<span>HNO2 =====> H+ + NO2-
</span>I<span>nitial concentration</span> = 0.311
<span>C = -x,x,x </span>
<span>E = 0.311-x,x,x
</span>KNO2 ====>K+ + NO2-
<span>Initial concentration = 0.189 </span>
<span>C= -0.189,0.189,0.189 </span>
E = 0,0.189,0.189
Answer:
The answer to your question is V2 = 66.7 ml
Explanation:
Data
Volume 1 = V1 = 400 ml
Pressure 1 = P1 = 1 atm
Volume 2 = V2 = ?
Pressure 2 = P2 = 6 atm
Process
1.- To solve this problem use Boyle's law
P1V1 = P2V2
-solve for V2
V2 = P1V1 / P2
-Substitution
V2 = (1)(400) / 6
-Simplification
V2 = 400 / 6
-Result
V2 = 66.7 ml
Complete question:
Write the condensed formula from left to right, starting with (CH3)x where x is a number.
See attached image for the structure formula of the compound
Answer:
(CH₃)₂CHC(CH₃)₃ named as 2,2,3-Trimethylbutane
Explanation:
If we number the longest chain of the carbon starting from the left, we will observe that there are four carbons in the straight chain as shown in the image.
Starting from first carbon from the left of the carbon chain, at carbon number number 2, there two alkyl group, that is two methyl (CH3 is two). Also at carbon number 3, there are three alkyl group, that is three methyl (CH3 is three).
The condensed formula will be written as;
(CH₃)₂CHC(CH₃)₃
This compound is named as 2,2,3-Trimethylbutane, an isomer of Heptane
The change is that the water will freeze to 0 or minus I don’t know as I’m not to sure
The notion <span>an empty balloon have precisely the same apparent weight on a scale as a balloon filled with air depends on the diameter of the balloon. The weight of the balloon filled with air is equal to the mass of the balloon and the mass of the air inside. The mass of air inside is equal to the density of air multiplied by the volume of the balloon. If the balloon is large, then the two masses are equal whereas if not, the mass of air inside the inflation is neglible</span>