Answer:
f. Sn^4+
c. second
e. Al^3+
d. third
Explanation:
This question comes from a quantitative analysis showing the flowchart of a common scheme for identifying cations.
Now, from the separation scheme, Let's assume that Sn⁴⁺ & Al³⁺ were given; Then, Yes, the separation will work.
However, there will be occurrence of precipitation after the 1st step1.
So, the <u>Sn⁴⁺</u> cation will precipitate after the <u>second </u>step. Then the <u>Al³⁺</u> cation will precipitate after the <u>third</u> step.
Answer:
.
Explanation
In HX , X is more electronegative than Y so HX will ionise more because of ionic bond between H and X . On the other hand H₂Y will be less polar as compared to HX so it will ionise to a lesser extent . Hence Ka will be more for HX . Ka represents the degree of ionisation of acid . Higher the ionisation , higher is the value of Ka . H₂Y which is less polar will ionise less and hence it will have lesser value of Ka .
Hence H₂Y will have value of 10⁻⁷ and HX will have value of ka equal to 10⁹ .
Moles of methanol = 9.27x10^24/6.02x10^23 = 15.398 moles.
Mass of methanol = moles of methanol x molar mass of methanol
= 15.398 x 32.042
= 493.38 grams.
Hope this helps!
Answer:
X = 2
Explanation:
As you know, the rate of a first-order reaction depends linearly on the concentration of a single reactant. The rate of a first-order reaction that takes the form
Answer:
Explanation:
In weight/volume (w/v) terms,
1 ppm = 1g m-3 = 1 mg L-1 = 1 μg mL-1
200 mL = 0.2 L
15 / 0.2 mg L-1 =75 ppm
