Lemonade mix and water
Solvent: Water
Solute: Lemonade mix
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.
If one or more nucleotide pairs are deleted from a DNA strand, this is known as a frameshift mutation
<h3>
Define Frameshift Mutation</h3>
Insertions or deletions in the genome that are not multiples of three nucleotides are referred to as frameshift mutations. They are a particular class of insertion-deletion (indel) alterations that are present in polypeptides' coding sequences. Here, there are no multiples of three in the number of nucleotides that are added to or subtracted from the coding sequence. They may result from really basic alterations like the insertion or deletion of a single nucleotide.
<h3>
Frameshift mutations' effects</h3>
One of the most harmful modifications to a protein's coding sequence is a frameshift mutation. They are quite prone to produce non-functional proteins that frequently interfere with a cell's metabolic processes and result in significant alterations to polypeptide length and chemical makeup. Frameshift mutations can cause the mRNA to stop translating too soon and create an extended polypeptide.
Learn more about Frameshift mutations here:-
brainly.com/question/12732356
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Answer:
1,033.56 grams of carbon dioxide was emitted into the atmosphere.
Explanation:
Energy absorbed by pork,E =
(assuming)
Total energy produced by barbecue = Q
Percentage of energy absorbed by pork = 10%


Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.
Q = 
Moles of propane burnt to produce Q energy =n


According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:
carbon dioxide gas.
Mass of 23.49 moles of carbon dioxide gas:
23.49 mol × 44 g/mol =1,033.56 g
1,033.56 grams of carbon dioxide was emitted into the atmosphere.