conversion of minus 1 coulomb meter in debye

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:

141.89 dm^3

Explanation:

The equation of the reaction is;

2H2O2(aq) --------->2H2O(l) + O2(g)

Now , we are told that the mass of hydrogen peroxide decomposed was 315g. Number of moles of hydrogen peroxide in 315g of the substance is given by;

Number of moles= mass/molar mass

Molar mass of hydrogen peroxide= 34.0147 g/mol

Number of moles= 315g/34.0147 g/mol = 9.26 moles of hydrogen hydrogen peroxide.

From the reaction equation;

2 moles of hydrogen peroxide yields 1 mole of oxygen

9.26 moles of hydrogen peroxide yields 9.26 ×1/2 = 4.63 moles of oxygen

From the ideal gas equation;

Volume of the gas V= the unknown

Pressure of the gas P= 0.792 atm

Temperature of the gas= 23°C +273 = 296 K

Number of moles of oxygen = 4.63 moles of oxygen

R= 0.082atmdm^3K-1mol-1

Hence, from PV=nRT

V= nRT/P

V= 4.63 × 0.082 × 296/0.792 = 141.89 dm^3

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: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:

even I have the same dought

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:

[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:

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:

less becuase that moisture adds weight

Explanation:

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: 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.

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:

C.No, because bromine has lower activity than iodine

Explanation:

Hello,

In this case, for answering the given question we should remember that the higher the period the higher the activity of an element, therefore, since iodine is in period 6 and bromine is in period 5, we can say that the described reaction is not possible due to the fact that bromine is less active. For that reason answer is C.No, because bromine has lower activity than iodine.

Best regards.

Answer:

Yes, because bromine has lower activity than iodine

Explanation:

The internal energy of 2 moles of argon gas (Ar) at 298 K is 186.07 J.

By increasing the no.of moles and temperature the internal energy can be increased.

The internal energy of a system is its overall energy associated with its random motions. Thermodynamically it is a state function because the change in its values depends only on the initial and final states of the system and does not depend upon the path.

For an ideal gas, the internal energy is calculated using the following equation:

U = n C T

here, n is the no.of moles, C be the molar heat capacity at constant volume and T be the temperature in K. C for Ar is 0.3122 J/(Kg K) and temperature and no.of moles are given 298 K and 2 moles respectively.

Thus U = 2 mols × 0.3122 J/ (Kg K) × 298 K

            = 186.07 J

By increasing the amount of gas and heating the system, the internal energy can be increased.  By increasing the temperature by 20 K will raise the internal energy by 10 J.

Hence, by these two ways the internal energy of a gaseous system can be increased.

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Answer:

Na + Cl ⇒ NaCl

Explanation:

Skeleton equation: Na + Cl ⇒ NaCl

This is already balanced so that is the answer.

HERE IS THE COMPLETE QUESTION

what is the calculated value of the cell potential at 298 k for an electrochemical cell with the following reaction, when the Cl2 pressure is 1.31 atm, the cl- concentration is at  1.28M, and the Ni2+ concentration is 1.24M

[tex]Cl_{2(g)} + Ni _{(s)} \to 2Cl^-_{aq} + Ni^{2+}_{aq}[/tex]

Answer:

[tex]\mathbf{E_{cell} = +1.4962 \ V}[/tex]

Explanation:

From the given question;

We can see that Nickel is oxidized and Chlorine is reduced. We also known that Oxidation occurs at the anode while reduction occurs at the cathode.

SO the Oxidation and the reduction reaction can be expressed as shown below.

Oxidation :      [tex]Ni_{s} \to Ni ^{2+}_{aq} + 2e^- \ \ \ \ E^0 = -0.26[/tex]

Reduction :      [tex]Cl_{2(g)} + 2e^- \to 2Cl^- _{(aq)} \ \ \ \ E^0} = 1.36[/tex]

                        [tex]Cl_{2(g)} + Ni _{(s)} \to 2Cl^-_{aq} + Ni^{2+}_{aq}[/tex]

[tex]E^0_{cell} = E^0_{cathode}-E^0_{anode}[/tex]

[tex]E^0_{cell} = E^0_{Cl_2/Cl^-}-E^0_{Ni^{2+}/Ni}[/tex]

[tex]E^0_{cell} = +1.36 - (-0.26)[/tex]

[tex]E^0_{cell} = +1.62 \ V[/tex]

By applying Nernst Equation; we have :

[tex]E_{cell} = E^0_{cell}- (\dfrac{0.0591}{n}) \ log Q[/tex]

where

n= 2 moles

[tex]Q= \dfrac{[Cl^-]^2[Ni^{2+}]}{P_{Cl_2}}[/tex]

[tex]E_{cell} = +1.62 \ V}- (\dfrac{0.0591}{2}) \ log \dfrac{[Cl^-]^2[Ni^{2+}]}{P_{Cl_2}}[/tex]

