Answer:
The Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol
Explanation:
We are given that
Volume of H2 at STP=52.8mL
Mass of magnesium metal ,M(Mg)=0.055g
We have to find the Experimental Molar Volume in L/mol of the Hydrogen gas.
Molar mass of Mg=24.305 g/mol
Number of moles=[tex]\frac{given\;mass}{molar\;mass}[/tex]
Using the formula
Number of moles of Mg=[tex]\frac{0.055}{24.305}[/tex]moles
Number of moles of Mg=0.00226moles
Number of moles of Mg=Number of moles of H2
Number of moles of H2=0.00226moles
Molar volume of Hydrogen gas (H2)=[tex]\frac{volume\;at\;STP}{No\;of\;moles\;H_2}[/tex]
Molar volume of Hydrogen gas (H2)=[tex]\frac{52.8}{0.00226}mL/mol[/tex]
Molar volume of Hydrogen gas (H2)=[tex]\frac{52.8}{0.00226}\times 10^{-3}L/mol[/tex]
[tex]1L=1000mL[/tex]
Molar volume of Hydrogen gas (H2)=23.36L/mol
Hence, the Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol
Give the number of lone pairs around the central atom and the molecular geometry of SCl2. Multiple Choice 3 lone pairs, linear 1 lone pair, bent 3 lone pairs, bent 0 lone pairs, linear 2 lone pairs, bent
Answer:
2 lone pairs, bent
Explanation:
According to the Valence Shell Electron Pair Repulsion Theory, the number of electron pairs on the valence shell of the central atom in a molecule influences the shape of the molecule.
The presence of lone pairs on the valence shell of the central atom causes the observed molecular geometry to deviate from the ideal geometry predicted on the basis of the valence shell electron pair repulsion theory.
SCl2 has four regions of electron density. This means that its electron domain geometry is tetrahedral. However, there are two lone pairs on the valence shell of the central atom hence the observed molecular geometry is bent.
What mass of NaOH would need to be dissolved in 500.0 mL of water to produce a solution with a pH of 12.40
Answer:
0.5024 g
Explanation:
Step 1: Calculate the concentration of H⁺
We will use the definition of pH.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -12.40 = 3.981 × 10⁻¹³ M
Step 2: Calculate the concentration of OH⁻
We will use the ionic product of water expression.
[H⁺] [OH⁻] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴/[H⁺] = 10⁻¹⁴/3.981 × 10⁻¹³ = 0.02512 M
Step 3: Calculate the initial concentration of NaOH
NaOH is a strong base and the molar ratio of NaOH to OH⁻is 1:1. Thus, the initial concentration of NaOH is 1/1 × 0.02512 M = 0.02512 M.
Step 4: Calculate the moles of NaOH
We will use the definition of molarity.
M = moles of NaOH/liters of solution
moles of NaOH = M × liters of solution
moles of NaOH = 0.02512 mol/L × 0.5000 L = 0.01256 mol
Step 5: Calculate the mass of 0.01256 moles of NaOH
The molar mass of NaOH is 40.00 g/mol.
0.01256 mol × 40.00 g/mol = 0.5024 g
Draw a Lewis structure for thiocyanic acid, HSCN, adding charges and lone electron pairs to the appropriate atoms.
Answer:
See explanation and image attached
Explanation:
Thiocyanic acid is made made up of hydrogen, sulphur, carbon and nitrogen atoms. Carbon is the central atom in the molecule.
The molecule has a total of sixteen valence electrons as shown in the image attached. There are no formal charges in the structure of the molecule as shown.
The molecule is linear in shape.
The time required to pass one Faraday of electricity through a solution with a current of 0.3A is
Answer:
89.35 hour
Explanation:
Recall :
Charge on 1 electron = 1.6 × 10^-19 C
1 mole contains = 6.023 × 10^23
Therefore, the charge on 1 mole of electron will be :
Charge per electron × 1 mole :
(1.6 × 10^-19) * (6.023 * 10^23) = 96500 C = 1 Farad
1 Farad = 96500 C
Using the formula :
Q = Current(I) * time(t)
Q = I*t
t = Q/I
Current, I = 0.3 A
t = 96500 / 0.3
t = 321666.66 second
t = 321666.66 / 3600 = 89.35 hour
Calculate the average atomic mass for X
Answer:
39.0229 amu
Explanation:
Hello there!
