Answer:
B. SONAR (Sound Navigation Ranging)
Explanation:
Just took the test
Answer:
1.0 ° C
Explanation:
The molar mass for Sodium Nitrate NaNO₃ = (23+14+(16×3)) = 85
Number of moles of NaNO₃ = mass of NaNO₃ /molar mass of NaNO₃
⇒ 17/85 = 1.38 moles
Since 1 mole of NaNO₃ dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed.
when 1.38 mole of NaNO₃ dissolved in 1 cubic decimeter of water, x kJ of heat energy is absorbed..
Then; x kJ of 1.38 mole of NaNo₃ = 1.38 × 40 kJ =55.2 kJ of heat absorbed.
Using the relation : Q = mcΔT to determine the temperature drop ; we get:
55.2 = 17 × 4 (ΔT)
55.2 = 68 ΔT
ΔT= 0.8 ° C
ΔT ≅ 1.0 ° C
Therefore, the drop in temperature when 17.0g of sodium nitrate is dissolved in 1 cubic decimeter of water is 1.0 ° C
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.
The answer is 4.
Gases have low densities, because of the increased space between hight-energy particles.