[tex]E_{cell} = +1.62 \ V}- (\dfrac{0.0591}{2}) \ log \dfrac{[1.28]^2[1.24]}{1.31*10^{-4}}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) \ log \dfrac{[1.6384][1.24]}{1.31*10^{-4}}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) \ log \dfrac{[1.6384*1.24]}{1.31}*10^{4}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) \ log \dfrac{[2.031616]}{1.31}*10^{4}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ log (1.55085*10^{4})[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ log (1.55085)+ log \ 10^{4}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ 0.1905 + 4[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ 4.1905[/tex]

[tex]E_{cell} = +1.62 \ V}- 0.1238[/tex]

[tex]\mathbf{E_{cell} = +1.4962 \ V}[/tex]

The question is incomplete; the complete question is;

Remembering that Sn2 reactions go with 100% inversion of configuration, while Sn1 reactions lead to racemization, explain why the reaction of (R)-2-butanol as in this experiment gives a mixture of about 75% (S)- 2 - bromobutane and about 25% (R)-2-bromobutane

Answer:

The reaction occurs mostly by SN2 mechanism

Explanation:

The two major routes for the synthesis of an alkyl halide involves the use of an alkene or an alkanol reacting with a hydrogen halide in both cases as the starting materials. The both routes involves the formation of a planar carbocation intermediate which is flat and planar.

Note that (R)-2-butanol is a secondary alcohol. Recall that nucleophillic substitution at a secondary carbon atom occurs mostly by SN2 mechanism. This implies that reaction of this alcohol is expected to proceed mostly by SN2 mechanism. SN2 mechanism involves the inversion of configuration, hence, we expect to have an excess of the product (S)- 2 - bromobutane in which there is an inversion of configuration compared with 25% (R)-2-bromobutane in which there is retention of configuration. This goes a long way to show us that the reaction occurs mostly by SN2 mechanism.

Secondly , steric factors also favour the formation of 75% (S)- 2 - bromobutane over 25% (R)-2-bromobutane. There is less steric hindrance in the transition state leading to the formation of (S)- 2 - bromobutane than that leading to the formation of (R)-2-bromobutane.

Answer:

K = 4.07x10⁻³

Explanation:

Based on the reaction:

NH₄I(s) ⇄ NH₃(g) + HI(g)

You can define K of equilibrium as the ratio of concentrations of reactants and products, thus:

K = [NH₃] [HI] / [NH₄I]

But, as NH₄I is a solid, is not taken into account in the equilibrium, that means K expression is:

K = [NH₃] [HI]

As the concentrations in equilibrium of the gases is:

[NH₃] = 4.34x10⁻²M

[HI] = 9.39x10⁻²M

Equilibrium constant, K, is:

K = 4.34x10⁻²M * 9.39x10⁻²M

Answer:

[tex]2.32(10^{-22})[/tex] atoms of Carbon

Explanation:

[tex]3.85 mol(\frac{6.02(10^{-22})atoms}{1 mol} )[/tex]

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:

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:

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:

(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:

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:

7L

Explanation:

using charles law

V1/T1 = V1/T1

Answer:

7l

Explanation:

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:

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:

Explanation:

Phosphoric acid H₃PO₄ is produced from reaction of water and tetraphosphorus decoxide  P₄O₁₀ as follows

P₄O₁₀ + 6 H₂O = 4 H₃PO₄

In this reaction 6 molecules of H₂O and one molecule of  phosphorus compound P₄O₁₀ is needed to produce phosphoric acid , ie the conversion factor of water to acid is 6 / 1

This ratio is called mole ratio of water to H₃PO₄.

So the required ratio is 6 : 1 .

Answer:

A = 0.7871g

B = 0.8417g

C = 0.3192g

D = 0.3897g

Explanation:

Hello,

The number of grams of sodium (Na) in 2g of NaCl is

Molar mass of NaCl = 58.44g/mol

Molar mass of Na = 23g

23g of Na = 58.44g of NaCl

x g of Na = 2g of NaCl

x = (2 × 23) / 58.44

x = 0.7871g

Therefore, 0.7871g of Na is present in 2g of NaCl

2.

2g of Na₃PO₄

Molar mass of Na₃PO₄ = 163.94g/mol

Molar mass of Na = 23g

(3 × 23)g of Na = 163.94 g of Na₃PO₄

x g of Na = 2g of Na₃PO₄

x = (2 × 69) / 163.94

x = 0.8417g

0.8417g of Na is present in 2g of Na₃PO₄

3.

2g of NaC₇H₅O₂

Molar mass of NaC₇H₅O₂ = 144.103g/mol

Molar mass of Na = 23g

23g of Na = 144.103g of NaC₇H₅O₂

x g of Na = 2g of NaC₇H₅O₂

x = (2 × 23) / 144.103

x = 0.3192g

0.3192g of Na is present in 2g of NaC₇H₅O₂

4. 2g of Na₂C₆H₆O₇

Molar mass of Na₂C₆H₆O₇ = 236.08g/mol

Molar mass of Na = 23g

(2 × 23)g of Na = 236.08g of Na₂C₆H₆O₇

x g = 2g of Na₂C₆H₆O₇

x = (46 × 2) / 236.08

x = 0.3897g

0.3897g is present in 2g of Na₂C₆H₆O₇