Name four types of salts​

Answer:

Volume of the calcium hydroxide solution used is 0.0235 mL.

Explanation:

Moles of KHP =

According to reaction, 2 moles of KHP  with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;

of calcium hydroxide

Molarity of the calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

Volume of the calcium hydroxide solution used is 0.0235 mL.

Answer:

The rate would have doubled

Explanation:

Answer:

Ca²⁺ and Cl⁻

Explanation:

In a chemical reaction, spectator ions are ions that are not involved in the reaction, that means are the same before and after the reaction.

In water, calcium hydroxide, Ca(OH)₂ is dissociated in Ca²⁺ and OH⁻. Also, hydrochloric acid, HCl, dissociates in H⁺ and Cl⁻. The reaction is:

Ca²⁺ + 2OH⁻ + 2H⁺ + 2Cl⁻ → 2H₂O + Ca²⁺ + 2Cl⁻

The ions that react are H⁺ and OH⁻ (Acid and base producing water)

And the ions that are not reacting, spectator ions, are:

Answer:

4.46 mol of NH3

Explanation:

The equation of he reaction is given as;

2N + 3H2 --> 2NH3

From the stochiometry of the reaction, 1 mol of Nitrogen produces 2 mol of Ammonia.

Mass of Nitrogen = 31.2g

Molar mass of Nitrogen = 14g/mol

Number of moles = Mass / Molar mass = 31.2 / 14 = 2.23 mol

Since 1 mol of N = 2 mol of NH3;

2.23 mol of N2 would produce x

x = 2.23 * 2 = 4.46 mol of NH3

Answer:

I hope it works

Explanation:

As we know that

w=m*g

given m=0.145 , g=9.8

hence we get

w= (9.8)*(0.145)

w=1.421 m/sec 2

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Answer:

n = 0.207 mole

Explanation:

We have,

P = 1 atm

V = 5 liter

R = 0.0821 L.atm/mol.K

T = 293 K

We need to find the value of n. The relation is as follows :

PV = nRT

Solving for n,

[tex]n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 5}{0.0821 \times 293}\\\\n=0.207\ \text{mol}[/tex]

So, the value of n is 0.207 mol.

Answer:

Gay lussacs law

Explanation:

Answer:

[tex]\Delta G=-359\frac{kJ}{mol}[/tex]

Explanation:

Hello,

In this case, we must remember that the Gibbs free energy is defined in terms of the enthalpy, temperature and entropy as shown below:

[tex]\Delta G=\Delta H -T\Delta S\\[/tex]

In such a way, for the given data, we obtain it, considering the conversion from J to kJ for the entropy in order to conserve the proper units:

[tex]\Delta G=-292\frac{kJ}{mol} -(298)(224\frac{J}{mol}*\frac{1kJ}{1000J} )\\\\\Delta G=-359\frac{kJ}{mol}[/tex]

Best regards.

Answer:

B- 358 kj

Explanation: I took the test

Answer:

the answer is 2,000 nickels.

Explanation:

we multiplied 100 by 100, because there are 100 cents in a dollar, and we divided 10,000 by 5, because there are 5 cents in a nickel.

Explanation:

The spontaneity of a system is deduced by the sign of the gibbs free energy value. If it is negative, it means the process / reaction is spontaneous however a positive value indicates the such process is not spontaneous.

Gibbs free energy, enthalpy and entropy are related by the following equation;

ΔG = ΔH - TΔS

A positive value of enthalpy, H and entropy, S means that G would always be a negative value at all temperatures.

Answer:

d. carboxyl

Explanation:

The presence of carbonyl group (>C=O)) and a hydroxyl group ( (−OH) on the same carbon atom is called a "carboxyl" group. A carboxyl group is represented as COOH and acts as the functional group part of carboxylic acids.

For example:

Hence, the correct option is "d. carboxyl ".

