Calculate the average atomic mass for element X(please help)

Wt Avg At Mass = 39.0229 amu

Explanation:

The Weight Average Atomic Mass in atomic mass units (amu) is the sum of the Weight Average Mass contributions of each isotopic mass. That is,

Wt. Avg. At. Mass (amu) = ∑ Wt. Avg. Contributions of each isotopic mass

Wt. Avg. Contribution = fractional abundance x isotopic mass(amu)

For this problem data:

Isotope   %Abundance     fractional         Isotopic             Wt Avg Isotopic

                                        abundance      mass(amu)        Contribution (amu)

   1                  9.67%           0.0967               38                      3.6746 amu

   2               78.68%           0.7868               39                    30.6852 amu

   3                 11.34%           0.1134                 40                      4.5360 amu

   4                  0.31%           0.0031                41                       0.1271  amu

Wt. Avg. Atomic Mass (amu) = ∑ Wt. Avg. Contributions  =  39.0229 amu

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