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3 years ago

A 300.0 mL saturated solution of copper(II) peroidate, Cu(IO4)2, contains .30 grams of dissolved salt. Calculate Ksp.

Chemistry

1 answer:

Sphinxa [80]3 years ago

8 0

Answer:

$Ksp=4.9 \times 10^{-8}$

Explanation:

balanced chemical reaction:

$Cu(IO_4)_2 \rightarrow CU^{2+} +2 IO_4^-$

concentration or molarity of Cu(IO4)2= mole/Volume in L

$mole=\frac{mass\, given}{molecular\,weight}$

$mole=\frac{0.30}{445.4}$

$mole=6.74 \times 10^{-4}$

volume= 300 ml=0.3 L

$molarity =\frac{mole}{volume\, in\, L}$

$molarity=2.3\times 10^{-3}$

Solubility means maximum salt which can be dissolved.

lets solubility=S

$[CU^{2+}]=S;$

$[IO_4^-] =2S$

$S=2.3\times 10^{-3}$

$Ksp=S\times(2S)^2$

$Ksp=4.9 \times 10^{-8}$

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How did Ernest Rutherford change the atomic model?

nadya68 [22]

Answer:

D. He showed that most of an atom's mass was located in the atom's

nucleus.

Explanation:

Ernest Rutherford changed the atomic model because of his experiment which was the gold foil experiment. A beam of alpha particles was aimed at a piece of gold foil, most particles passed through but some were scattered backward which showed that the middle of an atom (nucleus) is the where most of the mass is located.

6 0

3 years ago

Select the correct answer from each drop-down menu. At chemical equilibrium, the amount of because .

Elena-2011 [213]

Answer:

The answer that completes the question are in BOLD:

At chemical equilibrium, the amount of PRODUCT AND REACTANT REMAIN CONSTANT because the RATES OF THE FORWARD AND REVERSE REACTIONS ARE EQUAL.

Explanation:

In a reversible chemical reaction, an equilibrium is said to be achieved when the rates of the forward reaction is equal to that of the reverse reaction. A reversible reaction is one in which products are formed from reactants simultaneously with the formation of reactants from products.

The combination of two or more substances called REACTANTS gives rise to another substance called PRODUCT, which can in turn give rise to Reactants again. With time, the rate at which the reactants give rise the products, which is called the FORWARD REACTION will be equal to the rate at which the products give rise to the reactants, which is called REVERSE REACTION. At this point, the chemical reaction is said to be in a STATE OF EQUILIBRIUM.

When the rate at which both reaction occurs becomes equal i.e. at an equilibrium state, the concentration of both the reactants and the products becomes constant i.e. no longer changes. Hence, the amount of the reactants forming the products is the same as the amount of products forming the reactants.

N.B: At chemical equilibrium, the amount of the reactants and products does not necessarily equals zero (0). It simply means that there is no net change in the concentration/amount of both reactants and products.

3 0

3 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

3 years ago

if a sample of gas at 25.2 c has a volume of 536mL at 637 torr, what will its volume be if the pressure is increased to 712 torr

nignag [31]

Considering ideal gas:
PV= RTn

T= 25.2°C = 298.2 K

P1= 637 torr = 0.8382 atm

V1= 536 mL = 0.536 L

:. R=0.082 atm.L/K.mol

:. n= (P1V1)/(RT) = ((0.8382 atm) x (0.536 L))/
((0.082 atmL/Kmol) x (298.2K))

:. n= O.0184 mol

Then,
P2= 712 torr = 0.936842 atm

V2 = RTn/P2 = [(0.082atmL/
Kmol) x (298.2K) x (0.0184mol) ]/(0.936842atm)

:.V2 = 0.4796 L
OR
V2 = 479.6 ml

6 0

3 years ago

Ethanol, C2H6O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is

weeeeeb [17]

Answer:

This means 463 grams of ethanol would provide less amount of energy

Explanation:

Step 1: Data given

Heat of combustion of ethanol = 326.7 kcal/mol

The heat of combustion of octane = 1.308*10³ kcal/mol

Mass of octane = 463 grams

Molar mass octane = 114.23 g/mol

Molar mass ethanol = 46.07 g/mol

Step 2: Calculate moles octane

Moles octane = mass octane / molar mass octane

Moles octane = 463 grams / 114.23 g/mol

Moles octane = 4.05 moles

Step 3: Calculate energy of combustion of 4.05 moles octane

Combustion of 1 mol octane gives us: 1.308 * 10³ kcal/mol

Combustion of 4.05 moles octane gives us 4.05 * 1.308 * 10³ kcal/mol = 5.30 * 10³ kcal

This means the combustion reaction of 463 grams of octane gives us 5.30 * 10³ kcal

Step 4:

Heat of combustion of ethanol = 326.7 kcal/mol

OR in words: combustion of 1 mol ethanol gives us 326.7 kcal energy

Moles ethanol = 463 grams / 46.07 g/mol

Moles ethanol = 10.05 moles

Since combustion of 1 mol ethanol gives us 326.7 kcal

10.05 moles ethanol will give us = 10.05 * 326.7 = 3283.3 kcal = 3.28 * 10³ kcal

5.30 * 10³ kcal > 3.28 * 10³ kcal

This means 463 grams of ethanol would provide less amount of energy

3 0

3 years ago