Calculate how many moles of NO2 form from 3.0 mol N2O5 2N2O5(g)→4NO2(g)+O2(g)Express your answer using two significant figures.

Answer:

B - Making an Observation

Explanation:

Making an observation best describes the act of using senses or tools to gather information. Therefore, option B is correct.

The five senses—sight, taste, touch, hearing, and smell—gather data about our surroundings that the brain interprets. Based on prior experience (and subsequent learning), as well as by combining the data from each sensor, we make sense of this information.

Information gleaned from your five senses is referred to as an observation. These are smell, taste, touch, hearing, and sight. When you see a bird or hear it sing, you notice it.

The term observation, which is also used to sense five aspects of the world including vision, taste, touch, smell, and hearing, is used to describe utilizing the senses to examine the world, employing tools to take measurements, and looking at prior research findings.

Thus, option B is correct.

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Answer:

(a) The empirical formula of the compound is

m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O).

(b) The grams of O2 that were used in the reaction is 1.146 g

(c) The amount of O2 that would have been required for complete combustion is 1.401 g.

Explanation:

a. m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O)

(b) Using law of conservation of mass from above

m(O2) = m(CO) + m(CO2) + m(H2O) - m(CxHy)

m(O2) = 0.446 + 0.700 + 0.430 - 0.430

m(O2) = 1.146 g

The grams of O2 that were used in the reaction is 1.146 g

(c) for complete combustion, we need to oxidized CO to CO2

Then, 2CO +O2 = 2CO2

m(add)(O2) = M(O2)*¢(O2)/2 = M(O2) * {(m(CO))/(2M(CO))}

m(add)(O2) = 32 * {(0.446)/(2*28)} = 0.255 g

Note; Molar mass of O2 = 32, CO = 28

m(total)(O2) = m(O2) + m(add)(O2)

m(total)(O2) = 1.146 + 0.255 = 1.401 g

The amount of that grams would have been required for complete combustion is 1.401 g.

Note (add) and (total) were used subscript to "m"

Answer:

As K >>> 1, the reaction will shift to the products

Explanation:

To know the direction of any reaction you must calculate the equilibrium constant, K. If K is < 1, the reaction will shift to the reactants and if k > 1 the reaction will shift to the products.

With the reactions:

HSO₄⁻ ⇄ SO₄²⁻ + H⁺ Ka = 1.26x10⁻²

And:

NH₄⁺ ⇄ NH₃⁺ + H⁺ Ka = 5.6x10⁻¹⁰

The inverse reaction:

NH₃⁺ + H⁺ ⇄ NH₄⁺ 1/Ka = 1.8x10⁹

The sum of the reactions:

HSO₄⁻ + NH₃⁺ + H⁺ ⇄ NH₄⁺ + SO₄²⁻ + H⁺ K = 1.26x10⁻² ₓ 1.8x10⁹ = 2.3x10⁷

Answer:

I think it would be the Dead Sea

Explanation:

Because the dead sea is already usually in the warmer temperatures, the boiling point of the water would be lower than the rest.

Answer:

Option A. LR = N2O4, 45.7g N2 formed

Explanation:

The balanced equation for the reaction is given below:

N2O4(l) + 2N2H4(l) → 3N2(g) + 4H2O(g)

Next, we shall determine the masses of N2O4 and N2H4 that reacted and mass of N2 produced from the balanced equation. This is illustrated below:

Molar mass of N2O4 = 92.02 g/mol

Mass of N2O4 from the balanced equation = 1 x 92.02 = 92.02 g

Molar mass of N2H4 = 32.05 g/mol

Mass of N2H4 from the balanced equation = 2 x 32.05 = 64.1g

Molar mass of N2 = 2x14.01 = 28.02g/mol

Mass of N2 from the balanced equation = 3 x 28.02 = 84.06g

Summary:

From the balanced equation above,

92.02g of N2O4 reacted with 64.1g of N2H4 to produce 84.06g of N2.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

92.02g of N2O4 reacted with 64.1g of N2H4.

