What is the molarity of HCl if 36.99 mL of a solution of HCl contain 0.3219 g of HCl?

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Answer:

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to draw this Lewis dot structure by firstly realizing P has five valence electrons and F has seven because they are in groups VA and VIIA respectively, which means that the central atom is P with three F atoms around it as shown on the attached.

Also notice each fluorine atom has three lone pairs as their atoms just need one bond to complete the octet and the P atom has one lone pair as it needs three bonds to complete it.

Regards!

Answer:

80g

Explanation:

mass oxygen present in 1 mole of p4010

16×10=160gm

similarly

for 0.5 moles of p4010 160/2= 80gm

The number of atoms present in 0.45 moles of P₄O₁₀ is 1.08 x 10²³ atoms.

To determine the number of atoms, we use Avogadro's number, which states that there are approximately 6.022 x 10²³ particles (atoms, molecules, or formula units) in one mole of a substance.

In this case, we are given 0.45 moles of P₄O₁₀. To calculate the number of atoms, we multiply the number of moles by Avogadro's number:

Number of atoms = 0.45 moles P₄O₁₀ x (6.022 x 10²³ atoms / 1 mole)

Number of atoms = 2.7139 x 10²³ atoms

Rounding to three significant figures, the number of atoms present in 0.45 moles of P₄O₁₀ is approximately 1.08 x 10²³ atoms.

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Answer:

O2 is a covalent substance while NaCl is an ionic substance

Explanation:

In O2 molecule, the bond is between 2 oxygen atoms which are non - metals. Thus, this is a covalent bond since it involves 2 non metals.

Whereas, for the NaCl molecule, the bond is between a metal sodium (Na) and a non metal Chloride(Cl) and thus we can say this is an ionic bond.

Thus the difference is that O2 is a covalent substance while NaCl is an ionic substance.

Answer:

1) 0.00498 mol Cu.

2) 0.00000374 mol CO₂

Explanation:

Question 1)

We want to convert 3.00 * 10²¹ copper atoms into moles. Note that 3.00 is three significant figures.

Recall that by definition, one mole of a substance has exactly 6.022 * 10²³ amount of that substance. In other words, we have the ratio:

[tex]\displaystyle \frac{1\text{ mol}}{6.022\times 10^{23} \text{ Cu}}[/tex]

We are given 3.00 * 10²¹ Cu. To cancel out the Cu, we can multiply it by our above ratio with Cu in the denominator. Hence:

[tex]\displaystyle 3.00 \times 10^{21} \text{ Cu} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} \text{ Cu}}[/tex]

Cancel like terms:

[tex]=\displaystyle 3\times 10^{21} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} }[/tex]

Simplify:

[tex]\displaystyle = \frac{3\text{ mol Cu}}{6.022 \times 10^{2}}[/tex]

Use a calculator:

[tex]= 0.004981... \text{ mol Cu}[/tex]

Since the resulting answer must have three significant figures:

[tex]= 0.00498\text{ mol Cu}[/tex]

So, 3.00 * 10²¹ copper atoms is equivalent to approximately 0.00498 moles of copper.

Question 2)

We want to convert 2.25 * 10¹⁸ molecules of carbon dioxide into moles. Note that 2.25 is three significant digits.

By definition, there will be 6.022 * 10²³ carbon dioxide molecules in one mole of carbon dioxide. Hence:

[tex]\displaystyle \frac{6.022 \times 10^{23} \text{ CO$_2$}}{1\text{ mol CO$_2$}}[/tex]

To cancel the carbon dioxide from 2.25 * 10¹⁸, we can multiply it by the above ratio with the carbon dioxide in the denominator. Hence:

[tex]\displaystyle 2.25\times 10^{18} \text{ CO$_2$} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23} \text{ CO$_2$}}[/tex]

Cancel like terms:

[tex]\displaystyle= 2.25\times 10^{18} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23}}[/tex]

Simplify:

[tex]\displaystyle = \frac{2.25 \text{ mol CO$_2$}}{6.022\times 10^5}}[/tex]

Use a calculator:

[tex]=0.000003736...\text{ mol CO$_2$}[/tex]

Since the resulting answer must have three significant figures:

[tex]= 0.00000374\text{ mol CO$_2$}[/tex]

So, 2.25 * 10¹⁸ molecules of carbon dioxide is equivalent to approximately 0.00000374 moles of carbon dioxide.

