During the formation of an ionic compound

1) One atom losses electrons and the second gains electrons
2) Two atoms lose electrons
3) Two atoms gain electrons

Answer:

a. 1 mole of acid is equal to one equivalent.

b. 1.00 moles of HCl are found.

c. 1L of 2.00M NaOH is needed to reach the equivalence point

Explanation:

HCl reacts with NaOH as follows:

HCl + NaOH → NaCl + H2O

Where 1 mole of HCl reacts with 1 mole of NaOH. The reaction is 1:1

a. As the reaction is 1:1, 1 mole of acid is equal to one equivalent

b. The initial moles of HCl are:

1.00L * (2.00moles HCl / 1L) = 2.00 moles of HCl

At the halfway point, the moles of HCl are the half, that is:

1.00 moles of HCl are found

c. At equivalence point, we need to add the moles of NaOH needed for a complete reaction of the moles of HCl. As the moles of HCl are 2.00 and the reaction is 1:1, we need to add 2.00 moles of NaOH, that is:

2.00moles NaOH * (1L / 2.00mol) =

1L of 2.00M NaOH is needed to reach the equivalence point

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

            H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i)          0.115M                      0                  0

ΔC              -x                        +x                  +x

C(eq)    0.115M - x                   x                    x

            ≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷  => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion  concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M  value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

The concentration of hydroxide ion of given solution is 4.3 x 10⁻¹¹M.

We can calculate the concentration of hydroxide ions as follow:

[OH⁻][H⁺] = 10⁻¹⁴

Given chemical reaction with ICE table shown as below:

                H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq)

Initial:             0.115                      0                    0

Change:           -x                        +x                  +x

Equilibrium:  0.115-x                   +x                  +x

Given that, Ka = 4.3 x 10⁻⁷

Equilibrium constant for this reaction is written as:

Ka = [H⁺][HCO₃⁻]/[H₂CO₃]

4.3 x 10⁻⁷ = x² / 0.115

x = 2.32 x 10⁻⁴M = [H⁺]

Now we calculate the concentration of hydroxide ion as:

[OH⁻][H⁺] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 2.32 x 10⁻⁴ = 4.3 x 10⁻¹¹M

Hence, value of [OH⁻] is 4.3 x 10⁻¹¹M.

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Answer:

Zero-Order

Explanation:

The exothermic decaying of nitrous oxide at 575° C will lead to [tex]N_{2} and O_{2}[/tex] as follows:

[tex]2N_{2}O[/tex] → [tex]2N_{2}(g) + O_{2} (g)[/tex]

Hot platinum wire in the above reaction would function as a catalyst in the zero-order. However, if the reaction is considered in the gaseous phase, it will be more inclined towards second-order.

In the given scenario([tex]2N_{2}O[/tex] → [tex]2N_{2}(g) + O_{2} (g)[/tex]), the reactant molecules of Nitrous oxide are restricted to the ones which have linked themselves to the catalyst's surface. Once this limited surface is filled, the extra molecules of gas would remain vacant until the previously attached molecules with the surface are decayed entirely.

Answer:

D

Explanation:

We have to bear in mind that the acid is a weak acid. A weak acid does not dissociate completely in solution. We will have more concentration of undissociated acid than A^- and H3O^+ and OH^- in the system at equilibrium.

Being a weak acid, there is maximum concentration of water molecules followed by that of undissiociated acid.

Hence, for this solution, the concentration of ions in solution follows the order;

[H2O] > [HA] > [A-] ~ [H3O ] > [OH-]

Answer:

I think its A maybe am not sure

Explanation:

here's the answer to your question

Answer:

[tex]T_2= 36.7 \textdegree C[/tex]

Explanation:

Mass of Water [tex]m_w=6.90kg[/tex]

Temperature [tex]T=34.7 degrees[/tex]

Heat Flow [tex]H=57.1kJ[/tex]

Specific heat capacity of water [tex]\mu= 4.18J.g^(-1).K^(-1)[/tex]

Generally the equation for Final Temperature is mathematically given by

[tex]M*\mu *T_1 + Q = M*\mu *T_2[/tex]

[tex]T_2=\frac{M*\mu *T_1 + Q }{M*\mu}[/tex]

Therefore

[tex]T_2=\frac{6.90*4.18*34.7 + 57.1}{6.90*4.18}[/tex]

[tex]T_2= 36.7 \textdegree C[/tex]

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:

26mL

Explanation:

NaOH+HCl= NaCl+H2O

nHCl=0.0243*2=0.0486

nNaOh=nHCl

VNaOH=0.0486/1.87=0.026l=26ml

Answer:

26.0 mL

Explanation:

Answer:

I say Star S has radial motion

Explanation:

I'm not sure if it right but let me know if you have any other questions

Answer:

[tex]0.175\; \rm mol \cdot L^{-1}[/tex].

Explanation:

Magnesium chloride and silver nitrate reacts at a [tex]2:1[/tex] ratio:

[tex]\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)[/tex].

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

[tex]\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}[/tex].

The precipitate silver chloride [tex]\rm AgCl[/tex] is insoluble in water and barely ionizes. Hence, [tex]\rm AgCl\![/tex] isn't rewritten as ions.

Net ionic equation:

[tex]\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}[/tex].

Calculate the initial quantity of nitrate ions in the mixture.

[tex]\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}[/tex].

Since nitrate ions [tex]\rm {NO_3}^{-}[/tex] do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

[tex]n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol[/tex].

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be [tex](1/2)[/tex] of the concentration in the original solution.

[tex]\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

Answer:

19.2 g

Explanation:

Step 1: Write the balanced equation

P₂O₅(s) + 3 H₂O(l) ⇒ 2 H₃PO₄

Step 2: Calculate the moles corresponding to 5.29 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

5.29 g × 1 mol/18.02 g = 0.294 mol

Step 3: Calculate the theoretical yield of phosphoric acid, in moles

The molar ratio of H₂O to H₃PO₄ is 3:2. The theoretical yield of H₃PO₄ is 2/3 × 0.294 mol = 0.196 mol

Step 4: Calculate the mass corresponding to 0.196 moles of H₃PO₄

The molar mass of H₃PO₄ is 97.99 g/mol.

0.196 mol × 97.99 g/mol = 19.2 g

Answer:

is this question or you just asking I can't understand.

Answer:

a. -58 millivolts

Explanation:

The given Nernst equation is:

[tex]E_{ion} = 58 millivolts /z \Big[ log_{10} \Big( \dfrac{[ion]_{out}}{[ion]_{in}}\Big) \Big]}[/tex]

The equilibrium potential given by the Nernst equation can be determined by using the formula:

[tex]E_{Cl^-} = \dfrac{2.303*R*T}{ZF} \times log \dfrac{[Cl^-]_{out}} {[Cl^-]_{in}}[/tex]

where:

gas constant(R) = 8.314 J/K/mol

Temperature (T) = (20+273)K

= 298K

Faraday constant F = 96485 C/mol

Number of electron on Cl = -1

[tex]E_{Cl^-} = \dfrac{2.303*8.314*298} {(-1)*(96845)} \times log \dfrac{100} {10}[/tex]

[tex]E_{Cl^-} = - 0.05814 \ volts[/tex]

[tex]\mathsf{E_{Cl^-} = - 0.05814 \times 1000 \ milli volts}[/tex]

[tex]\mathsf{E_{Cl^-} \simeq - 58\ milli volts}[/tex]

Answer:

790 years

Explanation:

Given that;

0.693/t1/2 = 2.303/t log [A]o/[A]

So;

t1/2 =half life of carbon-14

t= age of the sample

[A]o= activity of the living sampoke

[A] = activity at time t

0.693/5.73 ×10^3 = 2.303/t log 85/77

1.21 × 10^-4 = 2.303/t log 1.1

1.21 × 10^-4 = 0.0953/t

t= 0.0953/1.21 × 10^-4

t= 790 years (to 2sf)

Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.

At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.

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Answer:

The molar mass of benzoic acid is 122.1 g/mol

The molar mass of hydrochloric acid = 36.5 g/mol

Explanation:

Benzoyl chloride is an organic compound with the molecular formula C₆H₅COCl. It is an acyl chloride since is it an organic derivative of a carboxylic acid. Acyl chlorides have the general molecular formula, R-COCl, where R is a side chain.

The R group of benzoyl chloride is the benzyl group C₆H₅. It reacts with water (hydrolysis) to produce hydrochloric acid and benzoic acid. The equation of the reaction is given below:

C₆H₅COCl + H₂O → C₆H₅CO₂H + HCl

The molar mass of benzoic acid as well as of hydrochloric acid is calculated from the sum of the masses of the atoms of the elements present in the compound thus:

Molar mass of carbon = 12.0107 g

Molar mass of hydrogen = 1.00784 g

Molar mass of oxygen = 15.999 g

Molar mass of chlorine = 35.453 g

Molar mass of benzoic acid, C₆H₅CO₂H containing 7 moles of atoms of carbon, 6 moles of atoms of hydrogen and 2 moles of atoms of oxygen = 7 × 12.0107 + 6 × 1.00784 + 2 × 15.999 = 122.1 g

Therefore, the molar mass of benzoic acid is 122.1 g/mol

Molar mass of hydrochloric acid, HCl, containing 1 mole of atoms of hydrogen and 1 mole of atoms of chlorine = 1 × 1.00784 + 1 × 35.453 = 36.5 g

Therefore, the molar mass of hydrochloric acid = 36.5 g/mol

Answer:

自分の仕事をする translate to english

Explanation:

Answer:

6.684g

Explanation:

Here, we can use the mole ratio of the gases to calculate.

We know that the mole ratio of the gases equate to their number of moles.

Firstly, we calculate the number of moles of the oxygen gas. The number of moles of the oxygen gas is equal to the mass of the oxygen gas divided by the molar mass of the oxygen gas. The molar mass of the oxygen gas is 32g/mol

Thus, the number of moles produced is 5.98/32 = 0.186875

Where do we move from here?

We know that if we place the partial pressure of oxygen over the total pressure, this would be equal to the number of moles of oxygen divided by the total number of moles. Now let’s do this.

449/851 = 0.186875/n

n =(0.186875 * 851)/449

n = 0.3542

Now we do the same for argon to get the number of moles of argon.

Firstly, we use dalton’s partial pressure law to get the partial pressure of argon. In the simplest form, the partial pressure of argon is the total pressure minus the partial pressure of oxygen.

P = 851 - 449 = 402 mmHg

We now use the mole ratio relation.

402/851 = n/0.3542

n = (402 * 0.3542) / 851

n = 0.1673

Since we now know the number of moles of argon, we can use this multiplied by the atomic mass of argon to get the mass.

The atomic mass of argon is 39.948 amu

The mass is thus 39.948 * 0.1673 = 6.684g

Answer:

COOHCOOH + 2OH⁻ ⇄  C₂O₄²⁻ + 2H₂O

Explanation:

The reaction of oxalic acid with a strong base like sodium hydroxide is the following:

COOHCOOH + OH⁻ ⇄ COOHCOO⁻ + H₂O    (1)

In this first reaction, the oxalic acid loses one proton. In a second reaction with NaOH, the ion COOHCOO⁻ loses its second proton to form ion oxalate as follows:

COOHCOO⁻ + OH⁻ ⇄ C₂O₄²⁻ + H₂O      (2)  

The general reaction between oxalic acid and NaOH is (eq 1 + eq 2):

COOHCOOH + 2OH⁻ ⇄  C₂O₄²⁻ + 2H₂O              

I hope it helps you!  

Answer:

D. [CO₂]

Explanation:

Let's consider the following equation at equilibrium.

CaCO₃(s) ⇄ CaO(s) + CO₂(g)

The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients. It only includes gases and aqueous species.

Kc = [CO₂]

Answer:

The various uses of water :

1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.

2. Water is used as a universal solvent.

3. water maintains the temperature of our body.

4. Water helps in digestion in our body.

5 .water is used in factories and industries.

6. Water is used to grow plants , vegetables and crops.

Answer:

The various use of water are;

I) Cooking.

ii) Drinking

III) Bathing

iv) Generating hydro- electricity

v) Construction work etc

Answer:

21.5mL of a 0.100M HCl are required

Explanation:

The sodium phenoxide reacts with HCl to produce phenol and NaCl in a 1:1 reaction.

To solve this question we need to find the moles of sodium phenoxide. These moles = Moles of HCl required to reach equivalence point and, with the concentration, we can find the needed volume as follows:

Mass NaC6H5O:

1.000g * 25% = 0.250g NaC6H5O

Moles NaC6H5O -116.09g/mol-

0.250g NaC6H5O * (1mol/116.09g) = 2.154x10⁻³ moles = Moles of HCl required

Volume 0.100M HCl:

2.154x10⁻³ moles HCl * (1L/0.100mol) = 0.0215L =

Answer

A

Explanation:

due to high specific heat capacity it loses heat and has low temperature

Answer:

1.28 M

Explanation:

Step 1: Given data

Mass of luminol (solute): 17.0 g

Volume of water: 75.0 mL (this is also the volume of solution)

Step 2: Calculate the moles corresponding to 17.0 g of luminol

The molar mass of luminol is 177.16 g/mol.

17.0 g × 1 mol/177.16 g = 0.0960 mol

Step 3: Calculate the molarity of the solution

We will use the definition of molarity

M = moles of solute / liters of solution

M = 0.0960 mol / 0.0750 L = 1.28 M

Answer:

LiOH(aq) → Li⁺(aq) + OH⁻(aq). 

Answer:

hello? are you still here? reply if you are

Answer:

35

Explanation:

is the answer for your question

The nature of attack on sites in a molecule depends on the nature of such sites. The following are the nature of the sites mentioned in the question:

1) The indicated carbon in compound A is electrophilic.

2) The indicated bond in compound B is nucleophilic.

3) The indicated carbon in compound C is electrophilic.

4) The indicated carbon in compound D is neither electrophilic nor nucleophilic.

5) The indicated oxygen in compound E is nucleophilic.

The terms "electrophilic" and "nucleophilic" are very common in chemistry.

An electrophilic center is usually positively charged, has a positive dipole or is electron deficient hence it attacks negative centers. The term itself means "electron loving". That actually means that it has an affinity for negative charges.

The -I inductive effect of the bromine atom in the carbon in compound A makes that carbon atom to be electrophilic. Also, the carbonyl bond and the O C H 3 attached to the carbon in compound C also makes it electrophilic.

The term "nucleophilic" literately means "nucleus loving". That means a specie that has a high affinity for positive charges. This specie must be electron rich.

The carbon atom in compound B has a double bond which is electron rich and can attack any positive center hence it is nucleophilic. Also, the oxygen atom in E bears two lone pairs of electrons which can attack any positive center in a molecule hence the oxygen atom is also nucleophilic.

In compound D, the carbon atom is bonded to a methyl, two hydrogens and a carbon. There is a nitrogen atom two bonds away. There is no +I or -I inductive effect on this carbon atom because the nitrogen atom is far away. Therefore, the indicated carbon in compound D is neither electrophilic nor nucleophilic.

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