Therefore, 50g of N2O4 will react with = (50 x 64.1)/92.02 = 34.83g of N2H4.

From the calculations made above, we can see that only 34.83g out 45g of N2H4 is required to react completely with 50g of N2O4.

Therefore, N2O4 is the limiting reactant and N2H4 is the excess reactant.

Finally, we shall determine the mass of N2 produced from the reaction.

In this case the limiting reactant will be used as it will produce the maximum yield of N2 since all of it is used up in the reaction.

The limiting reactant is N2O4 and the mass N2 produced can be obtained as illustrated below:

From the balanced equation above,

92.02g of N2O4 reacted to produce 84.06g of N2.

Therefore 50g of N2O4 will react to produce = (50 x 84.06)/92.02 = 45.7g of N2.

Therefore, 45.7g of N2 were produced from the reaction.

At the end of the day,

The limiting reactant is N2O4 and 45.7g of N2 were produced from the reaction.

Answer:

(+)-cholesterol = 71.88%

(-)-cholesterol = 28.12%

Explanation:

Asuming 1 gram of sample is dissolved in 1mL of water and the sample cell was 1dm long.

Enantiomeric excess is defined as the amount of pure enantiomer in a sample. The formula is:

ee = [α]mixture / [α]pure enantiomer.

Replacing:

ee = 14° / 32°×100 = 43.75%

As the sample is 14°, There is an excess of (+)-cholesterol and 56.25% is a 1:1 mixture of enantiomers.

That means percent composition of enantiomers is:

Answer:

5.56 × 10⁻⁸

Explanation:

Step 1: Given data

Step 2: Calculate the concentration of H⁺

We will use the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

[tex]Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}[/tex]

Answer:

pressure is = 54.9802atm

Explanation:

using ideal gas equation

PV=nRT

Answer:

Francium

Explanation:

The atomic radius increases from top to bottom in a group, and decreases from left to right across a period.

so francium (Fr) is the largest atom or has highest radii.

Hope this helps & please mark as brainiest!

Answer:

Francium has the largest atomic radius.

The general trend for atomic radii is increasing from top to bottom and decreasing from left to right so the one with the largest atomic radius will be in the bottom left of the periodic table.

Answer:

[tex]\Delta S=6045.8\frac{J}{K}[/tex]

Explanation:

Hello,

In this case, we can compute the change in the entropy for vaporization processes in term of the enthalpy of vaporization as shown below:

[tex]\Delta S=\frac{m*\Delta H}{T}[/tex]

Whereas the temperature is in Kelvins. In such a way, the entropy results:

[tex]\Delta S=\frac{1.00kg*2256x10^3\frac{J}{kg} }{(100+273.15)K}\\\\\Delta S=6045.8\frac{J}{K}[/tex]

Best regards.

Answer:

8.33mL or .0083L

Explanation:

Use m1 * V1 = m2 * V2

6.00M(x) = 0.100M(500mL)

solve for x

x= (.1 * 500) / 6

x=8.333 mL

Answer:

Explanation:

attached here is the diagram representing the structure

Answer:

Option A - When |ΔHsolute| > |ΔHhydration|

Explanation:

A solution is defined as a homogeneous mixture of 2 or more substances that can either be in the gas phase, liquid phase, solid phase.

The enthalpy of solution can either be positive (endothermic) or negative (exothermic).

Now, we know that enthalpy is amount of heat released or absorbed during the dissolving process at constant pressure.

Now, the first step in thus process involves breaking up of the solute. This involves breaking up all the intermolecular forces holding the solute together. This means that the solute molecules are separate from each other and the process is always endothermic because it requires energy to break interaction. Thus;

The enthalpy ΔH1 > 0.

Thus, the enthalpy of the solute has to be greater than the enthalpy of hydration.

An endothermic ΔHsolution occurs when |ΔH solute| < |ΔH hydration|.

A substance dissolves in water when the solute - solvent interaction exceeds the solute - solute solute interaction. The energy required to break the bonds between solutes is the ΔHsolute and the energy released when solute - solvent interaction take place is called the ΔHhydration.

