Diethyl ether (C2H5 )2O vaporizes at room temperature. If the vapor exerts a pressure of 233 mm Hg in a flask at 25 °C, what is the density of the vapor?​

Answer:

1/2 f1 will cross

Explanation:

answer it

Answer:

salt and sand

Explanation:it is what it is

Answer:

"1.4 mL" is the appropriate solution.

Explanation:

According to the question,

Now,

Increase in volume will be:

⇒ [tex]\Delta V = \alpha\times v_0\times \Delta \epsilon[/tex]

By putting the given values, we get

           [tex]=1.12\times 10^{-4}\times 500\times 25[/tex]

           [tex]=1.12\times 10^{-4}\times 12500[/tex]

           [tex]=1.4 \ mL[/tex]

Answer:

The correct option is D (KBr)

Explanation:

Potassium bromide (KBr) is a typical example of an IONIC CRYSTAL. Positive and negative ions are arranged in a regular pattern to give a giant crystal lattice in an ionic solid.

Ionic crystals are hard and have high melting points because the electrostatic forces holding the ions are strong. In the solid state, ionic compounds are poor conductors of electricity because the ions are held rigidly in place and so cannot moves about.

When melted or when dissolved in water, the ions are free to move about and the ionic substances become good conductors of electricity.

Ionic crystals are also BRITTLE. When struck, they tend to shatter because as planes of ions slip by one another, they pass from a condition of mutual attraction to one of mutual repulsion.

Answer:

When SO

2

is passed through an acidified solution of H

2

S, sulphur is precipitated out according to the reaction.

2H

2

S+SO

2

→2H

2

O+3S

Answer:

19.7 g.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to realize this problem can be solved by using a molecules-moles-mass relationship, starting with the given molecules, using the Avogadro's number and the molar mass of nitric acid (63.01 g/mol):

[tex]18.8x10^{22}molec*\frac{1mol}{6.022x10^{23}molec}* \frac{63.01g}{1mol} \\\\=19.7g[/tex]

Regards!

Explanation:

is responsible for interrupting the interaction between pectins and solvent molecules

Answer: The product of the given reaction is [tex]HNO_{3}[/tex] and the solution at equilibrium will be acidic.

Explanation:

When two or more chemical substances react together then it forms new substances and these new substances are called products.

For example, [tex]3N_{2}O_{5} + 3H_{2}O \rightarrow 6HNO_{3}[/tex]

This shows that nitric acid [tex](HNO_{3})[/tex] is the product formed and it is an acidic substance.

Hence, the solution at equilibrium will be acidic in nature.

Thus, we can conclude that the product of the given reaction is [tex]HNO_{3}[/tex] and the solution at equilibrium will be acidic.

Answer:The green growing on the penny of copper and the rust forming on the nail of iron are chemical changes. Boiling away salt water, scraping iron filings from a mixture of sand with a magnet, and breaking a rock with a hammer, are physical changes.

Explanation:

Answer: Sugar is added to water and initially completely dissolves, but eventually solid sugar collects on the bottom of the container. Sugar and water are both partially miscible. This produces a dynamic equilibrium. Ethanol (a liquid) is added to water and only a single layer is observed no matter how much ethanol is added. Ethanol and water are miscible.

Explanation:

When a substance (solute) dissolves partially in a solvent then it is known as partially miscible in the solvent. In such cases, a small amount of solute remains at the bottom of solution.

If a solute dissolves completely in solvent like water such that only one layer is seen in the solution then it means that the solute is miscible in solvent.

Thus, we can conclude that sugar is added to water and initially completely dissolves, but eventually solid sugar collects on the bottom of the container. Sugar and water are both partially miscible. This produces a dynamic equilibrium. Ethanol (a liquid) is added to water and only a single layer is observed no matter how much ethanol is added. Ethanol and water are miscible.

Answer:

acidic titration in its stable form

Explanation:

Methyl orange can change its color in titration solution. The yellow color is towards alkaline solution and red color is towards acidic solution. The Ph value of solution will change during this chemical process.

Answer:

Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:

element & mass %

phosphorus & 39.18%

sulfur & 60.82%

Write the molecular formula of X.

