Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have reacted.

Answer:is increase and more collisions :)

Explanation:

Answer:A medication break can ease side effects. A lack of appetite, weight loss, sleep troubles, headaches, and stomach pain are common side effects of ADHD medication.

Explanation: It may boost your child’s growth. Some ADHD medications can slow a child’s growth in height, especially during the first 2 years of taking it. While height delays are temporary and kids typically catch up later, going off medication may lead to fewer growth delays.

It won’t hurt your child. Taking a child off ADHD medication may cause their ADHD symptoms to reappear. But it won’t make them sick or cause other side effects.

Answer:

Mn is manganese and CO₃ is carbonate. Since the charge for CO₃ is -2 and the subscript is 2, the charge of Mn must be +4 so the answer is manganese (IV) carbonate.

Manganese (IV) carbonate is the name of Mn(CO[tex]_3[/tex])[tex]_2[/tex]. The only names used to identify salts are those of the cation or the anion.

The chemical formula of the anion (such as chloride or acetate) comes first in the name of a salt, which is followed with the identity of the cation (such as sodium or ammonium). They are created when acids and bases react, and they are always composed of either metal cations or cations made of ammonium. Manganese is Mn, and carbonate is CO[tex]_3[/tex]. The solution equals manganese (IV) carbonate since the charge for CO[tex]_3[/tex] is -2 but the subscript is 2, meaning that the charge of Mn has to be +4.

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Answer:

The answer is " 10.39"

Explanation:

Calculating acid moles:

[tex]= 0.02000 \ L \times 0.1000 \ M \\\\= 0.002000[/tex]

Calculating NaOH moles:

[tex]= 0.02012 \ L \times 0.1000 \ M \\\\= 0.002012[/tex]

calculating excess in OH-  Moles:

[tex]= 0.002012 - 0.002000\\\\=0.000012[/tex]

calculating total volume:

[tex]= 20.00 + 20.12\\\\ = 40.12 mL \\\\= 0.04012 L[/tex]

[tex][OH-]= \frac{0.000012} { 0.0472}[/tex]

           [tex]=0.00025 M[/tex]

[tex]pOH = - \log 0.00025[/tex]

        = 3.6

[tex]pH = 14 - pOH[/tex]

      = 10.39

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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Answer:

To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).

Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.

In this way, your reasoning is correct and it is probably a mistake in the book.

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:

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 catalyst cannot change an exothermic reaction into an endothermic reaction or vice versa.

Explanation:

Catalyst is basically a substance that enables a chemical reaction to occur at a faster rate as compared to the reaction without catalysis. It lowers the activation energy and temperature for a chemical reaction and a catalyst itself does not goes through any permanent chemical change. This means it does not get used in the process.

Exothermic and endothermic are the chemical reaction. Exothermic reactions absorb energy. This energy is absorbed in the form of heat. When the energy is released in the form of heat then this reaction is called endothermic. So one absorbs the heat and the other  releases it.

As we know that the catalyst does not undergo change at the end of the reaction so the energy or heat whether is absorbed or emitted or you can say whether the reaction is exothermic or endothermic, the total energy stays unchanged during the reaction. So with and without a catalyst, if both have same reactants and products and the difference in enthalpy between products and reactants will be the same.

Answer:

C.

Explanation:

Since the mass number is the number of protons and neutrons added together, the answer is 35. Since the questions are respectively electron quantity and mass number, the only answer choice with 35 as the second choice is C, so that is the correct answer.

Answer:

pressure is = 54.9802atm

Explanation:

using ideal gas equation

PV=nRT

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:

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:

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!

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:

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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:

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:

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:

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:

[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:

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:

4 and 6 would work for this

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:

Uranium-238 undergoes alpha decay to form Thorium-234 as daughter product.

Explanation:

Alpha decay is indicative of loss of the equivalents of a helium particle emission. The reaction equation for this reaction is shown below:

[tex]_{92} ^{238} U_{}[/tex]→ [tex]_{90} ^{234} Th_{} + _{2} ^{4} He_{}[/tex]

I hope this explanation is clear and explanatory.

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.

Answer:

158.7 g

Its the right answer

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:

(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:

[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.