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Mazyrski [523]
2 years ago
10

Three test tubes contain white crystalline organic solids A, B, and C, each of which melts at 149-150 degrees C. A 50-50 mixture

of A and B melts at 130-139 degrees C. A 50-50 mixture of A and C melts at 149-150 degrees C. In what range would a 50-50 mixture of B and C probably melt? What can you say about the identities of A, B, and C?
Chemistry
1 answer:
Digiron [165]2 years ago
3 0

Answer:

The mixture of B and C will melt as 130 - 139 ºC.

Explanation:

If the melting point (130 - 139 ºC) of a mixture of A and B is lower than the pure substances that is 149 - 150 ºC that means that one of these susbtances is an impurity because reduces the melting point.

If the melting point of the mixture of A and C is the same as the pure substances, we can deduce that they are the same substances, also because the melting point when a substance is pure is just 1 or 2 ºC like in this case.

So in a Mixture of B and C is going to be like the first case of the mixture among A and B, because the B substance is the impurity.

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m(Br₂) = 1,250 g.
m(BrₓOy) = 1,876 g.
n(Br₂) = m(Br₂) ÷ M(Br₂).
n(Br₂) = 1,25 g ÷ 159,81 g/mol.
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n(Br) = 2 · 0,0078 mol = 0,0156 mol.
m(O₃) = 1,876 g - 1,25 g = 0,626 g.
n(O₃) = 0,626 g ÷ 48 g/mol = 0,013 mol.
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n(Br) : n(O) = 1 : 2,5.
5 0
2 years ago
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How does 0.5 m sucrose (molecular mass 342) solution compare to 0.5 m glucose (molecular mass 180) solution?
mash [69]

Answer : Both solutions contain 3.011 X 10^{23} molecules.

Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain 3.011 x 10^{23} molecules.

Avogadro's Number is  N_{A} =  6.022 X 10^{23} which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.

Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.

Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.

Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.

We can calculate the number of molecules for each;

Number of molecules  = N_{A} X M;

∴  Number of molecules =  6.022 X 10^{23} X 0.5 mol/L X 1 L which will be  = 3.011 X 10^{23}

Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.

7 0
2 years ago
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The dipole moment (μ) of HBr (a polar covalent molecule) is 0.838D (debye), and its percent ionic character is 12.4 % . Estimate
myrzilka [38]

<span>When two electrical charges, of opposite sign and equal magnitude, are separated by a distance, a dipole is established. The size of a dipole is measured by its dipole moment (</span>μμ). Dipole moment is measured in Debye units, which is equal to the distance between the charges multiplied by the charge (1 Debye equals 3.34×10−30Cm3.34×10−30Cm). The dipole moment of a molecule can be calculated by Equation 1.11.1:

μ = qr

where

<span> <span>μ⃗ μ→ is the dipole moment vector</span> <span>qiqi is the magnitude of the ithith charge, and</span> <span>r⃗ ir→i is the vector representing the position of ithith charge.</span> </span>

 

r = μ/q

<span>r = [0.838D(3.34×10−30 C⋅m/ 1D)]/ (1.6×10−19 C) *0.124
</span> r = 1.41 x10^-10 m

 

7 0
2 years ago
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If fluorine 20 undergoes beta decay , what will it become ?
e-lub [12.9K]
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Ok now the reaction.

 20       20        0
F  -> Ne    +     e
 9         10       -1

Remember the atomic number determines the nature of the element ( i.e what elemnt it is).
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7 0
2 years ago
According to the following balanced reaction, how many moles of HNO3 are formed from 8.44 moles of NO2 if there is plenty of wat
kotegsom [21]

Answer:

5.63 mol.

Explanation:

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<em>3NO₂(s) + H₂O(l) → 2HNO₃(aq) + NO(g), </em>

It is clear that 3 mol of NO₂ reacts with 1 mol of H₂O to produce 2 mol of HNO₃ and 1 mol of NO.

<em>Water is present as an excess reactant and NO₂ is limiting reactant.</em>

<em></em>

  • To find the no. of moles of HNO₃ produced:

3 mol of NO₂ produces → 2 mol of HNO₃, from stichiometry.

8.44 mol of NO₂ produces → ??? mol of HNO₃.

∴ The no. of moles of HNO₃ are formed = (8.44 mol)(2 mol)/(3 mol) = 5.63 mol.

3 0
2 years ago
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