Combustion reactions are a notable source of carbon dioxide in the environment. Using the following balanced equation, how many grams of carbon dioxide are formed when 100.00 g of propane is burned? Express your answer to the correct number of significant figures. Be sure to show all steps completed to arrive at the answer. Equation: C3H8 + 5O2 ->>>>>>> 3CO2 + 4H2O

Answer:

a

Explanation:

answer is a on edg

Answer:

0.5 mol MgCl₂

Explanation:

Step 1: Write the balanced equation

Mg + 2 HCl → MgCl₂ + H₂

In words, 1 mole of Mg reacts with 2 moles of HCl to form 1 mole of MgCl₂ and 1 mole of H₂.

Step 2: Establish the appropriate molar ratio

The molar ratio of HCl to MgCl₂ is 2:1.

Step 3: Calculate the moles of MgCl₂ produced from 1 mole of HCl

1 mol HCl × (1 mol MgCl₂/2 mol HCl) = 0.5 mol MgCl₂

Answer:

it is 2.0, the above one is wrong

Explanation:

I did the test :

Answer:

[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]

Explanation:

Hello,

In this case, since water and oil are immiscible due to the oil's nonpolarity and water's polarity, when mixed, the oil remains on the water since it is less dense than water. In such a way, for a plastic sunk in the oil and floating on the water (in middle of them) we can conclude that the plastic have a mid density, therefore, the required organization is:

[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]

Best regards.

from

1moles=iatom

Mole=mass÷avogardos

Where

Avogadro's= 6.02×10²³

So moles = 65.0÷6.02×10²³

Atoms of zinc = 391.6 ×10²³

The number of atoms present in the given mass of Zinc that is 65.0gm is [tex]5.99\times10^{ 23}[/tex].

Atoms are the basic building blocks of matter. They are the smallest units of an element that retain the chemical properties of that element.

Now, to determine the number of atoms in a given number of moles, we can use Avogadro's number, which is approximately  [tex]6.022 \times10^{23}[/tex]atoms per mole.

First, we calculate the number of moles of zinc in 65.0g by dividing the given mass by the molar mass of zinc. The molar mass of zinc (Zn) is 65.38 g/mol.

Number of moles = Mass / Molar mass

Number of moles = 65.0g / 65.38 g/mol ≈ 0.9942 mol

Next, multiply the number of moles by Avogadro's number to find the number of atoms.

Number of atoms =[tex]Number of moles \times Avogadro's number[/tex]

Number of atoms = [tex]0.9942[/tex]mol × [tex]6.022 \times10^{23}[/tex] atoms/mol

Therefore, approximately [tex]5.99\times10^{ 23}[/tex] atoms are present in 65.0g of zinc.

Learn more about atoms here:

https://brainly.com/question/1566330

#SPJ2

Answer:

The pH of the solution will be 3

Explanation:

The strength of acids is determined by their ability to dissociate into ions in aqueous solution. A strong acid is any compound capable of completely and irreversibly releasing protons or hydrogen ions, H⁺. That is, an acid is said to be strong if it is fully dissociated into hydrogen ions and anions in solution.

Being pH=- log [H⁺] or pH= - log [H₃O⁺] and being a strong acid, all the HClO₃ dissociates:

HClO₄      +    H₂O        →      H₃O⁺      +      ClO₄-  

So: [HCLO₄]= [H₃O⁺]

The molar concentration is:

[tex]molar concentration=\frac{number of moles of solute}{volume solution}[/tex]

The molar mass of HClO₄ being 100 g / mole, then if 100 grams of the compound are present in 1 mole, 0.025 grams in how many moles are present?

[tex]moles of HClO_{4} =\frac{0.025 grams*1 mole}{100 grams}[/tex]

moles of HClO₄= 0.00025

Then:

[tex][HClO_{4}]=\frac{0.00025 moles}{0.25 L}[/tex]

[tex][HClO_{4}]=0.001 \frac{ moles}{ L}[/tex]

Being [HCLO₄]= [H₃O⁺]:

pH= - log 0.001

pH= 3

The pH of the solution will be 3

Answer:

The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.  

Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams

The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,  

Moles of water = weight of water/molecular weight

= 42.1 grams / 18 = 2.3

The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,  

= 214.2 * 0.2190 = 46.91 grams

The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,

= 214.2 * 0.4314 = 92.40 grams

The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,  

= 214.2 * 0.3497 = 74.91 grams

The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17

The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,  

92.40/78.96 = 1.17

The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,  

74.91/15.999 = 4.68.  

Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O

Answer:

CH3CH=NH2+>CH3CH2NH3 +

Explanation:

There are certain structural features that determine the stability of cationic species. These features that lead to the stability and higher acid strength of cations are those features that stabilize the cation.

CH3CH=NH2+ is more acidic than CH3CH2NH3 + owing to the fact that CH3CH=NH2+ contains a double bond in close proximity with the hydrogen that can be lost as a proton. Electron withdrawal by the double bond (greater s character) makes it easier for this hydrogen to be lost as a proton compared to CH3CH2NH3 +.

Density measures how tightly packed particles are.

If particles are tightly packed together, they will be more dense.

If they are loosely together, they will be less dense.

However, a common mistake is thinking that if something

is more dense it means that it's heavier.

However, that's not the case.

It has to do with how particles are packed in an object.

Answer:

After 47.9 days, will remain 14.5mg of the isotope

Explanation:

The radioactive decay follows always first-order kinetics where its general law is:  

Ln[A] = -Kt + ln[A]₀

Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.

We can find rate constant from half-life as follows:

[tex]t_{1/2} = \frac{ln2}{K}[/tex]

K = ln 2 / 27.7 days

K = 0.025 days⁻¹

Replacing, initial amount of isotope is 48.0mg = [A]₀ , K is 0.025 days⁻¹ and t = 47.9 days:

Ln[A] = -Kt + ln[A]₀

Ln[A] = -0.025 days⁻¹*47.9 days + ln (48.0mg)

ln [A] = 2.6726

[A] = e^ (2.6726)

[A] = 14.5mg

Answer:

The answer is "[tex]\bold{\log \frac{[0] mole}{[R]mole}}[/tex]"

Explanation:

[tex]E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}\\[/tex]

In the above-given equation, we can see from [tex]E_{ceu}[/tex], of both oxidant [tex]conc^n[/tex]as well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor  and each other suspend

[tex]\to \log \frac{\frac{\text{oxidating moles}}{25 \ ml}}{\frac{\text{moles of reduction}}{25 ml}} \ \ = \ \ \log \frac{\frac{\text{oxidating moles}}{30 \ ml}}{\frac{\text{moles of reduction}}{30 ml}} \\\\\\[/tex]

[tex]\to {\log \frac{[0] mole}{[R]mole}}[/tex]

The cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.

The cell potential has been given as the difference in the potential of the two half cells in the electrochemical reaction.

The two cells has been set with the concentration of solutions in the oxidation and reduction half cells.

The cell potential has been changed when there has been a change in the potential of the half cells.

The volume of 30 mL to the solution has been, resulting in the cell potential difference of x.

With the volume of 25 mL, there has been the difference in the potential being similar to the 30 mL solution, i.e. x.

Thus, the cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.

Learn more about cell potential, here:

https://brainly.com/question/1313684

Answer:

461.85 degrees Celsius

Answer:

0.14%

Explanation:

The computation of % is shown below:

As we know that

     HClO <=> H+ + ClO-

I         0.015          0      0

C          -a            +a     +a

E      0.015-a        a       a

Now

[tex]Ka = \frac{[H+][ClO-]}{[HClO]}[/tex]

[tex]= \frac{a^{2}}{(0.015 - a)} \\\\= 3.0 \times 10^{-8}[/tex]

[tex]a^{2} + 3.0 \times 10^{-8}a - 4.5 \times 10^{-10} = 0[/tex]

Now Solves the quadratic equation i.e.

[tex]a = 2.120 \times 10^{-5}[/tex]

[tex][H+] = a = 2.120 \times 10^{-5} M[/tex]

So,

% ionization is

[tex]= \frac{[H+]}{[HClO]}_{initial} \times 100\%\\\\= 2.120 \times 10^{-5}\div0.015 \times 100\%[/tex]

= 0.14%

Hence, the percentage of hypochlorous ionization is 0.14%

Answer:

Atomic no = 12 = Mg

Explanation:

It is given that,

The atomic number of two elements that are represented by letter Q and R are 9 and 12.

