Explain whether areas B and D show the same season.

A saturated solution of Pb(IO3)2 in pure water has a lead ion concentration of 5.0 x 10-5 Molar. What is the Ksp value of Pb(IO3

Orlov [11]

Answer:

Option (E) is correct

Explanation:

Solubility equilibrium of $Pb(IO_{3})_{2}$ is given as follows-

$Pb(IO_{3})_{2}\rightleftharpoons Pb^{2+}+2IO_{3}^{-}$

Hence, if solubility of $Pb(IO_{3})_{2}$ is S (M) then-

$[Pb^{2+}]=S(M)$ and $[IO_{3}^{-}]=2S(M)$

Where species under third bracket represent equilibrium concentrations

So, solubility product of $Pb(IO_{3})_{2}$ , $K_{sp}=[Pb^{2+}][IO_{3}^{-}]^{2}$

Here, $[Pb^{2+}]=S(M)=5.0\times 10^{-5}M$

So, $[IO_{3}^{-}]=2S(M)=(2\times 5.0\times 10^{-5})M=1.0\times 10^{-4}M$

So, $K_{sp}=(5.0\times 10^{-5})\times (1.0\times 10^{-4})^{2}=5.0\times 10^{-13}$

Hence option (E) is correct

7 0

2 years ago

How many moles of SnF₂ will be produced along with 48 grams of H₂? *

KIM [24]

Answer: 24 moles of $SnF_2$ are produced.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

${\text{Moles of} H_2}=\frac{48g}{2g/mol}=24moles$

$Sn+2HF\rightarrow SnF_2+H_2$

According to stoichiometry :

1 mole of $H_2$ is accompanied with = 1 mole of $SnF_2$

Thus 24 moles of $H_2$ is accompanied with = $\frac{1}{1}\times 24=24moles$ of $SnF_2$

Thus 24 moles of $SnF_2$ are produced.

8 0

2 years ago

Analysis of the water content of a lake found in the desert showed that it contained 17.5 percent chloride ion, and has a densit

tankabanditka [31]

Answer:

The molarity is 6.0 M.

Explanation:

The concentration of a solution is the amount of solute present in a given amount of solvent, or a given amount of solution.

In this problem, we are asked to convert percent by mass to molarity, which are two different concentration units.

The percent by mass is the ratio of the mass of a solute to the mass of the solution, multiplied by 100 percent. On the other hand, molarity (M) is defined as the number of moles of solute per liter of solution.

If we have 17.15 percent chloride ion, that means that we have 17.15 grams of chloride ion in 100 grams of solution. The density provided is the density of the solution, so we calculate how many mL correspond to 100 grams of solution:

1.23 g ----------- 1 mL

100 g ------------ x= 81.3 mL

Therefore, 17.5 grams of chloride ion are contained in 81.3 mL. Now we need to convert these grams into moles.

The atomic mass of chlorine is 35.5 g/mol:

35.5 g --------------- 1 mol

17.5 g --------------- x= 0. 49 mol

This 0.49 moles are contained in 81.13 mL, the definition of molarity says that this moles are contained in a liter (or what is the same, 1000 mL):

81.3 mL --------------- 0.49 mol

1000 mL ------------- x= 6.0 M

The molarity of the chloride ion will be 6.0 M.

5 0

2 years ago

Which of the following MOST accurately describes the arctic tundra?

Dmitrij [34]

cold

less plants

less rain

dry winds

permanent frozen ground called permafrost

3 0

2 years ago

Calculate the pH during the titration of 30.00 mL of 0.1000 M HCOOH(aq) with 0.1000 M NaOH(aq) after 29.3 mL of the base have be

pashok25 [27]

Answer:

3.336.

Explanation:

Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).

So, concentration of excess acid = [(NV)acid - (NV)base]/V total = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = 1.18 x 10⁻³ M.

For weak acids; [H⁺] = √Ka.C = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = 4.61 x 10⁻⁴ M.

∵ pH = - log[H⁺].

∴ pH = - log(4.61 x 10⁻⁴) = 3.336.

7 0

2 years ago