The balanced chemical reaction of potassium chloride involves 2 moles of potassium to react with 1 mole of chlorine and other are explained below.
<h3>What is balanced equation?</h3>
Balanced chemical equations are those equations in which all entities are present in same amount on reactant as well as on the product side of the reaction.
Given chemical reactions are:
- First equation is a balanced equation in which nitrogen reacts with hydrogen to form ammonia because all are present in same quantity on both sides.
- Second equation is not a balances equation as number of chlorine atoms are not present in same quantity, so balanced equation is:
2K + Cl₂ → 2KCl
- Balanced equation for the third reaction is:
CF₄ + 2Br₂ → CBr₄ + 2F₂
- Balanced equation for the fourth reaction is:
GaF₃ + 3Cs → 3CsF + Ga
- Fifth reaction between barium sulfide and platinum di fluoride is balanced as all atoms are present in same amount on both sides.
- Balanced equation for the last reaction is:
MgF₂ + Li₂CO₃ → MgCO₃ + 2LiF
Hence in the formation of potassium chloride balanced equation involves 2 moles of potassium to react with 1 mole of chlorine.
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<em>ANSWER 10 MOLES OF CARBON DIOXIDE</em>
2C0 + O2 → 2CO2
2 moles of carbon monoxide forming 2 moles of carbon dioxide
10 moles of carbon monoxide will form 10 moles of carbon dioxide
A triple bond is a type of covalent bond. It involves a sigma bond and two pi bonds (each bond involves 2 electrons). A carbon atom is capable of 4 bonds...a carbon-carbon triple bond is found in all alkynes.
Both transition metals and alkali metals are good conductors of heat and electricity, react with water, and are easily oxidized.
<h3>What are alkali metals and transition metals?</h3>
The alkali metals are elements of group 1 which are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs), and francium (Fr). They are also known as the s-block elements because they have their outermost electron in an s-orbital.
The alkali metals are shiny, soft, highly reactive metals and readily lose their outermost electron to create cations with charge +1. They can tarnish rapidly in the air due to oxidation by atmospheric moisture and oxygen.
Transition elements or transition metals are elements that have partially filled d-orbitals. An element having a d-subshell that is partially filled with electrons or can form stable cations with an incompletely filled d orbital.
Any element present in the d-block of the modern periodic table which consists of groups 3 to 12, is considered to be a transition element. For example, the mercury in the +2 oxidation state, corresponds to an electronic configuration of (n-1)d¹⁰. Many paramagnetic compounds are formed by transition metals because they have unpaired electrons in the d orbital.
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In balancing equations, we aim to get equal numbers of every type of atom on both sides of the equation, in order to satisfy the law of conservation of mass (which states that in a chemical reaction, every atom in the reactants is reorganised to form products, without exception). Therefore, let me walk you through question a:
<span>_Fe + _ H2SO4 --> _Fe2 (SO4)3 + _H2
First, take a stock-check of exactly what we currently have on each side (assuming that each _ represents a 1):
LHS: Fe = 1, H = 2, S = 1, O = 4
RHS: Fe = 2, H = 2, S = 3, O = 12,
There are two things to note here. Firstly, H2 (it should be subscript in reality) represents two hydrogen atoms bonded together as part of the ionic compound H2SO4 (sulphuric acid) - this two only applies to the symbol which is directly before it. Hence, H2SO4 only contains 1 sulphur atom, because the 2 applies to the hydrogen and the 4 applies to the oxygen. Secondly, the bracket before the 3 (which should also be subscript) means that there is 3 of everything within the bracket - (SO4)3 contains 3 sulphur atoms and 12 oxygen atoms (4 * 3 = 12).
Now let's start balancing. As a prerequisite, you must keep in mind that we can only add numbers in front of whole molecules, whereas it is not scientifically correct to change the little numbers (we could have two sulphuric acids instead of one, represented by 2H2SO4 (where the 2 would be a normal-sized 2 when written down), but we couldn't change H2SO4 to H3SO4).
The iron atoms can be balanced by having two iron atoms on the left-hand side instead of one:
2Fe </span>+ _ H2SO4 --> _Fe2 (SO4)3 + _H2
Now let's balance the sulphur atoms, by multiplying H2SO4 by 3:
2Fe + 3H2SO4 --> _Fe2 (SO4)3 + _H2
This has the added bonus of automatically balancing the oxygens too. This is because SO4- is an ion, which stays the same in a displacement reaction (which this one is). Take another stock check:
LHS: Fe = 2, H = 6, S = 3, O = 12
RHS: Fe = 2, H = 2, S = 3, O = 12
The only mismatch now is in the hydrogen atoms. This is simple to rectify because H2 appears on its own on the right-hand side. Just multiply H2 by 3 to finish off, and fill the third gap with a 1 because it has not been multiplied up. Alternatively, you can omit the 1 entirely:
2Fe + 3H2SO4 --> Fe2 (SO4)3 + 3H2
This is the balanced symbol equation for the displacement of hydrogen with iron in sulphuric acid.
For question b, I will just show you the stages without the explanation (I take the 3 before B2 to be a mistake, because it makes no sense to use 3B2Br6 when B2Br6 balances fine):
<span>B2 Br6 + _ HNO 3 -->_B(NO3)3 +_HBr
B2Br6 + _HNO3 --> _B(NO3)3 + 6HBr
B2Br6 + 6HNO3 --> _B(NO3)3 + 6HBr</span>
<span><span>B2Br6 + 6HNO3 --> 2B(NO3)3 + 6HBr</span>
Hopefully you can get the others now yourself. I hope this helped
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