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zhenek [66]
3 years ago
8

As student used a pen to draw a line across a piece of chromatography paper. he then placed a sample of dye on the drawn line fo

r analysis . is the student doing the right thing? why?​
Chemistry
1 answer:
AleksAgata [21]3 years ago
3 0

Answer is in the photo. I can only upload it to a file hosting service. link below!

tinyurl.com/wpazsebu

You might be interested in
What best describes the outer shell of the atoms in group d on this illustration of the periodic table?.
Lubov Fominskaja [6]

The outer shell of the atoms in group D of the periodic table consists of elements whose outermost shell is full

Their outermost shell consist of elements with valence electrons

The group D elements are found in between S and P blocks of the periodic table

<h3>What is an element?</h3>

An element is a substance which cannot be split into simpler units by ordinary process.

Few examples of elements which belongs to group D include the following:

  • Scandium
  • Titanium
  • Vanadium
  • Chromium
  • Manganese
  • Iron
  • Nickel
  • Copper
  • Zinc

Learn more about elements:

brainly.com/question/14514242

5 0
3 years ago
Which of the elements Magnesium, Calcium, Serum, Barium, looses electron readily​
anzhelika [568]

Answer:

mg. ca . ba. loses very easily

Explanation:

ok

6 0
3 years ago
Calculate the pH at the equivalence point when 22.0 mL of 0.200 M hydroxylamine, HONH2, is titrated with 0.15 M HCl. (Kb for HON
solniwko [45]

Answer:

pH = 3.513

Explanation:

Hello there!

In this case, since this titration is carried out via the following neutralization reaction:

HONH_2+HCl\rightarrow HONH_3^+Cl^-

We can see the 1:1 mole ratio of the acid to the base and also to the resulting acidic salt as it comes from the strong HCl and the weak hydroxylamine. Thus, we first compute the required volume of HCl as shown below:

V_{HCl}=\frac{22.0mL*0.200M}{0.15M}=29.3mL

Now, we can see that the moles of acid, base and acidic salt are all:

0.0220L*0.200mol/L=0.0044mol

And therefore the concentration of the salt at the equivalence point is:

[HONH_3^+Cl^-]=\frac{0.0044mol}{0.022L+0.0293L} =0.0858M

Next, for the calculation of the pH, we need to write the ionization of the weak part of the salt as it is able to form some hydroxylamine as it is the weak base:

HONH_3^++H_2O\rightleftharpoons H_3O^++HONH_2

Whereas the equilibrium expression is:

Ka=\frac{[H_3O^+][HONH_2]}{[HONH_3^+]}

Whereas Ka is computed by considering Kw and Kb of hydroxylamine:

Ka=\frac{Kw}{Kb}=\frac{1x10^{-14}}{9.10x10^{-9}}  \\\\Ka=1.10x10^{-6}

So we can write:

1.10x10^{-6}=\frac{x^2}{0.0858-x}

And neglect the x on bottom to obtain:

1.10x10^{-6}=\frac{x^2}{0.0858}\\\\x=\sqrt{1.10x10^{-6}*0.0858}=3.07x10^{-4}M

And since x=[H3O+] we obtain the following pH:

pH=-log(3.07x10^{-4})\\\\pH=3.513

Regards!

4 0
3 years ago
If a 40 g sample of NaCl is composed of 39.34% Sodium, how many grams of<br> that sample are Sodium?
alina1380 [7]

Answer:

15.7g

Explanation:

mass percent=mass of element/mass of compound *100

39.34=unknown(x)/40g×100

39.34=5x/2

multiplying both sides by 2,we get

5x=78.68

devide by 5 on both sides

x=15.74g Na

4 0
3 years ago
HELP PLEASE!
Maslowich
I think you don’t need the help anymore so can I just take these real quick ?
8 0
3 years ago
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