Answer:4.48483134×10 to the power of 19 m to the power of 2/s
Explanation:
You should work it out some and see
Answer:
Mean = 0.503
Confidence interval = 0.513, 0.493
Explanation:
Further explanations and calculations are found in the attachment below
Answer is the last one, it will not ionize, it already has an octet of electrons in it's valance shell
The percent by mass of nitric acid in the mixture is 3.36 %
2 HNO3 + Ba(OH)2 --->Ba(NO3)2 + 2H2O
so 1 mole Ba(OH)2 neutralizes 2 mol HNO3
moles of Ba(OH)2 present = molarity of base* volume of base
= 0.229 M * 15.6 ml
= 3.5724 milli mols
mols of HNO3 neutralized = 2* mols of Ca(OH)2 used
= 2* 3.5724 milli mols = 7.1448 x10^-3 mols
molar mass of HNO3 = 63.01 g/mol
mass of HNO3 present = molar mass * mols of HNO3
= 63.01 g/mol * 7.1448 x10^-3 mols
=0.4502 g
mass of sample = 13.4 g
mass % = mass of HNO3/mass of sample * 100
= 0.4502/13.4 *100
= 3.36 %
Nitric acid is a colorless, fuming, and distinctly corrosive liquid that could be a not unusual laboratory reagent and an important commercial chemical for the manufacture of fertilizers and explosives.
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Answer:
-) 2-methylbut-2-ene
-) 2-methylbut-1-ene
-) 3-methylbut-1-ene
Explanation:
in this case, the hydration of alkenes is a <u>marknovnikov reaction</u>, this means that the "OH" group would be added in the <u>most substituted carbon</u> of the double bond. (Figure 1)
For 2-methylbut-2-ene the most substituted carbon is the <u>tertiary carbon</u> (the carbon in the right of the double bond), so we will obtain the desired molecule. In 2-methylbut-1-ene the most substituted carbon is again the <u>tertiary carbon</u> (the carbon in the bottom of the double bond), so we will obtain 2-methyl-2-butanol. Finally, for 3-methylbut-1-ene the carbocation would be formed on carbon 3, this is a secondary carbocation. We can obtain a most stable carbocation if we do a <u>hydride shift</u> (Figure 2). With this new molecule is possible to obtain 3-methylbut-1-ene.