Answer:
None
Explanation:
Cl₂ is above Br₂ in the activity series.
Bromine will not displace chlorine from its salts.
The reaction will not occur.
Answer:
The electronic configuration that are incorrectly written is 1s²2s³2p⁶, 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴.
Explanation:
The electronic configuration of the elements corresponds to how all the electrons of an element are arranged in energy levels and sub-levels.
There are 7 energy levels —from 1 to 7— whose sublevels are described as s, p, d and f.
All electronic configurations begin with the term "1s" —corresponding to the sublevel s of level 1— so 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴ are incorrectly written. In addition, 4s²3d¹⁰4p⁷ is written incorrectly because is impossible to jump from the sublevel "s" to the sublevel "d" —which is found from level 3 and up— without passing through the sublevel "p".
In the case of 1s²2s³2p⁶, the wrong thing is that the sublevel "s" can only hold two electrons, not three.
The other options are correctly written.
Stoichiometry is the relationship between the relative quantities of substances taking part in a reaction or forming a compound, typically a ratio of whole integers. Hoped this helped!!!!. Also if you are trying to look for the formulas it should be online just type in stoichometry formulas.
The answer is 64.907 amu.
The atomic mass of an element is the average of the atomic masses of its isotopes. The relative abundance of isotopes must be taken into consideration, therefore:
atomic mass of copper = atomic mass of isotope 1 * abundance 1 + atomic mass of isotope 2 * abundance 2
We know:
atomic mass of copper = 63.546 amu
The atomic mass of isotope 1 is: 62.939 amu
The abundance of isotope 1 is: 69.17% = 0.6917
The atomic mass of isotope 1 is: x
The abundance of isotope 2: 100% - 69.17% = 30.83% = 0.3083
Thus:
63.546 amu = 62.939 amu * 0.6917 + x * 0.3083
63.546 <span>amu = 43.535 amu + 0.3083x
</span>⇒ 63.546 amu - 43.535 amu = 0.3083x
⇒ 20.011 amu = 0.3083x
⇒ x = 20.011 amu ÷ 0.3083 = 64.907 amu
Explanation:
Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.
HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−
Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M
Writing the information from the ICE Table in Equation form yields
6.6×10−4=x20.3−x6.6×10−4=x20.3−x
Manipulating the equation to get everything on one side yields
0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4
Now this information is plugged into the quadratic formula to give
x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2
The quadratic formula yields that x=0.013745 and x=-0.014405
However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86