Determine the mass in grams of 7.68 x 10^21 atoms of carbon. (The
mass of one mole of carbon is 12.01 g.)

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:

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

At the equilibrium, we have:  

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

0.030 M - x                      x               x

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

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:

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

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).            

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

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

\( n_{b} = n_{i} - n_{HCl} \)     (6)

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

\( n_{a} = n_{HCl} \)   (7)

\( n_{a} = 0.025 mol/L*0.010 L = 2.5 \cdot 10^{-4} moles \)

The concentrations are given by:

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

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

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

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

x = 4.54x10⁻⁵ = [OH⁻]

Then, the pH is:

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

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):

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

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

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

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

x = 1.40x10⁻⁵ = [OH⁻]

Then, the pH is:  

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

So, the pH is 9.15.

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

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

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

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

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

x = 5.013x10⁻⁷ = [OH⁻]      

Then, the pH is:  

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

So, the pH is 7.70.

\( n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0 \)    

\( n_{a} = 0.025 mol/L*0.036 L = 9.0 \cdot 10^{-4} moles \)

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

\( Ka = \frac{x^{2}}{Ca - x} \)  

\( Ka(Ca - x) - x^{2} = 0 \)   (10)          

We can find the acid constant as follows:

\( Kw = Ka*Kb \)

Where Kw is the constant of water = 10⁻¹⁴

\( Ka = \frac{1\cdot 10^{-14}}{1.76 \cdot 10^{-5}} = 5.68 \cdot 10^{-10} \)  

The concentration of NH₄⁺ is:

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

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

x = 2.78x10⁻⁶ = [H₃O⁺]

Then, the pH is:  

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

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).

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

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

Therefore, the pH is 3.43.

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Answer:

The coloring and possibly testing it

Potassium chloride is a medication that is used to treat or prevent low potassium levels in the body.Potassium can be removed from the body by certain diseases, illnesses, and drugs.

The addition of NaCl to a KCl mixture reduces the unpleasant side tastes associated with KCl, such as bitter, chemical, and metallic.

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Answer:

What is the question? I don't see it.

Answer:

0 degrees Celsius

Explanation:

Answer:

Water changes from a liquid to a solid at 0° Celsius

Carbon atoms may thus form bonds to as many as four other atoms. For example, in methane (CH 4​start subscript, 4, end subscript), carbon forms covalent bonds with four hydrogen atoms

__________________________________________________________

Earthquakes and volcanoes are often found along or near boundaries of continents and oceans.

The Ring of Fire is an area around the Pacific Ocean where a large number of earthquakes and volcanic eruptions occur due to the movement and interaction of tectonic plates. The Ring of Fire is also known as the Circum-Pacific Belt.

1. Seismologist - Scientists who study earthquakes

2. Faults - Are breaks or cracks in the rocks that make up Earth's crust.

3. Most earthquakes and volcanoes occur in Bands

4. Earthquakes and volcanoes are often Along or near boundaries of continents and oceans.

5. A fault is a break or a crack in the rocks of Earth's crust where an earthquake might take place.

7. An earthquake might cause a landslide when the ground shakes.

8. Landforms that are usually found along the boundary between plates are Mountains and Ocean trenches.

9. An earthquake is usually caused by movement along a fault.

10. According to the map, most earthquakes occur around the edges of the ocean and continental plate boundaries.

11. The Ring of Fire is an area around the Pacific Ocean where a large number of earthquakes and volcanic eruptions occur

There are many active volcanoes in the area of the Ring of Fire.

There are many seismic faults in the area of the Ring of Fire.

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Answer:

A. It will shift to the left

Explanation:

In the equilibrium:

C(s) ⇄ CO2(g)2CO(g)

The system will shift to the right if any change stimulate the production of gas -LeChatelier's principle-; in the same way, if a change doesn't favors the production of gas the system will shift to the left producing less gas.

The changes that increasing the pressure of the system, doesn't favors the gas production doing the system shift to the left.

A gas that is heated expands itsellf doing the pressure increases.

In the same way, if you compress the gas, the gas increases its pressure.

Thus, both changes increase pressure of the gas doing the system shift to the left.

Answer:

a negative ion, and an isotope.

Explanation:

The perfect atom consists of an equal balance in all 3. If the neutrons are not even with the protons, it is an isotope since it is like another version of the so called (but not actually) 'perfect' atom. If there is more electrons, the charge is negative, making it an ion.

