Answer:
0.0069
Explanation:
When converting between scientific notation try to remember that moving the decimal place to the left is positve and to the right is negative
Since you are changing from scientific to standard you will need to move to the left from the decimal point, making your answer 0.0069
Answer:
0.43 grams is the maximum mass of sodium sulfate that could be produced by the chemical reaction.
Explanation:
Mass of sulfuric acid = 0.98 g
Moles of sulfuric acid = 
Mass of sodium hydroxide = 0.240 g
Moles of sodium hydroxide = 

According to reaction, 2 moles of sodium hydroxide reacts with 1 mole of sulfuric acid , then 0.0060 moles of sodium hydroxide will react with :
of sulfuric acid
As we can see that we have 0.010 moles of sulfuric acid but only 0.0030 moles of sulfuric acid will react which indicates that it is in excessive amount where as sodium hydroxide is in limiting amount.
So, amount of sodium sulfate to be formed will depend upon moles of sodium hydroxide.
According to reaction, 2 moles of sodium hydroxide gives with 1 mole of sodium sulfate , then 0.0060 moles of sodium hydroxide will give :
of sodium sulfate
Mass of 0.0030 moles of sodium sulfate :
0.0030 mol × 142 g/mol = 0.426 g ≈ 0.43 g
0.43 grams is the maximum mass of sodium sulfate that could be produced by the chemical reaction.
Calculate the H positive from the pH equation: pH equals -log (H positive). This would be 10 to the -6.49. Let's call the acid HA. To calculate Ka in this equation, Ka equals H positive times A- over HA. HA is going to be the 0 0121. So, Ka=(10^-6.49)^2/0.0121. This equals 1.05*10^-13/0.0121. Ka then equals 8.65*10^-12.
We have that the Complete Expanded Structure of (CH3)2CHCH2OCH2CH3 is given in the attachment below
From the Question
(CH3)2CHCH2OCH2CH3
Generally for the condensed formula (CH3)2CHCH2OCH2CH3
We consider that this is a single bond connecting them
We consider
Hydrogen H(1)
Oxygen(8)
Carbon(6)
In conclusion
The Complete Expanded Structure of (CH3)2CHCH2OCH2CH3 is given in the attachment below.
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