A compound is found to contain 26.94 % nitrogen and 73.06 % fluorine by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 104.02 g/mol. What is the molecular formula for this compound?

Answers

Answer 1

Answer:

THE EMPIRICAL FORMULA OF THE COMPOUND IS NF2

THE MOLECULAR FORMULA OF THE COMPOUND IS N2F4

Explanation:

To calculate the empirical formula for the compound, we:

1. Write out the percentage weight of each elements

N = 26.94%

F = 73.06 %

2. Divide each by its atomic mass

( N= 14, F = 19)

N = 26.94 / 14 = 1.924

F = 73.06 / 19 = 3.845

3. Divide each by the smaller of the values

N = 1.924 / 1.924 = 1

F = 3.845 / 1.924 = 1,998

4. Round up to a whole number and write the empirical formula

N= 1

F = 2

So the empirical formula of the compound is N F2

To calculate the molecular formula, we:

(N F2 )n = molecular weight

( 14 + 19*2) n = 104.02

52 n = 104.02

n = 2.000

The molecular formula of the compound will be:

(N F2)2 = N2F4

In conclusion, the empirical formula of the compound is NF2 and the molecular formula of the compound is N2F4


Related Questions

The compound sodium hydroxide is a strong electrolyte. Write the transformation that occurs when solid sodium hydroxide dissolves in water. Use the pull-down boxes to specify states such as (aq) or (s).

Answers

Answer:

Solid sodium hydroxide dissolves in water to form an aqueous solution of ions.

NaOH(s) ⇌ Na+(aq) + OH–(aq) ΔH1 = ?

Solid sodium hydroxide reacts with aqueous hydrochloric acid to form water and an aqueous solution of sodium chloride.

NaOH(s) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH2 = ?

Solutions of aqueous sodium hydroxide and hydrochloric acid react to form water and aqueous sodium chloride.

Na+(aq) + OH–(aq) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH3 = ?

Need help with chemistry questions

Answers

Answer:

1. oxidation

2. reduction

3. oxidation

4. oxidation

Explanation:

Oxidation and Reduction in terms of hydrogen

Oxidation and Reduction with respect to Hydrogen Transfer. Oxidation is the loss of hydrogen. Reduction is the gain of hydrogen.

Oxidation and Reduction in terms of Oxygen

Oxidation and Reduction with respect to Oxygen Transfer. Oxidation is the gain of Oxygen. Reduction is the loss of Oxygen.

For each of the following reactions calculate the mass (in grams) of both the reactants that are required to form 15.39g of the following products.
a. 2K(s) + Cl2(g) → 2Cl(aq)
b. 4Cr(s) + 302(g) → 2Cr2O3(s)
c. 35r(s) + N2(g) → SraNa(s)

Answers

Answer:

a.

[tex]m_K=8.056gK\\ \\m_{Cl_2}=4.028gCl_2[/tex]

b.

[tex]m_{Cr}=10.51gCr\\ \\m_{O_2}=4.851gO_2[/tex]

c.

[tex]m_{Sr}=13.88gSr\\\\m_{N_2}=1.479gN_2[/tex]

Explanation:

Hello,

In this case, we proceed via stoichiometry in order to compute the masses of all the reactants as shown below:

a. [tex]2K+Cl_2\rightarrow 2KCl[/tex]

[tex]m_K=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{2molK}{2molKCl}* \frac{39.1gK}{1molK}=8.056gK\\ \\m_{Cl_2}=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{1molCl_2}{2molKCl}* \frac{70.9gCl_2}{1molCl_2}=4.028gCl_2[/tex]

b. [tex]4Cr+ 3O_2\rightarrow 2Cr_2O_3[/tex]

[tex]m_{Cr}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{4molCr}{2molCr_2O_3}* \frac{52gCr}{1molCr_2O_3}=10.51gCr\\ \\m_{O_2}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{3molO_2}{2molCr_2O_3}* \frac{32gO_2}{1molCr_2O_3}=4.851gO_2[/tex]

c. [tex]3Sr(s) + N_2(g) \rightarrow Sr_3N_2[/tex]

[tex]m_{Sr}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{3molSr}{1molSr_3N_2}* \frac{87.62gSr}{1molSr}=13.88gSr\\\\m_{N_2}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{1molN_2}{1molSr_3N_2}* \frac{28gN_2}{1molN_2}=1.479gN_2[/tex]

Regards.

