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2 years ago

Pls help :)

Chemistry

1 answer:

prisoha [69]2 years ago

6 0

Answer:

0.484 mole

Explanation:

1 mole of glucose reacts with 6 moles of O2, producing :

6 moles of CO2

6 moles of H2O

678 kcal

Using rule of three you have:

1 mole of glucose -> 678 kcal

x mole of glucose -> 328 kcal

x = 328/678 * 1 =0.484 mole

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Write the balanced Ka and Kb reactions for HSO3- in water. Be sure to include the physical states of each species involved in th

Hunter-Best [27]

Answer:

Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]

Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]

Explanation:

An amphoteric substance as HSO₃⁻ is a substance that act as either an acid or a base. When acid:

HSO₃⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₃²⁻(aq)

And Ka, the acid dissociation constant is:

When base:

HSO₃⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + H₂SO₃(aq)

And kb, base dissociation constant is:

6 0

1 year ago

Find the volume of a balloon of a gas at 842 mm Hg and -23 celcius if it’s volume is 915 milliliters at a pressure of 1170 mm Hg

Ronch [10]

The volume of a balloon f a gas at 842 mm Hg and -23 celsius if it’s volume is 915 milliliters at a pressure of 1170 mm Hg And a temperature of 24 celsius is 0.22 litres

Explanation:

Data given:

Initial volume of the balloon having gas V1= 915ml OR 0.195 L

initial pressure of the gas P1= 1170 mm Hg OR 1.53 atm

initial temperature of the gas T1 = 24 celsius or 273.15 + 24 = 297.15 K

Final pressure of the gas P2 = 842 mm Hg or 1.10 atm

final temperature of the gas T2 = -23 degrees or 273.15 - 23 = 250.15 K

Final volume at final temperature and pressure V2=?

The formula used is of Gas Law:

$\frac{P1V1}{T1}$ = $\frac{P2V2}{T2}$

V2 = $\frac{P1V1T2}{T1P2}$

putting the values in the equation:

V2 = $\frac{1.53 X 0.195 X 250.15}{297.15 X 1.10}$

V2 = 0.22 litres is the volume

The volume is 0.22 litres at a pressure of 1170 mmHg and temperature of -23 degrees.

5 0

2 years ago

If the 3.90 m solution from Part A boils at 103.45 ∘C, what is the actual value of the van't Hoff factor, i? The boiling point o

erma4kov [3.2K]

<span>Some of the solutions exhibit colligative properties. These properties depend on the amount of solute dissolved in a solvent. For boiling point elevation, we calculate the increase in temperature by the equation:

</span><span>ΔT(boiling point) = (Kb)mi

where Kb is a constant, m is the molality of the solution, i is the van't Hoff factor.

From the given data, we can easily calculate for i as follows:

</span>ΔT(boiling point) = (Kb)mi
103.45 - 100 = (0.512)3.90i
i = 1.73 <-------van't Hoff factor

7 0

2 years ago

You’re at the zoo and have a big red 1.80 L helium balloon. The barometric pressure today is 785 mmHg. Then you hear the roar of

nadezda [96]

Answer:

0.62 L

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 1.80 L

Initial pressure (P1) = 785 mmHg

Final pressure (P2) = 3.00 atm

Final volume (V2) =?

Step 2:

Conversion of the pressure in mmHg to atm.

It is important to express the initial and the final pressure in the same unit. Either express both in atm or in mmHg. What ever the case is, we'll still arrive at same answer. Here, we shall be converting from mmHg to atm. This is illustrated below:

760mmHg = 1atm

Therefore, 785 mmHg = 785/760 = 1.03 atm

Step 3:

Determination of the final volume. This is illustrated below.

We shall be applying the Boyle's law equation since the temperature is constant.

P1V1 = P2V2

Initial Volume (V1) = 1.80 L

Initial pressure (P1) = 1.03 atm

Final pressure (P2) = 3.00 atm

Final volume (V2) =?

P1V1 = P2V2

1.03 x 1.8 = 3 x V2

Divide both side by 3

V2 = (1.03 x 1.8) /3

V2 = 0.62 L

Therefore, the new volume of the balloon is 0.62 L

5 0

2 years ago

Read 2 more answers

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

2 years ago