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nalin [4]
3 years ago
6

A base has a molarity of 1.5M w.R.T the hydroxyl ion concentration. If 7.35 cm3 of this base is taken and diluted to 147 cm3, th

en what is the concentration of the hydroxyl ion.
Chemistry
1 answer:
daser333 [38]3 years ago
4 0

Answer:

0.077M is the concentration of the hydroxyl ion

Explanation:

Dilution factor is the ratio between the aliquot that is taken of a solution and the total volume of the diluted solution.

For the problem, dilution factor is:

7.53cm³ / 147cm³ =<em> 0.05122</em>

To obtain molarity of a diluted solution you must multiply dilution factor and initial molarity of the solution, thus:

1.5 M × 0.05122 = <em>0.077M is the concentration of the hydroxyl ion</em>

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the density of a black hole is 22.61 g/cm3 i think

7 0
3 years ago
Read 2 more answers
The pK1, pK2, and pKR of the amino acid lysine are 2.2, 9.1, and 10.5, respectively. The pK1, pK2, and pKR of the amino acid arg
almond37 [142]

Answer:

pH 9,8 is likely to work best for this separation

Explanation:

Ion exchange chromatography is a chemical process where molecules are separated by affinity to an ion exchange resin. To separate different aminoacids you must use the isoelectric point (That is the pH where the aminoacid will be in its neutral form).

For lysine, PI is:

pH = \frac{1}{2} (9,1+10,5) = 9,8

For arginine:

pH = \frac{1}{2} (9,0+12,5) = 10,75

At pH = 9,8 lysine will be in its neutral form and will not be retain in the column but arginine will be in +1 charge being retained by the ion exchange resin.

Thus, <em>pH 9,8 is likely to work best for this separation</em>

<em></em>

I hope it helps!

5 0
3 years ago
Consider the reaction: CH4 + 2O2 = CO2 + 2H2O
crimeas [40]

Answer:

The answer to your question is:

a)  80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction                             CH4   +   2O2   ⇒   CO2   +   2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

                               16 g of CH4 ----------------  2(32) g of O2

                               20 g              --------------    x

                               x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .  

                                CH4   +   2O2   ⇒   CO2   +   2H2O

                                15 g         22 g

                                16 g of CH4 ----------------  64 g of O2

                                15 g of CH4  ---------------   x

                               x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

                                 CH4   +   2O2   ⇒   CO2   +   2H2O

                        64 g of O2 ------------------  44 g of CO2

                        22 g of O2 ------------------   x

                        x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________  

                           CH4   +   2O2   ⇒   CO2   +   2H2O                              

                           10.31 g                     20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0
2 years ago
The formation of covalent bonds is...
Evgen [1.6K]
A covalent compound is made when two or more nonmetal atoms bond by sharing valence electrons. The shared valence electrons between two nonmetal atoms is called a covalent bond. Covalent bonds are formed when two atoms begin sharing electrons.
4 0
2 years ago
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

3 0
2 years ago
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