We need to write the electronic configuration of R. Atomic number shows the number of protons in atom.

For R, atomic number = 12

Its electronic configuration is : 2,8,2

It has two valance electrons in its outermost shell. The element is Magnesium (Mg).

Answer:

[tex]Speed = 3.30 \frac{km}{hr}[/tex]

Explanation:

Given

Distance = 1 mile

Time = 8.92 minutes

Required

Calculate Speed in km/hr

Speed is calculated as thus;

[tex]Speed = \frac{Distance}{Time}[/tex]

Substitute 1 mile for distance and 8.92 minutes for time

[tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex]

Convert Miles to Kilometres

If

[tex]1\ mile = 1609\ m[/tex];

Then

[tex]1\ mile = \frac{1609\ km}{1000}[/tex]

[tex]1\ mile = 1.609\ km[/tex] --- (1)

Convert minutes to hour

[tex]1\ minutes = 0.0167\ hour[/tex]

Multiply both sides by 8.92

[tex]8.92 * 1\ minutes = 0.0167\ hour * 8.92[/tex]

[tex]8.92 \ minutes = 0.148964\ hour[/tex] ---- (2)

By substituting (1) and (2) in [tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex], we have

[tex]Speed = \frac{1.609\ km}{0.48694\ hour}[/tex]

[tex]Speed = 3.304309 \frac{km}{hr}[/tex]

[tex]Speed = 3.30 \frac{km}{hr}[/tex] -- Approximated

Answer:

[tex]Ksp=1.2[/tex]

Explanation:

Hello,

In this case, as the saturated solution has 7.1 grams of sodium carbonate, the solubility product is computed by firstly computing the molar solubility by using its molar mass (106 g/mol):

[tex]Molar \ solubility=\frac{7.1gNa_2CO_3}{0.1L}*\frac{1molNa_2CO_3}{106gNa_2CO_3}=0.67M[/tex]

Next, as its dissociation reaction is:

[tex]Na_2CO_3(s)\rightleftharpoons 2Na^+(aq)+CO_3^{2-}(aq)[/tex]

The equilibrium expression is:

[tex]Ksp=[Na^+]^2[CO_3^{2-}][/tex]

And the concentrations are related with the molar solubility (2:1 mole ratio between ionic species):

[tex]Ksp=(2*0.67)^2*(0.67)\\\\Ksp=1.2[/tex]

Best regards.

Answer:

C) formaldehyde, H2C=O.

Explanation:

Hello,

In this case, given that the hydrogen bondings are known as partial intermolecular interactions between a lone pair on an electron rich donor atom, particularly oxygen, and the antibonding molecular orbital of a bond between hydrogen and a more electronegative atom or group. Thus, among the options, C) formaldehyde, H2C=O, will exhibit hydrogen bonding since the lone pair of electrons of the oxygen at the carbonyl group, are able to interact with hydrogen (in the form of water).

Best regards.

Answer:

C. electronegativity

Explanation:

THE ELEMENT WITH THE GREATE ELECTRONEGATIVITY IS ASSIGNED A NEGATIVE OXIDATION NUMBER EQUAL TIO ITS CHARGE

Answer:

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

Answer:

[tex]V_{HCl}=0.208L=208mL[/tex]

Explanation:

Hello,

In this case, since the chemical reaction is:

[tex]2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O[/tex]

We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:

[tex]n_{HCl}=2*n_{Mg(OH)_2}[/tex]

In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:

[tex]n_{Mg(OH)_2}=500mg*\frac{1g}{1000mg}*\frac{1mol}{58.3g} =0.00858mol[/tex]

Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:

[tex][H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M[/tex]

Then, since the concentration and the volume define the moles, we can write:

[tex][HCl]*V_{HCl}=2*n_{Mg(OH)_2}[/tex]

Therefore, the neutralized volume turns out:

[tex]V_{HCl}=\frac{2*0.00858mol}{0.0562\frac{mol}{L} }\\ \\V_{HCl}=0.208L=208mL[/tex]

Best regards.