At a certain temperature the equilibrium constant, Kc, equals 0.110 for the reaction: 2 ICl (g) ⇌ I2 (g) +Cl2 (g) What is the equilibrium concentration of ICl if 0.750 mol of I2 and 0.750 mol of Cl2 are initially mixed in a 1.00-L flask?

Answers

Answer:

The equilibrium concentration of ICl is 2.26 M

Explanation:

Chemical equilibrium is a state in which no changes are observed as time passes, despite the fact that the substances present continue to react. This is because chemical equilibrium is established when the forward and reverse reaction take place simultaneously at the same rate.

For the study of chemical equilibrium, the so-called equilibrium constant Kc is useful. Being:

aA + bB ⇔ cC + dD

the equilibrium constant Kc is:

[tex]Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }[/tex]

That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

For the reaction:

2 ICl (g) ⇌ I₂ (g) +Cl₂ (g)

the constant Kc is:

[tex]Kc=\frac{[I_{2} ]*[Cl_{2} ]}{[ICl]^{2} }[/tex]

Being Kc =0.110 and the concentration being the amount of moles of solute that appear dissolved in each liter of the mixture and being calculated by dividing the moles of the solute by the liters of the solution:

[tex][I_{2} ]=\frac{0.750 moles}{1 L} =0.750 \frac{moles}{L}[/tex][tex][Cl_{2} ]=\frac{0.750 moles}{1 L} =0.750 \frac{moles}{L}[/tex]

and replacing in the constant we get:

[tex]0.110=\frac{0.750*0.750}{[ICl]^{2} }[/tex]

Solving, you get the ICl concentration at equilibrium:

[tex][ICl]^{2} =\frac{0.750*0.750}{0.110 }[/tex]

[tex][ICl] =\sqrt{\frac{0.750*0.750}{0.110 }}[/tex]

[ICl]= 2.26 M

The equilibrium concentration of ICl is 2.26 M

Calculate ΔS∘rxn for the balanced chemical equation 2H2S(g)+3O2(g)→2H2O(g)+2SO2(g) Express the entropy change to four significant figures and include the appropriate units.

Answers

Answer:

-170.65

188.8+ 256.8-205.8-(2x205.2)

-170.65 is the entropy change.

What is Entropy Change?

Entropy trade is the phenomenon that is the measure of change of disorder or randomness in a thermodynamic gadget. It is associated with the conversion of heat or enthalpy completed in work. A thermodynamic device that has extra randomness means it has high entropy.

Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each extended by using their suitable stoichiometric coefficients, to reap ΔS° for the reaction.

Learn more about entropy change here: https://brainly.com/question/14257064

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1.) A sample of neon gas at a pressure of 0.646 atm and a temperature of 242 °C, occupies a volume of 515 mL. If the gas is cooled at constant pressure until its volume is 407 mL, the temperature of the gas sample will be ________°C.
2.) A sample of argon gas at a pressure of 0.633 atm and a temperature of 261 °C, occupies a volume of 694 mL. If the gas is heated at constant pressure until its volume is 796 mL, the temperature of the gas sample will be___________°C.
3.) 0.962 mol sample of carbon dioxide gas at a temperature of 20.0 °C is found to occupy a volume of 21.5 liters. The pressure of this gas sample ismm ____________ Hg.

Answers

Answer:1 )T2=134°C   2) T2=339.48°C. 3)

P=817.59 mmHg.

Explanation:

1.Given ;

pressure, P1 of neon gas = 0.646 atm

temperature, T1 =242oC + 273=515oC

Volume, V1 =515ml

Volume V2= 407ml

temperature , T 2= ?

