65) The molecules methane (CH4) and ammonia (NH3) both have four regions of high

Question

65) The molecules methane (CH4) and ammonia (NH3) both have four regions of high electron density around their central atoms, and yet these two molecules have different molecular geometries. Why are their molecule geometries different? ANSWER ON ATTACHED SHEETS 66) The chemical formula for octane is C8H18. How many moles of octane are present in a 1 gram sample of octane? you MuST sHow youR WORK ON ATTACHED sHEETS PLACE ANSWER HERE 67) How many grams of the compound dinitrogen pentoxide are needed to have 50.0 grams ofN? you MUST SHOW YOUR WORK ON ATTACHED SHEETS PLACE ANSWER HERE 68) How many grams of sulfur are present in 22.5 grams of disulfide carbon you MUST SHOW YOUR WORK ON ATTACHED SHEETS PLACE ANSWER HERE 69) The chemical formula for caffeine is C8H10N402. How many grams of carbon are present in a 12.46 gram sample of caffeine? you MUST sHow yoUR WORK ON ATTACHED sHEETs PLACE ANSWER HERE 70 What mass of Fe contains the same number of atoms as 15.83 grams of N? you MUST SHOW YOUR WORK ON ATTACHED SHEETS PLACE ANSWER HERE 71 80) ill in the blanks with the correct coefficient to balance the chemical equations below and on the next page. Fractional coefficients will not be accepted. If possible, coefficients must be reduced to the lowest whole number. Leave the blank empty if the coefficient should be one. H20 (D B203 (so H20 (l CO2 (g O2 (g) 2115 TAKE HOME QUIZ 2A. CHEM 121 FLOWERS

Explanation / Answer

Solution :-

Q65 ) The CH4 molecule have 4 bonding pair of electrons around the central carbon atom while in case of the NH3 It has 3 bonding pair of electrons and 1 lone pair of electrons around the central N atom.

The lone pair of the electrons repels the bonding pair of electrons in the NH3 molecules therefore the CH4 molecule have molecular geometry tetrahedral while in the NH3 molecule the the molecule geometry is trigonal pyramidal.

Q66) 12.4 g octane (C8H18)

Moles of octane = ?

Formula to calculate the moles is as follows

Moles = mass in gram / molar mass

Molar mass of the octane is 114.23 g per mol

Moles of octane = 12.4 g / 114.23 g per mol

                              = 0.109 mol octane

So moles of octane present are 0.109 mol octane

Q67) dinitrogen pentaoxide = N2O5

Mass of N2O5 = ?

Mass of N = 50.0 g

Lets first calculate the moles of the Nitrogen

Moles of N = 50.0 g / 14.007 g per mol = 3.5696 mol N

Now using the mole ratio lets calculate the moles of N2O5

3.5696 mol N * 1 mol N2O5 / 2 mol N = 1.7848 mol N2O5

Now lets convert the moles of N2O5 to its mass

Mass = moles * molar mass

Molar mass of N2O5 = 108.01 g per mol

Mass of N2O5 = 1.7848 mol * 108.01 g per mol

                          = 193 g N2O5

So we need 193 g N2O5 to get 50.0 g N

Q68) carbon disulfide = CS2

CS2 = 22.5 g

Mass of S= ?

Using the mole ratio of the CS2 and S we can calculate the mass of the S

(22.5 g CS2 * 1 mol / 76.139 g)*(2 mol S / 1 mol CS2)*(32.07 g / 1 mol S) = 19.0 g S

So it contains 19.0 g S

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.