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Yakvenalex [24]

3 years ago

12

To calculate the Ksp of 0.34 g of cesium sulfate (Cs2SO4) in 1.00 L of aqueous solution, which equilibrium equation will you use

?
A: Cs2SO4(s) <> 2Cs+(aq) + SO4^-2(aq)

1 answer:

FinnZ [79.3K]3 years ago

5 0

Answer:C

Explanation:

You might be interested in

An aqueous solution containing 17.5 g of an unknown molecular compound in 100.g water was found to have a freezing point of-1.8°

Zigmanuir [339]

Answer:

The answer to your question is 178.6 g

Explanation:

Data

ΔT = 1.8 °C

mass = 17.5 g

mass of water = 100 g

Kc = 1.86

Process

1.- Calculate the molality using the following formula

ΔTc = mKc

solve for m

m = ΔTc/Kc

substitution

m = 1.8/1.86

result

m = 0.968

2.- Calculate the number of moles

m = # of moles/kg of solvent

kg of solvent = 0.1 kg

# of moles = m x kg of solvent

# of moles = 0.968 x 0.1

# of moles = 0.0968

3.- Calculate the molar mass

x g molar mass --------------------- 1 mol

17.5 g --------------------- 0.0968 moles

x = (1 x 17.5)/0.0968

x = 178.6 g

4 0

3 years ago

a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat

Nostrana [21]

Answer:- $\frac{382.69J}{0C}$ .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, $\Delta T$ for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

$q=mc\Delta T$

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

$q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})$

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

$\Delta T$ for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

$q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)$

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, $q=C.\Delta T$

$C=\frac{q}{\Delta T}$

$C=\frac{1044.746J}{2.73^0C}$

$C=\frac{382.69J}{0C}$

So, the heat capacity of calorimeter is $\frac{382.69J}{0C}$ .

4 0

3 years ago

The perfect glass of sweet tea is 15% sugar. Elena is making a one gallon container of sweet tea. How many cups of sugar are in

Arte-miy333 [17]

There are 2.4 cups of sugar in the gallon of sweet tea.

<h3>What is the conversion of fluid measurement?</h3>

The conversion of fluid measure takes a standard analysis where the assumed container of a particular fluid is related to another container.

For example, we have two containers here:

The glass cup
The gallon

We need to know how many glass cups that are present in a gallon. From standard conversion rates:

1 gallon = 16 cups

However,

We learned that 1 cup contains 15% of sugar; Therefore, 16 cups will contain:

= 16 × 0.15

= 2.4 cups of sugar

Learn more about measurement conversion here:

brainly.com/question/777464

7 0

2 years ago

Gold forms a substitutional solid solution with silver. Calculate the number of gold atoms per cubic centimeter (in atoms/cm3) f

Leokris [45]

Answer:

Gold: 1.1 x 10²² atoms/cm³

Silver: 4.8 x 10²² atoms/cm³

Explanation:

100 g of the alloy will have 29 g of Au and 71 g of Ag.

19.32 g Au ____ 1 cm³

29 g Au ______ x

x = 1.5 cm³

10.49 g Ag ____ 1 cm³

71 g Ag _______ y

y = 6.8 cm³

The total volume of 100g of the alloy is x+y = 8.3 cm³.

Gold:

196.97 g Au____ 6.022 x 10²³ atoms Au

29 g Au _______ w

w = 8.9 x 10²² atoms Au

8.9 x 10²² atoms Au ____ 8.3 cm³

A ____ 1 cm³

A = 1.1 x 10²² atoms Au

Silver:

107.87 g Ag____ 6.022 x 10²³ atoms Ag

71 g Ag _______ w

w = 4.0 x 10²³ atoms Ag

4.0 x 10²³ atoms Ag ____ 8.3 cm³

B ____ 1 cm³

B = 4.8 x 10²² atoms Ag

6 0

3 years ago

How many grams of Fe will be produced if 705.0g of H20 are produced?

fredd [130]

<h3>Answer:</h3>

2265 g Fe₃O₄

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Fe₃O₄ + 4H₂ → 3Fe + 4H₂O

[Given] 705.0 g H₂O

<u>Step 2: Identify Conversions</u>

[RxN] 4 mol H₂O → 1 mol Fe₃O₄

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Fe - 55.85 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Molar Mass of Fe₃O₄ - 3(55.85) + 4(16.00) = 231.55 g/mol

<u>Step 3: Convert</u>

Set up stoich: $\displaystyle 705.0 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{1 \ mol \ Fe_3O_4}{4 \ mol \ H_2O})(\frac{231.55 \ g \ Fe_3O_4}{1 \ mol \ Fe_3O_4})$
Multiply/Divide/Cancel units: $\displaystyle 2264.74 \ g \ Fe_3O_4$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

2264.74 g Fe₃O₄ ≈ 2265 g Fe₃O₄

5 0

3 years ago