Answer:
Explanation:
AgCl ⇄ Ag⁺ + Cl⁻
m m m
If x mole of AgCl be dissolved in one litre .
[ Ag⁺ ] [ Cl⁻ ] = 1.6 x 10⁻¹⁰
m² = 1.6 x 10⁻¹⁰
m = 1.26 x 10⁻⁵ moles
So solubility of AgCl is 1.26 x 10⁻⁵ moles / L
The density of an object is defined as its mass divided by its volume. Mathematically, density = Mass / Volume. The unit of density is kilogram per cubic meter, kg / m^3 or g /cm^3.
For the question given above: the
Mass = 200.0 g
Volume = 100.0 cm^3
Therefore, Density = Mass / Volume = 200 / 100 = 2
Thus, the density of the object is 2 g /cm^3.
Answer:
A. 0.143 M
B. 0.0523 M
Explanation:
A.
Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).
KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄
The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:
1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol
The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.
5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:
M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M
B.
Let's consider the neutralization of potassium hydroxide and perchloric acid.
KOH + HClO₄ → KClO₄ + H₂O
When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.
![M_{A} \times V_{A} = M_{B} \times V_{B}\\M_{A} = \frac{M_{B} \times V_{B}}{V_{A}} \\M_{A} = \frac{0.143 M \times 10.1mL}{27.6mL}\\M_{A} =0.0523 M](https://tex.z-dn.net/?f=M_%7BA%7D%20%5Ctimes%20V_%7BA%7D%20%3D%20M_%7BB%7D%20%5Ctimes%20V_%7BB%7D%5C%5CM_%7BA%7D%20%3D%20%5Cfrac%7BM_%7BB%7D%20%5Ctimes%20V_%7BB%7D%7D%7BV_%7BA%7D%7D%20%5C%5CM_%7BA%7D%20%3D%20%5Cfrac%7B0.143%20M%20%5Ctimes%2010.1mL%7D%7B27.6mL%7D%5C%5CM_%7BA%7D%20%3D0.0523%20M)
The phosphate group of one nucleotide bonds covalently with the sugar molecule of the next nucleotide, and so on, forming a long polymer of nucleotide monomers. The sugar–phosphate groups line up in a “backbone” for each single strand of DNA, and the nucleotide bases stick out from this backbone. The carbon atoms of the five-carbon sugar are numbered clockwise from the oxygen as 1′, 2′, 3′, 4′, and 5′ (1′ is read as “one prime”). The phosphate group is attached to the 5′ carbon of one nucleotide and the 3′ carbon of the next nucleotide. In its natural state, each DNA molecule is actually composed of two single strands held together along their length with hydrogen bonds between the bases.
Answer:
![\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B1.447%20%5Ctimes%2010%5E%7B23%7D%5Ctext%7B%20molecules%20Cu%28OH%29%7D_%7B2%20%7D%7D)
Explanation:
1. Calculate the moles of copper(II) hydroxide
![\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20Cu%28OH%29%7D_%7B2%7D%20%3D%20%5Ctext%7B23.45%20g%20Cu%28OH%29%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20Cu%28OH%29%7D_%7B2%7D%7D%7B%5Ctext%7B97.562%20g%20Cu%28OH%29%7D_%7B2%7D%7D%20%3D%20%5C%5C%5C%5C%5Ctext%7B0.240%2036%20mol%20Cu%28OH%29%7D_%7B2%7D)
2. Calculate the molecules of copper(II) hydroxide