In this case, according to given information, the idea here is to multiply the percent abundance by the mass number of each isotope and then add them all together as shown below:
[tex]=0.0967*38+0.7868*39+0.1134*40+0.0031*41\\\\=3.6746+30.6852+4.536+0.1271\\\\=39.0229amu[/tex]
Regards!
If mercury barometer is replaced by water barometer, height of water column
i. will be less than that of Hg Column
ii. will be greater than that of Hg column iii. will be equal to that of Hg column
iv. will be none of these
Answer:
answer is first one 1 will be less then that of hg coloumn
If mass of an empty 9.4 mL pycnometer is 10.3 and the mass of the same pycnometer with an unknown liquid is 20.1. Determine the density of the unknown liquid to the correct number of significant figures in g/mL
Answer:
1.04 g/mL
Explanation:
Applying,
D = (m-m')/V................. Equation 1
Where D = Density of the unknown liquid, m = mass of the pycnometer when filled with unkwon liquid, m' = mass of the empty pycnometer, V = volume of the empty pycnometer
From the question,
Assuming the mass are in grams
Given: m = 20.1 g, m' = 10.3 g, V = 9.4 mL
Substitute these values into equation 1
D = (20.1-10.3)/9.4
D = 9.8/9.4
D = 1.04 g/mL
What is the
energy
2) The energy transition from n = 1 →n= 3 in hydrogen is 12.09 eV (1.6022 X 10J= 1 eV)-
of light emitted from n = 3 →n=1?
E e fiz
Answer:
energy is the ability to do work
7.23 One equivalent of sodium methanethiolate is added to an electrophile that has two leaving groups. Which product will be major
The question is incomplete, the complete question is shown in the image attached to this answer
Answer:
A
Explanation:
We can see from the conditions of the reaction that the incoming nucleophile is -SCH3 and there are two possible leaving groups in the substrate.
First of all, we have to look at the conditions of the reaction. We can see that the reaction is taking place in DMF, a polar aprotoc solvent. This condition favours the SN2 synchronous mechanism over the SN1 ionic mechanism.
Hence, the nucleophile at the 1-position is preferentially substituted owing to the conditions of the reaction.
Thus, option A is the major product of the reaction.
What volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example
The question is incomplete, the complete question is:
What volume (mL) of the sweetened tea described in Example 3.14 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example. The example is attached below.
Answer: 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink
Explanation:
We first calculate the number of moles of soft drink in a volume of 10 mL
The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}[/tex] .....(1)
Taking the concentration of soft drink from the example be = 0.375 M
Volume of solution = 10 mL
Putting values in equation 1, we get:
[tex]0.375=\frac{\text{Moles of sugar in soft drink}\times 1000}{10}\\\\\text{Moles of sugar in soft drink}=\frac{0.375\times 10}{1000}=0.00375mol[/tex]
Calculating volume of sweetened tea:
Moles of sugar = 0.00375 mol
Molarity of sweetened tea = 0.05 M
Putting values in equation 1, we get:
[tex]0.05=\frac{0.00375\times 1000}{\text{Volume of sweetened tea}}\\\\\text{Volume of sweetened tea}=\frac{0.00375\times 1000}{0.05}=75mL[/tex]
Hence, 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink
A sealed vessel initially contains 100 g of chlorine gas and 90 g of hydrogen gas. The two gases undergo reaction to form HCl. Which of the following statements is true?
a. 100 g HCl is produced
b. 190 g HCl is produced
c. less than 90 g HCl is produced
d. between 100 and 190 g of HCl is produced
Answer:
d.
Explanation:
H2 + Cl2 = 2HCL
From the equation 2g hydrogen combine with 71g of chlorine.
