KOH. Option B is correct
Explanations:A chemical compound are made up of more than one element combined together. According to the question, we need to determine the formula that represents a compound.
The compound there is KOH since it contains three elements (Potassium, Oxygen and Hydrogen)
3.0 x 10 ^23 atoms of calcium to moles
Answer:
[tex]0.5\text{ mole}[/tex]Explanation:
Here, we want to get the number of moles
Mathematically:
[tex]1\text{ mole has 6.02 }\times\text{ 10}^{23}\text{ atoms}[/tex]The number of moles in the question would be:
[tex]\frac{3.0\times10^{23}\text{ atoms}}{1}\times\frac{1\text{ mole}}{6.02\times10^{23}\text{ atoms}}\text{ = 0.5 mole}[/tex]What is the difference between concentrated and dilute with strong and weak?
The strength is a measure of the degree of ionization of a substance in an aqueous solution.
Some substances are strong and they dissociate completely (this is the case of hydrochloric acid, HCl, in water), some are weak and they do not dissociate completely (like acetic acid, CH3COOH).
The concentration is a measure of the amount of the substance in a solution and it is usually explessed in moles or percentage. It is a ratio of the substance to the solvent. A concentrated solution is a solution with a high percentage of the substance (solute) in the solution, and a dilute one is a solution with a low percentage of solute.
What is the density of hydrogen sulfide (H2S) at 0.2 atm and 311 K?Answer in units of g/L
Answer
Density = 0.267 g/L
Explanation
Given:
Pressure of H2S = 0.2 atm
Temperature = 311 K
We know:
The molar mass of H2S = 34,1 g/mol
R constant = 0.08206 L.atm/K.mol
Solution:
From the ideal gas law:
PV = nRT
We know that:
density = m/V
n = m/M
Therefore we can use the following equation to solve for density of H2S
[tex]\begin{gathered} density\text{ = }\frac{PM}{RT} \\ density\text{ = }\frac{(0.2\text{ atm x 34,1 g/mol\rparen}}{(0.08206\text{ }L.atm/K.mol\text{ x 311 K\rparen}} \\ \\ density\text{ = 0.267 g/L} \end{gathered}[/tex]Can you help show me the conversion works for this problem in my online homework please?
There are 2.04*10^-4 moles of hydrogen.
To start the calculation it is necessary to use the molar mass of glycine:
- Glycine molar mass: 75.08 g/mol.
From the formula of glycine, we know that in 75.08g of glycine, there are 5 moles of hydrogen. So, in 3.06*10^-3g of glycine will be another amount of hydrogen, and we can calculate it with a mthematical Rule of Three:
[tex]\begin{gathered} 75.08\text{gGlycine}-5\text{molHydrogen} \\ 3.06\cdot10^{-3}\text{gGlycine}-x=\frac{3.06\cdot10^{-3}\text{gGlycine}\cdot5\text{molHydrogen}}{75.08\text{gGlycine}} \\ x=2.04\cdot10^{-4}\text{molHydrogen} \end{gathered}[/tex]So, there are 2.04*10^-4 moles of hydrogen in 3.06*10^-3g of glycine.
What is the limiting reactant if 43.4 g of NH3 react with 30 g of NO? The balanced equation is 4NH3 + 6NO --> 5N2 + 6H2O
Answer
NO is the limiting reactant
Explanation
Given that:
The mass of NH3 that reacted = 43.4 g
The mass of NO that reacted = 30 g
The equation for the reaction is: 4NH3 + 6NO --> 5N2 + 6H2O
What to find:
The limiting reactant.
Step-by-step solution:
The first step is to convert the given mass of the reactants to moles.
Using the mole formula and the molar masses of (NH3 = 17.0 g/mol and NO = 30.0 g/mol)
[tex]\begin{gathered} Moles=\frac{Mass}{Molar\text{ }mass} \\ \\ Moles\text{ }of\text{ }NH_3=\frac{43.4g}{17.031g\text{/}mol}=2.55\text{ }mol \\ \\ Moles\text{ }of\text{ }NO=\frac{30g}{30.0g\text{/}mol}=1.0\text{ }mol \end{gathered}[/tex]The final step is to determine the limiting reactant by comparing the mole ratio from the given equation with the mole ratio in the step above.
