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andrey2020 [161]
3 years ago
7

Exactly 25.0 mL of an aqueous solution of barium hydroxide required 32.0 mL of 0.200 M nitric acid to neutralize it. What is the

concentration of the barium hydroxide solution?
Give only your numerical answer and do not include the M. For example, 0.262 or 1.25.​

Chemistry
2 answers:
Brrunno [24]3 years ago
7 0

Answer:

2000m the example 25.0\0.200

notka56 [123]3 years ago
6 0

Answer:

\frac{0.200  \times 320}{57}  = .112

Hope this helps

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An alkene X undergoes ozonolysis and gives two compounds Y and Z of molecular formula CaHO. Y and Z are functional isomers of ea
lys-0071 [83]

i. The given compound X is 2-methyl pent-2-ene. When it is reacted with ozone it forms an ozonide in the first step. In the second step the reduce to forms acetone and propanal.

ii. The structural formula of Y is CH_3-CO-CH_3 and Z is CH_3-CH_2-CHO.

iii. Alkenes, upon catalytic hydrogenation, form alkanes. This will occur in the presence of Nickel as the catalyst.

iv. The process of ozonolysis is useful in the field of pharmaceutics.

v. The test of unsaturation can be performed by passing a compound through Bromine water.

<h3>What is ozonolysis?</h3>

Ozonolysis is a reaction used in organic chemistry to determine the position of a carbon-carbon double bond in unsaturated compounds.

i. The given alkene X, that is subject to ozonolysis would be 2-methyl-2-pentene. Upon exposure to ozone, an ozonide is initially formed, after which it is broken down into 2 products - acetone and propanal, both with the molecular formula C₃H₆O.

The given compound X is 2-methyl pent-2-ene. When it is reacted with ozone it forms an ozonide in the first step. In the second step the reduce to forms acetone and propanal.

ii. The formulas of Y is CH_3-CO-CH_3 and Z is CH_3-CH_2-CHO. They are functional isomers as they have the same molecular formula but different functional groups - ketone and aldehyde.

iii. When alkenes undergo catalytic hydrogenation, they form alkanes, X will form 2 methyl petane on reaction with hydrogen gas in presence of Ni.

iv. The ozonolysis is used for the industrial-scale synthesis of pharmaceuticals.

v. The unsaturation of compound X can be proved by the bromine water test. As on reaction with it, the brown colour of bromine water becomes colourless due to the formation of dibromo alkane.

Learn more about the ozonolysis here:

brainly.com/question/14356308

#SPJ1

5 0
2 years ago
A student finds a rock on the way to school. In the laboratory he determines that the volume of the rock by placing the rock in
9966 [12]

Answer:

1.76 g/mL

Explanation:

You need to find the volume.  You can do this by subtracting the volume of the water and the rock by the volume of the water.

72.7 mL  -  50 mL  =  22.7 mL

Now that you have volume, divide the mass by the volume to find the density.

39.943 g/22.7 mL = 1.76 g/mL

7 0
3 years ago
Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemogl
Alex73 [517]

Without wasting much of our time, Here is the correct question.

Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemoglobin(aq) + O2(aq) -------> hemoglobin O2(aq) is first order in hemoglobin and first order in dissolved oxygen, with a rate constant of 4 × 10⁷ L mol⁻¹ s⁻¹. Calculate the initial rate at which oxygen will be bound to hemoglobin if the concentration of hemoglobin is 2 × 10⁻⁹ M and that of oxygen is 5 × 10⁻⁵M.

Answer:

4 × 10⁻⁶ M s⁻¹

Explanation:

The equation for the reaction between Hemoglobin molecules in blood that binds with oxygen molecule can be represent by:

hemoglobin_{(aq)  +  O_{2(aq)   ---------> hemoglobin.O_{2(aq)

Now, we are also being told to calculate only!, the  initial rate at which oxygen will be bound to hemoglobin.

So, If it is first order in hemoglobin and also first order in Oxygen molecule at the initial rate of the the reaction, therefore, the rate  for the reaction can be expressed as :

rate = k [hemoglobin_{(aq)}][O_{2(aq)}]

Let's not forget that we are so given some parameters;

where

k (rate constant) = 4 × 10⁷ L mol⁻¹ s⁻¹

[ hemoglobin_{(aq) ] = 2 × 10⁻⁹ M

[  O_{2(aq)  ]  =  5 × 10⁻⁵ M

Substituting our data given into the above rate formula, we have:

rate = (4 × 10⁷ L mol⁻¹ s⁻¹) × (2 × 10⁻⁹ M) × (5 × 10⁻⁵ M)

rate = 4 × 10⁻⁶ M s⁻¹     ( given that 1 M = 1 mol L⁻¹ )

∴ the initial rate at which oxygen will be bound to hemoglobin = 4 × 10⁻⁶ M s⁻¹

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