How many liters of O2 are required to produce 3.99 miles of CO2?

What is the source of carbon in a glucose molecule produced by photosynthesis?

Kazeer [188]

Answer:

The source of carbon in a glucose molecule produced by photosynthesis is carbon dioxide

4 0

3 years ago

Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

Mazyrski [523]

Answer:

-767,2kJ

Explanation:

It is possible to sum enthalpies of half-reactions to obtain the enthalpy of a global reaction using Hess's law. For the reactions:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁= −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂= +361.7 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl (g) ΔH₃= −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄= − 607.9 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅= − 44.0 kJ

The sum of (4) - (2) produce:

6) XCl₄(s) + O₂(g) ⟶ XO₂(s) + 2Cl₂(g) ΔH₆ = ΔH₄ - ΔH₂ = -969,6 kJ

(6) + 4×(3):

7) XCl₄(s) + 2H₂(g) + O₂(g) ⟶ XO₂(s) + 4HCl(g) ΔH₇ = ΔH₆ + 4ΔH₃= -1338,8 kJ

(7) - 2×(1):

8) XCl₄(s) + 2H₂O(g) ⟶ XO₂(s) + 4HCl(g) ΔH₈ = ΔH₇ - 2ΔH₁= -855,2kJ

(8) - 2×(5):

9) XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g) ΔH₉ = ΔH₈ - 2ΔH₅= -767,2kJ

I hope it helps!

6 0

3 years ago

A 14.1% by mass HClO3 solution has a density of 1.1690 g/mL. What is the concentration of this solution in molarity?

Scilla [17]

Answer:

[ HClO₃] = 1.93M

Explanation:

X % by mass, means that in 100 g of solution, we have X g of solute.

In this case, 14.1 g of HClO₃ are contained in 100 g of solution.

Density always referrs to solution. Let's calculate the volume of solution.

Solution density = Solution mass / Solution volume

1.1690 g/mL = 100 g / Solution volume

Solution volume = 100 g /1.1690 g/mL → 86.2 mL.

For molarity we must get moles of solute and volume of solution (L), because molarity is mol/L

Let's convert the mL of solution in L

86.2 mL . 1L / 1000mL = 0.0862 L

Now, let's determine the moles of solute. (Mass / Molar mass)

14.1 g / 84.45 g/mol = 0.167 moles

Molarity is mol/L → 0.167 moles / 0.0862 L = 1.93M

6 0

3 years ago

Suppose you are working in a chemistry lab and trying to make a solution of a substance called Sodium Chloride (which is a solid

Alenkasestr [34]

Answer:

a weighing balance, a measuring cylinder, a spatula, a beaker/flask, and a stirrer

Explanation:

The lab apparatus that would be needed to prepare a solution of sodium chloride would be a weighing balance, a measuring cylinder, a spatula, a beaker/flask, and a stirrer.

The weighing balance would be used to weigh out the required amount of sodium chloride. The beaker or flask would be placed on the weighing balance and its weight zeroed. The spatula would then be sued to take out the sodium chloride from its container into the beaker till the required amount is reached. The measuring cylinder would then be used to measure out the required volume of water which would be added to the salt in the beaker. The stirrer would then be used to stir the mixture in order for the salt to dissolve.

7 0

3 years ago

what is the volume of the air in a balloon that occupies 0.730 L at 28.0 c if the temperature is lowered to 0.00 C

svetoff [14.1K]

Answer:

The volume of the air is 0.662 L

Explanation:

Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:

$\frac{V}{T}=k$

If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

$\frac{V1}{T1}=\frac{V2}{T2}$

In this case:

V1= 0.730 L
T1= 28 °C= 301 °K (0°C= 273°K)
V2= ?
T2= 0°C= 273 °K

Replacing:

$\frac{0.730 L}{301K}=\frac{V2}{273K}$

Solving:

$V2=273K*\frac{0.730L}{301K}$

V2=0.662 L

The volume of the air is 0.662 L

6 0

3 years ago