Solution;

And at constant pressure, the volume cools at V2=407 mL at T2=?

From ideal gas equation, PV=nRT

V/T=constant

therefore

V1/V2=T1/T2 = T2=(V2 xT1)/V1

T2=(407 mL x 515 K)/515 mL= 407K.

T2= 407K -273= 134°C.   recall 0°C=273 K)

2..Given ;

pressure, P1 of neon gas = 0.633 atm

temperature, T1 =261oC + 273=534oC

Volume, V1 =694ml

Volume V2= 796ml

temperature , T 2= ?

Solution;

And at constant pressure, the volume expands  at V2=796mL at T2=?

From ideal gas equation, PV=nRT

V/T=constant

therefore

V1/V2=T1/T2 = T2=(V2 xT1)/V1

T2=(796 mL x 534 K)/694mL= 612.48K.

T2= 612.48K -273= 339.48°C. recall 0°C=273 K

3

Given;

moles of CO2= n=0.962 mol,

temperature T=20°C=20+273 K =293 K,

volume V=21.5 L,

gas constant R at L·mmHg/mol·K= 62.3637 L mmHg mol^-1 K^-1

Using  ideal gas equation PV=nRT

P=nRT/V

P=(0.962 mol)x(62.3637mmHg mol^-1 K^-1)x(293 K)/(21.5L)

P=817.59 mmHg.

The mole is a counting number that allows scientists to describe how individual molecules and atoms react. If one mole of atoms or molecules is equal to 6.022 x 10^32 atoms or molecules, how many molecules are in 23.45 g sample of copper (II) hydroxide, Cu(OH)2? Express your answer to the correct number of significant figures. (MM of Cu(OH)2 is 97.562g/mol. Be sure to show all steps completed to arrive at the answer.

Answers

Answer:

[tex]\large \boxed{1.503 \times 10^{23}\text{ molecules of Cu(OH)}_{2}}$}[/tex]

Explanation:

You must calculate the moles of Cu(OH)₂, then convert to molecules of Cu(OH)₂.

1. Moles of Cu(OH)₂[tex]\text{Moles of Cu(OH)}_{2} = \text{24.35 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \text{0.2496 mol Cu(OH)}_{2}[/tex]

2. Molecules of Cu(OH)₂[tex]\text{No. of molecules} = \text{0.2496 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= \mathbf{1.503 \times 10^{23}}\textbf{ molecules Cu(OH)}_{2}\\\text{There are $\large \boxed{\mathbf{1.503 \times 10^{23}}\textbf{ molecules of Cu(OH)}_{2}}$}[/tex]

The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].

According to the Avogadro number, the number of molecules in a mole of atom has been equivalent to the Avogadro constant. The value of Avogadro constant has been [tex]\rm 6.023\;\times\;10^2^3[/tex].

The moles of a compound has been given as:

[tex]\rm Moles=\dfrac{Mass}{Molar\;mass}[/tex]

The moles in 23.45 g copper (II) hydroxide has been:

[tex]\rm Moles=\dfrac{23.45}{97.562} \\Moles=0.24\;mol[/tex]

The moles of copper (II) hydroxide has been 0.24 mol.

The number of molecules in 0.24 mol sample has been driven by:

[tex]\rm 1\;mol=6.023\;\times\;10^2^3\;molecules\\0.24\;mol=0.24\;\times\;10^2^3\;molecules\\0.24\;mol=1.45\;\times\;10^2^3\;molecules[/tex]

The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].

For more information about molecules in a mole of sample, refer to the link:

https://brainly.com/question/24577356

Why can gasses change volume?
A. The forces holding the gas particles together are
stronger than gravity.
B. The gas particles have no mass, so they can change volume.
C. Gravity has no effect on gas particles, so they can float away.
O D. There are no forces holding the gas particles together.

Answers

Answer:

There are no forces holding the gas particles together.