Explanation:

The given molecule of phosphorus and sulfur has molar mass --- 316.25 g.

Empirical formula calculation:                      

element:              phosphorus                       sulfur

co9mposition:      39.185%                            60.82%

divide with

atomic mass:          39.185/31.0 g/mol           60.82/32.0g/mol

                              =1.26mol                           1.90mol

smallest mole ratio:   1.26mol/1.26mol =1      1.90mol/1.26 mol =1.50

multiply with 2:          2                                         3

Hence, the empirical formula is:

P2S3.

Mass of empirical formula is:

158.0g/mol

Given, molecule has molar mass --- 316.25 g/mol

Hence, the ratio is:

316.25g/mol/158.0 =2

Hence, the molecular formula of the compound is :

2 x (P2S3)

=[tex]P_4S_6[/tex]

Answer:

1) incorrect

2) incorrect

3) correct

4) correct

5) correct

6) incorrect

Explanation:

The correct electronic configuration of chromium is; [Ar] 3d⁵ 4s¹

The correct electronic configuration for Zr is; [Kr] 4d² 5s²

The correct electronic configuration of Cu^+ is; [Ar] 3d¹⁰

The electronic configuration of an atom refers to the arrangement of electrons in the atoms of such element.

The appropriate number of electrons and its properly written electronic configuration is clearly shown in this answer.

Answer:

A) habitat

Explanation:

a habitat is essentially the organisms "home". also known as a "niche"

Answer:

Explanation:

C₁₂H₂₂O₁₁

1 )

Molar mass = 12 x 12 + 22 x 1 + 11 x 16

= 144 + 22 + 176

= 342 g

2 )

100 mL of 1.0 M will contain 1.0 x0.100 = .1 mole of sucrose

0.1 mole of sucrose = 0.1 x 342 g = 34.2 g of sucrose.

So , mass of sucrose required is 34.2 g .

3 )

100 mL of .5 M sucrose = .100 x .5 mole of sucrose

= .05 mole of sucrose

.05 mole of sucrose = .05 x 342 g = 17.1 g of sucrose .

So , mass of sucrose required is 17.1 g .

Answer:

sp3 hybrid orbitals

Explanation:

The formula of a molecule gives us an idea of its structure and the nature of hybrid orbitals that are involved in the formation of the molecule.

AX4 corresponds to tetrahedral geometry. If a molecule is in tetrahedral geometry, it is most likely sp3 hybridized as usual.

Hence, a molecule with the formula AX 4 uses sp3 hybrid orbitals to form its bonds

Answer:

a)10.87

b)9.66

c)9.15

d)7.71

e) 5.56

f) 3.43

Explanation:

tep 1: Data given

Volume of 0.030 M NH3 solution = 30 mL = 0.030 L

Molarity of the HCl solution = 0.025 M

Step 2: Adding 0 mL of HCl

The reaction:    NH3 + H2O ⇔ NH4+ + OH-

The initial concentration:  

[NH3] = 0.030M    [NH4+] = 0M    [OH-] = OM

The concentration at the equilibrium:

[NH3] = 0.030 - XM

[NH4+] = [OH-] = XM

Kb = ([NH4+][OH-])/[NH3]

1.8*10^-5 = x² / 0.030-x

1.8*10^-5 = x² / 0.030

x = 7.35 * 10^-4 = [OH-]

pOH = -log [7.35 * 10^-4]

pOH = 3.13

pH = 14-3.13 = 10.87

Step 3: After adding 10 mL of HCl

The reaction:

NH3 + HCl ⇔ NH4+ + Cl-

NH3 + H3O+ ⇔ NH4+ + H2O

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.010 L = 0.00025 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00025 =0.00065 moles

Moles HCl = 0

Moles NH4+ = 0.00025 moles

Concentration at the equilibrium:

[NH3]= 0.00065 moles / 0.040 L = 0.01625M

[NH4+] = 0.00625 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.00625/0.01625)

pOH = 4.34

pH = 9.66

Step 3: Adding 20 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.020 L = 0.00050 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00050 =0.00040 moles

Moles HCl = 0

Moles NH4+ = 0.00050 moles

Concentration at the equilibrium:

