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xxMikexx [17]
3 years ago
14

A student placed 11.0 g of glucose (C6H12O6 ) in a volumetric flask, added enough water to dissolve the glucose by swirling, the

n carefully added additional water until the 100. mL mark on the neck of the flask was reached.
The flask was then shaken until the solution was uniform. A 55.0 mL sample of this glucose solution was diluted to 0.500 L .

How many grams of glucose are in 100. mL of the final solution?
Chemistry
1 answer:
Sedaia [141]3 years ago
4 0

Answer:

We have 12.1 grams of glucose in 100 mL of solution

Explanation:

Step 1: Data given

Mass of glucose = 11.0 grams

Volume = 100 mL

A 55.0 mL sample of this glucose solution was diluted to 0.500 L .

Step 2: Calculate concentration of glucose

Initially, amount of glucose available in 100 ml of the solution = 11.0 grams

Concentration of glucose = mass/ volume

Concentration=11.0 /0.100

= 110 g/L

We take away, 0.055 L of that solution, creating another solution with it but with 0.500 L of volume

Step 3: calculate concentration of the new volume

⇒ with C1 = Concentration of the first solution

⇒ with V1 = Volume of the first solution

⇒ with C2 = Concentration of the second solution 

⇒ with V2 = Volume of the second solution

C1V1= C2V2

110 * 0.055= C2 * 0.500

= 12.1 g/L -= the concentration of the new solution

Step 4: Calculate the mass in 0.100 L

Concentration = mass/ volume

12.1 = mass/0.100

=1.21 grams

We have 12.1 grams of glucose in 100 mL of solution

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Four oranges were placed in a series circuit. Approximately what voltage did this produce?
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Four times the original amount if only one orange was used

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3 0
3 years ago
The molar heat capacity of ethane is represented in the temperature range 298 K to 400 K by the empirical expression Cp,m in J K
BabaBlast [244]

Answer:

-88.66 kJ/mol

Explanation:

The expressions of heat capacity (Cp,m) for C(s) and for H₂(g) are:

C(s):  Cp,m/(J K-1 mol-1) = 16.86 + (4.77T/10³) - (8.54x10⁵/T²)

H₂(g): Cp,m/(J K-1 mol-1) = 27.28 + (3.26T/10³) + (0.50x10⁵/T²)

Cp = A + BT + CT⁻²

For the Kirchoff's Law:

ΔHf = ΔH°f + \int\limits^{T2}_{T1} {DCp(T)} \, dT

Where ΔH°f is the enthalpy at 298 K, T1 is 298 K, T2 is the temperature given (373 K), and DCp is the variation of Cp (products less reactants). ΔH°f  for ethene is -84.68 kJ/mol and the reaction is:

2C(s) + 3H₂(g) → C₂H₆

So, DCp:

dA = A(C₂H₆) - [2xA(C) + 3xA(H₂)] = 14.73 - [2x16.86 + 3x27.28] = -100.83

dB = B(C₂H₆) - [2xB(C) + 3xB(H₂)] = 0.1272 - [2x4.77x10⁻³ + 3x3.26x10⁻³] = 0.10788

dC = C(C₂H₆) - [2xC(C) + 3xC(H₂)] = 0 - (2x(-8.54x10⁵) + 3x0.50x10⁵) = 15.58x10⁵

dCp = -100.83 + 0.10788T + 15.58x10⁵T⁻²

\int\limits^{373}_{298} {-100.83 + 0.10788T + 15.58x10^5T^{-2}} \, dT = -3796.48 J/mol = -3.80 kJ/mol (solved by a graphic calculator)

ΔHf = -84.68 - 3.80

ΔHf = -88.66 kJ/mol

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