Explanation:

If 11.2 g of naphthalene, C10H8, is dissolved in 107.8 g of chloroform, CHCl3, what is the molality of the solution

Answers

Answer:

CHC12

Explanation:

i am not really sure i am onna do a quick research 4 u tho

Aqueous ammonia is added to a mixture of silver chloride and water. Given that Kf for the reaction between Ag+ and NH3 is large, which of the following are true?
A) The free ions are favored over the complex ion.
B) The complex ion is favored over solid silver chloride.
C) The free Ag+ ion is unstable.
D) More silver chloride will precipitate.

Answers

Answer:

B) The complex ion is favored over solid silver chloride

C) The free Ag+ ion is unstable.

Explanation:

Hello,

In this case, since the dissociation of solid silver chloride occurs at equilibrium with a neglectable solubility product (very small Ksp), which means that the solid tends to remain undissolved:

[tex]AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)[/tex]

By the addition of ammonia, the following reaction is favored:

[tex]Ag^+(aq)+2NH_3(aq)\rightleftharpoons [Ag(NH_3)_2]^+(aq)[/tex]

Which has a large equilibrium constant, which means that the formation of the complex is assured. In such a way, by addition of more ammonia, more complex will be formed, therefore B) The complex ion is favored over solid silver chloride is true. Moreover, C) The free Ag+ ion is unstable, since they tend to form the complex once they are formed by the solid silver chloride so it readily reacts.

Best regards.

How many atoms of oxygen are in one molecule of water (H2O)? one two four three

Answers

Answer:

there is one atom of oxygen and two atoms of hydrogen

Explanation:

One atom is in oxygen of water

The substance formed on addition of water to an aldehyde or ketone is called a hydrate or a/an:_______
A) vicinal diol
B) geminal diol
C) acetal
D) ketal

Answers

Answer:

B) geminal diol

Explanation:

Hello,

In this case, considering the attached picture, you can see that the substance resulting from the hydrolysis of an aldehyde or a ketone is a geminal diol since the two hydroxyl groups are in the same carbon. Such hydrolysis could be carried out in either acidic or basic conditions depending upon the equilibrium constant.

Regards.

Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)
Find the volume if the density is 2.6 g/mL and the mass is 9.7 g.(5 pts)
Find the mass if the density is 1.6 g/cm3 and the volume is 4.1 cm3 (5 pts)
Find the density if the initial volume of water is 12.8 mL, the final volume is 24.6 mL and the mass of the object is 4.3 g. Make a drawing to show the water displacement using a graduated cylinder. (gdoc, gdraw)

Answers

Answer:

[tex]\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}[/tex]

Explanation:

1. Density from mass and volume

[tex]\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}[/tex]

2. Volume from density and mass

[tex]V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}[/tex]

3. Mass from density and volume

[tex]\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}[/tex]

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water                = 12.8 mL

Volume of object               = 11.8 mL

[tex]\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}[/tex]

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

 

Ammonium phosphate is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid with ammonia . What mass of ammonium phosphate is produced by the reaction of of ammonia?

Answers

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.

The mass of ammonium phosphate produced is 14.3g

Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppose 0.963 g of methane is mixed with 7.5 g of oxygen. Calculate the minimum mass of methane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answers

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

[tex]n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4[/tex]

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

[tex]left\ over=0g[/tex]

Regards.

Calculate the concentration of H3O+ in a solution that contains 5.5 × 10-5 M OH- at 25°C. Identify the solution as acidic, basic, or neutral.

Answers

Explanation:

To calculate [H3O+] in the solution we must first find the pH from the [ OH-]

That's

pH + pOH = 14

pH = 14 - pOH

To calculate the pOH we use the formula

pOH = - log [OH-]

And [OH-] = 5.5 × 10^-5 M

So we have

pOH = - log 5.5 × 10^ - 5

pOH = 4.26

Since we've found the pOH we can now find the pH

That's

pH = 14 - 4.26

pH = 9.74

Now we can find the concentration of H3O+ in the solution using the formula

pH = - log H3O+

9.74 = - log H3O+

Find the antilog of both sides

H3O+ = 1.8 × 10^ - 10 M

The solution is basic since it's pH lies in the basic region.