From the equation, 4 moles of NH3 reacted with 6 moles of NO
So 2.55 moles of NH3 is expected to react with (2.55 x 6)/4 = 3.825 moles of NO
1.0 mol NO is less than 3.825 mol NO, hence, NO is the limiting reactant because it will be the reactant to be completely used up first.
If 1495 J of heat is needed to raise the temperature of a 315 g sample of a metal from 55.0°C to 66.0°C, what is the specific heat capacity of the metal?
The specific heat capacity of the metal that needs 1495 J of heat to raise the temperature is 0.43J/g°C.
How to calculate specific heat capacity?Specific heat capacity is the heat capacity per unit mass of a substance. It can be calculated by using the following formula:
Q = mc∆T
Where;
Q = quantity of heat absorbed or releasedm = mass of substancec = specific heat capacity∆T = change in temperatureAccording to this question, 1495 J of heat is needed to raise the temperature of a 315 g sample of a metal from 55.0°C to 66.0°C. The specific heat capacity can be calculated as follows;
1495 = 315 × c × {66 - 55}
1495 = 3465c
c = 1495/3465
c = 0.43J/g°C
Therefore, 0.43J/g°C is the specific heat capacity of the metal that requires a heat of 1495J.
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What wavelength (in nm) of light is emitted when an electron transitions from the n
= 8 to the n = 2 state?
The wavelength in nm of light emitted when an electron transitions from n = 8 to n = 2 state is 38900nm.
It is given that the electron transition took place from n = 8 to n = 2. To find the wavelength, the following formula is to be used:
1/λ = R(1/n²₁ - 1/n²₂)
Here, n₁ = 2 n₂ = 8 and R = 109677. Therefore,
1/λ = 109677(1/2² - 1/8²)
1/λ = 109677(1/4 - 1/64)
1/λ = 109677((16-1)/64)
1/λ = 109677((15)/64)
1/λ = 0.234375 x 109677
1/λ = 25,705.5469
λ = 1/25,705.5469
λ = 3.89 x 10⁻⁵cm
λ = 38900 x 10⁻⁹m
λ = 38900nm
Therefore, the wavelength in nm from n=8 to n=2 transition is 38900nm
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Which type of radiation would you consider ionizing radiation?MicrowavesVisible lightSound wavesGamma rays
Ionization radiation is a type of energy with such a high value that it is capable of removing electrons from atoms and converting them into ions. This type of energy is extremely dangerous. The types of energy that fall into this category are X-rays and gamma rays.
So, the answer will be Gamma rays
Match each colored row or group on the table to the family of elements it represents.Green (2nd column)Red (1st column)Purple (two rows below the rest of table)Blue (3rd to 12th columns)?Alkali metals?Alkaline earth metals?Transition metals?Lanthanides and actinides
Green column - Alkaline earth metals
Red column - Alkali metals
Purple column - Lanthanides and actinides
Blue column - Transition metals
Explanations:
The red block are made up of group 1 elements and these elements are known as Alkali metals.
The green column are the group 2 elements in the periodic table and are known as the Alkali earth metals.
For the blue columns, they are element from group 3 to 10 and they are known as the transition metals.
For the purple column, they are Lanthanides and actinides
Answer the following questions assuming you have a 1.3M (molarity) solution ofHNO3 (5 points each, 15 points total)A. How many moles of HNO3 would you have in 0.6 liters of the above solution?B. How many grams of HNO3 would be present in 0.6 liters of the above solution?C. If you began with 0.6 liters of solution, and diluted the solution to a concentration of 0.45M, whatwould the final volume of the solution be?
A) 0.78 moles
B) 49.15grams
C) 1.73L
Explanations:The formula for calculating the molarity of a solution is expressed as;
[tex]\begin{gathered} molarity=\frac{moles}{volume} \\ moles=molarity\times volume \end{gathered}[/tex]Given the following parameters
molarity of HNO3 = 1.3M
volume of HNO3 = 0.6L
A) moles of HNO3 = 1.3 * 0.6
moles of HNO3 = 0.78moles
B) Mass of HNO3 = moles * molar mass
Mass of HNO3 = 0.78 * 63.01
Mass of HNO3 = 49.15grams
There are 49.15grams of HNO3 present in 0.6L of the solution
C) According to dilution formula
[tex]\begin{gathered} C_1V_1=C_2V_2 \\ V_2=\frac{C_1V_1}{C_2} \\ V_2=\frac{1.3\times0.6}{0.45} \\ V_2=\frac{0.78}{0.45} \\ V_2=1.73L \end{gathered}[/tex]Therefore the final volume of the solution will be 1.73L
PLEASE HELP ME!!!