So 35.5 g Cl2 combines with 1g of H2
There are 100g of Cl2 so this will, by proportion, react with 100/35.5 g hydrogen.
This is 2.8 g hydrogen so the mass of HCl formed = 102.8 g.
The true statement is that d. between 100 and 190 g of HCl is produced.
How do find the mass of HCL?To find mass of HCL:
H2 + Cl2 = 2HCL
From the equation, 2g of hydrogen combines with 71g of chlorine.
So 35.5 g Cl2 combines with 1g of H2
There are 100g of Cl2 so this will, by proportion, react with 100/35.5 g of hydrogen.
This is 2.8 g hydrogen so the mass of HCl formed = 102.8 g.
Hydrogen chloride may be formed by the direct combination of chlorine (Cl2) gas and hydrogen (H2) gas.
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A chemist combines 33 g of methane with 289 g of oxygen to from 189 g of carbon dioxide and 30 g of water. Determine if the results of the following word problem adheres to the Law of Conservation of Mass.
Answer:
The correct answer is - no not adhere to the law of mass conservation.
Explanation:
According to the law of mass conservation in an isolated system, the mass can not be created or destroyed and in a chemical or physical change, the mass of products should be always equal to the mass of reactants.
On the basis of the law the mass of the chemical reaction-
Mass of products = mass of reactants
33 g of methane + 289g of oxygen = 189g of carbon dioxide + 30g of water
322g ≠ 219 g
which means this reaction does not adhere to the law of conservation.
Please help help please
Answer: The correct answer is B.
Explanation: Segregate most organic acids from oxidizing mineral acids. Keep oxidizers away from other chemicals, especially flammables.
Answer:
Segregate most organic acids from oxidizing mineral acids. Keep oxidizers away from other chemicals, especially flammables, combustibles, and toxic materials. Keep corrosives away from substances that they may react with and release corrosive, toxic, or flammable vapors.
what following oxide react with both acids and bases to form salts.
Answer:
mainly metal oxide use to react with both acid and bases to form salts such as zinc, aluminum etc.
One of the purposes of this lab is to determine the order of the reaction with respect to the Allura Red dye by creating first and second-order graphs for all four trials. The correct order of the reaction is the one where the slopes of the graphs for the four trials are roughly the same. Why is this important when choosing the order of the reaction
Answer: Hello the options related to your question are attached below
The slope is related to the rate constant so all four trials should have the same slope since the reactions are all the same ( Option C )
Explanation:
It is important when choosing the order of the reaction because the concentration of the bleaches used in the four trials are in excess hence their slopes have to be roughly the same and also because the reactions are similar and they where done at the same temperature, hence the slope of the first and second-order graphs will be the same.
Nitrogen monoxide and ozone react to form nitrogen dioxide and oxygen, like this:
NO(g) + O3 → NO2 + O2
The reaction is exothermic. Suppose a mixture of NO, O3 NO and O2 has come to equilibrium in a closed reaction vessel. Predict the change.
Answer:
The pressure of O2 will increase. Shift to the right.
The pressure of O3 will increase . Dhift to the left
Explanation:
The perturbations are:
The temperature is lowered. The pressure of O2 will
The temperature is raised. The pressure of O3 will:
We can Apply LeCh's principle and see the heat, ΔH, as a product of the reaction:
NO(g) + O3(g) → NO2(g) + O2(g) + ΔH
If temperature is lowered, the system will shift to the right in order to produce more heat doing:
The pressure of O2 will increase
In the other way, if temperature is raised, the system will shift to the left in order to decrease the amount of heat produced.
The pressure of O3 will increase
atomaticity of chlorine 1) 2, 2)1, 3) 32 , 4) 4.
Answer:
ATOMICITY OF CHLORINE IS 2Explanation:
Atomicity is defined as the total number of atoms present in a molecule.
The products obtained from hydroboration-oxidation of cis-2-butene are identical to the products obtained from hydroboration-oxidation of trans-2-butene. Draw the products and explain why the configuration of the starting alkene is not relevant in this case.