Answer:

Explanation:

by definition, 1 mole contains 6.02 x 10^23 of atoms (for elements) or molecules (for compounds)

3.00 x 10^21 atoms of copper / 6.02 x 10^23 of atoms

= 0.004983 moles of copper

= 4.98 x 10^(-3) moles of copper

2.25 x 10^18 molecules of carbon dioxide / 6.02 x 10^23 of molecules

= 0.000003737 moles of carbon dioxide

= 3.74 x 10^(-6) moles of carbon dioxide

Answer:

0.59 mol O₂

Explanation:

The balanced chemical equation for the decomposition of potassium chlorate (KClO₃) to produce potassium chloride (KCl) and oxygen gas (O₂) is the following:

2 KClO₃ → 2 KCl + 3 O₂

According to the equation, 3 moles of O₂ are produced from 2 moles of KClO ⇒ conversion factor: 3 mol O₂/2 mol KClO₃

Now, we calculate the number of moles of KClO₃ there is in 48.1 g, by dividing the mass into the molecular weight (Mw) of O₂:

Mw(KClO₃) = 39.1 g/mol + 35.4 g/mol + (16 g/mol x 3) = 122.5 g/mol

moles KClO₃ = mass KClO₃/Mw(KClO₃) = 48.1 g/(122.5 g/mol) = 0.3926 mol KClO₃

Finally, we multiply the moles of KClO₃ by the conversion factor to calculate the moles of O₂ produced:

0.3926 mol KClO₃ x 3 mol O₂/2 mol KClO₃ = 0.59 mol O₂

Answer:

The molarity is "0.050 M".

Explanation:

The given values are:

M1 = 0.100 M

M2 = ?

V1 = 25.0 mL

V2 = 50.0 mL

As we know,

⇒ [tex]M1\times V1=M2\times V2[/tex]

Or,

⇒ [tex]M2=\frac{M1\times V1}{V2}[/tex]

By putting the values, we get

          [tex]=\frac{0.100\times 25}{50}[/tex]

          [tex]=\frac{2.5}{50}[/tex]

          [tex]=0.05 \ M[/tex]

Answer:

Explanation:

a) AgCl(s) in 0.025 M NaCl

Equation:  AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

Initial conc :    S            O               O

equili conc :    O            S                S

                  NaCl(s) ⇒ Na⁺ (aq) + Cl⁻ (aq)

Initial conc :  0.025      0           0

equili conc :     0          0.025    0.025

Therefore the concentration:  Ag⁺ = 6.4 * 10^-9 M,  Cl⁻  = 0.025 M

attached below is the detailed solution of the

Answer:

The Rf value change will be  >  0.5

Explanation:

Given that Change in Solvent is proportional to change in polarity of solvent system

The change from solvent composition of 60/40  to 30/70 will cause an increase in the polarity of the system .

and Increase in Polarity = Increase in Rf value  because the compound will move to a higher distance

Answer:

Density of mercury is 13600 kg

Answer:

See explanation

Explanation:

a) The balanced reaction equation is;

Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH + PbCO3

b) When we include states of matter;

Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH(aq) + PbCO3 (s)

c) Complete ionic equation;

2Na^+(aq) + CO3^2-(aq) + Pb^2+(aq) + 2OH^-(aq) ----> 2Na^+(aq) + 2OH^-(aq) + PbCO3(s)

Net Ionic equation;

Pb^2+(aq) + CO3^2-(aq) ----> PbCO3(s)

I need more explain?

Answer:

C) METALLIC IS THE CORRECT ANSWER

Explanation: I just did the exam

he says he doesnt know sorry

Answer:

Molar mass for eugenol is 161.3 g/mol

Explanation:

This question talks about freezing point depression:

Our solute is eugenol.

Our solvent is camphor.

Formula to state the freezing point depression difference is:

ΔT = Kf . m . i where

ΔT = Freezing T° of pure solvent - Freezing T° of solution

In this case ΔT = 0.62°C

Kf for camphor is: 37°C /m

As eugenol is an organic compund, i = 1. No ions are formed.

To state the molar mass, we need m (molal)

Molal are the moles of solute in 1kg of solvent. Let's replace data:

0.62°C = 40 °C/m . m . 1

0.62°C / 40 m/°C = 0.0155 m

We convert mass of camphor from g to kg = 100 g . 1kg / 1000g = 0.1 kg

0.0155 molal = moles of solute / 0.1 kg

0.0155 m/kg . 0.1 kg = 0.00155 moles

We know that these moles are contained in 250 mg, so the molar mass will be:

0.25 g / 0.00155 mol = 161.3 g/mol

Notice, we convert mg to g, for the units!

Answer:

B

Explanation:

liquids is produce therefore it will have the less Entropy

Answer:

B.

