Answer: the percent yield of the reaction is 30.7%
Explanation:
The question requires us to determine the percent yield of the reaction that produces copper(I) sulfide, given the theoretical yield and actual yield.
The following information was provided by the question:
Amount of Cu used = 0.0970 mol of Cu
Actual yield of reaction = 2.37 g of Cu2S
To solve this problem, we'll need to determine the theoretical yield of the reaction, considering the amount of copper (Cu) used and the balanced chemical equation for this reaction, and then calculate the percent yield of the reaction.
1) Determining the theoretical yield of reaction:
The reaction between copper and sulfur to produce copper (II) sulfide can be written as:
[tex]2Cu\text{ + S}\rightarrow Cu_2S[/tex]Considering that 0.0970 moles of Cu were used in excess of S, and the stoichiometry of the reaction, we can calculate the amount of Cu2S produced:
2 mol Cu ---------------------- 1 mol Cu2S
0.0970 mol Cu ------------ x
Solving for x, we have that 0.0485 moles of Cu2S should be obtained from the reaction.
2) Calculating the percent yield of the reaction:
The percent yield of a chemical reaction can be calculated using the following equation:
[tex]percent\text{ yield = }\frac{theoretical\text{ yield}}{actual\text{ yield}}\times100\%[/tex]Note that the theoretical and actual yield values must be used in the same units. Thus, we need to convert the mass given for actual yield (2.37g) to number of moles, considering the molar mass of copper (I) sulfide (159.16 g/mol):
[tex]\begin{gathered} number\text{ of moles = }\frac{mass\text{ of sample \lparen g\rparen}}{molar\text{ mass \lparen g/mol\rparen}} \\ \\ number\text{ of moles = }\frac{2.37g}{159.16g/mol}=0.0149\text{ mol} \end{gathered}[/tex]Therefore, the actual yield corresponds to 0.0149 moles of Cu2S.
Now, we can calculate the percent yield of the reaction:
[tex]percent\text{ yield = }\frac{0.0149\text{ mol}}{0.0485\text{ mol}}\times100\%=30.7\%[/tex]Therefore, the percent yield of the reaction is 30.7%.
Given the following reaction:2 Li (s) + 2 H2O (I) = 2 LiOH (aq) + 1 H2 (g)How many grams of solid lithium must be used in this reaction in order to obtain 6.00 L of hydrogen gas at 1.10 atm and 25.0 C ?
Answer:
3.73 g of Li.
Explanation:
What is given?
Volume of hydrogen gas, H2 (V) = 6.00 L.
Pressure (P) = 1.10 atm.
Temperature (T) = 25.0°C + 273 = 298 K.
R = 0.082 L*atm/mol*K.
Step-by-step solution:
First, we have to find the number of moles of hydrogen gas (H2) using the ideal gas formula:
[tex]PV=nRT,[/tex]where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant and T is the temperature on Kelvin scale. So let's solve for 'n' and replace the given values:
[tex]n=\frac{PV}{RT}=\frac{1.10\text{ atm}\cdot6.00\text{ L}}{0.082\text{ }\frac{L\cdot atm}{mol\cdot K}\cdot298K}=0.270\text{ moles.}[/tex]Based on this, we have 0.270 moles of H2.
You can see in the chemical equation that 2 moles of Li reacted produce 1 mol of H2, so we have to state a rule of three:
[tex]\begin{gathered} 2\text{ moles Li}\rightarrow1\text{ mol H}_2 \\ ?\text{ moles Li}\rightarrow0.270\text{ moles H}_2 \end{gathered}[/tex]The calculation will look like this:
[tex]0.270\text{ moles H}_2\cdot\frac{2\text{ moles Li}}{1\text{ mol H}_2}=0.540\text{ moles Li.}[/tex]The final step is to convert 0.540 moles of Li to grams using its molar mass that can be found in the periodic table, which is 6.9 g/mol:
[tex]0.540\text{ moles Li}\cdot\frac{6.9\text{ g Li}}{1\text{ mol Li}}=3.73\text{ g Li.}[/tex]The answer is that we must use 3.73 g of Li to produce 6.00 L of H2 at 1.10 atm and 25.0 °C.
