Answer:
The three primary colors used when mixing dyes or paints are red, yellow, and blue. Other colors are often a mixture of these three colors. Try running a chromatography test again with non-primary-color markers, like purple, brown, and orange.
Explanation:
<h3><em>Mixtures that are suitable for separation by chromatography include inks, dyes and colouring agents in food. ... As the solvent soaks up the paper, it carries the mixtures with it. Different components of the mixture will move at different rates. This separates the mixture out.</em></h3>
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The molecular weight of water is <span>18.01528 g/mol.
So in 2.92 grams there are 2.92/</span>18.01528 = 0.1621 mol of particles.
1 mol contains 6,02214 × 10^<span>23 particles by definition.
So the nr of H2O molecules is </span>0.1621 * 6,02214 × 10^23 = 0,9761 × 10^23.
Every molecule has 2 H atoms, so you have to double that.
2* 0,9761 × 10^23 = 1.952 × 10^23.
The equilibrium constant is a value which represents the equilibrium of a reaction. It is a reaction quotient when the reaction reached equilibrium. If Keq is greater than 1, the mixture contains mostly the products. On the other hand, if Keq is less than 1, the mixture contains the reactants. For this case, the mixture contains mostly products.
Answer: It will take 8.2 minutes until the concentration decreases to 0.055 M
Explanation:
The time after which 99.9% reactions gets completed is 40 minutes
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
b) Time taken for 0.085 M to decrease to 0.055 M
Thus it will take 8.2 minutes until the concentration decreases to 0.055 M
Silver chloride produced : = 46.149 g
Limiting reagent : CuCl2
Excess remains := 3.74 g
<h3>Further explanation</h3>
Reaction
silver nitrate + copper(II) chloride ⇒ silver chloride + copper(II) nitrate
Required
silver chloride produced
limiting reagent
excess remains
Solution
Balanced equation
2AgNO3 (aq) + CuCl2 (s) → 2AgCl(s) + Cu(NO3)2(aq)
mol AgNO3 :
= 58.5 : 169,87 g/mol
= 0.344
mol CuCl2 :
=21.7 : 134,45 g/mol
= 0.161
mol ratio : coefficient of AgNO3 : CuCl2 :
= 0.344/2 : 0.161/1
= 0.172 : 0.161
CuCl2 as a limiting reagent
mol AgCl :
= 2/1 x 0.161
= 0.322
Mass AgCl :
= 0.322 x 143,32 g/mol
= 46.149 g
mol remains(unreacted) for AgNO3 :
= 0.344-(2/1 x 0.161)
= 0.022
mass AgNO3 remains :
= 0.022 x 169,87 g/mol
= 3.74 g
