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2 years ago

How many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?

Chemistry

2 answers:

miv72 [106K]2 years ago

6 0

Answer: 6.75 moles

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n}{V_s}$

Given;

Molarity = 4.5 M

n= moles of solute (glucose) = ?

$V_s$ = volume of solution in L = 1.5 L

Putting in the values we get,

$4.5=\frac{n}{1.5}$

$n=6.75moles$

Thus moles of glucose are 6.75 moles in 1.5 liters of a 4.5 M $C_6H_{12}O_6$ solution.

zepelin [54]2 years ago

3 0

For the answer to the question above asking, how many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?
The answer to your question is the the third one among the given choices which is 6.8 mol.
moles glucose = 1.5 x 4.5 = 6.8

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Nitrogen dioxide gas is dark brown in color and remains in equilibrium with dinitrogen tetroxide gas, which is colorless.

sukhopar [10]

Answer:

This reaction is exothermic because the system shifted to the left on heating.

Explanation:

2NO₂ (g) ⇌ N₂O₄(g)

Reactant => NO₂ (dark brown in color)

Product => N₂O₄ (colorless)

From the question given above, we were told that when the reaction at equilibrium was moved from room temperature to a higher temperature, the mixture turned dark brown in color.

This simply means that the reaction does not like heat. Hence the reaction is exothermic reaction.

Also, we can see that when the temperature was increased, the reaction turned dark brown in color indicating that the increase in the temperature favors the backward reaction (i.e the equilibrium shift to the left) as NO₂ which is the reactant is dark brown in color. This again indicates that the reaction is exothermic because an increase in the temperature of an exothermic reaction will shift the equilibrium position to the left.

Therefore, we can conclude that:

The reaction is exothermic because the system shifted to the left on heating.

8 0

2 years ago

Analysis of a gaseous chlorofluorocarbon, CClxFy, shows that it contains 11.79% C and 69.57% Cl. In another experiment, you find

uranmaximum [27]

Answer:

The molecular formula = $C_2Cl_{4}F_2$

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

% of C = 11.79

Molar mass of C = 12.0107 g/mol

% moles of C = 11.79 / 12.0107 = 0.9816

% of Cl = 69.57

Molar mass of Cl = 35.453 g/mol

% moles of Cl = 69.57 / 35.453 = 1.9623

Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,

% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64

Molar mass of F = 18.998 g/mol

% moles of F = 18.64 / 18.998 = 0.9812

Taking the simplest ratio for C, Cl and F as:

0.9816 : 1.9623 : 0.9812

= 1 : 2 : 1

The empirical formula is = $CCl_2F$

Also, Given that:

Pressure = 21.3 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 21.3 / 760 = 0.02803 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

Volume = 458 mL = 0.458 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K

⇒n = 0.00052445 moles

Given that :

Amount = 0.107 g

Molar mass = ?

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.00052445= \frac{0.107\ g}{Molar\ mass}$

$Molar\ mass= 204.0233\ g/mol$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol

Molar mass = 204.0233 g/mol

So,

Molecular mass = n × Empirical mass

204.0233 = n × 101.9147

⇒ n = 2

The molecular formula = $C_2Cl_{4}F_2$

6 0

3 years ago