Calculate the percentage oxygen in dinitrogen pentoxide
So, To calculate the mass percentage of dinitrogen pentoxide (N2O5) we will first calculate the molar weight of the molecule. The mass percentage will be calculated with the following equation:
[tex]\text{Mass percentage = }\frac{\text{Molar Mass oxygen}}{MolarmassmoleculeN_2\text{O}_5}\times100[/tex]Molar mass will be:
Element Atomic Mass #Atoms Mass
N 14.0067 2 28.0134 g/mol
O 15.9994 5 79.997 g/mol
Total mass = 28.0134 + 79.997 = 108.0104 g/mol
[tex]\text{Mass percentage = }\frac{\text{7}9.997}{108.0104}\times100\text{ = }74.064\text{ percent}[/tex]2. In each pair, CIRCLE the larger species.a. N or N3-b. Ca or Ca2+c. Si or Asd. Al or B
Answer:
a. N3-
b. Ca
c. As
d. Al
Explanation:
a. N or N3-
In this case, the largest species is N3- because it has 3 additional electrons in contrast to the neutral atom.
b. Ca or Ca2+
In this case, the largest species is Ca, because Ca2+ has 2 fewer electrons than the neutral atom of Ca.
c. Si or As
To analyze these elements, is is necessary to look for them in the Periodic Table of elements:
Since the atomic radius increases from top to bottom and from right to left, the largest is As.
d. Al or B
In this case, we also have to look for the elements in the Periodic table, and following the atomic radius increase, the largest is Al.
I have 2.50 x 10²³ atoms of titanium. How many moles of titanium do I have?
Answer: 2.50 x 10^23 atoms of titanium correspond to 0.415 moles of titanium.
Explanation:
The question requires us to calculate the number of moles of titanium (Ti) that correponds to 2.50 x 10^23 atoms of Ti.
We can apply the Avogadro's number to solve this problem: according to this proportionality constant, there are 6.022 x 10^23 particles in 1 mol of any compound (particles can be atoms, ions, molecules etc).
Thus, considering the Avogadro's number, we can write:
6.022 x 10^23 atoms Ti -------------------- 1 mol Ti
2.50 x 10^23 atoms Ti ---------------------- x
Solving for x, we'll have:
[tex]x=\frac{(1\text{ mol Ti\rparen}\times(2.50\times10^{23}\text{ atoms Ti\rparen}}{(6.022\times10^{23}\text{ atoms Ti\rparen}}=0.415\text{ mol Ti}[/tex]Therefore, 2.50 x 10^23 atoms of titanium correspond to 0.415 moles of titanium.
An unknown gas is found to consist of 10.48 g of nitrogen and 11.96 g if oxygen. What is the percentage composition of this gas? * A. 53.3% N, 46.7% O B. 74.5% N, 25.5% O C. 28.6% N, 71.4 % O D. 46.7%N, 53.3%O
Answer:
A. 53.3% N, 46.7% O
Explanation:
1st) It is necessary to add both quantities (10.48g of nitrogen and 11.96g of oxygen) to obtain the total mass of the sample, which represents 100%:
10.48g + 11.96g = 22.44g
22.44g represents the 100% of the sample.
2nd) Now we can calculate the percentage composition of the gas, using a mathematical rule of three and the amount of nitrogen and oxygen:
• Nitrogen:
[tex]\begin{gathered} 22.44g-100\% \\ 10.48g-x=\frac{10.48g*100\%}{22.44g} \\ x=46.7\% \end{gathered}[/tex]• Oxygen:
[tex]\begin{gathered} 22.44g-100\% \\ 11.96g-x=\frac{11.96g*100\%}{22.44g} \\ x=53.3\% \end{gathered}[/tex]We can also calculate the oxygen percentage by subtracting 100% minus the nitrogen percentage (46.7%):
100% - 4607% = 53.3%
So, the percentage composition is 53.3% N and 46.7% O.
What affect does a chemical reaction have on atoms?
