Answer:
both
Explanation:
id say that it could occur but also not as much. the moon would be smaller and further from the earth to where we would barely be able to see it. if the full moon is barely visible then im sure the total solar eclipse wouldn't be as noticeable as it is now. but thats just my opinion
The final volume V₂=4.962 L
<h3>Further explanation</h3>
Given
T₁=20 + 273 = 293 K
P₁= 1 atm
V₁ = 4 L
T₂=100+273 = 373 K
P₂=780 torr=1,02632 atm
Required
The final volume
Solution
Combined gas law :
P₁V₁/T₁=P₂V₂/T₂
Input the value :
V₂=(P₁V₁T₂)/(P₂T₁)
V₂=(1 x 4 x 373)/(1.02632 x 293)
V₂=4.962 L
Answer:
0.03682 mL of mercury
Explanation:
We know the density of the mercury which is 13.58 g/mL
density = mass / volume
volume = mass / density
Now we can calculate the volume of 0.5 g of mercury:
volume = 0.5 / 13.58 = 0.03682 mL of mercury
One of the many ways in order to solve for the vapor pressure of pure components at a given temperature is through the Antoine's equation which is written below,
P = 10^(A - B/C+T)
where A, B, and C are constants and T is the temperature in °C and P is the vapor pressure in mm Hg.
For hexane,
A = 7.01
B = 1246.33
C = 232.988
Substituting the known values,
P = 10^(7.01 - 1246.33/232.988+25)
<em> P = 151.199 mm Hg</em>
Answer:
Explanation:
This is an example of a limiting reactant question, and is very common as a general chemistry problem.
We first see the balanced equation, that is:
2CuCl2+4KI→2CuI+4KCl+I2
We first need to find the limiting reactant
We see that 0.56 g of copper(II) chloride (CuCl2) reacts with 0.64 g of potassium iodide (KI) . So, let's convert those amounts into moles.
Copper(II) chloride has a molar mass of
134.45 g/mol . So in 0.56 g of copper(II) chloride, then there exist
0.56g134.45g/mol≈4.17⋅10−3 mol
Potassium iodide has a molar mass of
166 g/mol . So, in 0.64 g of potassium iodide, there exist
if it wrong i am sorry