Answer:

d. Oxidation and reduction

Explanation:

For this question we have to remember the definition of each type of reaction:

-) Hydrogenation

In this reaction, we have the addition of hydrogen to a molecule. Usually, an alkene or alkyne. In the example, molecular hydrogen is added to a double bond to produce an alkane.

-) Alkylation

In this reaction, we have the addition of a chain of carbon to another molecule. In the example, an ethyl group is added to a benzene ring.

-) Hydrolysis

In this reaction, we have the breaking of a bond by the action of water. In the example, a water molecule can break the C-O bond in the ester molecule.

-) Halogenation

In this reaction, we have the addition of a halogen (atoms on the VIIIA group). In the example, "Cl" is added to the butene.

-) Ammoniation

In this reaction, we have the addition of the ammonium ion ([tex]NH_4^+[/tex]). In the example, the ammonium ion is added to an acid.

-) Oxidation and reduction

In this reaction, we have opposite reactions. The oxidation is the loss of electrons and the reduction is the gain of electrons. For example:

[tex]Ag^+~+~e^-~->~Ag[/tex] Reduction

[tex]Al~->~Al^+^3~+~3e^-[/tex] Oxidation

Answer:

Silver.

Explanation:

Hello,

In this case, for the redox reaction:

[tex]Ni^0(s)+Ag^+(aq)\rightarrow Ni^{2+}+Ag^0(s)[/tex]

We can see the nickel is being oxidized as its oxidation state increases from 0 to 2+ whereas the oxidation state of silver decreases from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.

Best regards.

The oxidizing agent in the redox reaction represented by the following cell notation is Silver.

The redox reaction is

Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)

here the nickel is being oxidized since its oxidation state rises from 0 to 2+ while on the other hand,  the oxidation state of silver is reduced from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.

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Answer:

The correct answer is 1.60 Liters.

Explanation:

The given reaction:

CO (g) + H₂(g) ⇔ CH₃OH (g)

Based on the given reaction, two moles of H₂ reacts with one mole of CO and produce one mole of CH₃OH.

It is mentioned that 3.20 L of H₂ is reacted, therefore, there is a need to convert it into moles.

As 22.4 L at standard temperature and pressure is equivalent to 1 mole.

Therefore, 1 L at STP will be, 1/22.4 mole

Now 3.20 L at STP will be,

= 1/22.4 × 3.20

= 0.1428 mole

And as mentioned in the reaction that 2 moles of H₂ gives 1 mole of CH₃OH, therefore, 1 mole of H₂ will give 1/2 mole of CH₃OH

Now, 0.1428 mole of H₂ will give,

= 0.1428/2 = 0.071 mole of CH₃OH

= 0.071 × 22.4 = 1.60 L

The volume, in liters, of CH₃OH gas formed is 1.60 L

From the question,

We are to determine the volume of CH₃OH formed

The given chemical equation for the reaction is

CO(g)+ H₂(g) → CH₃OH

The balanced chemical equation for the reaction is

CO(g)+ 2H₂(g) → CH₃OH

This means

1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH

Now, we will determine the number of moles of H₂ present in the 3.20 L H₂ at STP

1 mol of an ideal gas has a volume of 22.4 L at STP

Then,

x mole of the H₂ gas will have a volume of 3.20 L at STP

x = [tex]\frac{3.20 \times 1}{22.4}[/tex]

x = 0.142857 mole

∴ The number of mole of H₂ present is 0.142857 mole

Since

2 moles of H₂ reacts to produce 1 mole of CH₃OH

Then,

0.142857 mole of H₂ will react to produce [tex]\frac{0.142857}{2}[/tex] mole of CH₃OH

[tex]\frac{0.142857}{2} = 0.0714285[/tex]

∴ The number of moles of CH₃OH produced = 0.0714285 mole

Now, for the volume of CH₃OH formed

Since

1 mol of an ideal gas has a volume of 22.4 L at STP

Then,

0.0714285 mol of CH₃OH will have a volume of 22.4 × 0.0714285  at STP

22.4 × 0.0714285 = 1.5999984 L ≅ 1.60 L

Hence, the volume of CH₃OH gas formed is 1.60 L

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Answer:

An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction

Answer:

Triacontyl palmitate

Explanation:

In this case, we have a reaction between an acid and an alcohol. When we put together these kind of compounds an ester is produced. This reaction is called "esterification".

