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3 years ago

What is the boiling point of cyclohexane at 620 mm hg?

Chemistry

1 answer:

Alchen [17]3 years ago

8 0

The right answer is 271.78 degrees for cyclohexane.

For each liquid there is a precise relationship between pressure and boiling temperature, we can define the boiling temperature as the moment when the first bubble of vapor appears on the surface of a liquid.

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What is the density of a gas with a volume of 5.6 FL oz. and a mass of 37 grams?

Assoli18 [71]

Answer:

6.60

Explanation:

4 0

2 years ago

Read 2 more answers

Determine molar mass of Mn (ClO3)3

goldenfox [79]

M(Mn(ClO3)3)=(54.938)+(35.45x3)+(15.999x9)
M(Mn(ClO3)3)=305.279 g/mol

8 0

2 years ago

Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39

nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ

2) 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39 kJ

3) Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3: Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

This equation we add to the second equation

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

ΔH° = 2*18 + (-39) = 36 - 39 = -3 kJ

This new equation we will divide by 3

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) → CO2(g) + 2FeO(s)

ΔH° =-3/3 = -1 kJ

Now we will substract this new equation from the first equation

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

The next equation we will divide by 2

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

CO(g) + FeO(s) → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

7 0

2 years ago

3/7 of a sum is Rs.1800. Find the sum

STatiana [176]

Answer:

the sum is Rs 4,200

Explanation:

Given that

3 by 7 of the sum is $1,800

We need to find out the sum

So,

Here we assume the sum be x

So the following equation should be made

3 ÷ 7 × x = 1800

3x = 1800 × 7

x = (1800 × 7) ÷ 3

= Rs 4,200

Hence, the sum is Rs 4,200

The same would be considered and relevant

6 0

2 years ago

An acid solution is 0.100 M in HCl and 0.210 M in H2SO4. What volume of a 0.150 M solution of KOH must be added to 500.0 mL of t

Dmitry_Shevchenko [17]

Answer:

When we add 0.33L of KOH only the HCl solution will be neutralized. When we add 1.4 L of KOH, all of the acid (the HCl and H2SO4 solution) will be neutralized.

Explanation:

Step 1: Data given

The solution has 0.100 M HCl and 0.210 M H2SO4

Molarity KOH = 0.150 M

Volume of acid solution = 500 mL = 0.5 L

Step 2: Calculate moles of HCl

Moles HCl = Molarity HCl * volume

Moles HCl = 0.100 M * 0.5 L

Moles HCl = 0.05 moles

Step 3: Calculate moles of H2SO4

Moles H2SO4 = 0.210 M * 0.5 L

Moles H2SO4 = 0.105 moles

Step 4: The balanced neutralization reaction of KOH with HCl can be written as:

KOH + HCl → KCl + H2O

The mole ratio is 1:1

This means to neutralize 0.05 moles HCl, we need 0.05 moles KOH

Step 5: The balanced neutralization reaction of KOH with H2SO4 can be written as:

2KOH + H2SO4 → K2SO4 + 2H2O

The mole ratio KOH: H2SO4 is 2:1

This means to neutralize 0.105 moles H2SO4 we need 0.210 moles KOH

Step 6: Calculate volume of KOH needed to neutralize the solution

To neutralize the HCl solution: 0.05 moles / 0.150 M = 0.33 L KOH needed

To neutralize the H2SO4 solution: 0.210 moles / 0.150 M = 1.4 L KOH needed

When we add 0.33L of KOH the HCl solution will be neutralized. When we add 1.4 L of KOH the HCl and H2SO4 solution will be neutralized.

4 0

2 years ago