A, a heterogeneous mixture is a mixture where all the particles are not made up of one uniform composition. you can distinguish the different particles
Answer:
q = 14049 J
Explanation:
q = m*c*(t2-t1)
q = 350 * 0.892 * (70-25) =
312.2 * 45 = 14049 J
I might be getting a little confused but I could be right.
Hope this helps!
To solve this we use the
equation,
<span> M1V1 = M2V2</span>
<span> where M1 is the
concentration of the stock solution, V1 is the volume of the stock solution, M2
is the concentration of the new solution and V2 is its volume.</span>
<span>2.0 M x V1 = 0.50 M x 200 mL</span>
<span>V1 = 50 mL needed</span>
Answer:
The answer to your question is: density = 4 g/cm³
Explanation:
Data
Volume = 100 cm³
Mass = 400 g
Density = ?
Formula
density = mass/volume
substitution
density = 400/100 = 4 g/cm³
Ok first, we have to create a balanced equation for the dissolution of nitrous acid.
HNO2 <-> H(+) + NO2(-)
Next, create an ICE table
HNO2 <--> H+ NO2-
[]i 0.139M 0M 0M
Δ[] -x +x +x
[]f 0.139-x x x
Then, using the concentration equation, you get
4.5x10^-4 = [H+][NO2-]/[HNO2]
4.5x10^-4 = x*x / .139 - x
However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,
assume 0.139-x ≈ 0.139
4.5x10^-4 = x^2/0.139
Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.
We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.
Then to find percent dissociation, you do final concentration/initial concentration.
0.007M/0.139M = .0503 or
≈5.03% dissociation.