The number of moles of calcium chloride in 250 ml of a 9.0M of solution is 2.25 moles.

The amount of moles in a substance can be calculated by multiplying the molarity of the solution by the volume as follows;

no of moles = molarity × volume

According to this question, 250mL of a calcium chloride solution has a molarity of 9M. The number of moles can be calculated as follows:

no of moles = 9.0M × 0.250L

no of moles = 2.25 moles

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that the number of collisions depends on the energy, of the number of reactant particles, not their orientation.

Energy is referred to by scientists as the capacity for work.People have figured out how to transform energy from one type to the other and use it to accomplish tasks, making modern civilization possible.

boosts employee confidence and enables new hires to settle in more quickly; results in a more efficient and successful staff; increases employee retention; andencourages dialogue between the new employee and the supervisor.

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Answer:

C3H7O + O2 → CO2 + H2O

Explanation:

a) The full symbol equation for the oxidation reaction of an isomer of C3H7O can be represented as:

C3H7O + O2 → CO2 + H2O

b) The proper reagent for this oxidation reaction is O2 (oxygen gas). The catalyst required for this reaction depends on the specific conditions. Common catalysts used for oxidation reactions include transition metals such as platinum (Pt), palladium (Pd), or copper (Cu).

c) There is no need to remove the product (CO2 and H2O) from the reaction vessel because they are typically in the gas or liquid phase and do not significantly interfere with the reaction. The product gases can be easily vented out of the vessel, while the liquid water can be left in the reaction mixture. Additionally, the product CO2 is a stable and inert gas, which does not pose any hazards in most cases. Therefore, it is often not necessary to remove the products after the reaction is complete.

The calculated formula masses to 133.5 amu, we find that the substance with a formula mass closest to 133.5 amu is (d) AlCl3. Therefore, the answer is option D.

To identify the substance with a formula mass of 133.5 amu, we need to calculate the formula mass of each given substance and compare it to 133.5 amu.

(a) MgCl2:

The formula mass of MgCl2 can be calculated by adding the atomic masses of magnesium (Mg) and chlorine (Cl).

Mg: atomic mass = 24.31 amu

Cl: atomic mass = 35.45 amu

Formula mass of MgCl2 = (24.31 amu) + 2(35.45 amu) = 95.21 amu

(b) SCl:

The formula mass of SCl can be calculated by adding the atomic masses of sulfur (S) and chlorine (Cl).

S: atomic mass = 32.07 amu

Cl: atomic mass = 35.45 amu

Formula mass of SCl = 32.07 amu + 35.45 amu = 67.52 amu

(c) BCl:

The formula mass of BCl can be calculated by adding the atomic mass of boron (B) and chlorine (Cl).

B: atomic mass = 10.81 amu

Cl: atomic mass = 35.45 amu

Formula mass of BCl = 10.81 amu + 35.45 amu = 46.26 amu

(d) AlCl3:

The formula mass of AlCl3 can be calculated by adding the atomic mass of aluminum (Al) and 3 times the atomic mass of chlorine (Cl).

Al: atomic mass = 26.98 amu

Cl: atomic mass = 35.45 amu

Formula mass of AlCl3 = 26.98 amu + 3(35.45 amu) = 133.78 amu. Option D

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The mass of chlorine gas product formed is 61.4 g.

What is the mass ?

First, we need to determine which reactant is limiting and which is in excess. To do this, we can use the given mass of HCl and convert it to moles:

63.2 g HCl × (1 mol HCl/36.46 g HCl) = 1.73 mol HCl

Next, we can use stoichiometry to determine how many moles of \(Cl_{2}\) will be produced from 1.73 mol of HCl:

1.73 mol HCl × (2 mol \(Cl_{2}\)/4 mol HCl) = 0.865 mol \(Cl_{2}\)

Finally, we can convert the moles of \(Cl_{2}\) to grams using its molar mass:

0.865 mol \(Cl_{2}\) × (70.91 g \(Cl_{2}\)/1 mol \(Cl_{2}\)) = 61.4 g \(Cl_{2}\)

Therefore, the mass of chlorine gas product formed is 61.4 g.

What is stoichiometry ?

Stoichiometry is a branch of chemistry that deals with the calculation of the quantities of reactants and products involved in a chemical reaction. It involves using balanced chemical equations to calculate the amounts of reactants required to produce a certain amount of product, or the amount of product that can be obtained from a given amount of reactant. Stoichiometry is an important tool for predicting the outcomes of chemical reactions and for designing chemical processes in industry.

