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3 years ago

What is the pOH of a 0.00037 M solution of barium hydroxide

Chemistry

1 answer:

ruslelena [56]3 years ago

5 0

Answer: pOH = 3.13

Ba(OH)2 is a very basic substance. The general formula for pOH is - log(OH)

Barium Hydroxide produces 2 moles of OH for every mole of Ba(OH)2 present in the solution.

0.00037 M = 3.7 * 10^-4 Ba(OH)2 will produce 2 *0.00037 M of OH-

OH- = 2* 0.00037 = 0.00074

pOH = - log(0.00074) = 3.13

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A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 60.00 gram sample of the alcohol p

marin [14]

Answer:

$C_2H_6O$

Explanation:

The first step is the calculation of the moles of $H_2O$ and $CO_2$ , so:

$114.6~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=2.6~mol~of~CO_2$

$70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O$

Now, in 1 mol of CO2 we have 1 mol of C and in 1 mol of $H_2O$ we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to calculate the grams of C and O and then do the substraction form the initial amount, so:

$2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C$

$3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H$

$Total~grams=~31.25~+~7.82=39.08~g$

$grams~of~O=60.00~g-~39.08~g=20.92~g~of~O$

Now we can convert the grams of O to moles, so:

$20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O$

The next step is to divide all the mol values by the smallest one:

$O=\frac{1.30~mol~O}{1.30~mol~O}=~1$

$C=\frac{2.6~mol~C}{1.30~mol~O}=~2$

$H=\frac{7.82~mol~H}{1.30~mol~O}=6$

Therefore the formula is $C_2H_6O$

6 0

3 years ago

When reacting a solution of the following secondary hydroxyl amine with 2-butene under very high temperatures, the desired produ

Naddika [18.5K]

Answer:

Mechanism for top reaction is combustion while the side product at bottom path oximes

Explanation:

Hydroxiamine is a reducing agent that requires high temperature to combust in it'd ageous state to produced oxime while water is eliminated. The reverse is the case because the production of water during combustion I does not yield the fight result as it can e very irritating.

Increased temperature stimulates the protons to become free to combust and react with 2 butene and other aldehyde during chemical reaction.

8 0

2 years ago

What mass of octane must be burned in order to liberate 5270 kj of heat? δhcomb = -5471 kj/mol?

katrin2010 [14]

As,
5471 kJ heat is given by = 1 mole of Octane
Then,
5310 kJ heat will be given by = X moles of Octane

Solving for X,
X = (5310 kJ × 1 mol) ÷ 5471 kJ

X = 0.970 moles of Ocatne

So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
Moles = mass / M.mass
Or,
Mass = Moles × M.mass
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Mass = 110.83 g of Octane

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2 years ago

What are atoms made of

Olegator [25]

Protons, neutrons, and electrons.

7 0

3 years ago

Read 2 more answers

Chemicals that are able to slow or stop the reaction rate are inhibitors. reactants. products. catalysts.

erma4kov [3.2K]

Answer:inhibitors

Explanation: