Three bonding pairs around a central atom results in a ________. A. octahedral B. trigonal planar compound C. tetrahedral compound D. linear compound

Answer:

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Explanation:

In water, a substance that ionizes completely in solution is called a weak electrolyte.

Answer: strong electrolyte.

Explanation: In water, a substance that ionizes completely in solution is called a. a. weak electrolyte.

The best option for dissolving PbBr₂ is option (2)

2) Add a solution of NaBr

The reason for choosing the above option is as follows;

Lead (II) bromide is an inorganic powdery substance that has a solubility in water of 0.973 g/100 mL at 20°C. It is insoluble in alcohol but is soluble in alkali, ammonia, NaBr, and KBr.

PbBr₂ is slightly soluble in ammonia, and it reacts with NaOH to produce Pb(OH)₂ and NaBr.

Taking the solubility product of PbBr₂ as [tex]K_{sp}[/tex] = 6.60 × 10⁻⁶, in a solution of 0.5 M NaBr, we have;

PbBr₂ → Pb⁺ + 2Br⁻

[tex]K_{sp}[/tex] = [Pb]·[2Br]²

Therefore, we get;

6.60 × 10⁻⁶ = [x]·[0.5]²

Where;

x = The number of moles of lead, Pb, in per liter of solution

∴ x = (6.60 × 10⁻⁶)/[(0.5 )²] = 2.64 × 10⁻⁵.

The molar solubility of PbBr₂ per liter of NaBr, x = 2.64 × 10⁻⁵ mol/L

PbBr₂ is more soluble in NaBr.

Given that ammonium ion NH₄Br  in water gives similar products to ammonia, NH₃, it is expected to be more suitable to dissolve PbBr₂ in NaBr.

Therefore, the best solution for dissolving PbBr₂(s) is NaBr, the correct option is option (2) add a solution of NaBr.

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Answer:

reactivaty

Explanation:

here you go for the answer

Explanation:

the force of the lone pairs from the bottom would cancel out the force of the lone pairs from the top. Thus, the molecule will be linear.

Answer:

14720000

Explanation:

1 kg = 1000000 mg

14.72 kg = 14.72 x 1000000

=14720000

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Answer:

"[tex]6.7\times 10^{-4} \ atm[/tex]" is the right answer.

Explanation:

Given:

Partial pressure of [tex]N_2[/tex],

= 0.20 atm

Partial pressure of [tex]H_2[/tex],

= 0.15 atm

[tex]K_p = 1.5\times 10^3[/tex] at [tex]400^{\circ} C[/tex]

As we know,

⇒ [tex]K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}[/tex]

By putting the values, we get

    [tex]1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}[/tex]

        [tex]pNH_3^2 = \frac{0.000675}{1.5\times 10^3}[/tex]

                    [tex]=6.7\times 10^{-4} \ atm[/tex]

The measurement of body fluid compartments can be achieved by the dilution of chemical compounds that only circulate and disperse in the region of selected areas in the body. The dilution process is dependent on how the concentration is defined.

Given that:

∴

Concentration C  = 0.5 mg/100

Concentration C = 0.005 mg/ml

Since the mass amount of the chemical compound(i.e. insulin) and the concentration is known.

The volume of the body fluid compartment can be calculated as:

[tex]\mathbf{volume = \dfrac{\text{mass of the marker }}{concentration }}[/tex]

[tex]Volume = \dfrac{80 \ mg}{0.005 \ mg/ml}[/tex]

Volume = 16000 ml

Thus, it is known that insulin is generally utilized for the measurement of the extracellular fluid volume and serves as a cell impermeant marker.

As a result;

The volume of the extracellular fluid compartment is 16000 ml.

However, the tritium-labeled water is a good marker for the entire body fluid compartment due to the fact that:

Given that:

plasma activity of tritium = 20 cpm/ml

i.e.

In 1 ml of plasma, 20 cpm of tritium is present.