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Answer:

When a wine bottle is uncorked, its contents are exposed to air, causing them to oxidize. ... Bacteria naturally present in grapes can turn either the sugars in grape juice or the alcohol in wine into acetic acid, giving it a vinegar taste (and eventually producing a wine vinegar).

Explanation:

Answer:

132 54Xe

Explanation:

From the question given above, the following were obtained:

Proton = 54

Neutron = 78

Atomic number =?

Atomic symbol =.?

Next, we shall determine the atomic number of the isotope. This can be obtained as follow:

Atomic number of an element is simply defined as the number of protons in the atom of the element.

Thus,

Atomic number = proton number

Proton = 54

Atomic number = proton number = 54

Atomic number = 54

Therefore, the atomic number of the isotope is 54

Next, we shall determine the mass number of the isotope. This can be obtained as follow:

Proton = 54

Neutron = 78

Mass number =.?

Mass number = Proton + Neutron

Mass number = 54 + 78

Mass number = 132

Therefore, the mass number of the isotope is 132.

From the calculations made above, the isotope have:

Atomic number = 54

Mass number = 132

Comparing the atomic number with those in the periodic table, the isotope is Xenon (Xe).

Thus, the atomic symbol for the isotope is 132 54Xe.

Answer:

Transmutation is the conversion of an atom of one element into an atom of a different element through nuclear changes.

An alpha particle is a particle composed of 2 protons and 3 neutrons. It results from transmutation because of the change in protons in a large nucleus.

A chemical equation is balanced accord to the number of atoms of each element before and after the change. This is also shows the Law of Conservation of Matter.

A nuclear equation is balanced according to mass number and charge (atomic number). These equations can’t be balanced like chemical equations because the identities of the atoms can change.  On the left of equation, the top number is the atomic mass and the bottom number is the atomic number. Take note that if you add the atomic mass (top number), you’ll get the the atomic mass of of the atom pre-transmutation. Same applies with atomic number. This shows a balanced nuclear equation.

The atomic number is the number of protons in a nucleus. The mass number is the sum of the protons and neutrons in a nucleus. Electrons have very small masses, so they are not accounted in atomic mass.

Explanation:

because there is 2 hydrogen atoms in the reactions side of the equation (you can tell because the H has a 2 in the subscript) you have to have 2 hydrogen atoms in the reactants side to help balance out the equation. and since the copper and oxygen atoms are already balanced there is no coefficient needed.

Answer:

The kinetic-molecular theory applies on to ideal gases.. Many gases behave nearly ideally if pressure is not very high and temperature is not very slow. ... Because the attractive forces between gas particles are insignificant, gas particles glide easily past one another.

F

Explanation:

Answer:

See below

Explanation:

2C2H6 + 7O2 = 4CO2 + 6H2O

2C2H2 + 5O2 = 4CO2 + 2H2O

A lot of this is trial and error.  I start from left to right.  For the first reaction, you had

C2H6 + O2  =  CO2  +  H2O

You have 2 carbons on the left so you can try adding a 2 in front of the carbons on the right and keep going from there.  Now you have

C2H6  +  O2  =  2CO2  + H2O

You now have 2 C on each side.  C2 is the same as 2C.

The next step is to look at the hydrogens on the left.  You have 6 so you need 6 on the right.  You can add a 3 in front of the H on the right.  The H already has a subscript of a 2 so having the 3 in front will now make it 6H.  You now have this

C2H6  +  O2  =  2CO2  +  3H2O

So your C are equal (2 on each side) and your H is equal (6 on each side).  All that is left is the O.  So far you have 2 on the left but you have 7 on the right (4 from the 2CO2 and 3 more from 3H2O).  Since you have an even number on one side and an odd number on the other, there is no way to make this work.  So it's time to start over.

My next step was to put a 2 in front of C2H6 and keep going as we did above.  Start with this

2C2H6   + O2  = CO2  + H2O

You have 4 C on the left.  So you need to add a 4 in front of CO2 to get 4 C on the right.  Now you have this

2C2H6  +  O2  =  4CO2  +  H2O

Now on the left you have 12 H (2 in front times the 6 subscript on the H).  You need to have 12 on the right.  We can get that by putting a 6 in front of the H2O ( again the 6 in front times the 2 subscript of the H is 12 total).  You have this now

2C2H6  +  O2  = 4CO2  +  6H2O

You now have 4 C on both sides, 12 H on both sides.  All that is left now is the O.

