Answer:

Explanation:
1. Concentration of SO₄²⁻
SrSO₄(s) ⇌ Sr²⁺(aq) +SO₄²⁻(aq); Ksp = 3.44 × 10⁻⁷
0.0150 x
![K_{sp} =\text{[Sr$^{2+}$][SO$_{4}^{2-}$]} = 0.0150x = 3.44 \times 10^{-7}\\x = \dfrac{3.44 \times 10^{-7}}{0.0150} = \mathbf{2.293 \times 10^{-5}} \textbf{ mol/L}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BSr%24%5E%7B2%2B%7D%24%5D%5BSO%24_%7B4%7D%5E%7B2-%7D%24%5D%7D%20%3D%200.0150x%20%3D%203.44%20%5Ctimes%2010%5E%7B-7%7D%5C%5Cx%20%3D%20%5Cdfrac%7B3.44%20%5Ctimes%2010%5E%7B-7%7D%7D%7B0.0150%7D%20%3D%20%5Cmathbf%7B2.293%20%5Ctimes%2010%5E%7B-5%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D)
2. Concentration of Pb²⁺
PbSO₄(s) ⇌ Pb²⁺(aq) + SO₄²⁻(aq); Ksp = 2.53 × 10⁻⁸
x 2.293 × 10⁻⁵
![K_{sp} =\text{[Pb$^{2+}$][SO$_{4}^{2-}$]} = x \times 2.293 \times 10^{-5} = 2.53 \times 10^{-8}\\\\x = \dfrac{2.53 \times 10^{-8}}{2.293 \times 10^{-5}} = \mathbf{1.10 \times 10^{-3}} \textbf{ mol/L}\\\\\text{The concentration of Pb$^{2+}$ is $\large \boxed{\mathbf{1.10 \times 10^{-3}}\textbf{ mol/L}}$}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BPb%24%5E%7B2%2B%7D%24%5D%5BSO%24_%7B4%7D%5E%7B2-%7D%24%5D%7D%20%3D%20x%20%5Ctimes%202.293%20%5Ctimes%2010%5E%7B-5%7D%20%3D%202.53%20%5Ctimes%2010%5E%7B-8%7D%5C%5C%5C%5Cx%20%3D%20%5Cdfrac%7B2.53%20%5Ctimes%2010%5E%7B-8%7D%7D%7B2.293%20%5Ctimes%2010%5E%7B-5%7D%7D%20%3D%20%5Cmathbf%7B1.10%20%5Ctimes%2010%5E%7B-3%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20concentration%20of%20Pb%24%5E%7B2%2B%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.10%20%5Ctimes%2010%5E%7B-3%7D%7D%5Ctextbf%7B%20mol%2FL%7D%7D%24%7D)
Explanation:
The - 3 degree C( carbon atom) 2p atomic orbital + methyl C-H sigma molecular orbital because one C-H bond has to dissolve its bond and provide the H that is sigma molecular orbital and the carbonation is type 3 degree sp2 carbon.
Hyperconjugation is the stabilizing effect arising from the electrons ' engagement in a π-bond (usually C-H or C-C) with a neighboring empty or partly filled p-orbital or π-orbital to provide an expanded molecular orbital that enhances system stability.
The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4
Evgesh-ka [11]
In NH3 , let oxidation number of N be x
x + (+1)3 = 0
x = -3
In HNO3 , let oxidation number of N be x
1 + x + (-2)3 = 0
x = +5
In NO2 , let oxidation number of N be x
x + (-2)2 = 0
x = +4
Answer:
124.91mL
Explanation:
Given parameters:
P₁ = 1.08atm
V₁ = 250mL
T₁ = 24°C
P₂ = 2.25atm
T₂ = 37.2°C
V₂ = ?
Solution:
To solve this problem, we are going to apply the combined gas law;

P, V and T represents pressure, volume and temperature
1 and 2 delineates initial and final states
Convert the temperature to kelvin;
T₁ = 24°C, T₁ = 24 + 273 = 297K
T₂ = 37.2°C , T₂ = 37.2 + 273 = 310.2K
Input the variables and solve for V₂

V₂ = 124.91mL
Answer:
see attached
Explanation:
Dimensional analysis is useful whenever dimensions are involved. Unless it is quite clear that all of the problem dimensions are consistent (for example, all speeds in miles per hour, or all angles in degrees), dimensional analysis can be useful for keeping the math straight.
Only units of the same dimensions can be added or subtracted. When numbers are multiplied or divided or raised to a power, dimensional analysis can help ensure that the appropriate operations are being used on appropriate numbers. It can also help ensure that dimensions are being combined properly to give appropriate derived dimensions.
__
Scientific notation is a way of writing very large or very small numbers compactly. It can also help with "order of magnitude" estimates. If an answer using SI prefixes is appropriate, or if a number can be conveniently expressed in standard form, then scientific notation is usually not required.
On the other hand, SI prefixes may not be appropriate in some cases, or a problem may specify that scientific notation be used for expressing results. In those instances, scientific notation should be used.