How many moles of pentane are in 1.48x1021

multiply the number of each atom with its molecular mass. (see the periodic table)

5*12 + 12*1 = 72 gram per mole.

Or more accurately: 5*12.01078 + 12*1.007947 = 72.149264 g/mole

1.48 * 1021 number of molecules of pentane = 0.0002458 mol

Given data:

Number of molecules of pentane = 1.48 * 1021

Number of moles = ?

Solution:

1 mole = 6.02 × 1023 molecules.

If

1 molecule = 1/6.02 × 1023

So

1.48 × 1021 molecules = (1/6.02 × 1023) × 1.48 × 1021

                                    = 2.458 × 10-3 mol

                                    = 0.0002458 mol

So there are 0.000245 moles of pentane, if the number of molecules are 1.48 * 1021. Because 1 mole mole contain Avogadro's number of molecules.

Which is equal to 6.02 × 1023 molecules.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol