<u>Answer:</u> The amount of water required to prepare given amount of salt is 398.4 mL
<u>Explanation:</u>
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

We are given:
Molarity of solution = 0.16 M
Given mass of manganese (II) nitrate tetrahydrate = 16 g
Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol
Putting values in above equation, we get:

Volume of water = Volume of solution = 398.4 mL
Hence, the amount of water required to prepare given amount of salt is 398.4 mL
Answer: Rate of decomposition of acetaldehyde in a solution is 
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
For a reaction : 
![Rate=k[A]^x](https://tex.z-dn.net/?f=Rate%3Dk%5BA%5D%5Ex)
k= rate constant
x = order of the reaction = 2


Thus rate of decomposition of acetaldehyde in a solution is
Answer:
The suitable equation for this reaction is
2CO + O₂ -----> 2CO₂
Here, we are given that we have 2 grams of O₂
From the equation, we can see that 2 * Moles of O₂ = Moles of CO₂
Moles of O₂:
2/32 = 1/16 moles
Therefore, the number of moles of CO₂ is twice the moles of O₂
Moles of CO₂ = 2 * 1/16
Moles of CO₂ formed = 1/8 moles
Mass of CO₂ formed = Molar mass of CO₂ * Moles of CO₂
Mass of CO₂ formed = 44 * 1/8
Mass of CO₂ formed = 5.5 grams
Hence, option B is correct
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Answer: 2 atoms
Explanation: 2 in Co2 means 2 atoms
Answer:
The normal amount of disaccharide would be produced, but fewer monosaccharides would be produced.
Explanation:
The first reaction, the conversion of starch into disaccharides, is catalyzed by the enzyme amylase. <u>Since amylase is present in a normal amount, a normal amount of disaccharides will be produced.</u>
In the second reaction, these disaccharides will be transformed into monosaccharides by a disaccharidase. However, since t<u>here is less disaccharidase, there will be fewer monosaccharides produced than if it was a normal amount of amylase.</u>