Hope this helps you

Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH4Cl with 200.0 mL of 0.12 M NH3. The Kb for NH3 is 1.8 × 10-5.

Answers

Answer:

The pH of the solution is 9.06.

Explanation:

The reaction of the dissociation of NH₃ in water is:

NH₃(aq) + H₂O(l)  ⇄  NH₄⁺(aq) + OH⁻(aq)     (1)

[NH₃] - x                     [NH₄⁺] + x     x  

The concentration of NH₃ and NH₄⁺ is:

[tex] [NH_{3}] = \frac{n_{NH_{3}}}{V_{T}} = \frac{C_{i}_{(NH_{3})}*Vi_{NH_{3}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.12 M*0.2 L}{0.2 L + 0.25 L} = 0.053 M [/tex]

[tex] [NH_{4}^{+}] = \frac{C_{i}_{(NH_{4}^{+})*V_{NH_{4}^{+}}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.15 M*0.25 L}{0.2 L + 0.25 L} = 0.083 M [/tex]

From equation (1) we have:

[tex]Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}[/tex]

[tex] 1.8 \cdot 10^{-5} = \frac{(0.083 + x)*x}{0.053 - x} [/tex]

[tex] 1.8 \cdot 10^{-5}(0.053 - x) - (0.083 + x)*x = 0 [/tex]

By solving the above equation for x we have:

x =  1.15x10⁻⁵ = [OH⁻]

The pH of the solution is:

[tex] pOH = -log([OH^{-}]) = -log(1.15 \cdot 10^{-5}) = 4.94 [/tex]

[tex] pH = 14 - pOH = 14 - 4.94 = 9.06 [/tex]

Therefore, the pH of the solution is 9.06.

I hope it helps you!

The density of concentrated nitric acid (HNO3) is 1.413 g/mL. What volume in liters would be occupied by a mass of 47.2 g?

Answers

Answer:

The volume that a mass of 47.2 g would occupy is 0.0334 L

Explanation:

Density is the property that matter, whether solid, liquid or gas, has to compress into a given space. Density is defined as the amount of mass it has per unit volume, that is, the ratio between the mass of a body and the volume it occupies:

[tex]Density=\frac{mass}{volume}[/tex]

This indicates that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.

In this case:

density= 1.413 g/mLmass= 47.2 gvolume=?

Replacing:

[tex]1.413 \frac{g}{mL}=\frac{47.2 g}{volume}[/tex]

Solving:

[tex]volume=\frac{47.2 g}{1.413\frac{g}{mL} }[/tex]

volume=33.40 mL

Being 1,000 mL= 1 L:

volume= 0.0334 L

The volume that a mass of 47.2 g would occupy is 0.0334 L

Arrange the following in order of increasing boiling point: CH4, CH3CH3, CH3CH2Cl, CH3CH2OH. Rank from lowest to highest. To rank items as equivalent, overlap them.

Answers

Answer:

In order from lowest to highest:

Methane < Ethane < Chloroethene < Methanol

i.e: CH4 < CH3CH3 < CH3CH2OH < CH3CH2Cl

Explanation:

Compounds with stronger molecular fore have higher boiling points, thus making the molecules more difficult to pull apart. The presence of chains also increases the molecular dispersion. The dipole force of ethanol makes it have a very high boiling point.

I'm positive this explanation would suffice. Best of luck.

The order of increasing boiling points of the substances listed is; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH.

Intermolecular interactions occur between molecules. The boiling point and melting points of substances depends on the nature and magnitude of intermolecular interaction between the molecules of the substance.

The order of increasing boiling points of the substances listed is as follows; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH. CH3CH2OH has the highest boiling point due to intermolecular hydrogen bonds in the molecule. Though CH4 and CH3CH3 are both alkanes, CH3CH3 has a higher molecular mass and consequently greater dispersion forces and a higher boiling point.