Answer:
a) Attached below
b) The presence of racemic mixture found as product in both cases shows that products are identical ( i.e. they have same configuration
Explanation:
Diagrams of the products obtained from hydroboration-oxidation of cis-2-butene , hydroboration-oxidation of trans-2-butene.
attached below
The presence of racemic mixture found as product in both cases shows that products are identical ( i.e. they have same configuration )
Which of the following elements is the largest in size
a. O
b. He
c. K
d. H
Answer:
C. K
i took this class before
Answer: The largest element is K
Explanation: As K has the largest radius among O,He and H
We can use bond-line formulas to represent alkenes in much the same way that we use them to represent alkanes. Consider the following alkene: h5ch5e4 How many carbon atoms are sp2−hybridized in this alkene?
Answer:
2
Explanation:
The number of carbon atoms that are sp²-hybridized in this alkene is 2
Because all the single bonded carbon atoms in the alkene are sp²-hybridized
There are three(3) single formed via sp² orbitals and one ( 1 ) PI bond formed via Pure-P-orbital
attached below is the some part of the solution
Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation you submit is balanced. (You can edit both sides of the equation to balance it, if you need to.)
______________ → BaBr2 + H2O
Answer:
Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O
Explanation:
We have the products of a reaction and we have to predict the reactants. Since the products are binary salt and water, this must be a neutralization reaction. In neutralizations, acids react with bases. The acid that gives place to Br⁻ is HBr, while the base the gives place to Ba²⁺ is Ba(OH)₂. The balanced chemical equation is:
Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O
What effect does hybridization have on chemical bond
How many moles are contained in .984 molecules F2?
Answer:
1.6 x 10⁻²⁴ moles
Explanation:
We have 0.984 molecules of F₂. We know that 1 mol is equal to 6.022 x 10²³ molecules. Thus, we have a conversion factor: 1 mol/6.022 x 10²³ molecules
So, we multiply the molecules of F₂ by the conversion factor to calculate the moles:
0.984 molecules x 1 mol/6.022 x 10²³ molecules = 1.6 x 10⁻²⁴ moles
g Calculate the theoretical yield (in grams) of your product if you start with 0.50 grams of E-stilbene. The molecular weight of E-stilbene is 180.25 g/mol, and the molecular weight of the product is 340.058 g/mol. 0.5109 grams 0.9433 g 0.2342 g 0.6312 g
Answer:
0.9433g
Explanation:
Theoretical yield is defined as the mass produced assuming all reactant reacts producing the product.
Assuming the reaction is 1:1, we need to find the moles of E-stilbene (Reactant). If all reactant reacts, the moles of E-stilbene = Moles of product.
Using the molar mass of the product we can find the theoretical yield as follows:
Moles E-stilbene:
0.50g * (1mol/180.25g) = 0.00277 moles = Moles Product
Mass Product = Theoretical yield:
0.00277 moles * (340.058g/mol) = 0.9433g
Question 16(Multiple Choice Worth 5 points)
(04.01 LC) Which statement is true about the total mass of the reactants during a chemical change?
O It is destroyed during chemical reaction.
O It is less than the total mass of the products. O It is equal to the total mass of the products.
O It is greater than the total mass of the products.
Answer:
It is equal to the total mass of the products.
Explanation:
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Given the amount of camphor (200mg) we are using in this experiment, please determine how many mg of sodium borohydride to use in this reaction. We would like you to use 5.2 molar equivalents of this reagent. This means 5.2 times the mmol of camphor we are using. As an example: for 110.0 mg of camphor,142 mg of NaBH4 would be used (see if you can confirm this result). For complete credit, your work needs to be clearly drawn out!
Answer:
Explanation:
From the given information:
Camphor may be reduced as readily in the presence of sodium borohydride(NaHB4). The resulting compound which is stereoselective requires 1 mole of sodium borohydride (NaHB4) to reduce 1 mole of camphor in this reaction. The reaction is shown below.