A)Nuclear fusion releases protons and neutrons, so the total number of protons
and neutrons in a star changes throughout its life.
B)Nuclear fusion conserves protons and neutrons, so the total number of protons
and neutrons in a star changes throughout its life.
C)Nuclear fusion releases protons and neutrons, so the total number of protons
and neutrons in a star remains the same throughout its life.
D)Nuclear fusion conserves protons and neutrons, so the total number of protons
and neutrons in a star remains the same throughout its life.
A)Nuclear fusion releases protons and neutrons, so the total number of protons and neutrons in a star changes throughout its life is the true statement.
What are some uses for nuclear fusion?A suggested method of producing energy would use heat from nuclear fusion processes to produce electricity. A heavier atomic nucleus is created by the fusion of two lighter ones, which also produces energy. Fusion reactors are devices created to use this energy.
Nuclear fusion is less risky than nuclear fission because the fuel rods produced by nuclear fission include dangerous radioactive waste that may be used in weapons and must be maintained carefully for thousands of years.
Powering the Sun and other stars are nuclear fusion processes. Two light nuclei combine to produce one heavy nucleus during a fusion process. The resultant single nucleus's total mass is less than the mass, which causes the process to release energy.
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4 Al(s) + 302(g)->2Al2O3(s)When 42.39 g of Al and 85.16 g of O2 were reacted, 6.67 grams of aluminum oxidewere obtained. What is the percent yield? (Hint: You need to determine which one isthe limiting reactant and then the theoretical yield).
The reaction presented to us is balanced since we have 4 aluminum atoms and 6 oxygen atoms on both sides of the reaction.
Now, we have a given mass of aluminum and oxygen, we must first determine the moles of each using their molar mass.
Moles of Al
[tex]\begin{gathered} \text{Mol of Al}=GivengAl\times\frac{1molAl}{MolarMass,gAl} \\ \text{Mol of Al}=42.39gAl\times\frac{1molAl}{26.98gAl}=1.57molAl \end{gathered}[/tex]Moles of O2
[tex]\begin{gathered} MolO_2=GivengO_2\times\frac{1molO_2}{MolarMass,gO_2} \\ MolO_2=85.16gO_2\times\frac{1molO_2}{31.998gO_2}=2.66molO_2 \end{gathered}[/tex]Now that we have the number of moles, we will calculate what the limiting reagent is, that is, the reagent that limits the reaction by its number of moles.
To find the limiting reactant we must compare the amount of product obtained with the given amount of reactant separately. The reactant that produces the least amount of product is the limiting reactant.
Using Al as limiting reactant
[tex]\begin{gathered} \text{MolAl}_2O_3=MolAl\times\frac{2molAl_2O_3}{4molAl} \\ \text{MolAl}_2O_3=1.57molAl\times\frac{2molAl_2O_3}{4molAl}=0.785molAl_2O_3 \end{gathered}[/tex]Using O2 as a limiting reactant
[tex]\begin{gathered} \text{MolAl}_2O_3=MolO_2\times\frac{2molAl_2O_3}{3molO_2} \\ \text{MolAl}_2O_3=2.66molO_2\times\frac{2molAl_2O_3}{3molO_2}=1.77molAl_2O_3 \end{gathered}[/tex]Aluminum is the reagent that produces the least amount of aluminum oxide, so the limiting reagent will be Al. And it will produce 0.785 moles of Al2O3. In grams this will be:
[tex]\text{gAl}_2O_3=0.75molAl_2O_3\times\frac{101.95gAl_2O_3}{1molAl_2O_3}=80.03gAl_2O_3[/tex]The percent yield will be:
[tex]\begin{gathered} \text{Percent yield=}\frac{\text{Actual yield}}{Theoretical\text{ yield}}\times100\% \\ \text{Percent yield=}\frac{\text{6}.67gA_{}l_2O_3}{80.03ggA_{}l_2O_3}\times100\%=8.33\% \end{gathered}[/tex]The percent yield will be 8.33%
Please help me with this question.
