Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described by the chemical equation

MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

How much MnO2(s) should be added to excess HCl(aq) to obtain 185 mL Cl2(g) at 25 °C and 745 Torr ?

mass of MnO2:

Answer:

B

Explanation:

Spontaneous in chemical reactions means without any external input.

Occurring without added energy. Hence, option B is correct.

A spontaneous reaction is a reaction that supports the formation of products under the conditions under which the reaction is happening.

Spontaneous Reaction- a reaction that favours the formation of products at the conditions under which the reaction is occurring.

A non-spontaneous reaction can be made spontaneous if it is inside a controlled environment, this is what happens in nuclear power plants that create atomic fusion and fission in chambers that are controlled to control different particles to create nuclear active rays.

Hence, option B is correct.

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Answer:

photoelectric effect, phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it.

Answer: The osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.

Explanation:

Given: Mass = 5.0 g

Volume = 1 L

Molar mass of sucrose = 342.3 g/mol

Moles are the mass of a substance divided by its molar mass. So, moles of sucrose are calculated as follows.

[tex]Moles = \frac{mass}{molarmass}\\= \frac{5.0 g}{342.3 g/mol}\\= 0.0146 mol[/tex]

Hence, concentration of sucrose is calculated as follows.

[tex]Concentration = \frac{moles}{Volume (in L)}\\= \frac{0.0146 mol}{1 L}\\= 0.0146 M[/tex]

Formula used to calculate osmotic pressure is as follows.

[tex]\pi = CRT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure

C = concentration

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

[tex]\pi = CRT\\= 0.0146 \times 0.0821 L atm/mol K \times 298 K\\= 0.357 atm (1 atm = 760 torr)\\= 271.32 torr[/tex]

Thus, we can conclude that the osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.

The question is incomplete, the complete question his;

Which straight chain alkane below would you predict to be the most viscous? Why? All are liquids exhibiting the general bonding pattern CH3-(CH2)n-CH3

C9H20

C10H22

C5H12

C6H14

C12H26

Answer:

C12H26

Explanation:

Generally, the viscosity of a liquid increases with increase in molecular mass of the substance.

Liquids of high molecular mass do not flow easily. This means that they posses high viscosity.

Thus, since C12H26 has the highest molecular mass among the options given in the question, C12H26 exhibits the greatest viscosity.

Answer:

The Rf value of ibuprofen increases

Explanation:

TLC involves the elution of a solute using a mobile phase(solvent). The stationary phase is made of an adsorbent such as silica.

The extent of interaction between the solute and the mobile phase affects the Rf value. The greater the interaction between the solute and the solvent, the greater the Rf value.

On the other hand, the polarity of the solvent and the solute also affects the Rf value. If the solvent is changed from t-butyl methyl ether to acetone, the Rf value for ibuprofen increases because ibuprofen is polar and acetone is also polar hence there is greater interaction between the solvent and solute.

Explanation:

Population density is defined as the number of people present per square kilometre. Population density of India according to 2011 census is 382 persons per square kilometres.

Answer:

M=0.549M

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to perform this calculation by firstly assuming we have 1 kg of water as the solvent so that we have 0.550 moles of NaCl as well. Moreover, we realize we have 1000 grams of water and the correct mass of the solution can be calculated by converting 0.550 moles of NaCl to grams by using its molar mass:

[tex]m_{solute}=0.550mol*\frac{58.44 g}{1mol}= 32.14g\\\\m_{solution}=1000g+32.14g=1032.14g[/tex]

And subsequently, the volume in liters by using the density and the correct conversion factor:

[tex]V_{solution}=1032.14g*\frac{1mL}{1.03g} *\frac{1L}{1000mL} =1.002L[/tex]

Finally, the molarity will be:

[tex]M=\frac{0.550mol}{1.002L} =0.549M[/tex]

Regards!

Answer:

[tex]_{19} ^{40} K[/tex] → [tex]_{20} ^{40} Ca[/tex] + [tex]_{-1} ^{0} e[/tex] + ¯v

Explanation:

[tex]_{19} ^{40} K[/tex] is an unstable potassium isotope which has the capacity to undergo beta decay. During the decay process, it transforms to calcium atom with a release of beta particle and an antineutrino.

