The density of aluminum is 2.70 g/cm^3? How many aluminum atoms are in a block of aluminum with a volume of 114.2 cm^3?

68.77×10²³ atoms

Explanation:

Given data:

Density of Al = 2.70 g/cm³

Volume of Al block = 114.2 cm³

Number of atoms of Al = ?

Solution:

First of all we will determine the mass of Al.

D = m/v

2.70 g/cm³ = m/114.2 cm³

m = 2.70 g/cm³ × 114.2 cm³

m = 308.34 g

Now we will calculate the number of moles of Al.

Number of moles = mass/molar mass

Number of moles = 308.34 g/27 g/mol

Number of moles = 11.42 mol

Number of atoms of Al:

1 mole = 6.022×10²³ atoms

11.42 mol × 6.022×10²³ atoms/1 mol

68.77×10²³ atoms

where is the following

explanation:

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