We know that when  |ΔH solute| < |ΔH hydration|, energy is required to break up the solute - solute interaction and  ΔHsolution is endothermic.

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The answer is Hydrogyn bonding. It keeps the water molocules bonded together and in a liquid state, without it it'd be in a gashious state.

Answer:Hydrogen bonds keep them together in room temperature, hope this helps!

Explanation:

<!> Brainliest is appreciated! <!>

Answer:

5727 years or 5730 (rounded to match 3 sig figs) whichever one your teacher prefers

Explanation:

First Order decay has a half life formula of Half Life = Ln (2) / k = 0.693/K

Half-life = 0.693/k = 0.693/1.21 x10-4 =  5727 years or 5730 (rounded to match 3 sig figs)

This should be correct because if you google the half-life of 14 C it is ~ 5700 years

Answer:

C. A positive and a negative ion

Explanation:

Acids and bases are made up of charged particles known as ions. The ions present in acids are oppositely charged and are held together by strong electrostatic forces. When acids or bases are dissolved in water, the electrostatic forces holding their individual molecules together are weakened and these ions are free to move apart in a process known as dissociation. Dissociation occurs because of the attraction between the positive and negative ions in the acid and bases and the negative and positive polarity of water.

For example, when an acid like hydrochloric acid is dissolved in water it dissociates into positive and negative ions as follows:

HCl(aq) -----> H+ + Cl-

When a base like sodium hydroxide is dissolved in water, it dissociates into positive and negative ions as follows:

NaOH(aq) ----> Na+ + OH-

Answer:

yeah C is correct

Explanation:

Answer:

8.59 g

2.25 g

Explanation:

According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-

Moles of Pb(OH)CL is

[tex]= \frac{Mass}{Molar\ mass}[/tex]

[tex]= \frac{10.0 g}{259.65g / mol}[/tex]

= 0.0385 mol

Mass of PbO needed is

[tex]= 0.385mol Pb(OH) Cl\times \frac{1 mol PbO}{1molpb (OH) cl} \times \frac{223.2g PbO}{1mol PbO}[/tex]

After solving the above equation we will get

= 8.59 g

Mass of NaCL needed is

[tex]= \frac{1mol\ NaCl}{1molPb\ (OH)Cl} \times \frac{58.45NaCl}{1mol NaCl}[/tex]

After solving the above equation we will get

= 2.25 g

Therefore we have applied the above formula.

Answer:

intermolecular dipole-dipole hydrogen bonds

Explanation:

Water is a polar molecule. Recall that the central atom in water is oxygen. The molecule is bent, hence it has an overall dipole moment directed towards the oxygen atom. Since it has a permanent dipole moment, we expect that it will show dipole-dipole interactions in the liquid state.

Similarly, water contains hydrogen and oxygen. Recall that hydrogen bonds are formed when hydrogen is covalently bonded to highly electronegative elements. Hence, water in the liquid state exhibits strong hydrogen bonding. The unique type of dipole-dipole interaction in liquid water is actually hydrogen bonding, hence the answer.

Answer:

[tex]1) NH_{4}IO_{3}\\2) Pb(IO_{3})_{4} \\3) NH_{4}(C_{2}H_{3}O_{2})\\4) Pb(C_{2}H_{3}O_{2})_{4}[/tex]

Explanation:

[tex]1) NH_{4}^{+}IO_{3}^{-} ---> NH_{4}IO_{3}\\2) Pb^{4+}(IO_{3}^{-})_{4} --->Pb(IO_{3})_{4} \\3) NH_{4}^{+}(C_{2}H_{3}O_{2})^{-} ---> NH_{4}(C_{2}H_{3}O_{2})\\4) Pb^{4+}(C_{2}H_{3}O_{2})^{-} _{4} --->Pb(C_{2}H_{3}O_{2})_{4}[/tex]

Answer:

The density would be larger than the correct value.

Explanation:

First off, the realtionship between denisty and volume is given in the equation below;

Density = Mass / Volume

From this equation, Density is inversely proportional to volume. This means as the volume increases, the density decreases and as the volume decreases the density increases.

