Answer:
1.2 ( two significant digits)
Explanation:
5.4 - 4.21 answer should only have the the number of significant digits as the number in the equation with the least number of significant digits... in this case 5.4 is only two significant digits....so the answer should only have two significant digits
5.4 - 4.21 = 1.19 round to two digits = 1.2
Answer:
density=1.43 g/L
Explanation:
Since the density formula is density = mass / volume, we need to find out the mass of the gas and the volume is that of the container.
The mass of the gas is 130.0318 g-129.6375 g=0.3943 g
The gas volume is 276mL*(1L/1000mL) 0.276 L
density = mass / volume=0.3943g/0.276L
density =1.43g/L
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,

pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
Answer:
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