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aleksley [76]
4 years ago
15

A column is filled with four different liquids of different densities. A red liquid, a blue liquid, a green liquid, and a purple

liquid are all added into the column. This chart shows the densities of the liquids.
A 2-column table with 4 rows titled densities of liquids. The first column titled Liquids has entries red, blue, green, purple The second column has entries 1.2 grams per centimeter cubed, 1.6 grams per centimeter cubed, 0.8 grams per centimeter cubed, 0.1 grams per centimeter cubed.

In what order would the liquids would arrange themselves, from top to bottom?

1.blue, red, purple, green.
2.purple, green, red, blue.
3.purple, green, blue, red.
4.blue, red, green, purple.
Chemistry
2 answers:
deff fn [24]4 years ago
8 0

Answer:

Blue, Red, Green, Purple

Explanation:

zhuklara [117]4 years ago
7 0
4. Blue, Red, Green, Purple

(Lowest densities on the top, highest densities on the bottom)
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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
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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!

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