Explanation:

if the # of molecules (of gas) on the product side is less than # of molecules on the reactant side = entropy is decreasing, and vice versa.

if the # of molecules on the reactant side is less than # of molecules on the product side = entropy is increasing (it is more disorderly, chaotic)

A. 2 → 2 + 1 entropy increases

B. 1 + 4 → 1 + 1 entropy decreases

C. 2 → 1 + 3 entropy increases

D. 2 + 1 → 2 + 2 entropy increases

Answer:

Step 1:

N

H

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3

C

H

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C

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K

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B.

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C.

K

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D.

K

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Step 2:

N

H

2

C

H

2

C

H

2

N

H

+

3

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K

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B.

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C.

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D.

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Acid-

Ethylenediammonium ion refers to an organic compound with the chemical formula C₂H₄(NH⁺₃)₃. From its ammonia complex, its characteristics include:

The dissociation of a chemical compound(here, it's Ethylenediammonium) is the disintegration of a compound into simpler compounds or elements that may typically be recombined mostly under distinct conditions.

The dissociation of ethylenediammonium ion occurs in two stages and can be represented as follows:

From above, the formation of hydrogen ion H⁺ indicates that ethylenediammonium ion is an acid.

Thus, in the first dissociation, the equilibrium constant can be represented as: [tex]\mathbf{Ka_1}[/tex]

[tex]\mathbf{NH_3^+ CH_2CH_2NH_3^+_{(aq)} \rightleftharpoons^{\mathbf{Ka_1}} NH_2CH_2CH_2NH_3^+_{(aq)}+H^+_{(aq)}}[/tex]

The second stage of the dissociation can now be expressed as:

[tex]\mathbf{NH_2 CH_2CH_2NH_3^+_{(aq)} \rightleftharpoons ^{\mathbf{Ka_2}} NH_2CH_2CH_2NH_2_{(aq)}+H^+_{(aq)}}[/tex]

From above, we will notice that the equilibrium constant   is [tex]\mathbf{Ka_2}[/tex]

Thus, from the above explanation, we can see the complete two acid dissociation reactions for ethylenediammonium ion and the correct symbol for the equilibrium constant for each reaction.

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Answer:

Percent yield  = 72.07 %

Explanation:

Our reaction is:

C₄H₁₀O + NaBr + H₂SO₄ → C₄H₉Br + NaHSO₄ + H₂O

It is correctly balanced.

Let's determine which is the limiting reagent:

45 g . 1 mol / 74 g = 0.608 moles of C₄H₁₀O

67.1 g . 1 mol / 102.9 g = 0.652 moles of NaBr

97 g . 1 mol / 98 g = 0.990 moles of sulfuric acid

Ratio is always 1:1, so for 1 mol of NaBr and 1 mol of sulfuric acid we need 1 mol of C₄H₁₀O. We have 0.652 moles of NaBr, we need the same amount of C₄H₁₀O and we have 0.990 moles of acid, we need the same amount of C₄H₁₀O; we only have 0.608 moles, that's why C₄H₁₀O is the limiting reactant, there's no enough C₄H₁₀O.

Ratio is also 1:1, between reactant and product.

1 mol of C₄H₁₀O produces 1 mol of C₄H₉Br

Then, 0.608 moles will produce 0.608 moles of C₄H₉Br

We convert moles to mass: 0.608 mol . 136.9 g/mol = 83.25 g

That's the 100 % yield reaction

Percent yield  = (Yield produced / Theoretical yield) . 100

Percent yield = (60 g / 83.25 g) . 100 = 72.07 %

Answer:

C2HBr

Explanation:

The empirical formula is like the simpliest form so divide all by 3 and get the above formula.

Solution :

[tex]$BF_3 (g) + BCl_3 (g) \rightarrow BF_2 Cl + BCl_F(g)$[/tex]

Explanation 1 :

Spontaneity of the reaction is based on two factors :

-- the tendency to acquire a state of minimum energy

-- the energy of a system to acquire a maximum randomness.

Now, since there isn't much difference in the bond enthalpies of B-F and B-Cl. So, we can say the major driving factor is tendency to acquire a state of maximum randomness.

Explanation 2 :

A system containing the [tex]\text{"chemically mixed"}[/tex] B halides has a [tex]\text{greater entropy}[/tex] than a system of [tex]$BCl_3$[/tex] and [tex]BF_3[/tex].

It has the same number of [tex]\text{gas phase molecules}[/tex], but more distinguishable kinds of [tex]\text{molecules}[/tex], hence, more microstates and higher entropy.

Answer:

Option e.

