Answer:
talk to them.
Explanation:
some people you can never get to like you but by talking to all types of people you find the people you can relate to and when you do it should feel natural for you.
Answer:
0.165 mol·L⁻¹
Explanation:
1. Write the <em>chemical equation</em> for the reaction.
HNO₃ + KOH ⟶ KNO₃ + H₂O
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2. Calculate the <em>moles of HNO₃</em>
c = n/V Multiply each side by V and transpose
n = Vc
V = 0.027 86 L
c = 0.1744 mol·L⁻¹ Calculate the moles of HNO₃
Moles of HNO₃ = 0.027 86 × 0.1744
Moles of HNO₃ = 4.859 × 10⁻³ mol HNO₃
===============
3. Calculate the <em>moles of KOH
</em>
1 mol KOH ≡ 1 mol HNO₃ Calculate the moles of KOH
Moles of KOH = 4.859 × 10⁻³× 1/1
Moles of KOH = 4.859 × 10⁻³ mol KOH
===============
4. Calculate the <em>molar concentration</em> of the KOH
V = 29.4 mL = 0.0294 L Calculate the concentration
c = 4.859 × 10⁻³/0.0294
c = 0.165 mol·L⁻¹
Answer:
See picture for answer
Explanation:
First to all, an aldehyde is a carbonated chain with a Carbonile within it chain. It's call aldehyde basically because the C = O is always at the end of the chain. When the C = O is on another position of the chain, is called a ketone.
Now, in this exercise we have an aldehyde with 5 carbons, so the first carbon is the C = O. The remaining four carbon belong to the chain. however, we need to have a branched chain in this molecular formula.
If this the case, this means that the longest chain cannot have 5 carbons. It should have 4 carbons as the longest chain. The remaining carbon, would one branched.
In this case, we only have two possible ways to have an aldehyde with a branched chain, and 4 carbons at max. One methyl in position 2, and the other in position 3.
The remaining aldehyde with branched chain, cannot have 4 carbons as longest, it should have 3 carbons with longest chain and 2 carbons as radicals (In this case, methyl). In this way, we just have all the aldehyde with this formula and at least one branched chain. The other possible ways would be conformers or isomers of the first three.
See picture for the structures of these 3 aldehydes, and their names.}
CI2 + 2Nal ---> 2NalCI + I2
Nal = 149.9 g/mol.
NaCI = 58.5 g/mol
2 mol Nal makes 2 mol NaCI
0.29 g Nal x ( 1 mol / 149.9 g) = 0.0019 mol NaCI
00.19 mol NaCI x ( 58.5 g / 1 mol ) 0.11 g NaCI ( 2 sig figs)