Answer:Chemical reactions involve breaking chemical bonds between reactant molecules (particles) and forming new bonds between atoms in product particles (molecules).Explanation:
What volume of O2 collected at 22.0 degrees Celsius and 728mmHg would be produced by the decomposition of 8.15g KClO3
The volume of O₂ collected at 22 degrees Celsius and 728 mmHg would be produced by the decomposition of 8.15 grams of KClO₃ is 2.52 L
It is given that the mass of KClO₃ is 8.15g and the temperature is 22°C and the pressure is 728mmHg. The number of moles of KClO₃ is given by,
No of moles of KClO₃ = 8.15/122.55
No of moles of KClO₃ = 0.066 mol
The reaction for decomposition is,
2KClO₃ -----------> 2KCl + 3O2
2 mol of KClO3 gives 3 moles of O₂
Then, 0.66mol of KClO3 gives, 3/2(0.66) mol of O₂ which is 0.0998
The given temperature is 22 + 273 = 295K
The given pressure is 728mmHg which is 728/760 which is 0.958 atm
We know that
PV = nRT
0.958 x V = 0.0998x 0.0821x295
V = 2.52L
Therefore, the volume of Oxygen collected is 2.52L
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Zinc metal reacts with HCl according to the balanced equation:Zn + 2HCl à ZnCl2 + H2When 0.103 g of Zn is combined with enough HCl to make 50 mL of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.5 C to 23.7 C. Find the heat of the reaction. (ans: -160 kJ/mol)
-163kJ/mol
Explanations:Given the reaction between the zinc metal reacts with HCl according to the balanced equation:
[tex]Zn+2\text{HCl}\rightarrow\text{ZnCl}_2+H_2[/tex]The required heat of reaction will be calculated using the formula below:
[tex]\triangle H_{\text{rxn}}=-\frac{Q}{n_{Zn}}[/tex]where:
• Q is the ,heat produced
,• nZn is the, number of moles that reacted
The formula for calculating the quantity of heat produced is expressed as:
[tex]Q=mc\triangle\theta[/tex]where:
• m is the ,mass, of the ,Zinc metal
,• c is the ,specific heat capacity, of zinc
,• △θ is th,e change in temperature
Get the mass of zinc
[tex]\begin{gathered} \text{mass}=\text{ }density\times\text{volume} \\ m=\rho\times v \end{gathered}[/tex]The quantity of heat becomes:
[tex]Q=\rho\cdot v\cdot C\triangle\theta[/tex]Substitute the given parameters to have:
[tex]\begin{gathered} Q=\frac{1.02\cancel{g}}{\cancel{mL}}\times50\cancel{mL}\times\frac{4.18J}{\cancel{g^oC^{}}}\times(23.7-22.5)\cancel{^oC} \\ Q=1.02\times50\times4.18\times1.2 \\ Q=255.8Joules \end{gathered}[/tex]Next is to get the number of moles of Zinc that reacted (nZn)
[tex]\begin{gathered} nZ_n=\frac{Mass}{\text{Molar mass}} \\ nZ_n=\frac{0.103g}{65.4g\text{/mol}} \\ nZ_n=0.00157mole \end{gathered}[/tex]Get the required heat of reaction of Zinc:
[tex]\begin{gathered} \triangle H_{\text{rxn}}=-\frac{Q}{nZ_n} \\ \triangle H_{\text{rxn}}=-\frac{255.8Joules}{0.00157moles} \\ \triangle H_{\text{rxn}}=-162,929.93J\text{/mol} \\ \triangle H_{\text{rxn}}\approx-163kJ\text{/mol} \end{gathered}[/tex]Hence the heat of the reaction of Zinc is approximately -163kJ/mol
The frequency of a wave is 1.2 x 10° Hz. What is its wavelength? What kind of wave is this?
Answer: wavelength= ............
Explanation:
A 14.577 g sample of CaCl2 was added to 12.226 g of K2CO3 and mixed in water. A 3.554 g yield of CaCO3 was obtained. Calculate the percent yield of CaCO3.
A 14.577 g sample of CaCl₂ was added to 12.226 g of K₂CO₃ and mixed in water. A 3.554 g yield of CaCO₃ was obtained then the percent yield of CaCO₃ is 13.25%
Here given data is
14.577 g CaCl₂ was added to 12.226 g of K₂CO₃ and mixed in water
3.554 g yield of CaCO₃ was obtained
We have to calculate the percent yield of CaCO₃ = ?
Percent yield = actual yield/threotical yield×100
Actual yield = 3.554 g
Theroetical yield = 14.577 g CaCl₂ + 12.226 g of K₂CO₃
Theroetical yield = 26.803g
Now Percent yield = actual yield/threotical yield×100
Percent yield = 3.554 g /26.803g×100
Percent yield = 13.25%
The percent yield of CaCO₃ is 13.25%
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What are the OSHA PEL-TWyA and PEL-STEL values for nitric acid?