Explanation:

Molarity is the concentration that indicates moles of solute in 1 L of solution.

We have another concentration, percent by mass.

Percent by mass indicates mass of solute in 100 g of solution.

Our solute is HNO₃, our solvent is water.

17.5 g of nitric acid is the mass of solute. We can convert them to moles:

17.5 g . 1mol / 63g = 0.278 moles

We do not have volume of solution. We assume the mass is 100 g because the percent by mass but we need density to state the volume.

Density = Mass / Volume

Mass / Density = Volume

Once we have the volume, we need to be sure the units is in L, to determine molarity

M = mol /L

Answer:

when naming ionic compounds — those are only used in naming covalent molecular compounds. Do NOT use prefixes to indicate how many of each element is present; this information is implied in the name of the compound. since iron can form more than one charge. Ionic Compounds Containing a Metal and a Polyatomic Ion.

Answer:

since H2O is an acid, by the Arrhenius definition, it would donate a proton. Thus, the conjugate base is OH~

Answer:

1,3-dibromobenzene

Explanation:

An image of the compound described in the question is attached to this answer.

We need to reiterate here the rules of IUPAC nomenclature. The substituents in a compound must be named in such a way that they have the lowest number.

The compound described may also be named as 1,5-dibromobenzene but this name is disallowed because it gives the substituents a higher number than 1,3-dibromobenzene.

Answer:

the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Explanation:

Given the data in the question;

Co = 53 or [ 53 wt% B-47 wt% A ]

W∝ = 0.5 = Wβ

Cβ = 92 or [ 92 wt% B-8 wt% A ]

Now, lets set up the Lever rule for W∝ as follows;

W∝ = [ Cβ - Co ] / [ Cβ - C∝ ]

so we substitute our given values into the expression;

0.5 = [ 92 - 53 ] / [ 92 - C∝ ]

0.5 = 39 /  [ 92 - C∝ ]

0.5[ 92 - C∝ ] = 39

46 - 0.5C∝  = 39

0.5C∝ = 46 - 39

0.5C∝ = 7

C∝ = 7 / 0.5

C∝ = 14  or [ 14 wt% B-86 wt% A ]

Therefore, the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Answer:

m = 1125.9 g.

Explanation:

Hey there!

In this case, according to the given information, it turns out possible for us to solve this problem by using the definition of density as mass divided by volume:

[tex]d=\frac{m}{V}[/tex]

Thus, we solve for the mass in the equation to obtain:

[tex]m=d*V[/tex]

Then, we plug in the values to obtain:

[tex]m=2.70g/mL*417mL\\\\m=1125.9g[/tex]

Regards!

Answer:

1.) Propanone (ketone)

2.) Ethanal( aldehyde)

3.) 3-phenyl-2-propenal (aldehyde)

4.) Butanone (ketone)

5.) Ethanol ( alcohol)

6.) 2-propanol (alcohol)

Explanation:

In organic chemistry, ALCOHOL ( also known as alkanol) are compounds in which hydroxyl groups are linked to alkyl groups. They can be considered as being derived from the corresponding alkanes by replacing the hydrogen atoms with hydroxyl groups. The hydroxyl group is the functional group of the alcohol as it is responsible for their characteristic chemical properties. A typical example of alcohol is ethanol and 2-propanol.

Alkanals or ALDEHYDES have the general formula RCHO while alkanones or KETONES have the general formula RR'CO where R and R' may be alkyl or aryl groups. The main similarity between these two classes of compounds is the presence of the carbonyl group. In aldehydes, there is a hydrogen atom attached to the carbon In the carbonyl group while there is none on the ketones.

Some common examples of ketones are Propanone, Butanone while examples of aldehydes are Ethanal and 3-phenyl-2-propenal

Answer:

1.58 g

Explanation:

Step 1: Write the balanced equation

2 NaHCO₃ ⇒ Na₂CO₃ + H₂O + CO₂

Step 2: Calculate the moles corresponding to 2.50 g of NaHCO₃

The molar mass of NaHCO₃ is 84.01 g/mol.

2.50 g × 1 mol/84.01 g = 0.0298 mol

Step 3: Calculate the moles of Na₂CO₃ produced

The molar ratio of NaHCO₃ to Na₂CO₃ is 2:1. The moles of Na₂CO₃ produced are 1/2 × 0.0298 mol = 0.0149 mol

Step 4: Calculate the mass corresponding to 0.0149 moles of Na₂CO₃

The molar mass of Na₂CO₃ is 105.99 g/mol.

0.0149 mol × 105.99 g/mol = 1.58 g