As such, in 960,000 counts per min (cpm) of tritium-labeled water, the volume of the whole body compartment is:

[tex]\mathbf{= \dfrac{960000}{20} ml \plasma}[/tex]

= 48000 ml of plasma

Therefore, we can conclude that the volumes of the body compartment that can be determined are:

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Answer:

dissolution is the process

Answer:

In metrology, measurement uncertainty is the expression of the statistical dispersion of the values attributed to a measured quantity.By international agreement, this uncertainty has a probabilistic basis and reflects incomplete knowledge of the quantity value. It is a non-negative parameter.

Answer:

d

Explanation:

Answer:

NaCl, Na⁺,Cl⁻.

MgCl₂, Mg²⁺, Cl⁻.

CaO, Ca²⁺, O²⁻.

Li₃P, Li⁺, P³⁻.

Al₂S₃, Al³⁺, S²⁻.

Ca₃N₂, Ca²⁺, N³⁻.

FeCl₃, Fe³⁺, Cl⁻.

FeO, Fe²⁺, O²⁻.

Cu₂S, Cu⁺, S²⁻.

Cu₃N₂, Cu²⁺, N³⁻.

ZnO, Zn²⁺, O²⁻.

Ag₂S, Ag⁺, S²⁻.

K₂CO₃, K⁺, CO₃²⁻.

NaNO₃, Na⁺, NO₃⁻.

Ca(HCO₃)₂, Ca²⁺, HCO₃⁻.

Al(OH)₃, Al³⁺,OH⁻.

Li₃PO₄, Li⁺, PO₄³⁻.

K₂SO₄, K⁺, SO₄²⁻.

Explanation:

Sodium chloride. NaCl, formed by the cation Na⁺ and the anion Cl⁻.

Magnesium chloride. MgCl₂, formed by the cation Mg²⁺ and the anion Cl⁻.

Calcium oxide. CaO, formed by the cation Ca²⁺ and the anion O²⁻.

Lithium phosphide. Li₃P, formed by the cation Li⁺ and the anion P³⁻.

Aluminum sulfide. Al₂S₃, formed by the cation Al³⁺ and the anion S²⁻.

Calcium nitride. Ca₃N₂, formed by the cation Ca²⁺ and the anion N³⁻.

Iron(III)chloride. FeCl₃, formed by the cation Fe³⁺ and the anion Cl⁻.

Iron(II)oxide. FeO, formed by the cation Fe²⁺ and the anion O²⁻.

Copper(I)sulfide. Cu₂S, formed by the cation Cu⁺ and the anion S²⁻.

Copper(II)nitride. Cu₃N₂, formed by the cation Cu²⁺ and the anion N³⁻.

Zinc oxide. ZnO, formed by the cation Zn²⁺ and the anion O²⁻.

Silver sulfide. Ag₂S, formed by the cation Ag⁺ and the anion S²⁻.

Potassium carbonate. K₂CO₃, formed by the cation K⁺ and the anion CO₃²⁻.

Sodium nitrate. NaNO₃, formed by the cation Na⁺ and the anion NO₃⁻.

Calcium bicarbonate. Ca(HCO₃)₂, formed by the cation Ca²⁺ and the anion HCO₃⁻.

Aluminum hydroxide. Al(OH)₃, formed by the cation Al³⁺ and the anion OH⁻.

Lithium phosphate. Li₃PO₄, formed by the cation Li⁺ and the anion PO₄³⁻.

Potassium sulfate. K₂SO₄, formed by the cation K⁺ and the anion SO₄²⁻.

Answer:

CaCl2 has ionic bond because here calcium gives its electron to the chlorine atom and becomes positivetly charged ion.

Answer:

hydrogen nitrate + sodium hydrochlorate- sodium nitrate+ water + co2 (acid base reaction)

silver nitrate + calcium chloride - silver chloride+ calcium nitrate ( double displacement reaction)

hydrogen + nitrogen - ammonia gas ( simple contact reaction)

hydrogen peroxide - water + oxygen ( single displacement reaction)

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Answer:

an alcohol is a Hydroxyl group due to the OH~ that is associated with it's molecules

The functional group found in an alcohol is Hydroxyl . Therefore, the correct option is option D.

A functional group in organic chemistry is a substituent and moiety inside a molecule that triggers the molecule's distinctive chemical processes. No matter how the rest of a molecule is made up, the very same functional group would experience the same or a similar set of chemical events.

