German scientist Döbereiner was one responsible for grouping elements into triads based on most notably atomic mass, many of which can be found in the periodic table to be in a pattern (for example <span><span>Iron </span><span>Cobalt </span><span>Nickel, elements 26, 27, 28)</span></span>
Most Favored Nation status is an economic position in which a country enjoys the best trade terms given by its trading partner. That means it receives the lowest tariffs, the fewest trade barriers, and the highest import quotas (or none at all).
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Answer:
No
Explanation:
One mole of P₄ react with six moles of I₂ and gives 4 moles of PI₃.
When one gram phosphorus and 6 gram of iodine react they gives 8.234 g
ram of PI₃ .
Given data:
Mass of phosphorus = 1 g
Mass of iodine = 6 g
Mass of PI₃ = ?
Solution:
Chemical equation:
P₄ + 6I₂ → 4PI₃
Number of moles of P₄:
Number of moles = Mass /molar mass
Number of mole = 1 g / 123.9 g/mol
Number of moles = 0.01 mol
Number of moles of I₂:
Number of moles = Mass /molar mass
Number of moles = 6 g / 253.8 g/mol
Number of moles = 0.024 mol
Now we will compare the moles of PI₃ with I₂ and P₄.
I₂ : PI₃
6 : 4
0.024 :
4/6×0.024 = 0.02
P₄ : PI₃
1 : 4
0.01 : 4 × 0.01 = 0.04 mol
The number of moles of PI₃ produced by I₂ are less it will be limiting reactant.
Mass of PI₃ = moles × molar mass
Mass of PI₃ = 0.02 mol × 411.7 g/mol
Mass of PI₃ = 8.234 g
Because when equilibrium is reached, the reaction is still occurring in both directions, it's just that rate(forward) =rate(reverse) so there is no net change in the concentrations of the reactants or products.
Answer:
1.634 molL-1
Explanation:
The mol ration between NH3 and HCl is 1 : 1
Using Ca Va / Cb Vb = Na / Nb where a = acid and b = base
Na = 1
Nb = 1
Ca = 0.208 molL-1
Cb = ?
Va = 19.64 mL
Vb = 25.00mL
Solving for Cb
Cb = Ca Va / Vb
Cb = 0.208 * 19.64 / 25.0
Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)
Using the dilution equation;
C1V1 = C2V2
Initial Concentration, C1 = ?
Initial Volume, V1 = 25.00 mL
Final Volume, V2 = 250 mL
Final Concentration, C2 = 0.1634 molL-1
Solving for C1;
C1 = C2 * V2 / V1
C1 = 0.1634 * 250 / 25.00
C1 = 1.634 molL-1
