Homogeneous because it completely dissolved so its now one. Heterogeneous would be when they don't dissolve and you see 2 separate substances like sand in water.
Answer:
this is the minutes and hours this is representing
Answer:
2.8087*10^-12 kJ per mole of reaction (2.8087*10^-12 kJ/mol).
Explanation:
To calculate the energy produced, we need to write a balanced equation for the reaction and determine the change in the masses of the reactants and products. Afterward, we can use the energy equation to determine the energy produced. The balanced equation for the nuclear reaction is shown below:
³₁H + ²₁H ⇒⁴₂He + ¹₀n
The masses of atoms are ³₁H is 3.01605 amu, ²₁H is 2.0140 amu, ⁴₂He is 4.00260 amu, and ¹₀n is 1.008665 amu.
change in mass Δm = (3.01605+2.0140) - (4.00260+1.008665) = 0.0188 amu
Energy produced, E = m*C^2
C is the speed of light = 3*10^8 m/s and 1 amu = 1.66*10^-27 kg
Therefore:
E = 0.0188*1.66*10^-27 * (3*10^8)^2 = 2.8087*10^-12 kJ per mole of reaction.
Therefore, in scientific notation, the energy released is 2.8087*10^-12 kJ/mol
Answer:
0,040 M
Explanation:
The global reaction of the problem is:
Al(OH) (s) + OH⁻ ⇄ Al(OH)₂⁻(aq) K= 40
The equation of equilibrium is:
K = ![\frac{[Al(OH)_{2} ^-]}{[Al(OH)][OH^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAl%28OH%29_%7B2%7D%20%5E-%5D%7D%7B%5BAl%28OH%29%5D%5BOH%5E-%5D%7D)
The concentration of OH⁻ is:
pOH = 14 - pH = <em>3</em>
pOH = -log [OH⁻]
[OH⁻] = 1x10⁻³
Thus:
40 = ![\frac{[Al(OH)_{2} ^-]}{[Al(OH)][1x10^{-3}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAl%28OH%29_%7B2%7D%20%5E-%5D%7D%7B%5BAl%28OH%29%5D%5B1x10%5E%7B-3%7D%5D%7D)
<em>0,04M = ![\frac{[Al(OH)_{2} ^-]}{[Al(OH)]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAl%28OH%29_%7B2%7D%20%5E-%5D%7D%7B%5BAl%28OH%29%5D%7D)
</em>
This means that 0,04 M are the number of moles that the solvent can dissolve in 1L, in other words, solubility.
I hope it helps!
First, make sure all three measurements have the same unit (I recommend meters because that's what the question calls for)
1.0m=1.0m
5.6cm=0.056m
2.1mm=0.0021m
Then, you just want to multiply the three together to get the total volume.
1.0X0.056X0.0021=0.0001176 or 0.00012, if your teacher wants you to use proper sig figs
