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gavmur [86]
3 years ago
5

An aqueous solution of sucrose (C12H22O11C12H22O11) is prepared by dissolving 7.6330 gg in sufficient deionized water to form a

25.00 mLmL solution. Calculate the molarity of the solution.
Chemistry
1 answer:
Mariana [72]3 years ago
6 0

Answer:

The answer to your question is 0.64 M

Explanation:

Data

Sucrose C₁₂H₂₂O₁₁ mass = 7.633 g

volume = 25 ml

Molarity = ?

Process

1.- Calculate the molar weight of Sucrose

C₁₂H₂₂O₁₁ = (12 x 12) +(22 x 1) + (11 x 16)

                = 144 + 22 + 176

                = 342 g

2.- Calculate the moles of sucrose

                   342 g ------------------ 1 mol

                    7.633 g ---------------  x

                     x = (7.633 x 1) / 342

                    x = 0.0223 moles

3.- Calculate the molarity

Molarity = moles / volume (L)

Molarity = 0.0223 / 0.035

Molarity = 0.64

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Ethylene oxide (EO) is prepared by the vapor-phase oxidation of ethylene. Its main uses are in the preparation of the antifreeze
Rashid [163]

Answer:

a. ΔH^0_{rxn} = -108.0\frac{kJ}{mol}

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate ΔH^0_{rxn} , for the gas-phase reaction )  using the Hess's Law.

ΔH^0_{rxn} =  E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}

Given from the question, the table below shows the corresponding  ΔH^0_{t}(kJ/mol) for each compound.

Compound                    H^0_{t}(kJ/mol)

Liquid EO                       -77.4

CH_4_(g_)                            -74.9                

CO_(g_)                              -110.5

If we incorporate our data into the above previous equation; we have:

ΔH^0_{rxn} = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

          =   -108.0 \frac{kJ}{mol}

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

T_{initial} = 93.0°C   &

the  enthalpy of vaporization  (ΔH^0_{vap}) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as =  \frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}

Let's not forget as well, that  ΔH^0_{vap} = \frac{q}{mass}

If we substitute  ΔH^0_{vap}  for  \frac{q}{mass} in the above equation, we have;

specific heat capacity (c) = \frac{deltaH^0_{vap}}{T_{final}-T_{initial}}

Making (T_{final}- T_{initial}) the subject of the formula; we have:

T_{final}- T_{initial}  = \frac{delat H^0_{vap}}{specificheat capacity}

(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}

T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C

         = 227.76°C +93.0°C

          = 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C                

7 0
3 years ago
Zn(s)+S(s)-ZnS(s)<br> What is the product and physical state
Yuki888 [10]
<span> The Reactants are Zinc (Zn) and Sulfur (S).
The Product is Zinc Sulfide (ZnS).
All of them are solids.
The combined masses of the reactants must be 14 grams, too. Later in Chemistry you'll learn that's not really true, but it is for now.
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4 0
3 years ago
How many mol of hydrogen in 2.8 mol of caffeine.
pochemuha
Empirical formula of caffeine is C₈H₁₀N₄O₂.

In 1 mol of caffeine we have 10 mol of hydrogen.
In 2,8 mol of caffeine we have x mol of hydrogen.

x = 2,8 mol * 10 mol / 1 mol = 28 mol

ANSWER: There are 28 mol of hydrogen.

:-) ;-)
5 0
3 years ago
Read 2 more answers
The chemical formula for artificial sweetener is C7H5NO3S. How many carbon atoms will be found in 5 molecules of the artificial
irga5000 [103]
35

Because there’s 7 carbon atoms in every molecule of artificial sweetener

And if you have 5 molecules of that

7x5 =35
4 0
3 years ago
What is the molarity (M) of the following solutions?
Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

                                      = 27 + 3(16 + 1)

                                      = 27 + 3(17) = 27 + 51

                                      = 78 g/mole

Al(OH)_3 = 78 g/mole

Given mass= 19.2 g/mole

Mole = \frac{Given\ mass}{Molar\ mass}

Mole = \frac{19.2}{78}

Moles = 0.246

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Volume = 280 mL = 0.280 L

Molarity = \frac{0.246}{0.280)}

Molarity  = 0.879 M

Molarity  = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr​

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

Mole = \frac{235.9}{119}

Moles = 1.98

Volume = 2.6 L

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Molarity = \frac{1.98}{2.6)}

Molarity = 0.762 M

Molarity = 0.76 M

4 0
3 years ago
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