On the right side, you have a total of 14 ( 8 from the 4CO2 and 6 from 6H2O).  So you'd need to add 7 in front of the O2 to make it 14.

You're done.  4 C on both sides, 12 H on both sides and 14 O on both sides.

The second equation I just did the same process.

The properties of a natural resource that make it useful for humans as a material or energy source is the ability to convert mass into energy and vice versa.

The expression natural resources make reference to all types of matter and energy extracted from nature that can be used to produce goods and services.

Some examples of natural resources include for example irreversible resources such as fossil fuels (i.e., oil, or coal, gas, minerals such as metals, rocks, etc) as well as those based on the use of reversible energy such as eolic air energy, solar radiation or sunlight, soil and hydric resources or water.

Therefore, with this data, we can see that natural resources can be defined as any material and or energy obtained from nature that may be irreversible or reversibly used to produce goods and services.

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The concentration of ammonia in the solution is calculated as 4.48*10⁻⁴ M.

In chemistry, the abundance of constituent divided by total volume of a mixture is known as concentration.

C₃H₇COOH + NH⁴⁺ ⇌ C₃H₇COOH⁻ + NH⁴⁺

The propanoic acid (C3H7COOH) is a weak acid with known Ka value (1.3*10^-5), so we use the Henderson-Hasselbalch equation to calculate concentration of the acetate ion (C3H7COO-) in the solution:

pKa = -log(Ka) = -log(1.3*10⁻⁵) = 4.89

pH = pKa + log([C₃H₇COO-]/[C₃H₇COOH])

[C₃H₇COOH] = 0.18 M

[C₃H₇COOH-] = Ka*[C₃H₇COOH]/[H+] = (1.310⁻⁵)(0.18)/10^-pH

Kb = [ NH⁴⁺][OH⁻]/[NH₃]

Kb = [NH⁴⁺][OH⁻]/[NH₃] ≈ [NH⁴⁺][OH⁻]/0

[OH⁻] = Kb*[NH₃]/[NH⁴⁺] = (1.7610⁻⁵)[NH3]/0.18

[H⁺][OH-] = 1.0*10⁻¹⁴

[H⁺] = 10^-pH

NH₃] = [OH⁻] = 1.010^-14/[H⁺] - 2.4910⁻⁷

[NH₃] = 4.48*10⁻⁴ M

Therefore, the concentration of ammonia in the solution is 4.48*10⁻⁴ M.

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answer:when an herbivore eats meat and a carnivore eats the herbivore energy from the eatin meat is passed indirectly to the carnivore.

Answer:

242 cm³.

Explanation:

The following data were obtained from the question:

Initial volume (V₁) = 484 cm³

Initial temperature (T₁) = 570 K

Final temperature (T₂) = 285 K

Pressure = constant

Final volume (V₂) =?

The new volume (i.e final volume) of the balloon can be obtained by using the Charles' law equation as illustrated below:

V₁/T₁ = V₂/T₂

484 / 570 = V₂ / 285

Cross

484 × 285 = 570 × V₂

137940 = 570 × V₂

Divide both side by 570

V₂ = 137940 / 570

V₂ = 242 cm³

Therefore, the new volume of the balloon is 242 cm³

Explanation:

MnO4−+ SO32− → Mn2++ SO42−

Splitting into half equations;

MnO4−  → Mn2+

SO32− → SO42−

Balancing the electrons;

2 MnO4−  + 10 e- → 2Mn2+

5SO32− → 5SO42− + 10 e-

In an acidic medium, it becomes;

2 MnO4−  +  8 H+ → 2 Mn2+ + 4 H2O

5 SO32−  + H2O → 5 SO42−  +  2 H+

Final equation is;

2 MnO4- + 5 SO32- + 6 H+ → 2 (Mn)2+ + 5 SO42- + 3 H2O

Coefficient of H+ = 6

Coefficient of H2O = 3

Coefficient of MnO4- = 2

Coefficient of SO32- = 5

Coefficient of (Mn)2+- = 2

Coefficient of SO42- = 5

Answer:

\(5SO_3^{2-}+2MnO_4^{-}+6H^+ \rightarrow 5SO_4^{2-}+ 2Mn^{2+}+3H_2O\)