Through the reduction process of camphor, the reducing agent can reach the carbonyl face with a one-carbon linkage. The product stereoisomer is known as borneol.
If the molecular weight of camphor = 152.24 g/mol
and it mass = 200 mg
The its no of moles = 200 mg/ 152.24 g/mol
= 1.3137 mmol
Now the amount of the required mmol for NaBH4 to be consumed in the reaction = 5.2 × 1.3137 mmol
= 6.831 mmol
since the molar mass of NaBH4 = 37.83 g/mol
Then, using the same formula:
No of moles = mass/molar mass
mass = No of moles × molar mass
mass = 6.831 mmol × 37.83 g/mol
mass of NaBH4 used = 258.42 mg
A gas mixture is made by combining 8.7 g each of Ar, Ne, and an unknown diatomic gas. At STP, the mixture occupies a volume of 17.28 L. What is the molar mass of the unknown gas
Answer: Molar mass of the unknown gas is 73.153 g/mol.
Explanation:
Given: Mass of each gas = 8.7 g
Volume = 17.28 L
Let us assume that the molar mass of gas is m g/mol.
Molar mass of Ar is 40 g/mol and Ne is 20 g/mol.
Hence, total moles of each gas are as follows.
[tex](\frac{8.7}{40} + \frac{8.7}{20} + \frac{8.7}{m}) mol[/tex]
At STP, the total volume of these gases is as follows.
[tex](\frac{8.7}{40} + \frac{8.7}{20} + \frac{8.7}{m}) mol \times 22.4 L = 17.28 L\\(\frac{8.7}{40} + \frac{8.7}{20})22.4 L + \frac{8.7}{m} \times 22.4 L = 17.28 L\\14.616 + \frac{8.7}{m} \times 22.4 L = 17.28 L\\\frac{8.7}{m} \times 22.4 L = (17.28 L - 14.616)\\\frac{8.7}{m} \times 22.4 L = 2.664 \\m = 73.153 g/mol[/tex]
Thus, we can conclude that molar mass of the unknown gas is 73.153 g/mol.
ELECTROLYSIS OF MOLTEN NaCl IS DONE IN A DOWNS CELL OPERATING AT 7.0 VOLTS AND 4.0X10^4A. HOW MUCH Na(s) AND Cl2(g) CAN BE PRODUCED IN 8 HOURS IN SUCH A CELL?
Answer:
Mass of sodium produced = 2.74 × 10⁵ g of Na
Mass of chlorine produce = 4.24 × 10⁵ g of Cl₂
Explanation:
In the electrolysis of molten NaCl as described above, the quantity of charge used is given by the formula, Q = I × t
Where I isnthe current passed in amperes and t is time in seconds.
Q = 4.0 × 10⁴ A × (8 × 60 × 60) s = 1.152 × 10⁹ C
Equation for the discharge of sodium is; Na+ + e- ---> Na (s)
One mole of electrons is required to discharge one mole of Na
One mole of electron = 1 faraday = 96500 C
One mole of Na has a mass of 23 g
96500 C produces 23 g of Na
1.152 × 10⁹ C will produce 23 g × 1.152 × 10⁹ C / 96500 C = 2.74 × 10⁵ g of Na
Equation for the discharge of chlorine gas is; 2 Cl- ---> Cl₂(g) + 2e-
Two mole of electrons are required to discharge one mole of chlorine gas
Two moles of electron = 2 faraday = 2 × 96500 C = 193000
One mole of Cl₂ has a mass of 71 g
193000 C produces 71 g of Cl₂
1.152 × 10⁹ C will produce 71 g × 1.152 × 10⁹ C / 193000 C = 4.24 × 10⁵ g of Cl₂
The amount of Na produced is 274551 g and the amount of Cl₂ produced is 423763.5 g.
Current passed through the cell = [tex]4.0\times10^4 A[/tex]
Time = 8 Hours
We have to calculate the amount of Na and [tex]Cl_2[/tex] produced in 8 hours in the downs cell.
