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yarga [219]
2 years ago
15

What is the mass, in grams, of solute in 255mL of a 0.0525M solution of KMnO4 (MM =

Chemistry
1 answer:
aliya0001 [1]2 years ago
7 0

Answer:

Mass = 2.12 g

Explanation:

Given data:

Volume of KMnO₄ = 255 mL (255/1000 = 0.255 L)

Molarity = 0.0525 M

Mass in gram = ?

Solution:

First of all we will calculate the number of moles.

<em>Molarity = number of moles of solute / volume in litter</em>

0.0525 M = number of moles of solute / 0.255 L

Number of moles of solute = 0.0525 M ×0.255 L

Number of moles of solute = 0.0134 mol

Mass in gram:

<em>Number of moles = mass/ molar mass</em>

Mass = moles × molar mass

Mass = 0.0134 mol × 158.04 g/mol

Mass = 2.12 g

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For the reaction CO2(g) + H2(g)CO(g) + H20(g)
Studentka2010 [4]

Answer:

The ΔG° is 29 kJ and the reaction is favored towards reactant.

Explanation:

Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,  

ΔG° = ΔH°rxn - TΔS°rxn

= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K

= 41.2 kJ - 12.2 kJ

= 29 kJ

As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.  

5 0
2 years ago
Copper sulfate can crystalize as an anhydrate (no water in the crystal) or as three different hydrates (varying amounts of water
Maurinko [17]

Answer:

Two factors that might have a affect of which copper sulphate mineral will occur at a given location  is:

A. Copper sulphate high solubility in water

B. Also it binds nicely with the sediments  or the crystal.

Explanation:

As it is mentioned here that copper sulphate can be crystallized as an anhydrate which means that their  is no waterin those crystals or can be as of those three different hydrates whose crystal structure varies with the amount of water present in it.

The four forms are also given of the copper sulphate are:

  • Bonatite
  • Boothite
  • Chalcanthite
  • Chalcocyanite

So, the two factors that might give an affect which type of copper sulphate  mineral willoccur at a given location is:

A. The copper sulphate high solubility in water.

B. It binds extremely nicely with the sediments or say to the crystal. It is  also regulated by plants.

8 0
3 years ago
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

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