For the answer to the question above, I can't help you directly because I don't have a calculator right now. But I'll show you how to solve this.
<span>use the freezing point depression formula for this one: delta T = i * m * K where K is a constant, m is the molality (mol solute/kg solvent), and i is the van'hoff factor the van hoff factor is the number of ions that your salt dissociates into. Since it's an ALKALI flouride salt, how many ions? k is just a constant, you get it from a table in your textbook somewhere So you have everything to solve for the molality of the solution, once you did that, multiplying it by the mass of water to find the mols of the salt. Take the mass of the salt and divide by this mols to figure out the molar mass, and then compare it with the periodic table to identify the salt.
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<u>Mole solute</u> x mass of Water = Mol solute<u>
</u>kg Solvent
then
Mass of solute x <u> 1 </u> = molar mass
mole of solute
Answer:
4.5moles
Explanation:
First, let us balance the equation given from the question. This is illustrated below:
KClO3 —> KCl + O2
There are 2 atoms of O on the right side and 3 atoms on the left. It can be balance by putting 2 in front of KClO3 and 3 in of O2 as shown below
2KClO3 —> KCl + 3O2
Now, we have 2 atoms each of K and Cl on the left side and 1atom each of K and Cl on the right. It can be balance by putting 2 in front of KCl as shown below:
2KClO3 —> 2KCl + 3O2
Now the equation is balanced.
From the balanced equation,
2 moles of KClO3 produced 3 moles of O2.
Therefore, 3 moles of KClO3 will produce = (3 x 3) /2 = 4.5moles of O2.
Therefore 3 moles of KClO3 will produce 4.5 moles of O2
Answer:
At Equilibrium
[COCl₂] = 0.226 M
[CO] = 0.054 M
[Cl₂] = 0.054 M
Explanation:
Given that;
equilibrium constant Kc = 1.29 × 10⁻² at 600k
the equilibrium concentrations of reactant and products = ?
when 0.280 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K. [COCl²]
Concentration of COCl₂ = 0.280 / 1.00 = 0.280 M
COCl₂(g) ----------> CO(g) + Cl₂(g)
0.280 0 0 ------------ Initial
-x x x
(0.280 - x) x x ----------- equilibrium
we know that; solid does not take part in equilibrium constant expression
so
KC = [CO][Cl₂] / COCl₂
we substitute
1.29 × 10⁻² = x² / (0.280 - x)
0.0129 (0.280 - x) = x²
x² = 0.003612 - 0.0129x
x² + 0.0129x - 0.003612 = 0
x = -b±√(b² - 4ac) / 2a
we substitute
x = [-(0.0129)±√((0.0129)² - 4×1×(-0.003612))] / [2 × 1 ]
x = [-0.0129 ± √( 0.00017 + 0.01445)] / 2
x = [-0.0129 ± 0.1209] / 2
Acceptable value of x =[ -0.0129 + 0.1209] / 2
x = 0.108 / 2
x = 0.054
At equilibrium
[COCl₂] = (0.280 - x) = 0.280 - 0.054 = 0.226 M
[CO] = 0.054 M
[Cl₂] = 0.054 M
Active metals are those metals in the group 1 of the periodic table.
Electronegativity is the trend to atract electrons.
Active metals have few valence electron, because their last shell is of the kind ns^1 or ns^2
Then, these atoms do not trend to attract electrons. The most electronegative atomos are those who have 7 valence elecfrons; this is their last shell is of the kind ns^7, because when they attract one electron to its valence shell they will complete 8 electrons which is the most stable configuration.
