CHEMISTRY HIGH SCHOOL

What is the molarity of 0.50 g of Na dissolved in a 1.5 L solution?

Answers

Answer 1

Answer:

Answer : The molarity of solution is, 0.0145 mole/L

Solution : Given,

Mass of solute, Na= 0.50 g

Volume of solution = 1.5 L

Molar mass of Na= 23 g/mole

Molarity : It is defined as the number of moles of solute present in the one liter of solution.

Formula used :

$Molarity=(w_b)/(M_b* V)$

where,

$w_b$ = mass of solute, Na

V = volume of solution in liters

$M_b$ = molar mass of solute, Na

Now put all the given values in the above formula, we get the molarity of a solution.

$Molarity=(0.50g)/(23g/mole* 1.5L)=0.0145mole/L$

Therefore, the molarity of solution is, 0.0145 mole/L

Answer 2

Answer: Molar mass Na = 23.0 g/mol

1 mole Na ------- 23.0 g
? mole Na ------ 0.50 g

moles = 0.50 * 1 / 23.0

moles = 0.50 / 23.0

= 0.0217 moles of Na

Molarity = number of moles / liters solution

M = 0.0217 / 1.5

= 0.014467 M

hope this helps!

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1. An unknown compound was found to have a percent composition of: 47.0%potassium, 14.5% carbon, and 38.5 % oxygen. What is its empirical formula? If
the true molar mass of the compound is 166.22 g/mol, what is its molecular
formula?

Answers

Answer:The empirical formula is KCO2

Explanation:

A compound that contains both potassium and oxygen formed when potassium metal was burned in oxygen gas. the mass of the compound was 7.11 grams. The mass of the potassium metal was 3.91 grams. What mass of oxygen was involved in this reaction? Justify your answer.

Answers

Answer:
3.2 g of O₂

Solution:
This reaction is for the formation of Potassium Superoxide, The reaction is as follow,

K + O₂ → KO₂

First let us confirm that either the given amount of Potassium produces the given amount of Potassium oxide or not,
So,
As,
39.098 g (1 mole) K produced = 71.098 g of K₂O
So,
3.91 g of K will produce = X g of K₂O

Solving for X,
X = (3.91 g × 71.098 g) ÷ 39.098

X = 7.11 g of K₂O

Hence, it is confirmed that we have selected the right equation,
So,
As,
39.098 g of K required = 32 g of O₂
So,
3.91 g of K will require = X g of O₂

Solving for X,
X = (3.91 g × 32 g) ÷ 39.098 g

X = 3.2 g of O₂

The mass of oxygen that was involved in the reaction is calculated as below

mass of the oxygen =mass of the compound- mass of the potassium

that is 7.11 grams -3.91 grams= 3.2 grams of oxygen is involved

This is because the mass of the compound composes the mass of potassium + mass of the oxygen

A compound has a molar mass of 90 grams per mole and the empirical formula CH2O. What is the molecular formula of this compound?

Answers

CH2O = 12 + 2 + 16 = 30
90 /30 = 3
3 x (CH2O) = C3H6O3

Final answer:

The molecular formula of a compound with an empirical formula of CH₂O and a molar mass of 90 g/mol is C₃H₆O₃. This is found by dividing the compound's molar mass by the molar mass of the empirical formula, and using the ratio to multiply the subscripts in the empirical formula.

Explanation:

The molecular formula of a compound can be determined using the molar mass and the empirical formula given. The empirical formula for the compound is CH₂O, which has a molar mass of approximately 30 g/mol (12 g for C, 1 g for H, and 16 g for O). If the molar mass of the compound is 90 g/mol, we can find the ratio of the molar mass of the compound to the empirical formula by dividing 90 g/mol by 30 g/mol, which gives us 3.

This means the molecular formula of this compound is three times the empirical formula. So, to obtain the molecular formula, you multiply each subscript in the empirical formula by 3. If you do this for CH₂O, you get C₃H₆O₃ which is the molecular formula of the compound.

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Directions: Write the name of the type of chemical reaction in the space provided.(Can be Synthesis Reaction, Decomposition Reaction, Combustion, Single Displacement Reaction or Double Displacement Reaction)

Answers

The given reactions are combination reaction, decomposition reaction, neutralization reaction, displacement reaction and double displacement reaction.

What is displacement reaction?

A displacement reaction is a chemical process during which a one more reactive element being replaced from its compound by a less reactive element.

What is combination reaction?

When A combination reaction occurs when one or more elements or compounds react together to generate a new chemical. Equations of the following kind are used to depict such reactions.

X+ Y→XY

6) $Fe}(s)+3{O}_(2)(g)$ → $2{Fe}_(2) {O}_(3)(s)$ , this reaction is a kind of combination reaction reaction.

7) ${Zn}_(2)(s)+2 {HCl} (aq)$ → ${ZnCl}_(2)(a q)+{H}_(2)(g)$

This reaction is a kind of displacement reaction.

8) ${MgCO}_(3)(a q)+2 {HCl}(a q)$ → ${MgCl}_(2)(a q)+{H}_(2) {O}(l)+{CO}_(2)(g)$ .

This reaction is a kind of combustion reaction.

