Suppose the concentration of the NaOH solution was 0.5 M instead of 0.1 M. Would this titration have required more, less, or the same amount of NaOH solution for a complete reaction? Explain

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:

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:

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:

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:

(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:

even I have the same dought

Answer: single replacement reaction

Explanation:

A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution. Thus one element should be different from another element.

A general single displacement reaction can be represented as :

[tex]X+YZ\rightarrow XZ+Y[/tex]

The reaction [tex]2AgNO_3(aq)+Zn(s)\rightarrow 2Ag(s)+Zn(NO_3)_2(aq)[/tex]

When zinc metal is added to aqueous silver nitrate, zinc being more reactive than silver displaces silver atom from its salt solution and lead to formation of zinc nitrate and silver metal.

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:

[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:

Molecular formula for the gas is: C₄H₁₀

Explanation:

Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g

At STP → 1 atm and 273.15K

1 atm . 0.0336 L = n . 0.082 . 273.15 K

n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)

n = 1.500 × 10⁻³ moles

Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m

Now we propose rules of three:

If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C

58 g of gas (1mol) would have:

(58 g . 0.480) / 0.580 = 48 g of C  

(58 g . 0.100) / 0.580 = 10 g of H

 48 g of C / 12 g/mol = 4 mol

 10 g of H / 1g/mol = 10 moles

The molecular formula of the compound is C4H10.

At STP;

P = 1 atm

T = 273 K

V = 33.6 mL or 0.0336 L

R = 0.082 atmLK-1mol-1

n = ?

Hence;

n = PV/RT

n = 1 atm × 0.0336 L/0.082 atmLK-1mol-1 × 273 K

n = 0.0015 moles

Number of moles = mass/molar mass

Molar mass= Mass/Number of moles

Molar mass = 0.087 g/0.0015 moles

Molar mass = 58 g/mol

Mass of carbon = (58 g × 0.480) / 0.580 = 48 g of C  

Mass of hydrogen = (58 g × 0.100) / 0.580 = 10 g of H

Number of moles of carbon = 48 g of C / 12 g/mol = 4 mol

Number of moles of hydrogen = 10 g of H / 1g/mol = 10 moles

Formula of the compound must then be C4H10.

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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 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:

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:

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:

the nucelolo is not part of the atom

Explanation:

The nucleolus is NOT a part of the atom. The nucleolus has its definition in biology as an element of the cell.

Let's remember that the atom is made up of three fundamental elements that are: electrons, protons and neutrons. Protons and neutrons are found in the nucleus while electrons are orbiting around the nucleus.

Answer:

core

Explanation:

there is no such thing as a core in an atom.

The middle is known as the nucleus

Answer:

[tex]\large \boxed{1.64\times 10^{-5}\text{ mol/L }}[/tex]

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

                           2x      0.007 50 + x

[tex]K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}[/tex]

The amount of the sample in space is referred to as concentration, in this the Maximum concentration of [tex]Ag^+[/tex] is [tex](3.28 \times 10^{-5}\ M)[/tex].

        Concentration of [tex]Na_2CO_3 = 0.00750 M[/tex]

           [tex](CO_3)^{2-} = Na_2CO_3 = 0.00750\ M\\\\Ksp \ \ Ag_2CO_3 =( Ag^{+})^2 (CO_3)^{2-}\\\\8.10 \times 10^{-12} = (Ag^+)^2 \times (0.00750\ M)\\\\(Ag^+)^2 = \frac{(8.10 \times 10^{-12})}{ (0.00750\ M)}\\\\(Ag^+)^2 = 1.08 \times 10^{-9}\ M^2\\\\(Ag^+) = (1.08 \times 10^{-9}\ M^2)^{\frac{1}{2}}\\\\\[(Ag^+)\] = (3.28 \times 10^{-5}\ M)\\\\[/tex]

So, the Maximum concentration of [tex]Ag^+[/tex] is [tex](3.28 \times 10^{-5}\ M)[/tex].

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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:

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:

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:

THE NEW FREEZING POINT IS -4.196 °C

Explanation:

ΔTf = 1 Kf m

molarity of MgCl2:

Molar mass = (24 + 35.5 *2) g/mol

molar mass = 95 g/mol

7.15 g of MgCl2 in 100 g of water

7.15 g = 100 g

(7.15 * 100 / 1000) = 1000 g or 1 L or 1 dm3

= 0.715 g /dm3

Molarity in mol/dm3 = molarity in g/dm3 / molar mass

= 0.715 g /dm3 / 95 g/mol

m = 0.00752 mol/ dm3

So therefore:

ΔTf = i Kf m

1 = 3 (1 Mg and 2 Cl)

Kf = 1.86 °C/m

M = 0.752 moles

So we have:

ΔTf = 3 * 1.86 * 0.752

ΔTf = 4.196 °C

The new freezing point therefore will be 0 °C - 4.196 °C which is equals to - 4.196 °C

Answer:

pressure is = 54.9802atm

Explanation:

using ideal gas equation

PV=nRT

Answer:

dimethoxy(phenyl)methanol

Explanation:

For this question, we have to remember the mechanism of the Grignard reaction. In this case, phenylmagnesium bromide is our nucleophile, a carbo-anion is produced (step 1). Then this carbo-anion can attack the carbonyl group in the dimethyl carbonate, the double bond is delocalized into the oxygen producing a negative charge (step 2). Finally, with the addition of the hydronium ion ([tex]H_3O^+[/tex]), the anion can be protonated to produce the alcohol (dimethoxy(phenyl)methanol) (step 3).

See figure 1

I hope it helps!

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:

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:

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:

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: -1,030 kJ

Explanation:

To calculate the number of moles we use the equation:

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar mass}}[/tex]     .....(1)

Putting values in equation 1, we get:

[tex]\text{Moles of methanol}=\frac{45.3g}{32g/mol}=1.42mol[/tex]

The balanced chemical reaction is:

[tex]CH_3OH(l)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)[/tex]  [tex]\Delta H=-726kJ[/tex]

Given :

Energy released when 1 mole of [tex]CH_3OH[/tex] combusts = 726k J

Thus Energy released when 1.42 moles of [tex]CH_3OH[/tex] combusts =  [tex]\frac{726kJ}{1}\times 1.42=1030J[/tex]

Thus 1030 kJ of heat is released and as [tex]\Delta H[/tex] is negative for exothermic reaction, [tex]\Delta H=-1030kJ[/tex]

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 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:

A

Explanation:

The products are the stuff on the right side of the arrow (yield sign). In this case that would be C₆H₁₂O₂ + 6O₂.