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castortr0y [4]
3 years ago
8

Honestly reporting experimental findings is an example of using good

Chemistry
1 answer:
GaryK [48]3 years ago
4 0
I would say that honestly reporting experimental findings is an example of using good science because science is definitely about honesty in accepting experimental findings and realizing that one has to face up to the consequences and develop things from there rather than wishing the outcome was different. 
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What is the molarity of a solution prepared from 85. 0 g cacl2 in 300. 0 g of solution? the density of the solution is 1. 05 g/m
Yuki888 [10]

Answer: 2.6 M

Explanation:

8 0
3 years ago
QUESTION 5
mrs_skeptik [129]
Answer is b. Hope it help
7 0
3 years ago
What important processes in your body depend on mitosis
PilotLPTM [1.2K]
Mitosis makes new cells and replaces the cells that are old, lost or damaged
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3 years ago
determine the freezing point depression of a solution that contains 30.7 g glycerin (c3h8o3, molar mass
Blababa [14]

The freezing point depression of a solution containing 30.7 g of glycerin  is  calculated as -1.65°C

Equating :

It is given that,

Given mass of glycerin is = 30.7 grams (Solute)

Volume of water = 376 mL

K_{f}or molar -freezing-depression point is = 1.86°C/m

Molar mass of glycerin = 92.09 g/mole

Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:

Mass = Density X Volume

Mass = 1.0 g/mL X 376 mL

Mass = 376 g or 0.376 Kg

Using the formula of melting point depression, the equation becomes:

             ΔT_{f} = i ×K_{f} ×m

T⁰-T_{s}  = i *K_{f} *\frac{mass of glycerin}{molar mass of glycerin * mass of water     in     kg}

in which,

ΔT_{f} = change in freezing point

ΔT_{s} = freezing point of solution that has to be find

ΔT° = freezing point of water ()

Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.

Substituting the values in the above equation:

0⁰C₋T_{s} = 1 ×1.86°C/m ×\frac{30.7}{92.09g/mol * 0.376kg}

T_{s} = -1.65°C

Thus, the freezing point depression of a solution is  -1.65°C

<h2 />

Freezing point depression

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute

Is melting point elevation or depression?

Boiling point elevation is that the raising of a solvent's boiling point due to the addition of a solute. Similarly, melting point depression is the lowering of a solvent's freezing point due to the addition of a solute. In fact, because the boiling point of a solvent increases, its melting point decreases

Learn more about freezing point depression :

brainly.com/question/26525184

#SPJ4

8 0
1 year ago
λmax for the π → π* transition in ethylene is 170 nm. Is the HOMO-LUMO energy difference in ethylene greater than or less than t
-Dominant- [34]

Answer:

the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene

Explanation:

The λmax is the wavelength of maximum absorption. We could use it to calculate the HOMO-LUMO energy difference as follows:

For ethylene

E= hc/λ= 6.63×10^-34×3×10^8/170×10^-9= 1.17×10^-18J

For cis,trans−1,3−cyclooctadiene

E= hc/λ=6.63×10^-34×3×10^8/230×10^-9=8.6×10^-19J

Therefore, the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene

8 0
3 years ago
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