Assuming all thing's being normal;

Mass = 2g

Volume = 10ml

Density = 2 / 10 = 0.2 g/ml

Second case scenario;

'your pipet was incorrectly calibrated so that it transferred less than 10.00 mL"

Lets have a value of 8ml for our volume. Mass remains constant.

Density = 2 / 8 = 0.25 g/ml

The density would be larger than the correct value.

Answer: The density would be larger than the correct value.

First off, the relationship between density and volume is given by:

Density = Mass / Volume

From this equation, Density is inversely proportional to volume. This means as the volume increases, the density decreases and as the volume decreases the density increases.

Assuming all thing's being normal;

Mass = 2g

Volume = 10ml

Density = [tex]\frac{2}{10}=0.2[/tex]   g/ml

Second case scenario;

'your pipet was incorrectly calibrated so that it transferred less than 10.00 mL"

Lets have a value of 8ml for our volume. Mass remains constant.

Density = [tex]\frac{2}{8}= 0.25[/tex]  g/ml

The density would be larger than the correct value.

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Answer:

4 and 6 would work for this

Answer:

158.7 g

Its the right answer

Answer:

Whenever a system in equilibrium gets disturbed, the adjustment of the system is done in such a manner that the effect of the change gets nullified, this is known as Le Chatelier's Principle. Let us consider that if a reaction present in an equilibrium gets disturbed by changing pressure, concentration, pressure, or other things, then the reaction will move in such a manner so that it can attain the equilibrium again.  

Based on the given question, the equation is:  

HF (aq) + H2O (l) ⇒ H3O+ (aq) + F- (aq)

a) When HCl is added, the dissociation of HCl takes place within the water to give rise to Cl- and H3O+ ions. One can witness an overall enhancement in the H3O+ ions concentration and the shifting of the equilibrium will take place in the backward direction based on the Le-Chatelier's principle. Thus, on adding HCl, the concentration of H3O+ and F- ions decreases, and the concentration of HF increases.  

b) When the addition of a strong electrolyte like KF is done, the dissociation of KF is done into the F- and K+ ions. Thus, with the overall enhancement in the F- ions concentration, the shifting of equilibrium will take place in the backward direction based on the Le-Chatelier's principle. Thus, with the addition of KF, the concentration of H3O+ and F- ions decreases, and HF increases.  

c) With the addition of strong electrolytes like NaCl in the solution, the dissociation of NaCl takes place into the Cl- and Na+ ions. With the addition of NaCl, the equilibrium is not disturbed as the ions exhibit no influence on the given equilibrium reaction. Thus, the concentration of H3O+, HF, and F- ions remains unmodified.  

d) With the addition of KOH, the dissociation of KOH takes place into K+ and OH- ions. Based on Le-Chatelier's principle, the equilibrium will shift in the forward direction as the produced OH- ions will consume the hydronium ions. Therefore, the concentration of H3O+ and F- ions increases, while the concentration of HF decreases.  

e) With the addition of HF, that is, a weak acid, the equilibrium will move in the forward direction to counter the change as the concentration of reactant increases. Therefore, the H3O+ and F- ions decreases, and the concentration of HF increases.  

On adding HCl & KOH to the equilibrium concentration of hydronium ion changes, on adding KF concentration of fluoride ion changes, on adding NaCl no change occur and on adding HF concentration of both ion increases.

According to the equilibrium law whenever any stress is applied at the equilbrium state of chemical reaction, then the equilibrium will move on that direction where the effect of the applied stress will decreases.

Given chemical reaction is:

HF (aq) + H₂O (l) ⇄ H₃O⁺ (aq) + F⁻ (aq)

Hence affect on the equilibrium was discussed above.

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Answer:

O2

Explanation:

Answer:

o2

Explanation:

edge

Answer:

See explanation

Explanation:

-) Given the molecules propane (C3H8) and n-butane (C4H10) n-butane has the higher boiling point mainly due to a larger chain of carbons.

