Answer:
water, carbon dioxide reacted in the presence of sun light and chlorophyll and form glucose and oxygen.
Explanation:
Photosynthesis:
It is the process in which in the presence of sun light and chlorophyll by using carbon dioxide and water plants produce the oxygen and glucose.
Carbon dioxide + water + energy → glucose + oxygen
water is supplied through the roots, carbon dioxide collected through stomata and sun light is capture by chloroplast.
Chemical equation:
6H₂O + 6CO₂ + energy → C₆H₁₂O₆ + 6O₂
it is known from balanced chemical equation that 6 moles of carbon dioxide react with the six moles of water so, the mass of water will be,
The glucose is converted into pyruvate and prod
Answer:
6,45mmol/L of NaOH you need to add to reach this pH.
Explanation:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ <em>pka = 4,74</em>
Henderson-Hasselbalch equation for acetate buffer is:
5,0 = 4,74 + log₁₀![\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOO%5E-%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
Solving:
1,82 =
<em>(1)</em>
As total concentration of acetate buffer is:
10 mM = [CH₃COOH] + [CH₃COO⁻] <em>(2)</em>
Replacing (2) in (1)
<em>[CH₃COOH] = 3,55 mM</em>
And
<em>[CH₃COO⁻] = 6,45 mM</em>
Knowing that:
<em>CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O</em>
Having in the first 10mmol/L of CH₃COOH, you need to add <em>6,45 mmol/L of NaOH. </em>to obtain in the last 6,45mmol/L of CH₃COO⁻ and 3,55mmol/L of CH₃<em>COOH </em>.
I hope it helps!
Answer:
D) Dissociates very little in solution.
Explanation:
Answer:
ΔH°r = -184.6 kJ
Explanation:
Let's consider the following balanced equation.
H₂(g) + Cl₂(g) ⇄ 2 HCl(g)
We can calculate the standard enthalpy of the reaction (ΔH°r) using the following expression:
ΔH°r = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)
where,
ni are the moles of reactants and products
ΔH°f(p) are the standard enthalpies of formation of reactants and products
By definition, the standard enthalpy of formation of a simple substance in its most stable state is zero. Then,
ΔH°r = 2 mol × ΔH°f(HCl(g)) - [1 mol × ΔH°f(H₂(g)) + 1 mol × ΔH°f(Cl₂(g))]
ΔH°r = 2 mol × (-92.3 kJ/mol) - [1 mol × 0 + 1 mol × 0]
ΔH°r = -184.6 kJ
<span>2.70x10^7 pm^3
We will assume that the sodium atom closely approximates a sphere. So the expression for the volume of a sphere is
V = 4/3 pi r^3
V = 4/3 pi (186 pm)^3
V = 4/3 pi 6434856 pm^3
V = 26954261.78 pm^3
Round to 3 significant figures, giving 2.70x10^7 pm^3</span>