Answer:
12.34 amu
Explanation:
Let the 1st isotope be A
Let the 2nd isotope be B
Let the 3rd isotope be C
From the question given above, the following data were obtained:
1st Isotope (A):
Mass of A = 12.32 amu
Abundance (A%) = 19.5%
2nd isotope (B):
Mass of B = 13.08 amu
Abundance (B%) = 26.23%
3rd isotope (C):
Mass of C = 11.99 amu
Abundance (C%) = 54.27%
Atomic mass of X =?
The atomic mass of the element X can be obtained as follow:
Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]
= [(12.32 × 19.5)/100] + [(13.08 × 26.23)/100] + [(11.99 × 54.27)/100]
= 2.402 + 3.431 + 6.507
= 12.34 amu
Thus, the atomic mass of the element X is 12.34 amu
The specific heat of the iron can is determined as 0.449 J/g⁰C.
<h3>Specific heat of the iron can</h3>
The specific heat of the iron can is calculated as follows;
Q = mcΔθ
c = Q/mΔθ
where;
- Q is quantity of heat
- Δθ is change in temperature
- m is mass
c = 256/(50 x 11.4)
c = 0.449 J/g⁰C
Thus, the specific heat of the iron can is determined as 0.449 J/g⁰C.
Learn more about specific heat here: brainly.com/question/16559442
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The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
Learn more about pH of buffer:
brainly.com/question/21881762
The formulas indicate the number of elements in each compound. Notice the difference in ratios:
NaCl is 1:1, while H2O is 2:1.