What is a down cell?The Downs process is an electrochemical method for the commercial preparation of metallic sodium, in which molten NaCl is electrolyzed in a special apparatus called the Downs cell.
The total charge passed through the cell is calculated by the given formula as
Charge(Q) = Current(I) × time(t)
Q = [tex]4.0\times10^4 A \times t[/tex]
[tex]t = 8 \times 60\times 60[/tex] sec
t = 28800 sec
Q = [tex]4.0\times10^4 A \times 28800[/tex] sec
Q = [tex]115200\times 10^4 A\ sec[/tex]
We know that, Ampere = Coulombs per sec
Q = [tex]115200\times 10^4\ C[/tex]
1 mol of electrons 96500 C charge
Therefore, the number of mols of electrons carries this [tex]115200\times 10^4\ C[/tex] charge = [tex]\frac{115200\times10^4}{96500}[/tex] = 1.1937 × 10⁴ = 11937 mol electrons
In the Down's cell
Half cell reactions are:
Reduction half-reaction: [tex]2Na^+(aq)+2e^- \to 2Na(s)[/tex]
Oxidation half-reaction: [tex]2Cl^-(aq)\to Cl_2(g) + 2e^-[/tex]
We know that no. of moles = [tex]\frac{given \ mass}{molar \ mass}[/tex]
Molar mass of Na = 23 g/mol
The mass of Na formed = 11937 mol × 23 g/mol = 274551 grams
The molar mass of Cl₂ = 71 g/mol
The mass of Cl₂ = [tex]\frac{11937}{2}\times 71 = 423763.5 \ grams[/tex]
Hence, the amount of Na produced is 274551 g and the amount of Cl₂ produced is 423763.5 g.
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Which substrate is used in the last step of glycolysis
Pyruvate Kinase
Pyruvate Kinase performs a substrate level phosphorylation on ADP to generate an ATP and pyruvate, the final product of glycolysis.
PK dificiency is transmitted in an autosomal recessive disorder in which both alleles must contain the mutated gene, PK-LR.
Hope it helps you! ＼(^ᴥ^)／
There are four containers: a 100-mL beaker, 250-mL Erlenmeyer flask, a 500-mL beaker, and a 1-L Florence flask. They contain coffee, tea, water, and milk, although not in that order. Use the following facts to identify the beverage in each container.
a. the 500-mL container has a beverage commonly associated with breakfast.
b. the largest container has a colorless liquid (i.e. neither yellow nor orange).
c. the beverage in the smallest container is opaque. (you cannot see through it).
d. One clear liquid is in a container half the volume of a colored liquid.
e. The only combustible liquid has exactly twice the volume of an opaque liquid.
Answer:
There are four containers: a 100-mL beaker, 250-mL Erlenmeyer flask, a 500-mL beaker, and a 1-L Florence flask. They contain coffee, tea, water, and milk, although not in that order. Use the following facts to identify the beverage in each container.
a. the 500-mL container has a beverage commonly associated with breakfast.
b. the largest container has a colorless liquid (i.e. neither yellow nor orange).
c. the beverage in the smallest container is opaque. (you cannot see through it).
d. One clear liquid is in a container half the volume of a colored liquid.
e. The only combustible liquid has exactly twice the volume of an opaque liquid.
Explanation:
a. The 500-mL container has a beverage commonly associated with breakfast is coffee.
b. The largest container has a colorless liquid (i.e. neither yellow nor orange) water.
c. The beverage in the smallest container is opaque. (you cannot see through it) milk.
d. One clear liquid is in a container half the volume of a colored liquid tea.
The 500-mL container has a beverage commonly associated with breakfast is coffee. (rest answers are as follows)
How to indentify beverages ?The Indentification of the beverages can be done by knowing the content and optical activity that uniquely identify the container.
The 500-mL container has a beverage commonly associated with breakfast is coffee.The largest container has a colorless liquid (i.e. neither yellow nor orange) water.The beverage in the smallest container is opaque. (you cannot see through it) milk.One clear liquid is in a container half the volume of a colored liquid tea.Learn more about optical activity here ;
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