[NH3]= 0.00040 moles / 0.050 L = 0.008M

[NH4+] = 0.01 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.01/0.008)

pOH = 4.85

pH = 14 - 4.85 = 9.15

Step 4: Adding 35 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.035 L = 0.000875 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000875 =0.000025 moles

Moles HCl = 0

Moles NH4+ = 0.000875 moles

Concentration at the equilibrium:

[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M

[NH4+] = 0.000875 M / 0.065 L = 0.0135 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.0135/3.85*10^-4)

pOH = 6.29

pH = 14 - 6.29 = 7.71

Step 5: adding 36 mL HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.036 L = 0.0009 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.0009 =0 moles

Moles HCl = 0

Moles NH4+ = 0.0009 moles

[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M

Kw = Ka * Kb

Ka = 10^-14 / 1.8*10^-5

Ka = 5.6 * 10^-10

Ka = [NH3][H3O+] / [NH4+]

Ka =5.6 * 10^-10 =  x² / 0.0136

x = 2.76 * 10^-6 = [H3O+]

pH = -log(2.76 * 10^-6)

pH = 5.56

Step 6: Adding 37 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.037 L = 0.000925 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000925 =0 moles

Moles HCl = 0.000025 moles

Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M

pH = -log 3.73*10^-4= 3.43

The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:

a) pH = 10.86

b) pH = 9.66

c) pH = 9.15

d) pH = 7.70

e) pH = 5.56

f) pH = 3.43          

Initially, the pH of the solution is given by the dissociation of NH₃ in water.  

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻     (1)

The constant of the above reaction is:

[tex] Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.76\cdot 10^{-5} [/tex]   (2)

At the equilibrium, we have:  

   NH₃    +    H₂O   ⇄   NH₄⁺    +    OH⁻     (3)  

0.030 M - x                      x               x

[tex] 1.76\cdot 10^{-5}*(0.030 - x) - x^{2} = 0 [/tex]

After solving for x and taking the positive value:

x = 7.18x10⁻⁴ = [OH⁻]  

Now, we can calculate the pH of the solution as follows:

[tex] pH = 14 - pOH = 14 + log(7.18\cdot 10^{-4}) = 10.86 [/tex]

Hence, the initial pH is 10.86.

After the addition of HCl, the following reaction takes place:

NH₃ + HCl ⇄ NH₄⁺ + Cl⁻  (4)  

We can calculate the pH of the solution from the equilibrium reaction (3).            

[tex] 1.76\cdot 10^{-5}(Cb - x) - (Ca + x)*x = 0 [/tex] (5)  

The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:

[tex] n_{b} = n_{i} - n_{HCl} [/tex]     (6)

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^{-4} moles [/tex]          

[tex] n_{a} = n_{HCl} [/tex]   (7)

[tex] n_{a} = 0.025 mol/L*0.010 L = 2.5 \cdot 10^{-4} moles [/tex]

The concentrations are given by:

[tex] Cb = \frac{6.5\cdot 10^{-4} moles}{(0.030 L + 0.010 L)} = 0.0163 M [/tex]   (8)

[tex] Ca = \frac{2.5 \cdot 10^{-4} mole}{(0.030 L + 0.010 L)} = 6.25 \cdot 10^{-3} M [/tex]      (9)

After entering the values of Ca and Cb into equation (5) and solving for x, we have:  

[tex] 1.76\cdot 10^{-5}(0.0163 - x) - (6.25 \cdot 10^{-3} + x)*x = 0 [/tex]

x = 4.54x10⁻⁵ = [OH⁻]

Then, the pH is:

[tex] pH = 14 + log(4.54\cdot 10^{-5}) = 9.66 [/tex]

Hence, the pH is 9.66.

We can find the pH of the solution from the reaction of equilibrium (3).

The concentrations are (eq 8 and 9):

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 8.0\cdot 10^{-3} M [/tex]    

[tex] Ca = \frac{0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 0.01 M [/tex]    

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(8.0\cdot 10^{-3} - x) - (0.01 + x)*x = 0 [/tex]

x = 1.40x10⁻⁵ = [OH⁻]

Then, the pH is:  

[tex] pH = 14 + log(1.40\cdot 10^{-5}) = 9.15 [/tex]

So, the pH is 9.15.