The required nuclear equation is:

                                              [tex]_{19} ^{40} K[/tex] → [tex]_{20} ^{40} Ca[/tex] + [tex]_{-1} ^{0} e[/tex] + ¯v

The calcium formed due to this process is a stable atom.

The question is incomplete, the common question is;

What intermolecular attractive force is primarily responsible for the solubility of chlorine, Cl2, in water?

a. dipole - dipole

b. hydrogen bonding

c. dipole-induced dipole

d. ion-dipole

e. ion-induced dipole

Answer:

dipole-induced dipole

Explanation:

We have to remember that water is a polar molecule hence it possesses a dipole moment. Its dipole moment is responsible for the ability of water to dissolve many substances.

On the other hand, Cl2 is a nonpolar molecule bound together by only weak dispersion forces.

Recall that dispersion forces involve transient appearance of a dipole in a molecule.

Water molecules can induce a dipole in Cl2 thereby causing the both molecules to interact and Cl2 dissolves in water.

Answer:

Scientific method.

Explanation:

Scientific method is the way taken to acquire scientific knowledge. It includes experiments, statistical analysis of existing data, and all kinds of observations of the world around us, while theoretical research is based on deriving certain theories about the world from basic principles, in a mathematical or logical way. The scientific method applies to both types of research, and emphasizes that scientific research is objective, that it can be verified by other scientists, and that knowledge is not acquired without context, but in a way that leads to a greater understanding of previous research and the world. we live in. To contribute to this, researchers are expected to clearly record both their findings and the methods they use to arrive at the results.

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.

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I hope it helps you!  

Answer:

2.9 × 10²⁴ atoms Cu; 3.3 × 10²¹ atoms C; 1.44 × 10²⁵ atoms Hg; 9.63 × 10²² atoms Na

Explanation:

To convert moles to atoms, we need a conversion factor: Avogadro's number. There are 6.02 × 10²³ atoms in 1 mole of atoms.

4.9 mol Cu × 6.02 × 10²³ atoms Cu/1 mol Cu = 2.9 × 10²⁴ atoms Cu

5.4 × 10⁻³ mol C × 6.02 × 10²³ atoms C/1 mol C = 3.3 × 10²¹ atoms C

24.0 mol Hg × 6.02 × 10²³ atoms Hg/1 mol Hg = 1.44 × 10²⁵ atoms Hg

0.160 mol Na × 6.02 × 10²³ atoms Na/1 mol Na = 9.63 × 10²² atoms Na

Explanation:

This is a decomposition reaction. Firstly, you will want to write the chemical equation out and balance it.

[tex]2Hg_2O->4Hg+O_2[/tex] (The -> is supposed to be an arrow, sorry!)

We see that there's only 1mol of Oxygen made in the products, we can do some simple math to solve for the amount of grams of Oxygen produced according to the amount of the reactant (Hg2O).

[tex]32.2gHg_2O*\frac{1molHg_2O}{417.18gHg_2O}*\frac{1molO_2}{2molHg_2O}*\frac{32gO_2}{1molO_2}[/tex]

I want to break this down, just in case:

The 417.18gHg2O is the molecular mass of the molecule (so I doubled Hg and added 16 to it to get this number).

As we can see in the chemical equation, 1mol Hg2O produces 2mol O because Oxygen is a diatomic molecule (so there will always be two of it when it's by itself).

And finally, in 1mol O2 there are 32g of O2.

** When you do math like this, always make sure that all of your units cancel out except for the units you're looking for. For example, here we're looking for the grams of Oxygen, so after everything else cancels out, we should only have grams O2.

So, 1.23gO2 should be your answer.

Answer:

Ni

Explanation:

An active metal is a highly reactive metal. Active metals are found high up in the activity series.

Active metals react with other metals that are lower than them in the activity thereby displacing the lower metals from a solution of their salts. This is what may have happened in the other two reactions.