Answer:1=B
2=A
Explanation:
how do i calculate the molar mass of 5 and 65. K2Cr2O76. C12H22O11
To calculate the molar mass of a compound, we need to find at the periodic table the atomic mass of each element that belongs to the compound and then we multiply for the quantity of atoms of it.
So, for:
5. K2Cr2O7:
K = 39 g/mol
Cr = 52 g/mol
O = 16 g/mol
(2x39) + (2x52) + (7x16) = 78 + 104 + 112 = 294 g/mol
Molar mass of K2Cr2O7 = 294 g/mol
6. C12H22O11
C = 12 g/mol
H = 1 g/mol
O = 16 g/mol
(12x12) + (22x1) + (11x16) = 144 + 22 + 176 = 342 g/mol
Molar mass of C12H22O11 = 342 g/mol
What is one way an atom can become more stable?
Unstable atoms usually are very large atoms with a big nucleus, this configuration sometimes gives the atoms this unstable characteristic, which is why they decay so fast (some will decay in a matter of milliseconds), and radioactive decay is a way that atoms can find to become more stable, in this type of process, particles (alpha and beta decay) and energy (gamma decay), can be emitted from an atom in order to reach stability, therefore the best option is the 3rd option
How many grams of CO are produced when 33.0 g of C reacts? Fe2O3(s)+3C(s)→2Fe(s)+3CO(g)
Answer
77.0 grams of CO are produced when 33.0 g of C reacts.
Explanation
Given:
Mass of C = 33.0 g
Equation: Fe₂O₃(s) + 3C(s) → 2Fe(s) + 3CO(g)
What to find;
The grams of CO that are produced when 33.0 g of C reacts.
Step-by-step solution:
Step 1: Convert 33.0 g of C to moles.
Using the formula for moles and the molar mass of C = 12.0 g/mol, the moles of C is
[tex]\begin{gathered} Moles=\frac{Mass}{Molar\text{ }mass} \\ \\ Moles\text{ }of\text{ }C=\frac{33.0\text{ }g}{12.0\text{ }g\text{/}mol}=2.75\text{ }mol \end{gathered}[/tex]Step 2: Determine the moles of CO produced.
Using the mole ratio of C and CO from the balance equation and the moles of C in step 1; the moles of CO produced is calculated as follows:
[tex]\begin{gathered} 3\text{ }mol\text{ }C=3\text{ }mol\text{ }CO \\ \\ 2.75\text{ }mol\text{ }C=x \\ \\ x=\frac{2.75\text{ }mol\text{ }C}{3\text{ }mol\text{ }C}\times3\text{ }mol\text{ }CO \\ \\ x=2.75\text{ }mol\text{ }CO \end{gathered}[/tex]Step 3: Convert the moles of CO produced to mass in grams.
The moles of CO produced in step above can be converted to mass in grams as shown below
[tex]\begin{gathered} Mass=Moles\times Molar\text{ }mass \\ \\ Molar\text{ }mass\text{ }of\text{ }CO=12.0+16.0=28.0\text{ }g\text{ /}mol \\ \\ Mass\text{ }of\text{ }CO\text{ }produced=2.75\text{ }mol\times28.0\text{ }g\text{/}mol \\ \\ Mass\text{ }of\text{ }CO\text{ }produced=77.0\text{ }g \end{gathered}[/tex]Hence, 77.0 grams of CO are produced when 33.0 g of C reacts.
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Isotopes of certain ___ elements that spontaneously emit ____ and ____ from the _____.
Isotopes of certain __radioactive__ elements that spontaneously emit __radiation__ and _rays___ from the __nuclei____
What causes isotopes to emit radiation ?Any of multiple species of the same chemical element with differing masses whose nuclei are unstable and spontaneously produce radiation in the form of alpha, beta, and gamma rays are known as radioactive isotopes, also known as radioisotopes, radionuclides, or radioactive nuclides.
Isotopes are identical elemental atoms with differing quantities of neutrons. Numerous elements have one or more radioactive isotopes. Due to the instability of their nuclei, they decay and release radiation.Many of the radioisotopes in the uranium, thorium, and actinium natural radiation decay series, as well as the naturally occurring radioisotopes potassium-40 and carbon-14, generate gamma radiation.Learn more about Isotopes here:
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Perform the followingmathematical operation, andreport the answer to thecorrect number of significantfigures.
