Answer:- 0.273 kg
Solution:- A double replacement reaction takes place. The balanced equation is:
We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.
= 63.8 g aluminum nitrate
From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.
We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.
= 
sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.
= 0.273 kg
So, 0.273 kg of 35% m/m sodium chlorate solution are required.
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<u>ANSWER</u></h3>
2. Neon, only
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<u>EXPLANATION</u></h3>
When Na and F combine they form an electron configuration of 2-8. Na electron configuration is 2-8-1 while F is 2-7, so when they form an ionic bond F will gain Na outermost electron to complete its valence shell due to having a higher negativity. Neon has an electron configuration of 2-8 but argon has 2-8-8.
Window cleaner is ammonia dissolved in water. This is an example of a _______.
Thin layer chromatography(TLC) works with the principle of separation through adsorption.
It is used in the isolation and extraction of lipids through the following steps:
- apply the lipid samples spots in the bottom of the plate.
- also apply sample solution to the marked spot
- pour the mobile phase into the TLC chamber and use a moist filter paper to cover it. this is done to maintain equal humidity.
- then place the plate in the TLC chamber and close it with a lid.
- the plate is immersed into the solvent (mobile phase) for its development. this is done, keeping in mind that the sample spot should be above the solvent.
- once the sample spots are developed, they are removed and dried.
- this is later viewed using the UV light chamber to see the isolation of the lipid sample.
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Answer:
Alphabet C :NADH is oxidized,CO2 is reduced
