Answer:
D (But as a heads up, you wrote barium instead of Bromine)
Explanation:
The Potassium atom will lose an electron since its valence shell only has one, while Bromine has 7 electrons in its valence shell. Potassium wants to get rid of its one electron, Bromine wants to gain that one electron to get a full shell.
Potassium will become a CATION with a positive charge (since it lost an electron), Bromine will become an ANION with a negative charge (since it gained an electron)
Answer:
Final pH: 9.49.
Round to two decimal places as in the question: 9.5.
Explanation:
The conjugate of B is a cation that contains one more proton than B. The conjugate of B is an acid. As a result, B is a weak base.
What's the pKb of base B?
Consider the Henderson-Hasselbalch equation for buffers of a weak base and its conjugate acid ion.
![\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctext%7BpOH%7D%20%3D%20%5Ctext%7BpK%7D_b%20%2B%20%5Clog%7B%5Cfrac%7B%5B%5Ctext%7BSalt%7D%5D%7D%7B%5B%5Ctext%7BBase%7D%5D%7D%7D)
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![\displaystyle \text{pK}_b = \text{pOH} -\log{\frac{[\text{Salt}]}{[\text{Base}]}}\\\phantom{\text{pK}_b} = 4.5 - \log{\frac{0.750}{1.05}} \\\phantom{\text{pK}_b} =4.64613](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctext%7BpK%7D_b%20%3D%20%5Ctext%7BpOH%7D%20-%5Clog%7B%5Cfrac%7B%5B%5Ctext%7BSalt%7D%5D%7D%7B%5B%5Ctext%7BBase%7D%5D%7D%7D%5C%5C%5Cphantom%7B%5Ctext%7BpK%7D_b%7D%20%3D%204.5%20-%20%5Clog%7B%5Cfrac%7B0.750%7D%7B1.05%7D%7D%20%5C%5C%5Cphantom%7B%5Ctext%7BpK%7D_b%7D%20%3D4.64613)
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What's the new salt-to-base ratio?
The 0.005 mol of HCl will convert 0.005 mol of base B to its conjugate acid ion BH⁺.
Initial:
;
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After adding the HCl:
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Assume that the volume is still 0.5 L:
![\displaystyle [\text{B}] = \frac{n}{V} = \frac{0.520}{0.5} = 1.04\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BB%7D%5D%20%3D%20%5Cfrac%7Bn%7D%7BV%7D%20%3D%20%5Cfrac%7B0.520%7D%7B0.5%7D%20%3D%201.04%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
.![\displaystyle [\text{BH}^{+}] = \frac{n}{V} = \frac{0.380}{0.5} = 0.760\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BBH%7D%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7Bn%7D%7BV%7D%20%3D%20%5Cfrac%7B0.380%7D%7B0.5%7D%20%3D%200.760%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
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What's will be the pH of the solution?
Apply the Henderson-Hasselbalch equation again:
![\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}} = 4.64613 + \log{\frac{0.760}{1.04}} = 4.50991](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctext%7BpOH%7D%20%3D%20%5Ctext%7BpK%7D_b%20%2B%20%5Clog%7B%5Cfrac%7B%5B%5Ctext%7BSalt%7D%5D%7D%7B%5B%5Ctext%7BBase%7D%5D%7D%7D%20%3D%204.64613%20%2B%20%5Clog%7B%5Cfrac%7B0.760%7D%7B1.04%7D%7D%20%3D%204.50991)
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The final pH is slightly smaller than the initial pH. That's expected due to the hydrochloric acid. However, the change is small due to the nature of buffer solutions: adding a small amount of acid or base won't significantly impact the pH of the solution.
Answer:
A. Refracted
Explanation:
Incident rays parallel to the optical axis are reflected from the mirror.
Given data:
Hydrogen (H) = 3.730 % by mass
Carbon (C) = 44.44%
Nitrogen (N) = 51.83 %
This means that if the sample weighs 100 g then:
Mass of H = 3.730 g
Mass of C = 44.44 g
Mass of N = 51.83 g
Now, calculate the # moles of each element:
# moles of H = 3.730 g/ 1 g.mole-1 = 3.730 moles
# moles of C = 44.44/12 = 3.703 moles
# moles of N = 51.83/14 = 3.702 moles
Divide by the lowest # moles:
H = 3.730/3.702 = 1
C = 3.703/3.702 = 1
N = 3.702/3.702 = 1
Empirical Formula = HCN
In your hand, the ball has higher potential energy than kinetic because it is still off of the ground but it isn't moving so there is no kinetic. As the ball rises, its potential and kinetic energy increases. At its peak, it has very high potential energy and very low kinetic energy. As it falls, the potential energy decreases but kinetic does not.