Ni is the most active metal listed in the question since it can react a compounds with Pb(NO3)2(aq) to liberate Pb metal.

a. Br, Cl, F

b. Na, K, Ba

c. He, Ar, Ne

d. Ca, Ba, Mg

Answer:

a. halogen : F ,Cl ,Br l ,At

b an alkali metal: Na,Li, Rb, Cs

c. a noble gas: He, Ne, Kr, Ar

d. an alkaline earth metal: Be,Mg,Ca, Sr

hope it helps

Answer:

MoClBr₂

Explanation:

First we calculate the mass of bromine in the compound:

Then we calculate the number of moles of each element, using their respective molar masses:

Now we divide those numbers of moles by the lowest number among them:

Meaning the empirical formula is MoClBr₂.

Answer:

The formula of Al³⁺ and SO₄²⁻ is aluminum sulfate.

Explanation:

The formula for aluminum sulfate is Al₂(SO₄)₃. If we say in terms of ions. The ions are Al³⁺. It is a positive ion or the cation. Other ion is SO₄²⁻. It is sulfate ion. It is anion.

Aluminum sulphate is used in water purification and as a mordant in dyeing and printing textiles.

Hence, the formula of Al³⁺ and SO₄²⁻ is aluminum sulfate.

Answer:

Hydrogen bonding - London dispersion forces - dipole-dipole interactions

Strongest ----> Weakest

Solution :

a). The separation of 4-methylbenzoic acid from 1,4-dimethoxybenzene will work but it will result in lower recovery.

In the reaction of acid-base to form a sodium 4 - methoxy benzoate, that is soluble in the water, 4-methoxy benzoic acid reacts with the sodium bicarbonate to give sodium 4-methoxybenzoate as well as carbonic acid.

b). The pKa for the 4-methoxybenzoic acid is [tex]4.46[/tex], and that of carbonic acid is [tex]6.37[/tex]

c). The Keq for the reaction is [tex]10(6.37 - 4.46) = 101.91[/tex]

Therefore, the equilibrium lies to the right  and also the reaction favors the products and the separation works.

But the recovery will be low when compared to the extraction with Sodium hydroxide as the strong base will drive the equilibrium further to the right position, thus neutralizing all the acids virtually. And the weak base will partially neutralize the acid.

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: The type of solvent that should be dried with calcium chloride is esters while magnesium sulfate is diethyl ether

Explanation:

Drying agents are mainly hygroscopic substances that has the ability to absorb water on exposure to the atmosphere but not enough to form solutions. They are used in desiccators. Examples of drying agents include:

--> CALCIUM CHLORIDE: This is a compound of calcium that is found in soil water and sea water. It is prepared by the action of dilute hydrochloric acid on calcium trioxocarbonate(IV). The anhydrous salt is used in drying a wide variety of solvent including esters.

--> MAGNESIUM SULFATE: This is a slightly acidic drying agent. It works well in solvents like diethyl ether. It is a fast drying agent because it comes as a fine powder with a large surface area.

molarity = moles of solute / volume of solution

moles of solute = molarity × volume of solution

moles of solute = 0.245 mol/L × 0.1500 L

moles of solute = 0.03675 mol

moles of solute = 0.0368 mol

-----------------------------------------------------------

Step 1: Calculate the molar mass of solute.

molar mass of solute = (40.08 g/mol × 1) + (35.45 g/mol × 2)

molar mass of solute = 110.98 g/mol

Step 2: Calculate the mass of solute.

mass of solute = moles of solute × molar mass of solute

mass of solute = 0.03675 mol × 110.98 g/mol

mass of solute = 4.08 g

Note: The volume of solution must be expressed in liters (L).

Answer:

[tex]\boxed {\sf \bold {0.0368 \ mol \ CaCl_2}}}}[/tex]

[tex]\boxed {\sf \bold {4.08 \ g \ CaCl_2}}}}}[/tex]

Explanation:

Molarity is a measure of concentration in moles per liter.

