SHOW ALL THE WORK 30. a) A new gaseous hydride of boron, B x H y, has been synth

Question

                                                    SHOW ALL THE WORK

30. a) A new gaseous hydride of boron, BxHy, has been synthesized and collected at 25.0 0 C. The sample is analyzed and found to contain 2.206g of boron and 0.344g of hydrogen. what is the empirical formula of this compound?

B) This new gas occupies 849 mL. the pressure in the laboratry is 745 mm Hg. what is the molar mass of this compound?

C) what is the molecular formula of this compound?

  

the following information for bromine trifluoride, BrF supply the fol i) Lewis structure cture ormat (3 points) b) Electron domain/pair geometry c) Molecular geometry (molecular shape) d) Type of hybridization around the central atom e) Is BrF, polar or nonpolar? 33, A 432-g quantity of NaOH is added to 50.0 grams of water at 2230C in a constant-pressure calorimeter. After the NaOH totally dissolve the temperature of the solution is 28.50 C Calculate the change in enthalpy, in kJ/mol, for this reaction. The specific heat of the solution is 399J/g C. (8 points) 34. When an aqueous solutions ofCoCh and Na2CO3 are mixed amage ta precipitate forms. a) Write the balanced molecular equation, including all products and phases/states for the reaction.

Explanation / Answer

A)

Let total mass be 100 gm.

B: 2.206 g

H: 0.344 g

Divide by molar mass to get number of moles of each:

B: 2.206/10.81 = 0.204

H: 0.344/1.008 = 0.341

Divide by smallest to get simplest whole number ratio:

B: 0.204/0.204 = 1

H: 0.341/0.204 = 1.67

MULTIPLY BOTH BY 3

B:1*3 = 3

H : 1.67*3 = 5

So empirical formula is:B3H5

B)

Given:

P= 745.0 mm Hg = (745.0/760) atm = 0.98 atm

V= 849.0 mL = (849.0/1000) L = 0.849 L

T = 25.0 oC = (25.0+273) K = 298 K

use:

P * V = n*R*T

0.98 atm * 0.849 L = n * 0.0821 atm.L/mol.K * 298.0 K

n = 3.402*10^-2 mol

now use:

number of mol = mass / molar mass

3.402*10^-2 mol = (2.206 + 0.344) g / MM

MM = 75 g/mol

C)

Molar mass of B3H5,

MM = 3*MM(B) + 5*MM(H)

= 3*10.81 + 5*1.008

= 37.47 g/mol

Now we have:

Molar mass = 75.0 g/mol

Empirical formula mass = 37.47 g/mol

Multiplying factor = molar mass / empirical formula mass

= 75.0/37.47

= 2

Hence the molecular formula is : B6H10

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