In our case, the alcohol is a structure with 30 carbon in which the "OH" group is bonded on carbon 1. The name of this compound is "n-triacontanol". The acid is a structure in which we have 16 carbon in which the "COOH" group is placed on carbon 1. The name of this compound is "palmitic acid". The ester produced by the acid and the alcohol is "Triacontyl palmitate".

See figure 1.

I hope it helps!

Answer:

The empirical formulae for the compound is CH3.

Answer:

226Ra88→222Rn86+4He2

Explanation:

An α-particle usually consists of a helium nucleus which indicates the type of decay that was undergone in this radioactive process.

During α-decay(alpha decay), an atomic nucleus emits an alpha particle.

Answer:

NaN3 = 47.2 g

Explanation:

Given:

2 NaN3 ⇒ 2 Na  + 3 N2

Find:

Amount of NaN3

Computation:

N2 moles = Product of N2 / molar mass of N2

N2 moles =30.5/28

N2 moles = 1.0893

2NaN3 makes 3(N2 )

So,

NaN3 moles = (2/3) moles of N2  

NaN3 moles = ( 2/3) × 1.0893

NaN3 moles = = 0.7262

NaN3 mass = 0.7262 x 65

NaN3 = 47.2 g

Answer:

NaN3 = 47.2 g

Explanation:

Given:

2 NaN3 ⇒ 2 Na  + 3 N2

Find:

Amount of NaN3

Computation:

N2 moles = Product of N2 / molar mass of N2

N2 moles =30.5/28

N2 moles = 1.0893

2NaN3 makes 3(N2 )

So,

NaN3 moles = (2/3) moles of N2  

NaN3 moles = ( 2/3) × 1.0893

NaN3 moles = = 0.7262

NaN3 mass = 0.7262 x 65

NaN3 = 47.2 g

Explanation:

Answer:

pH = 3.49

Explanation:

We have a buffer system formed by a weak acid (HNO₂) and its conjugate base (NO₂⁻ coming from KNO₂). We can calculate the pH  of a buffer ssytem using the Henderson-Hasselbach equation.

pH = pKa + log [base] / [acid]

pH = -log Ka + log [NO₂⁻] / [HNO₂]

pH = -log 4.50 × 10⁻⁴ + log 0.290 M / 0.210 M

pH = 3.49

The pH of the solution containing 0.210 M nitrous acid (HNO₂) and 0.290 M potassium nitrite (KNO₂) is 3.49

We'll begin by calculating the the pKa of acid. This can be obtained as follow:

Acid dissociation constant (Ka) = 4.50×10¯⁴

pKa = –Log Ka

pKa = –Log 4.50×10¯⁴

pKa = 3.35

Concentration of HNO₂, [HNO₂] = 0.210 M

Concentration of KNO₂, [KNO₂] = 0.290 M

The pH of the solution can obtain by using the Henderson-Hasselbach equation as illustrated below:

pH = pKa+ log [NO₂⁻] / [HNO₂]

pH = 3.35 + log (0.290 / 0.210)

Thus, the pH of the solution is 3.49

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Answer:

The frequency of the photon is 7.41*10¹⁶ Hz

Explanation:

Planck states that light is made up of photons, whose energy is directly proportional to the frequency of radiation, according to a constant of proportionality, h, which is called Planck's constant. This is expressed by:

E = h*v

where E is the energy, h the Planck constant (whose value is 6.63*10⁻³⁴ J.s) and v the frequency (Hz or s⁻¹).

