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Murrr4er [49]
4 years ago
13

The chemical equation given below represents the chemical reaction between lithium (Li) and sulfur (S). In the equation, why is

the number 2 present in front of lithium on the reactants side of the equation?
A. to show the number of lithium atoms involved in the reaction
B. to show the number of electrons gained by lithium in the reaction
C. to show the atomic number of lithium in the periodic table
D. to show the number of electrons lost by lithium in the reaction
Chemistry
2 answers:
algol [13]4 years ago
5 0

Answer:

A. to show the number of lithium atoms involved in the reaction

Explanation:

The reaction between lithium and sulfur produces lithium sulfide where two lithium and one sulfur atoms react.

Balanced equation: 2 Li + S = Li2S

The 2 present before Li on the reactant's side shows how many lithium atoms are supposed to be present per every 1 sulfur atom for this reaction to be balanced and true.

That is it! :)

Olegator [25]4 years ago
4 0

Answer: Option (A) is the correct answer.

Explanation:

The given chemical reaction equation is as follows.

         2Li + S \rightarrow Li_{2}S

As the number of atoms on both reactant and product side are the same. Hence, it is a balanced chemical reaction equation.

Here, number 2 before lithium atom represents that two atoms of lithium are taking part in the reaction. As there is no coefficient or number placed before the sulfur atom this means that only one atom of sulfur is reacting in the reaction.

Thus, we can conclude that the number 2 present in front of lithium on the reactants side of the equation shows the number of lithium atoms involved in the reaction.

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What is the name of the compound with the formula NH HCO3 ?
svet-max [94.6K]

Answer:

Acid ammonium carbonate // Ammonium bicarbonate.

Explanation:

Hello, a 4 is missing in the fist H, thus:

NH_4HCO_3

The presence of the hydrogen between the ammonium and the carbonates characterizes the salt as an acid salt, so you could name it as acid ammonium carbonate or ammonium bicarbonate (similar to the sodium bicarbonate which is NaHCO_3).

Best regards.

6 0
3 years ago
How can you write Mg(OH)2 in subscripts
Liula [17]
Mg(OH)2 is the correct chemical formula because it takes two Hydroxide ions to bond with Magnesium normally it would be written with a subscript number 2 rather than the parenthesis and number 2
7 0
3 years ago
Dogs use the same glycolysis and cellular processes as humans use to produce ATP. A young dog has never had much energy. He is b
scZoUnD [109]

Answer:

<h2>Dog's mitochondria lack the transport protein that transport  pyruvate ( end product of glycolysis) across the outer mitochondrial  membrane .</h2>

Explanation:

1. As given here that dog's mitochondria can use only fatty acids and also  amino acids for their respiration, and  as compared to others, Dong's cell produce more lactate then normal,  this indicate that his mitochondrial membrane is different then others.  

2. The aerobic phases of cellular respiration in eukaryotes occur within  mitochondria. These aerobic phases are the TCA Cycle and the electron transport chain. Glycolysis occurs in the cytoplasm and the products of glycolysis enter into the mitochondria to continue cellular respiration.

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3 0
3 years ago
A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi
Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of Na = 5.0 g

Mass of Cl_2 = 10.0 g

Molar mass of Na = 23 g/mol

Molar mass of Cl_2 = 71 g/mol

First we have to calculate the moles of Na and Cl_2.

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol

and,

\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}

\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

2Na+Cl_2\rightarrow 2NaCl

From the balanced reaction we conclude that

As, 2 mole of Na react with 1 mole of Cl_2

So, 0.217 moles of Na react with \frac{0.217}{2}=0.108 moles of Cl_2

From this we conclude that, Cl_2 is an excess reagent because the given moles are greater than the required moles and Na is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 0.217 mole of HCl react to give 0.217 mole of NaCl

Now we have to calculate the mass of NaCl

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

Molar mass of NaCl = 58.5 g/mole

\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g

Now we have to calculate the percent yield of this reaction.

Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = \frac{5.24g}{12.7g}\times 100

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0
3 years ago
In a 1.0-liter container there are, at equilibrium, 0.10 mole h2, 0.20 mole n2, and 0.40 mole nh3. what is the value of kc for t
bogdanovich [222]
The reaction involved here would be written as:

2N2 + 3H2 = 2NH3

The equilibrium constant of a reaction is the ratio of the concentrations of the products and the reactants when in equilibrium. The expression for the equilibrium constant of this reaction would be as follows:

Kc = [NH3]^2 / [N2]^2[H2]^3
Kc = 0.40^2 / (0.20)^2 (0.10)^3 
Kc = 4000
6 0
3 years ago
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