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Rainbow [258]
2 years ago
12

In the Acid-Base Chemistry Experiment the measurement of pH will be used to explore acid-base concepts, analyze an unknown, and

determine percent K and Fe in the iron salt. The pH is measured using pH paper, acid-base indicators, and a combination pH electrode. Quantitative measurements are made using a combination pH electrode interfaced with a computer in a pH titration using dilute NaOH. The molarity of a strong acid (HCl), equivalent weight and pKa of an unknown weak acid, and percent K and Fe in your iron salt are all determined in this way. Samples of the iron salt are run through a cation exchange column to prepare for titration. Choose either True or False for each of the following statements : A solution with a hydronium molarity of 0.00045 is acidic. pH is a way to express the hydronium concentration over a wide range.Percent K and Fe are determined by doing ion exchange then a pH titration.About 0.2M HCl is the titrant used for the pH titrations. A Lewis base is a species that can donate a proton to an acid
Chemistry
1 answer:
aleksley [76]2 years ago
5 0

Answer

a)  A solution with a hydronium molarity of 0.00045 is acidic. True

Doing the calculus of pH

pH= -Log [H^{+}] = -Log (0.00045)

b) pH is a way to express the hydronium concentration over a wide range. True

pH means –Log[H+] and this value is used to express a wide range of hydronium concentration sometimes obtaining pH minor than zero.

c) Percent K and Fe are determined by doing ion exchange then a pH titration. False

Usually, Fe is determined by redox titration with potassium permanganate due to it’s more accurate. On the other hand, K is determined usually by volumetric process which includes precipitation like potassium picrate precipitate

d) About 0.2M HCl is the reagent used for the pH titrations. False.

In order to do pH titration, it is possible to use a wide range of HCl concentrations and other acids as reagent if the analyte is a basic compound. Otherwise, if the analyte is an acid compound you should use a basic compound as reagent.

e) A Lewis base is specie that can donate a proton to an acid. False

A Lewis base is an electron pair donor.

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How much heat is required to raise the temperature of 670g of water from 25.7"C to 66,0°C? The specific heat
inna [77]

Answer:

Explanation:

q= mc theta

where,

Q = heat gained

m = mass of the substance = 670g

c = heat capacity of water= 4.1 J/g°C    

theta =Change in temperature=( 66-25.7)

Now put all the given values in the above formula, we get the amount of heat needed.

q= mctheta

q=670*4.1*(66-25.7)

  =670*4.1*40.3

=110704.1

8 0
2 years ago
Carl's chemistry teacher asked him to make a 1 M sodium chloride solution. Carl measured 58.44 grams NaCl, added it to a volumet
Nastasia [14]

Answer:

After measuring the solute, Carl should first dissolve the solid in a small amount of DI water before diluting to the total volume.

Explanation:

To ensure that all the solute dissolves in the solution, first dissolve the solid in less than the total volume of solution needed.

4 0
2 years ago
Read 2 more answers
the half life of the radioactive element strontium-90 is 29.1 years. If 16 grams of strontium-90 are initially present, how many
8_murik_8 [283]

Answer:

4 g after 58.2 years

0.0156 After 291 years

Explanation:

Given data:

Half-life of strontium-90 = 29.1 years

Initially present: 16g

mass present after 58.2 years =?

Mass present after 291 years =?

Solution:

Formula:

how much mass remains =1/ 2n (original mass) ……… (1)

Where “n” is the number of half lives

to find n

For 58.2 years

n = 58.2 years /29.1 years

n= 2

or  291 years

n = 291 years /29.1 years

n= 10

Put values in equation (1)

Mass after 58.2 years

mass remains =1/ 22 (16g)

mass remains =1/ 4 (16g)

 mass remains = 4g

Mass after 58.2 years

mass remains =1/ 210 (16g)

mass remains =1/ 1024 (16g)

mass remains = 0.0156g

8 0
2 years ago
Read 2 more answers
Consider a beaker of water sitting on the pan of an electronic scale that has been tared. A metal weight hanging from a string i
natka813 [3]

Answer:

See explanation below for answers

Explanation:

We know that the balance is tared, so the innitial weight would be zero. Now, let's answer this by parts.

a) mass of displaced water.

In this case all we need to do is to substract the 0.70 with the 0.13 g. so:

mW = 0.70 - 0.13

mW = 0.57 g of water

b) Volume of water.

In this case, we have the density of water, so we use the formula for density and solve for volume:

d = m/V

V = m/d

Replacing:

Vw = 0.57/0.9982

Vw = 0.5710 mL of water

c) volume of the metal weight

In this case the volume would be the volume displaced of water, which would be 0.5710 mL

d) the mass of the metal weight.

In this case, it would be the mass when the metal weight hits the bottom which is 0.70 g

e) density.

using the above formula of density we calculate the density of the metal

d = 0.70 / 0.5710

d = 1.2259 g/mL

4 0
2 years ago
One of the emission spectral lines for Be31 has a wavelength of 253.4 nm for an electronic transition that begins in the state w
max2010maxim [7]

Explanation:

The given data is as follows.

   \lambda = 253.4 nm = 253.4 \times 10^{-9}m      (as 1 nm = 10^{-9})

            n_{1} = 5,        n_{2} = ?

Relation between energy and wavelength is as follows.

                    E = \frac{hc}{\lambda}

                       = \frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{253.4 \times 10^{-9}}

                       = 0.0784 \times 10^{-17} J

                       = 7.84 \times 10^{-19} J

Hence, energy released is 7.84 \times 10^{-19} J.

Also, we known that change in energy will be as follows.

     \Delta E = -2.178 \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}

where, Z = atomic number of the given element

 7.84 \times 10^{-19} J = -2.178 \times (4)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}

    \frac{7.84 \times 10^{-19} J}{34.848} = \frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}

      0.02 + 0.04 = \frac{1}{n^{2}_{1}}

                      n_{1} = \sqrt{\frac{1}{0.06}}

                          = 4

Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.

4 0
2 years ago
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