So the frequency will be:

[tex]v=\frac{E}{h}[/tex]

Being E= 4.91*10⁻¹⁷ J and replacing:

[tex]v=\frac{4.91*10^{-17} J}{6.63*10^{-34} J.s}[/tex]

You can get:

v= 7.41*10¹⁶ [tex]\frac{1}{s}[/tex]= 7.41*10¹⁶ Hz

The frequency of the photon is 7.41*10¹⁶ Hz

Answer: Reaction (1) , (3) and (4) are accompanied by an increase in entropy.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

(1) [tex]2SO_2(g)+O_2(g)\rightarrow SO_3(g)[/tex]

3 moles of reactant are changing to 1 mole of product , thus the randomness is increasing. Thus the entropy also increases.

2) [tex]H_2O(l)\rightarrow H_2O(s)[/tex]

1 mole of Liquid reactant is changing to 1 mole of solid product , thus the randomness is decreasing. Thus the entropy also decreases.

3) [tex]Br_2(l)\rightarrow Br_2(g)[/tex]

1 mole of Liquid reactant is changing to 1 mole of gaseous product , thus the randomness is increasing. Thus the entropy also increases.

4)  [tex]H_2O_2(l)\rightarrow H_2O(l)+\frac{1}{2}O_2(g)[/tex]

1 mole of Liquid reactant is changing to half mole of gaseous product and 1 mole of liquid product, thus the randomness is increasing. Thus the entropy also increases.

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

[tex]MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-[/tex]

And the undergoing chemical reaction:

[tex]MgCl_2+2NaF\rightarrow MgF_2+2NaCl[/tex]

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

[tex]n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2[/tex]

Next, the moles of magnesium chloride consumed by the sodium fluoride:

[tex]n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2[/tex]

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

[tex]n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2[/tex]

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[tex][Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M[/tex]

[tex][F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M[/tex]

Thereby, the reaction quotient is:

[tex]Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}[/tex]

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

Answer:

12.5g

Explanation:

Half life = 2.4 Minutes.

The half life of a compound is the time it takes to decay to half of it's original concentration or mass.

Time lapsed= 7.2 minutes. This is equivalent to 3 half lives ( 3 * 2.4)

Initial mass = 100g

First half life;

100g --> 50g

Second half life;

50g --> 25g

Third half life;

25g --> 12.5g

The amount of Zn-71 that remains after 7.2 mins has elapsed is 12.5 g

We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 2.4 mins

Time (t) = 7.2 mins

[tex]n = \frac{t}{t_{1/2}} \\\\n = \frac{7.2}{2.4} \\\\[/tex]

Thus, 3 half-lives has elapsed.

Original amount (N₀) = 100 g

Number of half-lives (n) = 3

[tex]N = \frac{N_{0}}{2^{n}} \\\\N = \frac{100}{2^{3 }}\\\\N = \frac{100}{8}\\\\[/tex]

Thus, the amount of Zn-71 that remains after 7.2 mins is 12.5 g

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Answer:

[tex]x_{et}=0.6068[/tex]

Explanation:

Hello,

In this case, since the mole fraction of a compound, in this case ethanol, in a binary mixture, in this constituted by both water and ethanol, is mathematically defined as follows:

[tex]x_{et}=\frac{n_{et}}{n_{et}+n_{w}}[/tex]

Whereas [tex]n[/tex] accounts for the moles in the solution for each species, we must first compute the moles of both ethanol (density: 0.789 g/mL and molar mass: 46.07 g/mol) and water (density: 1g/mL and molar mass: 18.02 g/mol)

[tex]n_{et}=10.00mL\ et*\frac{0.789g\ et}{mL\ et} *\frac{1mol\ et}{46.07g\ et}=0.1713mol\ et\\ \\n_w=2.00mL\ w*\frac{1g\ w}{mL\ w} *\frac{1mol\ w}{18.02g\ w}=0.1110mol\ w[/tex]

Therefore, the mole fraction turns out:

[tex]x_{et}=\frac{0.1713mol}{0.1713mol+0.1110mol}\\\\x_{et}=0.6068[/tex]

Best regards.

