CO has a molar mass of 44.01 g/mol, which means that 1 mole of CO weighs 44.01 grams.
We want to convert 14.35 grams of ethanol to acetic acid, so we need to know how many moles of ethanol there are. Let's first multiply the given mass of ethanol by its molar mass:
14.35 (g) x 32.06 (g/mol) = 45.85 mol
So, we have 45.85 moles of methanol. If methanol reacts with CO to produce acetic acid, we need to know how many moles of CO we need to use, based on the stoichiometry of the reaction. The reaction equation gives us a 2:1 ratio for CO and methanol:
2 CH₃OH(l) + CO(g) → CH₃CO₂H(aq)
So, for every mole of CH₃OH we react, we will need 0.5 mole of CO. Therefore, we need 23.92 moles of CO to convert all the methanol to acetic acid.
Using the mass of 1 mole of CO, we can convert moles of CO to grams:
23.92 (mol) x 44.01 (g/mol) = 1050.93 (g)
So, we need 1051 grams of CO to completely convert 14.35 grams of methanol to acetic acid.
When a barium atom loses two electrons it becomes a positive ion and its radius decreases. Barium (Ba) has atomic number 56 so it has 2 electrons in first shell of an atom to become stable according to duplet rule. Then other 52 electrons revolve in the shells according to octet rules.
Another 2 electrons are in the outermost shell. To become stable electrons lose to form barium ions (Ba+2). Hence, by losing 2 electrons the outermost shell will be diminished so its radius decreases and by losing electrons it becomes positive ions.
Answer:
The correct answer is option 3, It becomes a positive ion and its radius decreases
Explanation:
As per the Octet rule, Barium has 2 electrons in its outermost shell. When it loses the two electron it gains two positive charge i.e Ba2+. As the barium loses the two electron from its outermost shell, the outermost shell becomes vacant and thus is no more considered as a part of atomic geometry of the barium atom and since the outermost shell is considered negligible the radius of barium atom reduces automatically.
gas ➡ solid
liquid ➡ gas
solid ➡ gas
A Brønsted-Lowry base is a species that accepts a proton (hydrogen ion) from another species. NH3, OH-, and even water itself are examples of Brønsted-Lowry bases, denoting they accept protons.
A Brønsted-Lowry base is a species that can accept a proton (a hydrogen ion) from another species. For instance, in a reaction between water and ammonia, NH3 is the Brønsted-Lowry base because it accepts a proton from water. This means that any species capable of accepting a proton, such as hydroxide ion (OH-), ammonia (NH3), or water itself can be considered a Brønsted-Lowry base.
For example, think about the dissociation of water:
H2O + H2O ⇌ H3O+ + OH-
In this reaction, water is acting as both a Brønsted-Lowry acid and a Brønsted-Lowry base. One water molecule donates a proton and becomes a hydroxide ion (the conjugate base), while the other accepts a proton to become hydronium (the conjugate acid).
Another example would be the ionization of ammonia in water:
NH3 + H2O ⇌ NH4+ + OH-
Here, ammonia (NH3) is the Brønsted-Lowry base as it accepts a proton from water to become ammonium (NH4+).
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Methane and ethane demonstrate the law of definite composition and the law of multiple proportions as these both compounds have the exact atoms but in different ratios.
As for methane, it can be seen that every methane molecule contains 4 hydrogen atoms and 1 carbon atom. Where masses of carbon and hydrogen are in ratio - 12 g C / 4 g H -- reduce to 3g C/ 1 g H .
Ethane contains 2 carbon atoms and 6 hydrogen atom and their mass ratio will be - 24 g C / 6 g H -- reduce to 4 g C/ 1 g H .
So both ethane and methane demonstrate the law of definite composition and the law of multiple proportions.
Law of definite composition: The elements which are present in the compound is combined in the same proportion by mass.
Law of multiple proportion: When two or more compounds are formed by the combination of two elements then the mass ratio of one element is combined with the fixed mass of the other element.
In case of methane, it consist of one carbon atoms and four hydrogen atoms implies that has definite composition. In terms of mass, in methane, carbon and hydrogen atoms are combined in a definite ratio i.e. 12 g C/ 4 g H. Thus, methane has definite composition.
Now, carbon and hydrogen combines to give a class i.e. hydrocarbon. In this case, for every constant mass of carbon the ratio of hydrogen will always reduce to 4/3 ratio for the formation of ethane (hydrocarbon). Thus, law of multiple proportion followed.
Similarly, in case of ethane, this compound also consist of exact atoms but in different ratios.
In case of ethane, it consist of two carbon atoms and six hydrogen atoms implies that has definite composition. In terms of mass, in ethane, carbon and hydrogen atoms are combined in a definite ratio i.e. 24 g C/ 6 g H. Thus, methane has definite composition.
Now, carbon and hydrogen combines to give a class i.e. hydrocarbon. In this case, for every constant mass of carbon the ratio of hydrogen will always reduce to a specific ratio for the formation of methane (hydrocarbon). Thus, law of multiple proportion followed.
(2) NaOH(aq) + HCl(aq) -->NaCl(aq) + H2O(l) + heat
(3) H2O(l) + heat --> H2O(g)
(4) H2O(l) + HCl(g) -->H3O+(aq) + Cl-(aq) + heat
The equation where the term heat represents the heat of fusion is NaCl(s) + heat -->NaCl(l).
The heat of fusion is the heat required to break the intermolecular interactions in a solid as it is converted to the liquid. Hence, the process of fusion is the process of converting a solid into a liquid.
Examining the equations shown in the options closely, we will discover that in the equation; NaCl(s) + heat -->NaCl(l), heat is supplied to sodium chloride solid and it is converted to liquid. The heat involved in this process is the heat of fusion.
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Answer:
The answer is 1
Explanation: