it small because that bond is not very polar notice that all of those arrows they cancel. because they point in opposite directions. that's why BH 3 is nonpolar if we were to draw the Lewis structure of bf3. ... so even though the boron fluorine bond is polar the molecule as a whole is nonpolar.
Answer:
Wt. Avg. Atomic Weight => 63.35457 amu
Explanation:
Given Isotopic %Abundance fractional Wt Avg
At. Mass (amu) abundance contribution
Cu-63 62.93 69.09 0.6909 43.4783
Cu-65 64.9278 20.0668 0.200668 20.0668
Wt Average of all isotopes = ∑Wt Avg Contributions
= 43.4783 amu + 20.0668 amu = 63.35457 amu
Answer:
When heat is added to a substance, the molecules and atoms vibrate faster. As atoms vibrate faster, the space between atoms increases. The motion and spacing of the particles determines the state of matter of the substance. The end result of increased molecular motion is that the object expands and takes up more space.
hope this helps!!!!!!
<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
