Potassium carbonate, K 2CO 3, sodium iodide, NaI, potassium bromide, KBr, methanol, CH 3OH, and ammonium chloride, NH 4Cl, are s
slava [35]
Answer:
Potassium carbonate (K₂CO₃)
Explanation:
The compounds dissociate into ions in water, as follows:
K₂CO₃ → 2 K⁺ + CO₃⁻ ⇒ 3 dissolved particles per mole
NaI → Na⁺ + I⁻ ⇒ 2 dissolved particles per mole
KBr → K⁺ + Br⁻ ⇒ 2 dissolved particles per mole
CH₃OH → CH₃O⁻ + H⁺ ⇒ 2 dissolved particles per mole
NH₄Cl → NH₄⁺ + Cl⁻ ⇒ 2 dissolved particles per mole
Therefore, the largest number of dissolved particles per mole of dissolved solute is produced by potassium carbonate (K₂CO₃).
Answer:
0.15 l of 4.0 m stock KCl solution should betaken
Explanation:
N1V1=N2V2
6*0.1=V2*4
V2=0.15L
CH4 + 2 O2 ---> CO2 + 2 H2O Q = 891,6 kJ / mol CH4
1 mol CH4 = 16 g
16 g ---- 891,6 kJ
x g ----- 272 kJ
x = 272 kJ × 16 g / 891,6 kJ = 4,88 g
You must burn 4,88 g of CH4.
:-) ;-)
212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.
So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then,
So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.