Carbon dioxide and Water are the products of cellular respiration.
Answer:
Saturated solution
We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.
Explanation:
Step 1: Calculate the mass of water
The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.

Step 2: Calculate the mass of glucose per 100 g of water
550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.

Step 3: Classify the solution
The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.
Take the attached picture of a periodic table as a guide. You are finding for a solid metal. Therefore, streamline your choices by looking at elements written in black bold letters, because they are all solid. Next, if you look at the center, the legend for metals are colors in orange, yellow, flesh, lavender, pink, and cyan blue. These region would be your choices. Next, you want to find a metal that is shiny and ductile. The shiny appearance is a common characteristic of luster by materials. Ductility is the ability of a metal to stretch when under tensile stress. These properties are best exhibited by metals in the transitions metals colored in pink. Therefore, the answer to your question would be any of the metal in the pink area. Examples are Titanium, Chromium, Gold, Silver, Platinum, Tungsten, etc.
Answer:
a. 0.182
b. 1.009
c. 1.819
Explanation:
Henderson-Hasselbach equation is:
pH = pKa + log [salt / acid]
Let's replace the formula by the given values.
a. 3 = 3.74 + log [salt / acid]
3 - 3.74 = log [salt / acid]
-0.74 = log [salt / acid]
10⁻⁰'⁷⁴ = 0.182
b. 3.744 = 3.74 + log [salt / acid]
3.744 - 3.74 = log [salt / acid]
0.004 = log [salt / acid]
10⁰'⁰⁰⁴ = 1.009
c. 4 = 3.74 + log [salt / acid]
4 - 3.74 = log [salt / acid]
0.26 = log [salt / acid]
10⁰'²⁶ = 1.819
A, O2 has to be a reactant for combustion to burn