[tex]molarity= \frac {moles \ of \ solute}{liters \ of \ solution}[/tex]

In this solution, there are 150.0 milliliters of solution and the molarity is 0.245 M CaCl₂ or 0.245 mol CaCl₂ per liter.

First, convert the milliliters to liters. There are 1000 milliliters in 1 liter.

Now, substitute the known values (molarity and liters of solution) into the formula. The moles of solution are unknown, so we can use x.

[tex]0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L}[/tex]

We are solving for x, so we must isolate this variable. It is being divided by 0.150 L. The inverse of divisions is multiplication, so we multiply both sides by 0.150 L.

[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L} * 0.150 L[/tex]

[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L=x[/tex]

The units of liters cancel.

[tex]0.150 *0.245 \ mol \ CaCl_2 =x[/tex]

[tex]0.03675 \ mol \ CaCl_2[/tex]

The original measurements have 3 significant figures, so our answer must have the same.

We should round to the ten thousandths place. The 5 to the right of this place tells us to round the 7 up to an 8.

[tex]\bold {0.0368 \ mol \ CaCl_2}[/tex]

We can convert mass to moles using the molar mass. These values are found on the Periodic Table. They are the same as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units.

The solute is calcium chloride: CaCl₂. Look up the molar masses of the individual elements.

Notice that chlorine has a subscript of 2. We must multiply the molar mass by 2.

Add calcium's molar mass.

Use the molar mass as a ratio.

[tex]\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]

Multiply the moles of calcium chloride we calculated above.

[tex]0.0368 \ mol \ CaCl_2 *\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]

The units of moles of calcium chloride cancel.

[tex]0.0368 *\frac {110.98 \ g\ CaCL_2}{ 1 }[/tex]

[tex]4.084064 \ g\ CaCl_2[/tex]

Round to 3 significant figures again. For this number, it is the hundredths place. The 4 in the thousandths place tells us to leave the 8.

[tex]\bold {4.08 \ g \ CaCl_2}[/tex]

the answer to the question is B.UV-B

Explanation:

The electron density number as well as the sublevel letter are used to describe valence electrons in an atom. The third total energy and subbasement p, for example, is denoted by 3p. The electron configuration of oxygen, for example, is 1s^2 2s^2 2p^4, which means the first two electrons will couple up in the 1s orbital, while the following two protons will pair up in the 2s orbital.

The sample atom is Carbon with electron configuration; 1s² 2s² 2p².

The principal energy level of an electron refers to the shellp in which the electron is located relative to the atom's nucleus. In this case only 2 energy levels exist in a carbon atom; which are energy level 1 and 2

The sublevels exist within a principal energy and the electron configuration of an atom is described with consideration of energy sublevels. The sublevels in a carbon atom are;

The orbitals in this configuration are: 1s 2s 2px 2py 2pz in which case; each orbital can accommodate 2 electrons each.

Ultimately, the location of an element on the periodic table with respect to group and period are used to determine the valency and no. of energy levels in the atom of that Element.

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The question is incomplete, the complete question is;

. A chemist determined by measurements that 0.015 moles of hydrogen chloride participate in a chemical reaction. Calculate the mass of hydrogen chloride that participates Round your answer to 2 significant digits. x S. ?

Answer:

0.54 g

Explanation:

Recall that;

Number of moles = mass/molar mass

Molar mass of HCl =36.5 g/mol

Mass= number of moles × molar mass

Mass= 0.015 moles × 36 g/mol

Mass= 0.54 g

Answer:

Calcium

Explanation:

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

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

1) To show that heat flows spontaneously from high temperature to low temperature

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

2) Entropy change for Decomposition of mercuric oxide

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg  -  S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Answer:

e. HCOOH and NaCHOO

Explanation:

For a buffer solution, both an acid and its conjugate base are required.

With the information above in mind, we can discard options a) and c), as those combinations are not of an acid and its conjugate base.

Now it is a matter of comparing the pKa (found in literature tables) of the acids of the remaining three acids:

The acid with the pKa closest to the desired pH is HCOOH, so the correct answer is e. HCOOH and NaCHOO