Answer:
1. C- Three.
2. A- Methionine
3. D- Translocation.
4. C- OH.
5. A - 5'
6. A - 3' carbon
7. A. adenine and guanine
Explanation:
1. A codon is a group of three nucleotide sequence that encodes or specifies an amino acid. This means that, during translation (second stage of gene expression), when a CODON is read, an amino acid is added to the growing peptide chain.
2. The codon that initiates the translation process is called a start codon. It has a sequence: AUG and it specifies Methionine amino acid. Hence, during translation where a tRNA binds to the mRNA codon to read it and add its corresponding amino acid, a tRNA with a complementary sequence of AUG (start codon) binds to it and carries Methionine amino acid.
3. Translocation is a process during translation whereby the mRNA-tRNA moeity moves forward in the ribosome to allow another codon to move into the vacant site for translation process to continue.
4. The sugar component of a nucelotide that makes up the nucleic acid (DNA or RNA) i.e. ribose or deoxyribose, contains an hydroxyll functional group (-OH).
5. A nucleotide consists of a pentose (five carbon) sugar, phosphate group and a nitrogenous base. The phosphate group (PO43-) is attached to the 5' carbon of the sugar molecule.
6. The free hydroxyll group (-OH) of the five carbon sugar molecule in DNA is attached to its 3' carbon.
7. Nitrogenous bases are the third component of a nucleotide, the other two being pentose sugar and phosphate group. The nitrogenous bases are four viz: Adenine, Guanine, Cytosine, and Thymine. These bases are classified into Purines and Pyrimidines based on the similarity in their structure. Adenine (A) and Guanine (G) are Purines because they possess have two carbon-nitrogen rings, as opposed to one possessed by Pyrimidines (Thymine and Cytosine).
3. The density of acetic anhydride is 1.08 g/mL. How many moles of acetic anhydride are used in this experiment
Answer:
Number of moles = 0.1058 mol
Explanation:
Density = 1.08 g/mL
Volume = 10ml
Density = Mass / Volume
Mass = Density * Volume
Mass = 1.08 * 10 = 10.8g
The molar mass of acetic anhydride = 102.09 g/mol
Molar mass = Mass / Moles
Upon solving for moles;
Moles = Mass / Molar mass
Moles = 10.8 / 102.09 = 0.1058 mol
The moles of acetic acid used in the experiment has been 0.1058 mol.
Density has been defined as mass of a substance per unit volume. The density has been expressed as:
[tex]\rm Density=\dfrac{Mass}{Volume} [/tex]
The moles have been the mass of the substance with respect to the molar mass. The moles of a substance has been given as:
[tex]\rm Moles=\dfrac{Mass}{Molar\;mass} [/tex]
Computation for the Moles of Acetic AcidThe given sample of acetic acid has density = 1.08 g/ml
The volume of the sample used in the experiment has been 10 ml.
Substituting the values for the mass of acetic acid:
[tex]\rm 1.08=\dfrac{Mass}{10}\\ Mass=1.08\;\times\;10\;g\\ Mass=10.8\;g[/tex]
The mass of the acetic acid used has been 10.8 g.
The molar mass of acetic acid has been 102.09 g/mol.
Substituting the values for the moles of acetic acid:
[tex]\rm Moles=\dfrac{10.8}{102.09} \\ Moles=0.1058\;mol[/tex]
The moles of acetic acid used in the experiment has been 0.1058 mol.
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Use the reaction I2(s) I2(g), H = 62.4 kJ/mol, S = 0.145 kJ/(molK)
At what temperature is the reaction at equilibrium?
A.157K
B.430K
C.0.002K
D.62K
Answer: B. 430 K
Explanation:
According to Gibb's equation:
[tex]\Delta G=\Delta H-T\Delta S[/tex]
[tex]\Delta G[/tex] = Gibbs free energy
[tex]\Delta H[/tex] = enthalpy change = +62.4 kJ/mol
[tex]\Delta S[/tex] = entropy change = +0.145 kJ/molK
T = temperature in Kelvin
[tex]\Delta G[/tex] = +ve, reaction is non spontaneous
[tex]\Delta G[/tex] = -ve, reaction is spontaneous
[tex]\Delta G[/tex] = 0, reaction is in equilibrium
[tex]\Delta H-T\Delta S=0[/tex] for reaction to be spontaneous
[tex]T=\frac{\Delta H}{\Delta S}[/tex]
[tex]T=\frac{62.4kJ/mol}{0.145kJ/molK}=430K[/tex]
Thus the Reaction is spontaneous when temperature is 430 K.
Answer:
430 K
Explanation:
i just took the test on a pex :)
8.1 moles of HCl is put into a beaker with enough water to make a total of 189.98 milliliters of
solution. What is the molarity of the resulting solution? (Round the answer to two decimal places
regardless of significant figures. Do not include units in the answer.)
Answer:
42.64 M
Explanation:
Step 1: Given data
Moles of HCl (solute): 8.1 molVolume of the solution: 189.98 mLStep 2: Convert the volume of the solution to liters
We will use the relationship 1 L = 1,000 mL.
[tex]189.98 mL \times \frac{1L}{1,000mL} = 0.18998L[/tex]
Step 3: Calculate the molarity of the solution
The molarity is equal to the moles of solute divided by the liters of solution.
[tex]M= \frac{8.1mol}{0.18998L} = 42.64 M[/tex]
what is required on a chemical label
Answer: What is required on a chemical label includes pictograms, a signal word, hazard and precautionary statements, the product identifier, and supplier identification.
