1 millimeter (ml) is equal to 0.1 centimeters (cm), so all you have to do is take however many mm you have, and multiply that by 0.1. Then you have your answer. So, 59 times 0.1 is 5.9
~Silver
Sika have more food choices because they eat both grasses and shrubs, compared to the white-tailed dear who only eats shrubs.
The question is incomplete, here is the complete question:
Calculate the mole fraction of the ionic species KCl in the solution A solution was prepared by dissolving 43.0 g of KCl in 225 g of water.
<u>Answer:</u> The mole fraction of KCl in the solution is 0.044
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of water = 225 g
Molar mass of water = 18 g/mol
Putting values in equation 1, we get:
Given mass of KCl = 43 g
Molar mass of KCl = 74.55 g/mol
Putting values in equation 1, we get:
Mole fraction of a substance is given by:
Moles of KCl = 0.577 moles
Total moles = [0.577 + 12.5] = 13.077 moles
Putting values in above equation, we get:
Hence, the mole fraction of KCl in the solution is 0.044
Answer:
The principle of superposition states that the oldest sedimentary rock units are at the bottom, and the youngest are at the top. Based on this, layer C is oldest, followed by B and A. hope that helps
Explanation:
A=top
B=middle
C=bottom
Answer:
1-(tert-butoxy)-2-methylpropane
Note: there is a mistake in formula, the correct formula is (CH₃)₂-CH-CH₂-O-C(CH₃)₃ not (CH₃)₂-CH-CH₂-O(CH₃)₃, because oxygen is a divalent compound.
Explanation:
<em>Structural formula is attached</em>
IUPAC naming rules
1. start numbering the chain from the functional group. In this compound we start from oxygen side.
2. Here we can see that at position 1 there is an oxy group along with a tertiary carbon having three methyl groups. So we write it as 1-tert-butoxy. Which means that there is a methoxy group at position 1 along with a tertiary carbon.
3. At position 2 we can see that there is a methyl group attached to the main chain, so we write it as 2-methyl.
4. Now we count the total number of carbons in the main chain. As we can see that there are 3 carbons in the remaining or parent chain, so we write it as propane
5. So the IUPAC name of the compound will be 1-(tert-butoxy)-2-methylpropane