The correct answer would be 3.) Location of an electron
The density would be the same for the whole bar as well as one half of the bar. Density is a identity I believe, by this I mean that it stays the same no matter how little or how much of the same substance you have. Since density = mass / volume, half the bar has half of the weight as well as half of the volume of the whole bar, making the density the same.
For example, a block weighs 10 grams and has a volume of 5 ml. the density would be d = 10/5 or, d = 2g/ml
Half of the block weighs 5 grams and has a volume of 2.5 ml. The density is d = 5/2.5, or, d = 2 g/ml.
See, although there are different amounts of the same substance, their density is the same.
<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61
<u>Explanation:</u>
We are given:
Initial moles of iodine gas = 0.100 moles
Initial moles of hydrogen gas = 0.100 moles
Volume of container = 1.00 L
Molarity of the solution is calculated by the equation:
Equilibrium concentration of iodine gas = 0.0210 M
The chemical equation for the reaction of iodine gas and hydrogen gas follows:
<u>Initial:</u> 0.1 0.1
<u>At eqllm:</u> 0.1-x 0.1-x 2x
Evaluating the value of 'x'
The expression of 
for above equation follows:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
![[HI]_{eq}=2x=(2\times 0.079)=0.158M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.079%29%3D0.158M)
![[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D%280.1-x%29%3D%280.1-0.079%29%3D0.0210M)
![[I_2]_{eq}=0.0210M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.0210M)
Putting values in above expression, we get:
Hence, the value of equilibrium constant for the given reaction is 56.61
Answer:
Vf = 1.22 mL
Explanation:
If we assume that the pressure is constant and the number of moles does not change, we can say that the volume of the gas is modified in a directly ratio, to the Absolute Temperature.
Let's convert the values:
91°C + 273 = 364K
0.9°C + 273 = 273.9K
Volume decreases if the temperature is decreases
Volume increases if the T° increases
V₁ / T₁ = V₂ / T₂ → 1.63mL /364K = V₂ / 273.9K
V₂ = (1.63mL /364K) . 273.9K → 1.22 mL