9) ${NiCl}_(2)(s)$ → ${Ni}(s)+{Cl}_(2)(g)$ .

This reaction is a kind of decomposition reaction.

10) ${C}(s)+6 {H}_(2)(g)+{O}_(2)(g)$ → $2 C}_(2) {H}_(6) {O}(s)$ .

This reaction is a kind of combination reaction.

11) $\mathrm{C}_(12) \mathrm{H}_(22) \mathrm{O}_(11)(s) \rightarrow 12 \mathrm{C}(s)+11 \mathrm{H}_(2) \mathrm{O}(\mathrm{g})$ .

This reaction is a kind of dissociation reaction.

12) $2 \mathrm{LiI}(a q)+\mathrm{Pb}\left(\mathrm{NO}_(3)\right)_(2)(a q) \rightarrow 2 \mathrm{LiNO}_(3)(a q)+\mathrm{PbI}_(2)(s)$ .

This reaction is a kind of double displacement reaction.

13) $\mathrm{CdCO}_(3)(s) \rightarrow \mathrm{CdO}(s)+\mathrm{CO}_(2)(g)$ .

This reaction is a kind of decomposition reaction

14) $\mathrm{Cl}_(2)(g)+2 \mathrm{KBr}(a q) \rightarrow 2 \mathrm{KCl}(a q)+\mathrm{Br}_(2)(g)$ .

This reaction is a kind of single displacement reaction.

15) $\mathrm{BaCl}_(2)(a q)+2 \mathrm{KIO}_(3)(a q) \rightarrow \mathrm{Ba}\left(\mathrm{IO}_(3)\right)_(2)(s)+2 \mathrm{KCl}(a q)$ .

This reaction is kind of double displacement reaction.

16) $2 \mathrm{Mg}(s)+\mathrm{O} 2(\mathrm{~g}) \rightarrow 2 \mathrm{MgO}(s)$ .

This reaction is a kind of combination reaction.

17) $AgNO_(3) (aq) +KI(aq)$ → $KI(aq) + KNO_(3)(aq)$ .

This reaction is a kind of double displacement reaction.

18) $2 \mathrm{Li}(s)+\mathrm{H}_(2) \mathrm{O}(b) \rightarrow 2 \mathrm{LiOH}(a q)+\mathrm{H}_(2)(g)$ .

This reaction is a kind of single displacement reaction.

19) $\mathrm{C}(\mathrm{s})+\mathrm{O}_(2)(\mathrm{~g}) \rightarrow \mathrm{CO}_(2)(\mathrm{~g})$ .

This reaction is a kind of combination reaction.

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combination reaction,decombinationreaction,displacementreaction,decombinatioreaction,neutrilizatioreaction

Determine how many moles of O2 are required to react completely with 6.7 moles C6H14.

Answers

I think *extra stress on the word think lol* it's 6.7 moles of O₂.
unless you have the full formula, giving me the amount of moles in C₆H₁₄ and O₂.

hopefully this helps. :/

When 8.00 × 1022 molecules of ammonia react with 7.00 × 1022 molecules of oxygen according to the chemical equation shown below, how many grams of nitrogen gas are produced?4 NH3(g) + 3 O2(g) → 2 N2(g) + 6 H2O(g)

Answers

Answer: 1.848 g

Explanation: To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadro's number}}$ ....(1)

For ammonia:

Putting values in above equation, we get:

$\text{Moles of ammonia}=(8.00* 10^(22))/(6.023* 10^(23))=0.132mol$

$\text{Moles of oxygen}=(7.00times 10^(22))/(6.023* 10^(23))=0.116mol$

For the reaction:

$4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)$

By Stoichiometry of the reaction,

4 moles of ammonia combine with 3 moles of Oxygen

Thus 0.132 moles of ammonia will combine with= $(3)/(4)* 0.132=0.009mol$ of oxygen

Thus ammonia is the limiting reagent as it limits the formation of product.

4 moles of ammonia produces 2 moles of nitrogen

0.132 moles of ammonia will produce= $(2)/(4)* 0.132=0.066 moles$ of nitrogen

Molar mass of nitrogen = 28 g/mol

Amount of nitrogen produced= ${text {no of moles}}* {text {molar mass}}=0.066* 28=1.848g$

Final answer:

In the given chemical reaction, 8.00 x 10²² molecules of ammonia would produce 37.3 grams of nitrogen gas.

Explanation:

In this chemical reaction, every 4 molecules of ammonia (NH₃) produce 2 molecules of nitrogen gas (N₂). Given you have 8.00 x 10²² molecules of ammonia, this would produce (8.00 x 10²² / 2) x 2 = 8.00 x 10²² molecules of nitrogen gas, according to the reaction stoichiometry.

One molecule of nitrogen gas (N₂) has a molar mass of 28 g/mol. To convert molecules to moles, we need Avogadro's number (6.022 x 10²³ molecules = 1 mol). So, the number of moles of nitrogen gas is (8.00 x 10²² molecules / 6.022 x 10²³ molecules/mol) = 1.33 mole.

Now, by using the molar mass of nitrogen gas, we can find the mass. So, the mass of nitrogen gas = molar mass x moles = 28 g/mol x 1.33 mol = 37.3 grams.

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