Answer:

7.4 × 10⁻⁹ M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for Li₃PO₄: 5.9 × 10⁻¹⁷

Concentration of lithium ion: 0.0020 M

Step 2: Write the reaction for the solution of Li₃PO₄

Li₃PO₄(s) ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)

Step 3: Calculate the phosphate concentration required for a precipitate to occur

The solubility product constant is:

Ksp = 5.9 × 10⁻¹⁷ = [Li⁺]³ × [PO₄³⁻]

[PO₄³⁻] = 5.9 × 10⁻¹⁷ / [Li⁺]³

[PO₄³⁻] = 5.9 × 10⁻¹⁷ / 0.0020³

[PO₄³⁻] = 7.4 × 10⁻⁹ M

Answer:

Molar solubility is 1.12x10⁻⁴M

Explanation:

The dissolution of magnesium hydroxide is:

Mg(OH)₂(s) ⇄ Mg²⁺ + 2OH⁻

The molar solubility represents the moles of the solid that the solution can dissolve, that could be written as:

Mg(OH)₂(s) ⇄ X + 2X

Where X is solubility.

If you obtained a [OH⁻] = 2.24x10⁻⁴M and you know [OH⁻] = 2X:

2X = 2.24x10⁻⁴M

X = 2.24x10⁻⁴M/2

X =1.12x10⁻⁴M

Answer:

See explanation

Explanation:

1)

Mg3N2(s) + 6H2O(l) ------------> 3Mg(OH)2 + 2NH3(g)

Number of moles of magnesium nitride= mass/molar mass= 4.86g/100.9494 g/mol = 0.048 moles

1 mole of magnesium nitride yields 3 moles of magnesium hydroxide

0.048 moles of magnesium nitride yields 0.048 moles × 3= 0.144 moles of magnesium hydroxide

Theoretical yield of magnesium hydroxide = 0.144 moles × 58.3197 g/mol = 8.398 g

Percent yield= actual yield/ theoretical yield × 100

Percent yield= 7.18/8.398 × 100/1 = 85.5%

2)

N2(g) + 3H2(g) -------> 2NH3(g)

Number of moles of hydrogen gas = mass/ molar mass = 6.41g/ 2gmol-1 = 3.205 moles of hydrogen gas.

From the balanced reaction equation;

3 moles of hydrogen gas yields 2 moles of ammonia

3.205 moles of hydrogen gas yields 3.205 × 2/3 = 2.1367 moles of ammonia

Theoretical yield of ammonia = 2.1367 moles × 17 gmol-1 = 36.3 g

Percent yield = actual yield/ theoretical yield ×100

Percent yield = 26.2/36.3 ×100

Percent yield = 72.2%

3)

3Cl2(g) + 3H2O(l) ------> HOCl3(aq) + 5HCl(aq)

Number of moles of water= mass/ molar mass = 3.79g/18 gmol-1 = 0.21 moles

Since

3 mole of water yields 5 mole of HCl

0.21 moles of water yields 0.21 × 5/3 = 0.35 moles of HCl

Theoretical yield of HCl = 0.35 moles × 36.5 gmol-1 = 12.775 g

Percent yield = actual yield/ theoretical yield × 100/1

Percent yield = 8.70/12.775 ×100

Percent yield = 68.1%

Answer:

3

Explanation:

Resonance is a valence bond concept put forward by Linus Pauling to explain the fact that the observed properties of a molecule may be as a result of the fact that its actual structure lie somewhere between a given number of structural extremes called canonical structures or resonance structures.

There are three resonance structures for SO3 that obey the octet rule. All the S-O bonds in SO3 are equivalent in these resonance structures.

Seven equivalent resonance structures for the molecular of SO3 can be drawn without breaking the octet rule.

We can arrive at this answer because:

In an SO3 molecule, electrons are shared between atoms. This sharing can be done with seven resonance structures.

These structures are shown in the figure below.

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