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.
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Answer:
Neutrons.
Explanation:
Isotopes can be defined as the atom of an element that has the same number of protons but different number of neutrons. This ultimately implies that, the isotopes of an element have the same atomic number (number of protons) but different atomic mass (number of nucleons).
The isotope of an element is denoted by
Where; X is the symbol of the element.
A is the atomic mass or number of nucleons.
Z is the atomic number or number of protons.
<em>Therefore, the number of neutrons = A - Z</em>
<em>Isotopes of carbon differ with respect to the number of neutrons.</em>
<em>Basically, there are three (3) Isotopes of Carbon and these are;</em>
<em>1. Carbon-12: it has an atomic mass of 12 with 6 numbers of proton and neutron respectively. </em>
<em>2. Carbon-13: it has an atomic mass of 13 with 6 numbers of proton and 7 numbers of neutron. </em>
<em>3. Carbon-14: it has an atomic mass of 14 with 6 numbers of proton and 8 numbers of neutron. </em>
Answer:
Explanation:
From the given information:
The equation for the reaction can be represented as:

The I.C.E table can be represented as:
2SO₂ O₂ 2SO₃
Initial: 14 2.6 0
Change: -2x -x +2x
Equilibrium: 14 - 2x 2.6 - x 2x
However, Since the amount of sulfur trioxide gas to be 1.6 mol.
SO₃ = 2x,
then x = 1.6/2
x = 0.8 mol
For 2SO₂; we have 14 - 2x
= 14 - 2(0.8)
= 14 - 1.6
= 12.4 mol
For O₂; we have 2.6 - x
= 2.6 - 1.6
= 1.0 mol
Thus;
[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,
[O₂] = 1/50 = 0.02 M ,
[SO₃] = 1.6/50 = 0.032 M
Kc = [SO₃]² / [SO₂]² [O₂]
= ( 0.032²) / ( 0.248² x 0.02)
= 0.8325
Recall that; the equilibrium constant for the reaction
= 0.8325;
If we want to find:

Then:


Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.