The OSHA limit for STEL-TWA is 2ppm and the PEL-STEL value is 4ppm to the exposure to nitric acid.
OSHA refers to Occupational Safety and Health Administration has adopted standard values for the exposure to harmful and hazardous chemicals so that the safety of workers and testing animals could be ensured.
Here, STEL-TWA stands for Short Term Exposure Limit - Time-Weighted average, and PEL-STEL stands for Permissible Exposure Limit- Short Term Exposure Limit.
This is usually set up in industries to ensure the safety of the workers and the maximum exposure short-term time limit is 15min and 2ppm for STEL-TWA and 4 ppm for PEL-STEL.
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Is chlorine malleable or brittle?
Chlorine is not malleable or ductile. This is so because nonmetals, like chlorine, are brittle.
What is a chlorine physical characteristic?The odor of chlorine is distinctively intrusive and upsetting. The liquid is pure amber, while the gas is a greenish yellow tint. The information on the physical characteristics of chlorine as discovered by various researchers reveals some variances.
What is brittle versus ductile or malleable?In essence, metals that are malleable or ductile can both deform in some ways through plastic deformation in response to stress. Brittle metals are those that fracture without significant plastic deformation. Brittle is the antithesis of ductile and malleable in this context.
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Turn the following word equation into a symbolic equation: potassium hydroxide and hydrobromic acid react to form potassium bromide and water.
The chemical formula of potassium hydroxide is KOH.
The chemical formula of hydrobromic acid is HBr.
The chemical formula of potassium bromide is KBr.
The chemical formula of water is H2O.
The symbolic equation for this reaction is:
[tex]KOH+HBr\rightarrow KBr+H_2O[/tex]It means that the correct answer is the first choice.
17. What is a bond?
Bond:
Bonds are forces that keep atoms together to make compounds or molecules.
In chemical bonds electrons are involved. And there are three kinds of bonds:
1. Covalent bonds: where the electrons are shared between the atoms.
2. Polar covalent bonds
3. Ionic bonds: happens when one atom gives to the other atom electrons.
14) How many mL of 5.724 M HNO3 are needed to prepare a 183.95 mL of 2.063 M HNO3?
Answer
66.30 mL
Explanation
Given parameters:
Initial concentration, C₁ = 5.724 M
Final volume, V₂ = 183.95 mL
Final volume, V₂ = 183.95 mLFinal concentration, C₂ = 2.063 M
What to find:
The initial volume, V₁
Step-by-step solution:
Using the dilution law:
[tex]C_1V_1=C_2V_2[/tex]Substitute the given parameters into the formula to get V₁
[tex]\begin{gathered} 5.724M\times V_1=2.063M\times183.95mL \\ \text{Divide both sides by 5.724M} \\ \frac{5.724M\times V_1}{5.724M}=\frac{2.063M\times183.95mL}{5.724M} \\ V_1=66.30\text{ mL} \end{gathered}[/tex]Hence, 66.30 mL of 5.724 M HNO3 are needed to prepare a 183.95 mL of 2.063 M HNO3?
Determine the mass of the sample based on the following data to two decimal places- mass of beaker and cover: 50.09- mass of beaker, cover, and sample before heating: 51.04- mass of beaker, cover, and sample after heating: 50.88
Step 1 - Understanding the problem
We need to determine the mass of a sample which probably contains water as an impurity. Therefore, if we just weight the sample, we won't be able to separate its own weight from that of water.
What can be done, then, is heating the sample first in order to evaporate all the water, and only then weighting it. That's exactly what the procedure in the exercise has done.
The idea here thus is that we just need to subtract two weights:
[tex]m_{\text{sample}}=m_{\text{sample}+\text{water}}-m_{\text{water}}[/tex]Step 2 - Discovering the mass of the sample
We can discover the mass of the sample by simply subtracting the mass of beaker and cover from the final mass (beaker + cover + sample after heating).
This will give us the mass of the sample because the water has evaporated after heating:
[tex]m_{\text{sample}}=50.88-50.09=0.79g[/tex]The mass of the sample is thus 0.79g.