This permits the design of synthetic chemistry as well as the methodical forecasting of chemical reactions as well as the behaviour of chemical molecules. Other functional groups close by can affect a functional group's reactivity. Retrosynthetic analysis can be used to design organic synthesis by using functional group interconversion. The functional group found in an alcohol is Hydroxyl .

Therefore, the correct option is option D.

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Answer: 1090°C

Explanation: According to combined gas laws

(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2

where P1 = initial pressure of gas = 80.0 kPa

V1 = initial volume of gas = 10.0 L

T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K

P2 = final pressure of gas = 107 kPa

V2 = final volume of gas = 20.0 L

T2 = final temperature of gas

Substituting the values,

(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2

T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)

T2 = 513 K × (1.3375) × (2)

T2 = 1372.275 K

T2 = (1372.275 - 273) °C

T2 = 1099 °C

1090 degree Celsius

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Answer:

The correct option is (c)

Answer:

the law of conservation of mass

[tex]\boxed{\sf {AlCl_3\atop Aluminium\:Chloride}+{H_2\atop Hydrogen}\longrightarrow {Al\atop Aluminium}+{HCl\atop Hydrochloric\:acid}}[/tex]

Balanced Equation:-

[tex]\boxed{\sf {2AlCl_3\atop Aluminium\:Chloride}+{3H_2\atop Hydrogen}\longrightarrow {2Al\atop Aluminium}+{6HCl\atop Hydrochloric\:acid}}[/tex]

Answer

Iridium(IV)Nitrite

The correct IUPAC name of the Ir(NO₂)₄ compound is Iridium(IV)Nitrite.

Whether it's in a continuous chain or just a ring, the largest chain of carbons joined by a single bond serves as the basis for IUPAC nomenclature.

A chemical compound would seem to be a substance that contains numerous similar molecules made of atoms from different elements joined by chemical bonds.

The given compound is Ir(NO₂)₄. It can be seen that 4 nitro group is attached with Ir and its coordination number is 4. Hence, the IUPAC name will be Iridium(IV)Nitrite.

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Answer:

vinegr and soda ..................

........

Answer:

acid + base = salt

so the answer is vinegar and soda

Explanation:

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Answer:

Mg+F2= Mgf2

Explanation:

F 2 is an oxidizing agent, Mg is a reducing agent. ; Pale-yellow to greenish gas with a pungent, irritating odor.

Answer:

Consequence

Answer:

1.12 mol/L.

Explanation:

From the question given above, the following data were obtained:

Mole of LiCl = 3.75 moles

Volume = 3.36 L

Molarity =?

Molarity is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is expressed as:

Molarity = mole / Volume

With the above formula, we can obtain the molarity of the solution as follow:

Mole of LiCl = 3.75 moles

Volume = 3.36 L

Molarity =?

Molarity = mole /Volume

Molarity = 3.75 / 3.36

Molarity = 1.12 mol/L

Thus, the molarity of the solution is 1.12 mol/L

Chemistry 11 Solutions

978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85

Amount in moles, n, of the NaCl(s):

NaCl

2.5 g

m

n

M

58.44 g

2

4.2778 10 m l

ol

o

/m

u

Molar concentration, c, of the NaCl(aq):

–2 4.2778 × 10 mol

0.100

0.42778 mol/L

0.43 mol

L

/L

n

c

V

The molar concentration of the saline solution is 0.43 mol/L.

Check Your Solution

The units are correct and the answer correctly shows two significant digits. The

dilution of the original concentrated solution is correct and the change to mol/L

seems reasonable.

Section 8.4 Preparing Solutions in the Laboratory

Solutions for Practice Problems

Student Edition page 386

51. Practice Problem (page 386)

Suppose that you are given a stock solution of 1.50 mol/L ammonium sulfate,

(NH4)2SO4(aq).

What volume of the stock solution do you need to use to prepare each of the

following solutions?

a. 50.0 mL of 1.00 mol/L (NH4)2SO4(aq)

b. 2 × 102 mL of 0.800 mol/L (NH4)2SO4(aq)

c. 250 mL of 0.300 mol/L NH4

+

(aq)

What Is Required?