We can find the pH of the solution from reaction (3).

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 3.85\cdot 10^{-4} M [/tex]      

[tex] Ca = \frac{0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 0.0135 M [/tex]      

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(3.85\cdot 10^{-4} - x) - (0.0135 + x)*x = 0 [/tex]

x = 5.013x10⁻⁷ = [OH⁻]      

Then, the pH is:  

[tex] pH = 14 + log(5.013\cdot 10^{-7}) = 7.70 [/tex]  

So, the pH is 7.70.

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0 [/tex]    

[tex] n_{a} = 0.025 mol/L*0.036 L = 9.0 \cdot 10^{-4} moles [/tex]

Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.

At the equilibrium, we have:                

NH₄⁺    +    H₂O   ⇄   NH₃    +    H₃O⁺

Ca - x                             x               x

[tex] Ka = \frac{x^{2}}{Ca - x} [/tex]  

[tex] Ka(Ca - x) - x^{2} = 0 [/tex]   (10)          

We can find the acid constant as follows:

[tex] Kw = Ka*Kb [/tex]

Where Kw is the constant of water = 10⁻¹⁴

[tex] Ka = \frac{1\cdot 10^{-14}}{1.76 \cdot 10^{-5}} = 5.68 \cdot 10^{-10} [/tex]  

The concentration of NH₄⁺ is:

[tex] Ca = \frac{9.0 \cdot 10^{-4} moles}{(0.030 L + 0.036 L)} = 0.0136 M [/tex]      

After solving the equation (10) for x, we have:

x = 2.78x10⁻⁶ = [H₃O⁺]

Then, the pH is:  

[tex] pH = -log(H_{3}O^{+}) = -log(2.78\cdot 10^{-6}) = 5.56 [/tex]

Hence, the pH is 5.56.

Now, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).

[tex] C_{HCl} = \frac{0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L}{(0.030 L + 0.037 L)} = 3.73 \cdot 10^{-4} M = [H_{3}O^{+}] [/tex]      

[tex] pH = -log(H_{3}O^{+}) = -log(3.73 \cdot 10^{-4}) = 3.43 [/tex]

Therefore, the pH is 3.43.

Find more about pH here:

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I hope it helps you!  

Answer:

d

Explanation:

Answer: A balanced equation for the given reaction is [tex]NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)[/tex].

Explanation:

The reaction equation will be as follows.

[tex]NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)[/tex]

Number of atoms on the reactant side is as follows.

Number of atoms on the product side is as follows.

Since number of atoms on both the reactant and product sides are equal. Hence, the reaction equation is balanced.

Thus, we can conclude that a balanced equation for the given reaction is [tex]NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)[/tex].

Answer:

The appropriate answer is "9.225 g".

Explanation:

Given:

Required level,

= 63 ppm

Initial concentration,

= 22 ppm

Now,

The amount of free SO₂ will be:

= [tex]Required \ level -Initial \ concentration[/tex]

= [tex]63-22[/tex]

= [tex]41 \ ppm[/tex]

The amount of free SO₂ to be added will be:

= [tex]41\times 225[/tex]

= [tex]9225 \ mg[/tex]

∵ 1000 mg = 1 g

So,

= [tex]9225\times \frac{1}{1000}[/tex]

= [tex]9.225[/tex]

Thus,

"9.225 g" should be added.

Answer:

heat rate= 1281W

length = 15.8m

Explanation:

we have this data to answer this question with

Tmi = 85 degrees

Tmo = 35 degrees

Ts = 25 dgrees

flow rate = 25 degrees

using engine oil property from table a-5

Tm = Tmo - TMi/2 = 333k

u =0.522x10⁻²

k = 0.26

pr = 51.3

cp = 2562 J/kg.k

mcp(Tmo-Tmi) =

0.01 x 2562(35-85)

= 1281 W

we find the change in Tim

= [(35-25)-(85-25)]/ln[(35-25)/(85-25)]