Explanation:

Hello,

In this case, given the reaction:

\(MnO_4^{-}+ SO_3^{2-} \rightarrow Mn^{2+}+ SO_4^{2-}\)

We first identify the oxidation state of both manganese and sulfur at each side:

\(Mn^{7+}O_4^{-}+ S^{4+}O_3^{2-} \rightarrow Mn^{2+}+ S^{6+}O_4^{2-}\)

So we have the oxidation and reduction half-reactions below, including the addition of water and hydronium as it is in acidic media:

\(S^{4+}O_3^{2-}+H_2O \rightarrow S^{6+}O_4^{2-}+2H^++2e^-\)

\(Mn^{7+}O_4^{-}+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\)

Next, we exchange the transferred electrons:

\(5*(S^{4+}O_3^{2-}+H_2O \rightarrow S^{6+}O_4^{2-}+2H^++2e^-)\\2*(Mn^{7+}O_4^{-}+8H^++5e^- \rightarrow Mn^{2+}+4H_2O)\\\\5S^{4+}O_3^{2-}+5H_2O \rightarrow 5S^{6+}O_4^{2-}+10H^++10e^-\\2Mn^{7+}O_4^{-}+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\)

Then we add the resulting half-reactions and simplify the transferred electrons:

\(5S^{4+}O_3^{2-}+5H_2O+2Mn^{7+}O_4^{-}+16H^+ \rightarrow 5S^{6+}O_4^{2-}+10H^++ 2Mn^{2+}+8H_2O\)

We rearrange the terms in order to simplify water and hydronium molecules:

\(5S^{4+}O_3^{2-}+2Mn^{7+}O_4^{-}+16H^+-10H^+ \rightarrow 5S^{6+}O_4^{2-}+ 2Mn^{2+}+8H_2O-5H_2O\\\\5S^{4+}O_3^{2-}+2Mn^{7+}O_4^{-}+6H^+ \rightarrow 5S^{6+}O_4^{2-}+ 2Mn^{2+}+3H_2O\)

Finally we write the balanced reaction in acidic media:

\(5SO_3^{2-}+2MnO_4^{-}+6H^+ \rightarrow 5SO_4^{2-}+ 2Mn^{2+}+3H_2O\)

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When you increase the amount of carbon dioxide in the atmosphere, you also increase the amount of carbonic acid in the ocean.

Carbon dioxide is absorbed from the atmosphere into the ocean through a process called "oceanic uptake," which is facilitated by the exchange of gases at the air-sea interface.

As more carbon dioxide is emitted into the atmosphere, the concentration of carbon dioxide in the ocean increases, leading to a phenomenon called "ocean acidification".

Ocean acidification can have a number of negative impacts on marine organisms, including reduced growth rates and weakened shells or skeletons.

Thus, we can conclude this increases the amount of carbonic acid in ocean.

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Explanation:

A nucleus consists o both electrons and proton. Therefore, an unstable nucleus undergoes radiation. And this causes a formation of a new atom. This is known as radioactive decay. Therefore, an unstable nucleus emits energy as it decays.

Answer:

The correct answer is B.herefore, an unstable nucleusherefore, an unstable nucleusR

Answer:

Protons

Explanation:

Metals, Nonmetals, and Metalloids group the elements on the periodic table. Protons is not one of these but rather a subatomic particle that all the elements have.

Answer:

protons

Explanation:

Answer:  the law that the product of the pressure and the volume of one gram molecule of an ideal gas is equal to the product of the absolute temperature of the gas and the universal gas constant.An ideal gas is a gas that conforms, in physical behaviour, to a particular, idealized relation between pressure, volume, and temperature called the ideal gas law

The growth of the new plant depicts the asexual reproduction characteristic. The characteristic that describes the growth of the new plant in Stephan's mother cutting a twig from a rose bush and planting it in the soil is asexual reproduction.

Asexual reproduction is the mode of reproduction by which organisms generate offspring that are identical to the parent's without the fusion of gametes. Asexual reproduction is a type of reproduction in which the offspring is produced from a single parent.

The offspring created are clones of the parent plant, meaning they are identical to the parent.The new plant in Stephan’s mother cutting a twig from a rose bush and planting it in the soil depicts the process of asexual reproduction, which is the ability of a plant to reproduce without seeds. In asexual reproduction, plants can reproduce vegetatively by cloning themselves using their roots, bulbs, or stems.

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