Answer:
construction means everything is built from
EXCEPT NATURE
Explanation:
MARK BRAINLIEST THANK U
HAVE A LOVELY DAY
Answer: true
sound waves need particles to move through
Answer:
Atoms are composed of a central nucleus which is surrounded by orbiting <u>Electrons</u>.
Explanation:
The word Atom is derived from atomos meaning indivisible particle. An atom is a very small sized particle and its size is approximately 100 picometers.
Atoms are composed of a Nucleus which contains protons and neutrons and the nucleus is surrounded by orbiting electrons. Also, the nucleus is positively charged due to +ve protons and the electrons revolving around nucleus are -vely charged making overall atom neutral in nature.
Atoms react with another atoms due to the presence of valence electrons present in the valence shell of an atom. The valence electrons would make a covalent bond by mutually sharing the electrons or it may form an ionic bond by gaining and loosing valence electron.
Answer:
282.7KPa
Explanation:
Step 1:
Data obtained from the question.
Number of mole of (n) = 1.5 mole
Volume (V) = 13L
Temperature (T) = 22°C = 22 + 273°C = 295K
Pressure (P) =..?
Gas constant (R) = 0.082atm.L/Kmol
Step 2:
Determination of the pressure exerted by the gas.
This can be obtained by using the ideal gas equation as follow:
PV = nRT
P = nRT /V
P = 1.5 x 0.082 x 295 / 13
P = 2.79atm.
Step 3:
Conversion of 2.79atm to KPa.
This is illustrated below:
1 atm = 101.325KPa
Therefore, 2.79atm = 2.79 x 101.325 = 282.7KPa
Therefore, the pressure exerted by the gas in KPa is 282.7KPa
Answer:
![\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D)
Less than the concentration of Pb2+(aq) in the solution in part ( a )
Explanation:
From the question:
A)
We assume that s to be the solubility of PbI₂.
The equation of the reaction is given as :
PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹
[Pb²⁺] = s
Then [I⁻] = 2s
![K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BPb%24%5E%7B2%2B%7D%24%5D%5BI%24%5E%7B-%7D%24%5D%7D%5E%7B2%7D%20%3D%20s%5Ctimes%20%282s%29%5E%7B2%7D%20%3D%20%204s%5E%7B3%7D%5C%5Cs%5E%7B3%7D%20%3D%20%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%5C%5C%5C%5Cs%20%3D%5Cmathbf%7B%20%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D%5C%5C%5C%5C%5Ctext%7BThe%20mathematical%20expressionthat%20can%20be%20used%20to%20determine%20the%20value%20of%20%20%7D%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D)
B)
The Concentration of Pb²⁺ in water is calculated as :
![\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D)
![\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7B7%2A10%5E%7B-9%7D%7D%7B4%7D%7D%7D)
![\mathbf{s} =\sqrt[3]{1.75*10^{-9}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%7D%20%3D%5Csqrt%5B3%5D%7B1.75%2A10%5E%7B-9%7D%7D)

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI




The equilibrium constant:
![K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5BPb%5E%7B2%2B%7D%7D%5D%5BI%5E-%5D%5E2%20%5C%5C%20%5C%5C%20K_%7Bsp%7D%20%3D%20s%2A%281.0%2A2s%29%5E2%20%3D7%2A1.0%5E%7B-9%7D%20%5C%5C%20%5C%5C%20s%20%3D%207%2A10%5E%7B-9%7D%20%5C%20%5C%20%20m%2FL)
It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.