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Answer:
177.277amu
Explanation:
the total occuring isotopes for Hafnium is =6.
First isotope had an atomic weight of 173.940amu
Second isotope =175.941amu
Third isotope =176.943amu
Fourth isotope=177.944amu
Fifth isotope. =178.946amu
sixth isotope .179.947amu
<em>Avera</em><em>ge</em><em> </em><em>ato</em><em>mic</em><em> </em><em>wei</em><em>ght</em><em> </em><em>of</em><em> </em><em>Haf</em><em>nium</em><em>=</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>the </em><em>atomi</em><em>c</em><em> </em><em>weights</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>iso</em><em>topes</em><em>/</em><em> </em><em>Tota</em><em>l</em><em> </em><em>occu</em><em>ring</em><em> </em><em>isotopes</em>
Thus, 173.940amu+175.941amu+176.943amu+177.944amu+178.946amu+179.947amu.= 1063.661amu
Average atomic weight= 1063.661amu /6 = 177.2768333amu
= 177.277amu to 3 decimal places.
Answer:
Explanation:
El hielo encierra la primavera antes de espolvorear la sal.
Cuando se rocía sal sobre el hielo, disminuye el punto de fusión del hielo 32 ° F a un poco por debajo de 32 °, por lo tanto, se acumula.
A medida que el hielo se vuelve a congelar, encierra la primavera
N = 4 moles of Ar2, P = 1.90 atm, V = ?
T = 50C = 273 + 50K = 323K
PV = nRT --> V = nRT/P
V = (4)(.0821)(323)/1.90
V = 106.07/ 1.9
V = 55.8 L
Explanation:
We have a molecule composed of 3" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; font-style: normal; font-variant: inherit; font-weight: normal; one; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">33 iron atoms, and 4" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: p; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">44 atoms of another element. We are given the following information: it has 2.36 g" role="presentation" style="box-sizing: inherit; margin: 0px; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">2.36 g2.36 g of iron for 3.26 g" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; font-style: normal; font-variant: inherit; font-weight: -wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">3.26 g3.26 g of molecule.
I want to find the molar mass of the compound, I have tried so far:
m=3.26 g=0.00326 kg" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; font-style: normal; font-variant: inherit; font-weight: normal; font-stretch: inherit; line-height: normal; font-family: inherit; font-size: 15px; vertical-align: baseline; display: inline; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">m=3.26 g=0.00326 kgm=3.26 g=0.00326 kg
Since it has 3" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; nt-variant: inherit; font-weight: normal; font-stretch: inherit; line-height: normal; font-family: inherit; font-size: 15px; vertical-align: baseline; display: inline; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">FeFe and 4" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; font-style: normal; font-variant: inherit; font-weight: normal; font-stretch: inherit; line-height: normal; ; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">44 atoms of an unknown substance, therefore:
3+4=7 atoms,1 mol=6.022⋅1023 atoms76.022⋅1023=1.16⋅10−23" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; font-style: normal; font-variant: : ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; width: 10000em !important; ; font-family: inherit; eight: none; min-width: 0px; min-height: 0px; position: relative;">0.003260.00326 by 1.16⋅10−23" role="presentation"; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">1.16⋅10−231.16⋅10−23 and I obtained 2.79429⋅1019" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; f inherit; font-size: 15px; vertical-align: baseline; display: inline; text-indent: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">2.79429⋅10
IT'S TOTAL ANSWER OF ITS AND THIS QUESTION IS IN MATHEMATION FINAL EXAM. PLEASE GIVE❤ AND MARK ME A BRAINLIST