = 17°C

Temperature (Kelvin) = 17°C + 273 = 290K.

Step 3:

Obtaining an expression for the density.

From the ideal gas equation PV = nRT, we can obtain an equation for the density as follow:

PV = nRT. (1)

But: number of mole(n) = mass (m)/Molar Mass(M) i.e

n = m/M

Substitute the value of n into equation 1

PV = nRT

PV = mRT/M

Divide both side by m

PV /m = RT/M

Divide both side by P

V/m = RT/MP

Invert the above equation

m/V = MP/RT (2)

Recall:

density (d) = mass(m) / volume(V) i.e

d = m/V

Replace m/V in equation 2 with d

m/V = MP/RT

d = MP/RT.

Step 4:

Determination of the density.

Temperature (T) = 290K

Pressure (P) = 1.02 atm

Molar Mass of helium (M) = 4g/mol

Gas constant (R) = 0.082atm.L/Kmol

Density (d) =?

d = MP/RT

d = 4 x 1.02 / 0.082 x 290

d = 0.172g/L

Therefore, the density of the helium gas is 0.172g/L

7 0
3 years ago
`I am holding a balloon containing 439 mL of gas over my fireplace. The temperature and pressure of the gas inside the balloon i
Andrew [12]

Answer:

- 0.07 °C

Explanation:

At constant pressure and number of moles, Using Charle's law  

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = 439 mL  = 0.439 L ( 1 L = 0.001 mL )

V₂ = 0.378 L

T₁ = 317.15 K

T₂ = ?

Using above equation as:

\frac{0.439}{317.15}=\frac{0.378}{T_2}

T_2=\frac{0.378\cdot \:317.15}{0.439}=273.08\ K

The conversion of T(K) to T( °C) is shown below:

T( °C) = T(K) - 273.15  

So, <u>T = 273.08 - 273.15 °C = - 0.07 °C</u>

3 0
3 years ago
In a tissue that metabolizes glucose via the pentose phosphate pathway, C-1 of glucose would be expected to end up principally i
Mnenie [13.5K]

Answer:

A.

Carbon dioxide

Explanation:

In a tissue that metabolizes glucose via the pentose phosphate pathway, C-1 of glucose would be expected to end up principally in Carbon dioxide

5 0
2 years ago
Pls help :(
Ilia_Sergeevich [38]

I would say only two

7 0
3 years ago
Read 2 more answers
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