To find out how many significant numbers there are in a number we can rely on a few rules:
1. Non-zero digits are always significant
2. Any zeros between non-zero digits are significant
3. Final zeros or trailing zeros are significant if they come after the non-zero numbers in a decimal number situation
Example: .500 and .632000
In this question, we have:
0.0016/0.849 = 0.0019, this is the correct value with the correct number of significant figures
In the following equation, how many grams of CO2 are produced if 2 moles of O2 react completely with methane (CH4)? CH4 + 2O2 -> CO2 + 2H2OHELPFUL INFO: molar mass of CO2=44g/mol, molar mass of O2=32g/mol, molar mass of CH4=16g/mol, and molar mass of water = 18g/mol
Step 1 - "Reading" the equation
The given chemical equation is:
[tex]CH_{4(g)}+2O_{2(g)}\to CO_{2(g)}+2H_2O_{(l)}[/tex]The bigger numbers, those that come before the formulas of the substances, indicate the quantity in moles of each substance required in this rection.
We can "read" this reaction thus as :
one mole of CH4 reacts with 2 moles of O2 thus producing one mole of CO2 and two moles of H2O
As the exercise is specifically asking about the proportion between O2 and CO2, we can further simplify this statement to:
two moles of O2 produce one mole of CO2
Step 2 - Obtaining the "recipe" for the reaction
In order to obtain a relation in grams, we need to multiply the number of moles of each substance by its respective molar mass. Let's remember that:
two moles of O2 produce one mole of CO2
Therefore, converting to grams (molar masses: 32 g/mol for O2; 44 g/mol for CO2)
[tex]\begin{gathered} O_2\to2\text{ moles }\times32\text{ g/mole = 64g} \\ CO2\to1\text{ mole }\times44\text{ g/mole = 44g} \end{gathered}[/tex]We have now obtained the "recipe" for this reaction:
64g of O2 produce 44g of CO2
Pretty much as we would with a cake recipe, we can use this to predict how much CO2 is formed.
Since the reaction of two moles of O2 is exactly what we have already calculated, 44g of CO2 would be formed in this reaction.
Sulfur burns in oxygen to form sulfur dioxide. Write a skeleton equation for this chemical reaction.
A skeletal chemical equation is a representation of a chemical reaction using chemical formulae of reactants and products.
Sulfur = S
Oxygen = O
Sulfur dioxide = SO2
The skeleton equation is:
[tex]S+O_2\text{ }\Rightarrow SO_2[/tex]Here the skeleton equation = balanced equation.
If a gas has a pressure of 6 atm , a temp of 53k and a volume of 78 liters but the temp changes to 17k and the pressure changes to 85 atm what is the new volume
answer and explanation
we are given initial values of temperature, pressure and volume as well as the final value of temperature and pressure and we are asked to find the final volume.
we can do this using the combined gas equation
P1V1/T1 = P2V2/T2
when we plug in the values we get:
(6atm x78L)/53K = (85atm x V2/17K
8.83 = 85atm x V2 / 17K
V2 = 1.77 Liters
A 51.72 g sample of a substance is initially at 23.5 °C. After absorbing 2547 J of heat, the temperature of the substance is 123.4 °C. What is the specific heat ( ) of the substance?
The heat absorbed or released by a substance is given by the following formula:
[tex]Q=m\cdot Cp\cdot(T2-T1)[/tex]Where Q is the heat absorbed or released, m is the mass of the substance, Cp is the specific heat, T2 is the final temperature and T1 is the initial one.
We know the values of Q, m, T2 and T1 because they are given by the question statement, and we have to find the value of Cp.
Solve the equation for Cp and use the given values to find its value:
[tex]\begin{gathered} Cp=\frac{Q}{m\cdot(T2-T1)} \\ Cp=\frac{2547J}{51.72g(123.4\degree C-23.5\degree C)} \\ Cp=\frac{0.49J}{g\degree C} \end{gathered}[/tex]The specific heat of the substance is 0.49J/g°C.
Which equation is balanced?
O 2Fe +02 → 2Fe2O3
O 3Fe +302 → 3Fe2O3
O 4Fe +302 → 2Fe2O3
O Fe +0₂ Fe₂O3
Answer:
C) 4Fe + 3O2 → 2Fe2O3
Explanation:
For this question, first you would go through all the options and make sure that there is an equal amount of elements on each side.