In this question, in propane, we have a chain of three carbons. In butane, we have fourth carbons. If we have more carbons we will have more interactions. If we have more interactions we have to give more energy to go from liquid to gas, therefore we will have a higher boiling point.

-) Given the molecules diethyl ether (CH3 CH2OCH2 CH3) and 1-butanol (CH3 CH2CH2 CH2OH) 1-butanol has a higher boiling point mainly due to hydrogen bonding.

In the case of butanol, we have the "OH" group. If we have a hydrogen bond to a heteroatom (O, S, P, or N) we will have the presence of this type of interaction between molecules. If we have more interactions we have to give more energy to go from liquid to gas, therefore we will have a higher boiling point.

I hope it helps!

Answer:is increase and more collisions :)

Explanation:

Answer:

five(5) hydrogen are formed

Explanation:

A metal ion is a type of atom compound that has an electric charge.

Such atoms willingly lose electrons in order to build positive ions called cations. The selected  Ions are :

[tex]1. Mn^2^+\\\ 2. Ca^2^+\\\ 3. Co^2^+\\\ 4. Fe^2^-\\\ 5. K^+[/tex]

Answer:

[tex]0.48~\frac{J}{g~^{\circ}C}[/tex]

Explanation:

In this question, we have to remember the relationship between Q (heat) and the specific heat (Cp) the change in temperature (ΔT), and the mass (m).

[tex]Q=m*Cp*ΔT[/tex]

The next step is to identify what values we have:

[tex]Q~=~1870~J[/tex]

[tex]m~=~85.01~g[/tex]

[tex]ΔT~=~45.2~^{\circ}C[/tex]

[tex]Cp~=~X[/tex]

Now, we can plug the values and solve for "Cp":

[tex]1870~J=~85.01~g~*Cp*45.2~^{\circ}C[/tex]

[tex]Cp=\frac{1870~J}{85.01~g~*45.2~^{\circ}C}[/tex]

[tex]Cp=0.48~\frac{J}{g~^{\circ}C}[/tex]

The unknow metal it has a specific value of [tex]0.48~\frac{J}{g~^{\circ}C}[/tex]

I hope it helps!

Answer: 1) Endothermic

2) Yes, absorbed.

3) 166.86 kJ will be absorbed.

Explanation:

1) To determine if a reaction is endothermic (heat is absorbed by the system) or exothermic (heat is released by the system), first calculate its change in Enthalpy, which is given by:

ΔH = [tex]H_{products} - H_{reagents}[/tex]

For the reaction 2Fe₂O₃(s) ⇒ 4FeO(s) + O₂(g):

Enthalpy of Reagent (Fe₂O₃(s))

Enthalpy of formation for Fe₂O₃(s) is - 822.2 kJ/mol

The reaction needs 2 mols of the molecule, so:

H = 2(-822.2)

H = - 1644.4

Enthalpy of Products (4FeO(s) + O₂(g))

Enthalpy of formation of O₂ is 0, because it is in its standard state.

Enthalpy of formation of FeO is - 272.04 kJ/mol

The reaction produces 4 mols of iron oxide, so:

H = 4(-272.04)

H = -1088.16

Change in Enthalpy:

ΔH = [tex]H_{products} - H_{reagents}[/tex]

ΔH = - 1088.16 - (-1644.4)

ΔH = + 556.2 kJ/mol

The change in enthalpy is positive, which means that the reaction is absorving heat. Then, the chemical reaction is Endothermic.

2) When Fe₂O₃(s) reacts, heat is absorbed because it is an endothermic reaction.

3) Calculate how many mols there is in 94.2 g of Fe₂O₃(s):

n = [tex]\frac{mass}{molar mass}[/tex]

n = [tex]\frac{94.2}{160}[/tex]

n = 0.6 mols

In the reaction, for 2 mols of Fe₂O₃(s), 556.2 kJ are absorbed. Then:

2 mols --------------- 556.2 kJ

0.6 mols ------------- x

x = [tex]\frac{0.6*556.2}{2}[/tex]

x = 167 kJ

It will be absorbed 167 kJ of energy, when 94.2 g of Fe₂O₃(s) reacts.