What length of this wire contains exactly 1.00 A gold wire has a diameter of 1.0

Question

What length of this wire contains exactly 1.00 A gold wire has a diameter of 1.00 mm. mole of gold? (density of Au 17.0 g/cm') 4. A) 2630 m B) 3.69 m C)251 m D)14.8m E) 62.7m A 1.375 g sample of mannitol, a sugar found in seaweed, is burned completely in oxygen to give 1.993 g of carbon dioxide and 0.9519 g of water. The empirical formula of mannitol is 5. A) CHO 6. Ammonia reacts with oxygen to form nitric oxide and water vapor 4NFb + 5 O2 4N0 + 6 H2O When 40.0 g NHs and 50.0 g Oz are allowed to react, which is the limiting reagent? A) NH B)0 C)NO D HO ENo reagent is limiting.

Explanation / Answer

4.

Molar mass of gold = 196.97 g/mol

Density = mass / volume

Volume = mass / density

= 196.97 / 17

= 11.586 cm³

Diameter = 1.00 mm

Radius = diameter / 2 = 1.00/2 mm= 0.5 mm

= 0.05 cm
Volume of wire = 3.14*r²*length
11.58 = 3.14*0.05²*length
length = 11.586/(3.14*0.05²)
length = 11.586/0.00785
length = 11.586/0.00785
length = 1,475.9 cm

= 14.759 m

= 14.8 m

5.

Given that mass of CO2 = 1.993 g

mass of H2O = 0.9519 g

now calculate the mole of C, H:

mass of CO2 = 1.993 g

dividing by the molar mass 44 = 1.993 g /44.0 g/ mole

=0.0453 moles of CO2
0.0453 moles of CO2 since there is 1 mole so C in CO2
= 0.0453 moles of C

0.0453 moles of C *12.01 g / mole =
0.544 g of C which is the mass of C in the compound

mass of H2O = 0.9519 g

0.9519 g/ 18.02 g/ mole= 0.0528 moles of H2O

since there are 2 moles of H in 1 moles of H2O =
0.1057 moles of H or
0.1057 moles of H *1.008 g/ mol=

0.1065 g of H which is the mass of H in the compound

mass of compound burnt = 1.3750 g

mass of C+H = 0.650 g
mass of O = 0.725 g by difference
moles of O = 0.04529

molar ratio of C : H :O = 0.0453 : 0.1057 : 0.0453

or after dividing by the smallest 0.0453 ….
molar ratio of C : H :O = 1.000 : 2.334 : 1.000
multiply by 3 to get whole numbers 3.000 7.001 3.000

empirical formula of mannitol is C3H7O3 ..

6.

4NH3 + 5O2 4NO + 6H2O

given 40.0 g NH3

50.0 g O2

Number of moles= amount in g / molar mass

= 40.0 g/ 17.03 g/ mole

= 2.35 mole NH3

Moles of O2 = 50 g /32.0 g/ mole

= 1.56 mole

Now calculate the amount of NH3 which is recated with this amount of O2:

1.56 mole O2 * 4 mole NH3/5 mole O2 = 1.248 mole NH3

Thus NH3 present in excess, O2 is limiting agent

The answer is B

The limiting agent has due to following properties:

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