Answer:
- <u><em>Yes, 200 ml of fluid can be transferred to a 1-quart container.</em></u>
Explanation:
You must compare the two volumes, 200 ml and 1 quart. If 200 ml is less than or equal to 1 quart, then 200 ml of fluid can be transferred to a 1-quart container, else it is not possible.
To compare, the two volumes must be on the same system of units.
Quarts is a measure of volume equivalent to 1/4 of gallon.
One gallon is approximately 3.785 liters.
3.785 liter = 3.785 liter × 1,000 ml/liter
Then, to convert 1 quart to ml use the unit cancellation method:
- (1/4)gallon × 3.785 liter/gallon × 1,000ml / liter = 946.25 ml
Thus, you get that a 1-quart container has volume of 946.25 ml, which allows that 200ml of fluid be transferred to it.
The correct answer is higher melting point, bound by metal metal bonds.
While alkali metals only have one valence electron, alkaline earth metals have two. Metal to metal connections hold the metals together. Alkaline earth metals have a stronger metallic connection and a higher melting point because they have two valence electrons.
the characteristics that Group 2 metals excel in over Group 1 metals.
- Initial Ionization Potential
- Group 2 items are more difficult than group 1 elements.
- Strong propensity to produce bivalent compounds
As a result, group 2 metals have stronger metallic bonding, which leads to increased cohesive energy and compact atom packing. This explains why group 2 metals are harder and have higher melting and boiling temperatures than group 1 metals.
To learn more about Group 2A(2) refer the link:
brainly.com/question/9431096
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I have done this question before and even though you didn’t provide the options. I think the correct option is elements
Answer:
-3.7771 × 10² kJ/mol
Explanation:
Let's consider the following equation.
3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
n: moles
ΔG°f(): standard Gibbs free energy of formation
p: products
r: reactants
ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))
ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)
ΔG° = -377.71 kJ = -3.7771 × 10² kJ
This is the standard Gibbs free energy per mole of reaction.
Answer:
6 oxygen atoms
Explanation:
Step 1: Given data
Number of dinitrogen monoxide molecules (N₂O): 6
Number of oxygen atoms (O): ?
Step 2: Calculate the appropriate ratio
The ratio of dinitrogen monoxide molecules to oxygen atoms is 1:1.
Step 3: Use the ratio to calculate the number of oxygen atoms
6 molecule N₂O × (1 atom O/1 molecule N₂O): 6 atom O