[OH﹘] = 6.5 x 10-5Mfind the pH and pOH
pH = 9.81
pOH = 4.19
Explanations:The formula for calculating the pOH of a solution is given as:
[tex]pOH=-log[OH^-][/tex]Given the following parameters
[tex][OH^-]=6.5\times10^{-5}M[/tex]Substitute
[tex]\begin{gathered} pOH=-log(6.5\times10^{-5}) \\ pOH=-(-4.19) \\ pOH=4.19 \end{gathered}[/tex]Determine the pH of the solution
[tex]\begin{gathered} pH+pOH=14 \\ pH=14-pOH \\ pH=14-4.19 \\ pH=9.81 \end{gathered}[/tex]Therefore the pH and pOH are 9.81 and 4.19 respectively
Determine the mole fraction of NaCl in a solution that has 0.589 mol NaCl and 0.625 mol water.
Step 1
A mole fraction is defined as:
Mole fraction (compound X) = moles of compound X/total moles
It has no units
-------------------------
Step 2
The solution is formed by solute and solvent. Therefore, the total moles will be:
0.589 moles NaCl + 0.625 moles water = 1.214 moles
------------------------
Step 3
Molar fraction (NaCl) = moles NaCl/total moles = 0.589 moles/1.214 moles = 0.49 approx.
Answer: Molar fraction of NaCl = 0.49
How must nuclear waste that is considered high-level waste to be stored
Answer: the best option to answer the question is the second one (letter B).
Explanation:
The question requires us to choose, among the options given, which one best corresponds to the correct form of disposal fo high-level nuclear waste.
High-level nuclear waste corresponds mainly to highly radioactive materials produced as byproducts of reactions that occur inside nuclear reactors. When they are not of use anymore, they are disposed in underground repositories, without any recycling. The place of disposal of this type of nuclear waste is usually placed in cannister which are placed underground and sealed with rocks. Since the time of decay of high-level nuclear waste can reach hundreds of thousands of years, this type of waste must be disposed of in a way that provides adequate protection for a long time.
Considering the information above, we can say that the best option to answer the question is the second one (letter B).
What molarity of a 53.96 mL magnesium hydroxide solution is required to neutralize 93.24mL of a 7.306 M solution of hydrochloric acid solution, which creates as products, magnesium chloride and water?
Answer
6.312 M
Explanation
Given:
Volume of Mg(OH)₂, Vb = 53.96 mL
Volume of HCl, Va = 93.24 mL
Molarity of HCl, Ma =7.306 M
What to find:
The molarity of Mg(OH)₂, Mb
Solution:
The first step is to write a balanced chemical equation for the reaction.
Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O
Mole ratio is 1:2; that is na = 2 and nb = 1
Hence, the molarity of Mg(OH)₂, Mb is calculated using the formula below:
[tex]\frac{M_aV_a}{n_a}=\frac{M_bV_b}{n_b}[/tex]Plugging the values of the given parameters into the formula, we have:
[tex]\begin{gathered} \frac{7.306M\times93.24mL}{2}=\frac{M_b\times53.96mL}{1} \\ \\ Cross\text{ }multiply \\ \\ M_b\times53.96mL\times2=7.306M\times93.24mL\times1 \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }53.96mL\times2 \\ \\ \frac{M_b\times53.96mL\times2}{53.96mL\times2}=\frac{7.306M\times93.24mL\times1}{53.96mL\times2} \\ \\ M_b=6.312\text{ }M \end{gathered}[/tex]Therefore, the molarity of a 53.96 mL magnesium hydroxide solution that is required to neutralize 93.24mL of a 7.306 M solution of the hydrochloric acid solution is 6.312 M
How many atoms (or molecules) are present in 1 mole?Question 19 options:A) 1B) 6.022×1020C) 6.022×1023D) 6.022
Answer:
The answer is C.
Explanation:
The number of particles (atoms/molecules) in 1 mole is referred to as the Avogadro's number. The Avogadro's number = 6.022 * 10 ^ 23. Therefore, the answer is C.
