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Crank
3 years ago
14

Rank the following aqueous solutions from highest to lowest freezing point: 0.1 m FeCl3, 0.30 m glucose (C6H12O6), 0.15 m CaCl2.

Assume complete dissociation.

Chemistry
1 answer:
Tresset [83]3 years ago
8 0

Answer:

Explanation: See images below for explanation

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A chemist prepares a solution of magnesium chloride MgCl2 by measuring out 49.mg of MgCl2 into a 100.mL volumetric flask and fil
Flura [38]

Answer:  Molarity of Cl^- anions in the chemist's solution is 0.0104 M

Explanation:

Molarity : It is defined as the number of moles of solute present per liter of the solution.

Formula used :

Molarity=\frac{n\times 1000}{V_s}

where,

n= moles of solute

Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.049g}{95g/mol}=5.2\times 10^{-4}moles  

V_s = volume of solution in ml = 100 ml

Now put all the given values in the formula of molarity, we get

Molarity=\frac{5.2\times 10^{-4}moles\times 1000}{100ml}=5.2\times 10^{-3}mole/L

Therefore, the molarity of solution will be 5.2\times 10^{-3}mole/L

MgCl_2\rightarrow Mg^{2+}+2Cl^-

As 1 mole of MgCl_2 gives 2 moles of Cl^-

Thus 5.2\times 10^{-3}  moles of MgCl_2 gives =\frac{2}{1}\times 5.2\times 10^{-3}=0.0104

Thus the molarity of Cl^- anions in the chemist's solution is 0.0104 M

6 0
3 years ago
How did Bohr describe the arrangement of particles within an atom?
alexandr1967 [171]

Answer: A

Explanation: Protons and neutrons form the nucleus of the atom, with electrons orbiting it.

4 0
3 years ago
Read 2 more answers
The number of electrons is not equal to number protons if...?
sineoko [7]

Answer:

the answer is not correct

3 0
2 years ago
The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I
ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:  

Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\

Equation:  

3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\

Calculating the amount of MnCO_3

= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\=  13.1 \ kg

For point b:

Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\  of\  Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\

=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg

Equation:  

3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\

Calculating the amount of MnO_2:  

= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\=  6516 \ g \\\\=6.52 \ kg\\\\

7 0
3 years ago
Given the equation representing a reaction at equilibrium: N₂ + 3H₂ ⇄ 2NH₃ What occurs when the concentration of H₂ is increased
ValentinkaMS [17]
That would cause the equation to shift right, and make more NH3 and decrease the amount of N2
4 0
3 years ago
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