For the first option: 2Fe +02 → 2Fe2O3
You have only 2Fe atoms and 2O atoms on your reactant side and 4Fe and 6O atoms on your product side. This is not balanced.
For option b: 3Fe +302 → 3Fe2O3
You have 3Fe and 6O atoms on your reactant side while on your product side you have 6Fe and 6O atoms. This is not balanced.
For option c: 4Fe +302 → 2Fe2O3
You have 4Fe and 6O atoms on your reactant side and 4Fe and 6O atoms on your product side. This is balanced!
For option d: Fe +0₂ → Fe₂O3
You have 1Fe and 2O atoms on your reactant side and 2Fe and 3O atoms on your product side. This is not balanced.
***Remember to multiply the subscript of an element by their coefficient (number in front of molecule) if there is any.
A bike rides at an average speed of 25 km/h. How many minutes will it take for this rider to ride a distance of 20 km?
Answer:
48 minutes
Explanation:
60 divided by 25 equals 2.4 then you times 2.4 by 20
Consider the balanced chemicalreaction below and determine thepercent yield for carbon dioxide if 4.50moles of propane yielded 7.64 moles ofcarbon dioxide.C3H8 +502 + 3CO2 + 4H2O-→
To solve this, we must know some terms:
% Yield = (Actual yield / Theoretical yield) x 100
Actual yield is already given = 7.64 moles
To obtain the theoretical yield, we use stoichiometry and proceed like this:
C3H8 + 5O2 => 3CO2 + 4H2O
1 mole C3H8 ------- 3 x 1 mole CO2
4.5 moles C3H8 ------ x
x = 13.5 moles
Therefore, our % yield:
% Yield = (7.64 moles/13.5 moles) x 100 = 57 % approximately
Answer: % yield = 57 %
What are the products of the neutralization reaction between HNO3 andCa(OH)2?OA. CaO, NO₂, and H₂OB. Ca3N₂ and H₂OOC. Ca(NO3)2 and H₂OOD. CaNO3 and H30+
Ca(NO3)2 and H₂O. Option C is correct
Explanations:What is a neutralization reaction?The combination of acid and base to form salt and water at the product is known as a neutralization reaction.
Hence the products of the neutralization reaction between HNO3 and Ca(OH)2 must be salt and water
From the given option, the equivalent product that are salt and water is Ca(NO3)2 and H₂O which gives the required product.
I need help with something please
Since both metals are in the same group in the periodic table and both non-metals are also in the same group in the periodic table, they have the same charge (metals = 1+; non-metals = 1-). So this makes the reaction easier for us:
NaCl + LiBr = LiCl + NaBr
The solubility in hexane of organic compound O is measured and found to be 0.520-that would contain 120. mg of O at this temperature.Be sure your answer has the correct unit symbol and 3 significant digits.00.0olo88at 25. °C. Calculate the volume of a saturated solution of O in hexanemLE
According to the explanation given in the previous session, now we have hexane in an organic compound with the solubility of 0.520 g/mL at 25°C. We need to find the value of volume at 120.0 mg
1 gram = 1000 mg
120.0 mg = 0.12 grams
Now we can calculate the volume:
0.520g = 1 mL
0.12g = x mL
0.520x = 0.12
x = 0.12/0.520
x = 0.231 mL is the volume for this compound
What is the concentration of A pill weighing 325mg containing 22mg Mg
Step 1 - Understanding percentual concentration
To express the concentration of a substance as a percentual concentration, we just need to divide its mass by the total mass:
[tex]concentration\text{ in \% = }\frac{m_{substance}}{m_{total}}\times100[/tex]Step 2 - Calculating the concentration in the pill
We want to calculate the concentration of Mg in the pill. Therefore, we'll divide its mass by the total mass of the pill:
[tex]concentration\text{ of Mg =}\frac{22mg}{325mg}\times100=6.8\text{ \%}[/tex]Answer: the concentration of Mg in the pill is 6.8%
What is neutralization reaction? Use an example to describe the components of neutralization reaction
1) Neutralization reaction.
In this type of reaction an acid and a base react to form a salt and water. In the course of the reaction H+ and OH- are produced. Afterward, they combine to produce water.
2) Example.
Acid: HCl
Base: NaOH
Salt: NaCl
[tex]\text{HCl}+\text{NaOH}\rightarrow NaCl+H_2O[/tex]