You need to calculate the initial volume, V1, of (NH4)2SO4(aq) stock solution

needed to prepare each given dilute solution.

The dilution gives the relationship between the molarity and the volume of the solution. The volume of stock solution with a molarity of 1.50 mol/L is 50 mL.

Dilution is said to be the addition of more volume to the concentrated solution to make it less in molar concentration. This tells about the inverse and indirect relationship between the volume and the molar concentration of the solution.

Given,

Initial volume = V₁

Initial molar concentration (M₁) = 1.50 mol/L

Final volume (V₂) = 125 mL = 0.125 L

Final molar concentration (M₂)= 0.60 mol/L

The dilution is calculated as:

M₁V₁ = M₂V₂

V₁ = M₂V₂ ÷ M₁

Substituting the values in the above formula as

V₁ = M₂V₂ ÷ M₁

V₁ = (0.60 mol/L × 0.125 L) ÷ 1.50 mol/ L

V₁ = 0.05 L

= 50 mL

Therefore, 50 mL of stock solution is needed to make a 0.60 mol/L solution.

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Answer:

B. 1.65 L

Explanation:

Step 1: Write the balanced equation

2 SO₂(g) + O₂(g) ⇒ 2 SO₃(g)

Step 2: Calculate the moles of SO₂

The pressure of the gas is 1.20 atm and the temperature 25 °C (298 K). We can calculate the moles using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.20 atm × 1.50 L / (0.0821 atm.L/mol.K) × 298 K = 0.0736 mol

Step 3: Calculate the moles of SO₃ produced

0.0736 mol SO₂ × 2 mol SO₃/2 mol SO₂ = 0.0736 mol SO₃

Step 4: Calculate the volume occupied by 0.0736 moles of SO₃ at STP

At STP, 1 mole of an ideal gas occupies 22.4 L.

0.0736 mol × 22.4 L/1 mol = 1.65 L

Fats may be either saturated or unsaturated. A saturated fat is a fat that consists of triglycerides whose carbon chains consist entirely of carbon-carbon single bonds. ... An unsaturated fat is a fat that consists of triglycerides whose carbon chains contain one or more carbon-carbon double bonds.

Answer:

C. sedimentary rock

Explanation:

Deposition is basically a process in which sediments are added to land.

B is the answer to your question.

C2H4 is not capable of hydrogen bonding because the H's are attached to the Carbon, and the charge is 0.

Although HCl cannot hydrogen bond, that aspect does not hinder it's ability to dissolve. Because HCl is polar and so is water, the positive side of H2O will be attracted to the negative side of HCl, thus "tearing" the molecule apart. (Like dissolves like - polar dissolves polar)

Based on the Solubility rule, KBr is soluble because it contains a group 1 metal.

Answer:

See explanation and image attached

Explanation:

Aromatic compounds undergo electrophilic aromatic substitution reactions in which the aromatic ring is maintained.

Substituted benzenes may be more or less reactive towards electrophilic aromatic substitution than benzene depending on the nature of the substituent present in the ring.

Substituents that activate the ring towards electrophilic substitution such as -OCH3 are ortho-para directing.

The major products of the bromination of anisole are p-bromoanisole and o-bromoanisole. The resonance structures leading to these products are shown in the image attached.

Answer:

See the answer below

Explanation:

From the original equation in the image, the mole ratio of C:CO2:CO is 1:1:2. This means that for every 1 mole of C and CO2, 2 moles of CO would be produced.

Now, looking at the simulation below the equation of the reaction, 3 moles of C and 8 moles of CO2 were supplied as input. Applying this to the original equation of reaction, C seems to be a limiting reagent for the reaction because the ratio of C to CO2 should 1:1.

Hence, taking all the 3 moles of C available means that only 3 moles out of the available 8 for CO2 would be needed. 3 moles c and 3 moles CO2 means that 6 moles CO would be produced (remember that the ratio remains 1:1:3 for C, CO2, and CO). This means that 5 moles CO2 would be leftover.

In other words, all the 3 moles C would be consumed, 3 out of 8 moles CO2 would be consumed, and 6 moles CO would be produced while 5 moles CO2 would be leftover.

Answer:

B. 0.0000005461m

I used the method of moving the decimal.