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Ilia_Sergeevich [38]
3 years ago
6

Why do we use ammonium formate as a hydrogen surrogate instead of using hydrogen gas in this experiment?

Chemistry
1 answer:
NemiM [27]3 years ago
7 0

Answer:

Ammonium formate can also be used in palladium on carbon (Pd/C) reduction of functional groups. In the presence of Pd/C, ammonium formate decomposes to hydrogen, carbon dioxide, and ammonia.

Explanation:

This hydrogen gas is adsorbed onto the surface of the palladium metal, where it can react with various functional groups.

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D.) It cannot be broken down into a simple substance through chemical means...
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In a pure metal, the electrons can be thought of as [ Select ] throughout the metal. Using molecular orbital theory, there [ Sel
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Explanation:

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3 years ago
What is the m of a solution where 0.500 moles of a salt are dissolved in 100.0 ml of solution? 25.0m 5.00m 50.0m o.500m 2.50m?
My name is Ann [436]
Molarity=moles/litre
molarity=0.5/0.1
molarity=5.00m
3 0
3 years ago
A person suffering from hyponatremia has a sodium ion concentration in the blood of 0.116 M and a total blood volume of 4.7 L .
klemol [59]

Explanation:

Normal moles of Na^{+} = volume × normal concentration

                              = 4.7 × 0.139 = 0.6533 mol

Moles of Na^{+} in hyponatremia blood = volume × hyponatremia concentration

                              = 4.7 × 0.116 = 0.5452 mol

Moles of NaCl to be added = moles of extra Na^{+} needed

                            = 0.6533 mol - 0.5452 mol = 0.1081 mol

Mass of NaCl = moles × molar mass of NaCl

                        = 0.1081 mol × 58.443

                        = 6.317g

                        = 6.32 g (approx)

Thus, we can conclude that mass of sodium chloride would need to be added to the blood is 6.32 g.                    

3 0
3 years ago
How many grams of ammonium chloride (gram formula mass= 53.5 g) are contained in .500 L of a 2.00 M solution?
Elza [17]

Answer:

53.5g of NH4Cl

Explanation:

First, we need to obtain the number of mole of NH4Cl. This is illustrated below:

Volume = 0.5L

Molarity = 2M

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 2 x 0.5

Mole = 1mole

Now, let us convert 1mole of NH4Cl to gram. This is illustrated below:

Molar Mass of NH4Cl = 53.5g/mol

Number of mole = 1

Mass =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass = 1 x 53.5

Mass = 53.5g

Therefore, 53.5g of NH4Cl is contained in the solution.

8 0
3 years ago
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