Word or symbol used to describe how data was measured
1 answer:
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Density of 3.52g/ml means for each milliliter of space there are 3.52 grams of diamond.
To set the stage:
Set the denominators equal:
Remove the denominators:
Isolate x by dividing both sides by 3.52:
To set the stage:
Set the denominators equal:
Remove the denominators:
Isolate x by dividing both sides by 3.52:
Answer:
9.0 g/cm³
Explanation:
Density can be computed with the formula:
Where:
D = Density
M = Mass
V = Volume
In your problem we are given:
84 cm³ = volume
760 g = mass
So we just plug in our given into the formula:
The answer would then be:
9.0 g/cm³
Answer:
ΔT=-747,13°C
Explanation:
Sensible heat is<em> the amount of thermal energy that is required to change the temperature of an object</em>, the equation for calculating the heat change is given by:
Q=msΔT
where:
- Q, heat that has been absorbed or realeased by the substance [J]
- m, mass of the substance [g]
- s, specific heat capacity [J/g°C] (
- ΔT, changes in the substance temperature [°C]
To solve the problem, we clear ΔT of the equation and then replace our data:
Q=msΔT
ΔT=Q/ms
Δ°C
<em>(Note that Q=-14900 J because there is a </em><u><em>LOST</em></u><em> of thermal energy)</em>
Thus, the change in temperature of the steel bar is -747,13°C, meaning that the temperature of the bar decreases.
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.
