Answer: Molarity of the prepared solution is 
Explanation:
Molarity is defined as the number of moles of solute dissolved per liter of the solution.
where,
n= moles of solute
= volume of solution in ml = 500 ml
moles of solute =
Now put all the given values in the formula of molarity, we get
Thus molarity of the prepared solution is
Answer:
The answer is 529.6 g
Explanation:
I took the test and somehow got it right :0 but yea this is the correct answer. I hope this helps :)
Answer:
Yes
Explanation:
Neither Calcium nor hydrogen nor oxygen are alkali metals
- Alkali metals are Group 1 elements
- They are very soft and mellable.
- H,Li,Na,K,Cs,Fr are included in it however H is not a actual member
Here
Ca(OH)_2 is the compound
Vant Hoff factor is 3
As it's greater than 1 it's a very strong base .
The pOH of this comes around 1.5 to 1.3
So very heavy ionic dissociation
Strong Alkali
A + B → AB: ✔ synthesis
AB → A + B: ✔ decomposition
Hydrocarbon + O2 → CO2 + H2O: <span>✔ combustion</span>
AB + CD → AD + CB: <span>✔ replacement</span>
The absorbance reported by the defective instrument was 0.3933.
Absorbance A = - log₁₀ T
Tm = transmittance measured by spectrophotometer
Tm = 0.44
Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654
True absorbance can be calculated by true transmittance, Tm = T+S(α-T)
S = fraction of stray light = 6%= 6/100 = 0.06
α= 1, ideal case
T = true transmittance of the sample
Tm = T+S(α-T)
now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233
therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)
i.e; 0.3933
To know more about transmittance click here:
brainly.com/question/17088180
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