I give brainliest

2 answers:

8 0

Answer: During winter the Northern Hemisphere leans away from the sun, there are fewer daylight hours and the sun hits us at an angle. This makes it appear lower in the sky. In equatorial regions the length of days and the directness of sunlight don't change as much.

Hope this helps :)

Citrus2011 [14]2 years ago

3 0

Answer:

What he said... more light gives more energy to plants, making them grow more.

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The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

7 0

2 years ago

Find the pH of a 0.100 molar H2C6O6 solution with ka, where KA is equal 8.0×10–5

lubasha [3.4K]

The pH of the solution is 2.54.

Explanation:

pH is the measure of acidity of the solution and Ka is the dissociation constant. Dissociation constant is the measure of concentration of hydrogen ion donated to the solution.

The solution of C₆H₂O₆ will get dissociated as C₆HO₆ and H+ ions. So the molar concentration of 0.1 M is present at the initial stage. Lets consider that the concentration of hydrogen ion released as x and the same amount of the base ion will also be released.

So the dissociation constant Kₐ can be written as the ratio of concentration of products to the concentration of reactants. As the concentration of reactants is given as 0.1 M and the concentration of products is considered as x for both hydrogen and base ion. Then the

$K_{a}=\frac{[H^{+}][HB] }{[reactant]}$