Learn more: https://brainly.com/question/865531

Which of the following compounds is more soluble in a 0.10 M NaCN solution than in pure neutral water? Ca3(PO4)2 AgBr CaCO3 Mg(OH)2 NH4ClO4

Answers

Answer:

AgBr

Explanation:

Silver bromide has a very low solubility product constant of about 7.7 ×10^-13 in pure water hence it is not quite soluble in pure water.

However, with NaCN, the AgBr forms the complex [Ag(CN)2]^2- which has a formation constant of about 5.6 ×10^8. This very high formation constant implies that the complex is easily formed leading to the dissolution of AgBr in NaCN.

The equation for the dissolution of AgBr in cyanide is shown below;

AgBr(s) + 2CN^-(aq) ----> [Ag(CN)2]^2-(aq) + Br^-(aq)



If you have 2.4L of SO2 gas (at STP) how many moles of sulfur dioxide do you have?

Answers

Answer:

0.107 mole of SO2.

Explanation:

1 mole of a gas occupy 22.4 L at standard temperature and pressure (STP).

With the above information, we can simply calculate the number of mole of SO2 that will occupy 2.4 L at STP.

This can be obtained as follow:

22.4 L contains 1 mole of SO2.

Therefore, 2.4 L will contain = 2.4/22.4 = 0.107 mole of SO2.

Therefore, 0.107 mole of SO2 is present in 2.4 L at STP.

Zinc is used as a coating for steel to protect the steel from environmental corrosion. If a piece of steel is submerged in an electrolysis bath for 24 minutes with a current of 6.5 Amps, how many grams of zinc will be plated out? The molecular weight of Zn is 65.38, and Zn+2 + 2e– → Zn. Question 7 options: A) 3.17 g of Zn B) 1.09 g of Zn C) 6.34 g of Zn D) 12.68 g of Zn

Answers

Answer:

A) 3.17 g of Zn

Explanation:

Let's consider the reduction of Zn(II) that occurs in an electrolysis bath.

Zn⁺²(aq) + 2e⁻ → Zn(s)

We can establish the following relations:

1 min = 60 s1 A = 1 C/sThe charge of 1 mole of electrons is 96,468 C (Faraday's constant).When 2 moles of electrons circulate, 1 mole of Zn is deposited.The molar mass of Zn is 65.38 g/mol

The mass of Zn deposited under these conditions is:

[tex]24min \times \frac{60s}{1min} \times \frac{6.5C}{s} \times \frac{1mol\ e^{-} }{96,468C} \times \frac{1molZn}{2mol\ e^{-}} \times \frac{65.38g}{1molZn} = 3.17 g[/tex]

Answer:

A.) 3.17

Explanation:

I got it right in class!

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Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that could be present in hard water. Assume fluoride is the only anion present that will precipitate calcium ion.

Answers

Answer:

[tex][F^-]_{max}=4x10{-3}\frac{molF^-}{L}[/tex]

Explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

[tex]Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)[/tex]

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

[tex][F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\[/tex]

[tex][F^-]_{max}=4x10{-3}\frac{molF^-}{L}[/tex]

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plsssss help!!! Deep Space 1 was a spacecraft powered by an engine that gave off xenon particles to change velocity. It had a mass of 500 kg. Which expression can be used to find the spacecraft’s acceleration if its engine created a net force of 0.10 N? A) 0.10 N 500 kg B) 500 kg · 0.10 N C) 500 kg 0.10 N D) 500 kg + 0.10 N

Answers

Answer:

Spacecraft’s acceleration (a) = 0.10 N / 500 kg

Explanation:

Given:

Mass of Spacecraft (M) = 500 Kg

Force generate by engine (F) = 0.10 N

Find:

Spacecraft’s acceleration (a)

Computation:

F = Ma

0.10 = 500 (a)

a = 0.10 / 500

Spacecraft’s acceleration (a) = 0.10 N / 500 kg

The expression which can be used to find the spacecraft’s acceleration if its engine created a net force of 0.10 N is  0.10 N/500Kg.