= -50/ln0.167

= -50/-1.78976

= 27.9°c

we finf the required reynold number

4x0.01/πx0.003x0.522x10⁻²

= 0.04/0.00004921

= 812.8

= 813

we find approximate correlation

NuD = hd/k

NuD = 3.66

3.66 = 0.003D/0.26

cross multiply

0.003D = 3.66x0.26

D = 3.66x0.26/0.003

= 317.2

As = 1281/317x27.9

= 0.145

As = πDL

L = As/πD

= 0.145/π0.003

= 0.145/0.009429

L = 15.378

Answer:

Explanation:

/ means divided by

* means multiply

1. formula is

partial pressure = no of moles(gas 1)/ no of moles(total)

0.30 mol CO/0.60 mol CO2 + 0.30 mol CO + 0.10 mol H20 ->

.3/(.6+.3+.1) =

.3/1 =

.3 =

partial pressure of CO

2.

.3 * .8 atm = .24

khanacademy

quizlet

The partial pressure of the CO is 0.24 atm if the total pressure of the mixture was 0.80 atm.

Dalton's Law of partial pressure states that the total pressure exerted by non reacting gaseous mixture at a constant temperature and given volume is equal to the sum of partial pressure of all gases.

Dalton's Law of partial pressure using mole fraction of gas

Partial pressure of carbon monoxide (CO) = Mole fraction of carbon monoxide (CO) × Total pressure

Now, we have to find the first mole fraction of CO

Mole fraction of carbon monoxide (CO) = [tex]\frac{\text{moles of solute}}{\text{total moles of solute}}[/tex]

                                                                  = [tex]\frac{\text{moles of CO}}{\text{moles of CO}_2 + \text{moles of CO} + \text{moles of H}_{2}O}[/tex]

                                                                  = [tex]\frac{0.30}{0.60 + 0.30 + 0.10}[/tex]

                                                                  = [tex]\frac{0.30}{1}[/tex]

                                                                  = 0.3

Now, put the value in above equation, we get that

Partial pressure of carbon monoxide (CO)

= Mole fraction of carbon monoxide (CO) × Total pressure

= 0.3 × 0.8

= 0.24 atm

Thus, the partial pressure of the CO is 0.24 atm is the total pressure of the mixture was 0.80 atm.

Learn more about the Dalton's Law of partial Pressure here: https://brainly.com/question/14119417

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Answer:

150.000 g

Explanation:

The law of conservation of mass states that the mass of reactants and products of a reaction must be equal to one another.

In other words, for this case:

We are given all required data to calculate the mass of the KOH solution:

Answer:

-46.67 J.

Explanation:

We are given;

Initial Pressure = 15atm = 15 × 10^(3) J

Final pressure = 1atm = 1 × 10^(3) J

Temperature = 300k

The pressures were converted to Joules.

Formula for the entropy change is;

∆S_system = ∆S_surrounding = -(dQ)/T

-(dQ)/T = (-(15 × 10^(3)) - (1 × 10^(3))/300)

= -46.67 J.

Answer: The reaction, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.

Explanation:

A chemical reaction in which one element of a compound is replaced by another element participating in the reaction.

For example, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex]

Here, the element manganese is replaced by carbon atom. As only one element gets replaced so, it is a single-displacement reaction.

Thus, we can conclude that [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.

Answer:

Calcium

Explanation:

Zinc cannot displace Ca because calcium is above it in the reactivity series

Answer:

green gemstone

Explanation:

hope this helps someone

Explanation:

The number of atoms in 1mol of every element can be represented by Avogadro's number, which is [tex]6.022*10^{23}[/tex].

Knowing this, now we can find the atoms in each of these molecules!

[tex]16.8gSr*\frac{1molSr}{87.62gSr} *\frac{6.022*10^{23}atomsSr}{1molSr} =[/tex]

1.15*10^23 atoms of Sr

[tex]26.5gFe*\frac{1molFe}{55.85gFe} *\frac{6.022*10^{23}atomsFe}{1molFe} =[/tex]

2.86*10^23 atoms of Fe

[tex]8.94gBi*\frac{1molBi}{208.98gBi} *\frac{6.022*10^{23}atomsBi}{1molBi} =[/tex]

2.58*10^22 atoms of Bi

[tex]40.0gP*\frac{1molP}{30.97gP} *\frac{6.022*10^{23}atomsP}{1molP}=[/tex]

7.78*10^23 atoms of P

Answer:

option a

hope helps you

have a great day