Answer:

The balanced chemical reaction is given as:

[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]

Explanation:

When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.

The balanced chemical reaction is given as:

[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]

According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.

Answer:

there are 37,8541 liters in 10 gallons

ℯ ℴ ℴ ℴℯℯ

it has a weak adhesion

Answer:

[tex][H^+]=0.000123M[/tex]

[tex]pH=3.91[/tex]

Explanation:

Hello,

In this case, dissociation reaction for acetic acid is:

[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]

For which the equilibrium expression is:

[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]

Which in terms of the reaction extent [tex]x[/tex] could be written as:

[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]

Thus, solving by using a solver or quadratic equation we obtain:

[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]

And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:

[tex][H^+]=0.000123M[/tex]

Now, the pH is computed as follows:

[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]

Best regards.

Answer:

Pentan-2-ol

Explanation:

On this reaction, we have a Grignard reagent (ethylmagnesium bromide), therefore we will have the production of a carbanion (step 1). Then this carbanion can attack the least substituted carbon in the epoxide in this case carbon 1 (step 2). In this step, the epoxide is open and a negative charge is generated in the oxygen. The next step, is the treatment with aqueous acid, when we add acid the hydronium ion ([tex]H^+[/tex])  would be produced, so in the reaction mechanism, we can put the hydronium ion. This ion would be attacked by the negative charge produced in the second step to produce the final molecule: "Pentan-2-ol".

See figure 1

I hope it helps!

Answer:

The mass of ammonium phosphate produced is 14.3g

Explanation:

Full question contains: "What mass of ammonium phosphate is produced by the reaction of 4.9g of ammonia"

Ammonium phosphate ((NH₄)₃PO₄) can be produced by the reaction of phosphoric acid (H₃PO₄) with ammonia (NH₃) as follows:

H₃PO₄ + 3NH₃ → (NH₄)₃PO₄

Where 1 mole of phosphoric acid reacts with 3 moles of ammonia producing 1 mole of ammonium phosphate.

To know how many grams of ammonium phosphate we need to find moles of ammonia that react, and, with the chemical equation we can find moles of ammonium phosphate and its mass as follows:

Moles ammonia (Molar mass: 17.031g/mol):

4.9g × (1mol / 17.031g) = 0.288 moles of ammonia you have in 4.9g

Moles of ammonium phosphate (149.09g/mol) and its mass:

As 0.288 moles of NH₃ are reacting and 3 moles of ammonia produce 1 mole of ammonium phosphate, moles produced are:

Moles (NH₄)₃PO₄:

0.288 moles NH₃ ₓ (1 mol (NH₄)₃PO₄  / 3 mol NH₃) = 0.0959 moles (NH₄)₃PO₄

These moles are, in grams:

0.0959 moles (NH₄)₃PO₄ ₓ (149.09g / mol) = 14.3g ammonium phosphate.

Answer:

[tex]V_2=1.17L[/tex]

Explanation:

Hello,

In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:

[tex]V_1C_1=V_2C_2[/tex]

Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:

[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]

Best regards.

Answer:

a. Cyclohexanone

Explanation:

The principle of IR technique is based on the vibration of the bonds by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is a specific energy that generates a specific vibration. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.

Now, we must remember that the lower the wavenumber we will have less energy. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.

If we look at the structure of all the molecules we will find that in the last three we have heteroatoms (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of resonance structures which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.

The molecule that fulfills this condition is the cyclohexanone.

See figure 1

I hope it helps!

Answer:

0.05 mole of Br2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

C2H4 + Br2 —› C2H4Br2

From the balanced equation above,

1 mole of C2H4 reacted with 1 mole of Br2 to produce 1 mole of C2H4 Br2.

Finally, we shall determine the number of moles bromine that will react with 0.05 mole of C2H2.

The number of mole of Br2 needed for the reaction can be obtained as follow:

From the balanced equation above,

1 mole of C2H4 reacted with 1 mole of Br2.

Therefore, 0.05 mole of C2H4 will also react with 0.05 mole of Br2.

Therefore, 0.05 mole of Br2 is needed for the reaction.