What is the coefficient of H2O in the balanced half reaction? (See picture)
Firstly we would determine the oxidation number of the elements to determine which is undergoing oxidation and which is undergoing reduction:
[tex]\begin{gathered} Mn\text{ }in\text{ }MnO_4^-:Mn+(-2\times4)=-1 \\ Mn\text{ }in\text{ }MnO_4^-:Mn-8=-1 \\ Mn\text{ }in\text{ }MnO_4^-:Mn=-1+8=+7 \\ \\ Mn\text{ }in\text{ }MnO_4^{2-}:Mn+(-2\times4)=-2 \\ Mn\text{ }in\text{ }MnO_4^{2-}:Mn=-2+8=+6 \end{gathered}[/tex]Mn oxidation number goes from +7 to +6 which means it is undergoing a reduction. A decrease in oxidation number means reduction.
[tex]\begin{gathered} S\text{ }in\text{ }HSO_3^-:1+(-2\times3)+S=-1 \\ S\text{ }in\text{ }HSO_3^-:1-6+S=-1 \\ S\text{ }in\text{ }HSO_3^-:S=-1+5=+4 \\ \\ S\text{ }in\text{ }SO_4^{2-}:S+(-2\times4)=-2 \\ S\text{ }in\text{ }SO_4^{2-}:S-8=-2 \\ S\text{ }in\text{ }SO_4^{2-}:S=+6 \end{gathered}[/tex]S oxidation goes from +4 to +6 meaning it is undergoing oxidation. There is an increase in oxidation number.
We will now balance the oxidation half reaction:
[tex]HSO_3^-+H_2O\rightarrow SO_4^{2-}+3H^++2e[/tex]The coefficient for the water molecule is 1 in the balnced half reaction.
Which of the following combinations will form a solution?Group of answer choicesOil and waterSand and waterSugar and waterSand and sugar
Answer: the combination "sugar and water" will form a solution.
Explanation:
The question requires us to determine which of the combinations given will form a solution.
To solve this question, let's consider this definition of solution: a homogenous mixture of two or more substances in relative amounts. This means that, to be considered a solution, two mixed substances must form a homogeneous mixture, i.e., without separation of phases.
Considering the options given by the question, we can say:
- oil and water do not mix (because they have different polarities), and separate in different phases when are put together;
- sand and water do not form a homogeneous mixture: sand is deposited at the bottom of the vessel and forms a separated phase of solid;
- sugar and water can form a homogeneous mixture, therefore they form a solution;
- sand and sugar form visible separated phases when mixed together (we can see the particles of sugar and sand separated).
Therefore, considering the points above, we can say that only sugar and water form a solution.
Are all covalent molecules soluble in water ?
What is the wavelength of light with a frequency of 3.215 x 10^15 s-¹?
A. 9.325 x 10^-8m
B. 3.110 x 10^-16m
C. 1.072 x 10^7 m
D. 9.639 x 10^23m
The wavelength of the light will be 9.63 × 10²³ m .The wavelength of the wave is calculated using the frequency formula.
The quantity of cycles finished in a certain amount of time is known as frequency. It also indicates the number of crests that pass past a specific position in a unit of time. It is often referred to as the reciprocal of time. Hertz are used to express frequency (Hz). The frequency of the wave is calculated using the frequency formula.
The total number of times a repeated event occurs in a unit of time is known as its frequency. Depending on the quantities known, various frequency formulas exist to calculate frequency. The terms frequency (f), time period (T), wave speed (V), and wavelength () are all found using the formula for the frequency of a wave.
We know that ,
Frequency= Speed of light/ wavelength
Wavelength= Frequency × Speed of light
= 3.215 ×10¹⁵ × 2.997× 10⁸
= 9.639 × 10²³ m
The wavelength will be obtained by frequency -wavelength formula a and it will come as 9.63 × 10²³ m .
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I need help getting the answer on all of theseFirst one
The Law of Conservation of Mass can be state in different ways. Essentially, this law states that if we are on a closed system, no mass can be created nor dstroyed. In chemical terms, this means that in a chemical reaction the number of atoms of each element has to be the same in the reactants and the products.
Given 20 grams of Na (s) and 10 grams of Cly, which reactant is in excess? And, how much NaCl should be produced from these amounts of the two reactants?