We know that force is the product of the mass a body and its acceleration. The result of motion is the action of an unbalanced force. We have the following information;

Mass of the spacecraft = 500 kg

Force on the engine =  0.10 N

From Newton's law;

F = ma

F = force

m = mass

a = acceleration

a = F/m

acceleration = 0.10 N/500Kg

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The solubility of calcium oxalate, CaC2O4, in pure water is 4.8 × 10‑5 moles per liter. Cal­culate the value of Ksp for silver carbonate from this data.

Answers

Answer:

2.3 × 10⁻⁹

Explanation:

Step 1: Write the reaction for the solution of calcium oxalate

CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)

Step 2: Make an ICE chart

We can relate the molar solubility (S) with the solubility product constant (Ksp) through an ICE chart.

        CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)

I                                0                 0

C                              +S               +S

E                                S                 S

The solubility product constant is:

Ksp = [Ca²⁺] × [C₂O₄²⁻] = S² = (4.8 × 10⁻⁵)² = 2.3 × 10⁻⁹

The electrolysis of molten AlCl 3 for 2.50 hr with an electrical current of 15.0 A produces ________ g of aluminum metal.

Answers

Can we actually get more information

Amino acids can be synthesized by reductive amination. Draw the structure of the organic compound that you would use to synthesize serine.

Answers

Answer:

Following are the answer to this question:

Explanation:

For the reductive amination of its carbonyl group, amino acids could be synthesized by reducing ammunition, which can be synthesized in the given attachment file:

please find the attachment:

Methanol liquid burns readily in air. One way to represent this equilibrium is: 2 CO2(g) + 4 H2O(g)2 CH3OH(l) + 3 O2(g) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above. 1) CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) K1 = 2) CO2(g) + 2 H2O(g) CH3OH(l) + 3/2 O2(g) K2 = 3) 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g)

Answers

Answer:

Answers are in the explanation

Explanation:

It is possible to obtain K of equilibrium of related reactions knowing the laws:

A + B ⇄ C K₁

C ⇄ A + B K = 1 /K₁

The inverse reaction has the inverse K equilibrium

2A + 2B ⇄ 2C K = K₁²

The multiplication of the coefficients of reaction produce a k powered to the number you are multiplying the coefficients

For the reaction:

2 CO2(g) + 4 H2O(g) ⇄ 2 CH3OH(l) + 3 O2(g) K

1) CH3OH(l) + 3/2 O2(g) ⇄ CO2(g) + 2 H2O(g)

This is the inverse reaction but also the coefficients are dividing in the half, that means:

[tex]K_1 = \frac{1}{k^{1/2}} = (1/K)^{1/2}[/tex]

2) CO2(g) + 2 H2O(g) ⇄ CH3OH(l) + 3/2 O2(g)

Here,the only change is the coefficients are the half of the original reaction:

[tex]K_2 = K^{1/2}[/tex]

3) 2CH3OH(l) + 3 O2(g) ⇄ 2 CO2(g) + 4 H2O(g)

This is the inverse reaction. Thus, you have the inverse K of equilibrium:

[tex]K_3 = \frac{1}{K}[/tex]

How many kg/hr of steam are produced by a 50HP boiler?

Answers

Answer:

Explanation:

50 HP = 50 x 746 watt

= 37300 watt

= 37300 J /s

heat produced in one hour = 60 x 60 x 37300 J

= 134280 x 10³ J

latent heat of vaporization = 2260 x 10³ J / kg .

for evaporation of water of mass 1 kg , 2260 x 10³ J of heat is required .

kg of water being evaporated by boiler per hour

= 134280 x 10³ / 2260 x 10³

= 59.41 kg

rate of production of steam

= 59.41 kg / hr .

Which part of an atom is mostly empty space?
A. nucleus
B. proton cloud
C. electron cloud
D. neutron

Answers

Answer:

C. Electron cloud

the electron is around 1/2000 times the size of the proton.

If you imagine the proton a a marble in the middle of a football field, the electrons will revolve around the last row

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