In order to answer this question we will need a few steps
First one will be to set up the properly balanced equation:
2 Na + Cl2 -> 2 NaCl
20 grams of Na
10 grams of Cl2
From this balanced equation we see that the molar ratio between Na and Cl2 is 2:1, that means that for every 2 moles of Na, we will need 1 mol of Cl2 in order to proceed with the reaction
Now we need to identify the limiting and excess reactant, we will do that by checking how many moles of each element we have and how much we should have in order to properly react with the other element, let's check it:
Molar mass for Na is 23g/mol
23g = 1 mol
20g = x moles
x = 0.89 moles of Na in 20 grams
So if we have 0.89 moles of Na, we should have 0.44 moles of Cl2, but we don't know if we have this amount of moles of Cl2 in 10 grams of it, let's check
Molar mass for Cl2 is 71g/mol
71g = 1 mol
10g = x moles
x = 0.14 moles
We have fewer moles than we actually needed, let's do the same step but now focusing on Cl2
Cl2 in 10 grams has 0.14 moles, therefore we will have 0.28 moles of Na, remember the molar ratio 2:1
23g = 1 mol
x grams = 0.28 moles
x = 6.44 grams of Na
If we have 10 grams of Cl2, we would only need 6.44 grams of Na, since we have more available, we can say that Na is in excess and Cl2 is the limiting reactant
To find out the mass of NaCl produced from 10 grams of Cl2 (we have to use the limiting reactant), NaCl has a molar mass of 58.44g/mol and the molar ratio will be again 2:1, one mol of Cl2 and 2 moles of NaCl, if we have 0.14 moles of Cl2, therefore we will have 0.28 moles of NaCl
58.44 g = 1 mol
x grams = 0.28 moles
x = 16.4 grams, the value in the options is a little higher, but not too much higher, that must be due some rounding up
Letter A
For the following reaction, 0.156 moles of sulfuric acid are mixed with 0.567 moles of calcium hydroxide.
The first step is to write the reaction using the molecular formulas of the substances:
[tex]H_2SO_4+Ca(OH)_2\rightarrow CaSO_4+2H_2O[/tex]Now that the equation is balanced, we have to divide the amount of each reactant by its stoichiometric coefficient, that in this case is 1 for both of them:
[tex]\begin{gathered} \frac{0.156mol}{1}=0.156mol \\ \frac{0.567mol}{1}=0.567mol \end{gathered}[/tex]The least result will indicate the limiting reagent, which in this case is sulfuric acid. It means that its formula is H2SO4.
Basing our calculations in this reagent, we can find the maximum amount of calcium sulfate that is produced:
[tex]0.156molH_2SO_4\cdot\frac{1molCaSO_4}{1molH_2SO_4}=0.156molCaSO_4[/tex]It means that the maximum amount of calcium sulfate that can be produced is 0.156 moles.
Consider the following reaction:
Mg²⁺(aq) + Cu(s) → Cu²⁺(aq) + Mg(s).
In this reaction, Mg²⁺(aq) is: ________
separate an oxidation-reduction equation into 2 half equations one for oxidation and one for reduction.
cu(s) ⇒ [tex]cu^{2+}[/tex](aq) + 2e- ........ oxidation
[tex]mg^{2+}[/tex](aq) + 2e- ⇒ mg(s) ...........reduction
as with all chemical equations one must balance mass and charge both.
An oxidizing agent causes another substance to be oxidized.The oxidizing agent is reduced.in other words to accept their electrons. Common oxidizing agents are oxygen, hydrogen peroxide and the halogens.
A reducing agent causes another substance to be reduced. the reducing agent is oxidised.for example hydrogen carbon hydrogen sulfide and Ammonia are reducing agents
learn more about the equation here;
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you discover the head of a match contains 3.75 g of sulfur. How many atoms of sulfur does the match contain? (hint: grams > moles > atoms)
As the question says, we are going to follow the path of the hint, first by finding the number of moles of Sulfur in 3.75 grams, and we can do that by using its molar mass, which is 32g/mol
32g = 1 mol
3.75g = x moles
32x = 3.75
x = 3.75/32
x = 0.12 moles of Sulfur in 3.75 grams
Now we have the number of moles, and to find the number of atoms, we need to use the Avogadro's constant number, which is the number of atoms in a single mol, this value is 6.02*10^23 atoms in 1 mol
1 mol = 6.02*10^23 atoms
0.12 moles = x atoms
x = 7.22*10^22 atoms of Sulfur in 3.75 grams