Explanation:
According to Occupational Safety and Health Administration (OSHA), the following are required on the label of a chemical;
pictograms a signal word hazard and precautionary statementsproduct identifiersupplier identificationAll these requirements on a chemical label helps to minimize the risk associated with such chemical substance especially if it has been marked as hazardous.
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Calculate the molar mass of (NH4)3AsO4.
Answer:
193.07 g / mol
Explanation:
Molar mass of (NH₄)₃ = 3 * (14.01 + 4 * 1.01) = 54.15
As = 74.92 and O₄ = 4 * 16 = 64
Answer is 54.15 + 74.92 + 64 = 193.07
The ΔHfus of silver is 11.30 kJ⋅mol−1 , and its ΔSfus is 9.150 J⋅mol−1⋅K−1 . What is the melting point of silver?
Answer:
[tex]T=1235K=962\°C[/tex]
Explanation:
Hello,
In this case, from thermodynamics, it is widely known that the entropy of fusion is defined in terms of the enthalpy of fusion at the fusion temperature (melting point) as shown below:
[tex]\Delta _fS=\frac{\Delta _fH}{T}[/tex]
Thus, solving for the melting point:
[tex]T=\frac{\Delta _fH}{\Delta _fS}[/tex]
Hence, for silver:
[tex]T=\frac{11.30kJ/(mol)*\frac{1000J}{1kJ} }{9.150J/(mol*K)} \\\\T=1235K=962\°C[/tex]
Which is a typical high temperature for a metal such as silver.
Best regards.
If 500 mL of 2.0 M HCl is diluted with water to a volume of 1 liter, what is the molarity of the new solution?
Answer: 1 mole per L
Explanation:
(2M)(500mL)=(M)(1000mL)
1000=(M)(1000mL)
M=1
Taking into account the definition of dilution, if 500 mL of 2.0 M HCl is diluted with water to a volume of 1 liter, the molarity of the new solution is 1 M.
DilutionWhen it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.
Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.
In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.
A dilution is mathematically expressed as:
Ci×Vi = Cf×Vfwhere
Ci: initial concentrationVi: initial volumeCf: final concentrationVf: final volumeFinal molarityIn this case, you know:
Ci= 2.0 MVi= 500 mLCf= ?Vf= 1 L= 1000 mLReplacing in the definition of dilution:
2.0 M× 500 mL= Cf× 1000 mL
Solving:
(2.0 M× 500 mL)÷ 1000 mL= Cf
1 M= Cf
In summary, if 500 mL of 2.0 M HCl is diluted with water to a volume of 1 liter, the molarity of the new solution is 1 M.
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Match the words in the below to the appropriate blanks in the sentences.
1. When comparing the two elements A s and S n , the more metallic element is ______based on periodic trends alone.
2. When comparing the two elements G e and S b , the more metallic element is ________ based on periodic trends alone.
A. Ge
B. Pb
C. Sb
D. impossible to determine
E. K
Answer:
Sn and Ge
Explanation:
To obtain the more metallic element, we compare the group in which both elements are. Generally the element with the lower ionzation energy is he more metallic one.
Ionization energy increases fro left to right across a period. Ionization energy decreases down the group.
1. When comparing the two elements A s and S n , the more metallic element is ______based on periodic trends alone.
Sn has a lower ionization energy so it is the more metallic one.
2. When comparing the two elements G e and S b , the more metallic element is ________ based on periodic trends alone.
Ge has a lower ionizaiton energy compared to Sb. So it is more metallic element than Sb.
A sample of oxygen is collected over water at a total pressure of 692.2 mmHg at 17°C. The vapor pressure of water at 17°C is 14.5 mmHg. The partial pressure of the O2
Answer:
677.7 mmHg
Explanation:
The first empirical study on the behaviour of a mixture of gases was carried out by John Dalton. He established the effects of mixing gases at different pressures in the same vessel.
Dalton's law states that,the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the individual gases present in the mixture of gases. When a gas is collected over water, the gas also contains some water vapour. The partial pressure of the gas will now be given as; total pressure of gas mixture - saturated vapour pressure of water (SVP) at that temperature.
Given that;
Total pressure of gas mixture = 692.2 mmHg
SVP of water at 17°C = 14.5 mmHg
Therefore, partial pressure of oxygen = 692.2-14.5
Partial pressure of oxygen = 677.7 mmHg
If 50 ml of 1.00 M of H2SO4 and 50 ml of 2.0 M KOH are mixed what is the concentration of the resulting solutes?
Answer: [H2SO4] = 0.5M;
[KOH] = 1M
Explanation: Molarity is the solution concentration defined by:
molarity = [tex]\frac{mol}{L}[/tex] or M
To determine the concentration of the mixture, find how many mols of each compound there are in the mixture:
50 mL = 0.05L
H2SO4
1 mol/L * 0.05L = 0.05mol
KOH
2mol/L * 0.05L = 0.1 mol
The mixture has a total volume of:
V = 50 + 50 = 100 mL = 0.1 L
The concentration of the resullting solutes:
[H2SO4] = [tex]\frac{0.05}{0.1}[/tex] = 0.5 M
[KOH] = [tex]\frac{0.1}{0.1}[/tex] = 1 M
Concentration of H2SO4 is 0.5M while for KOH is 1M.
The force that opposes drag and is powered by combustion reactions in the
engine is
Answer:
Thrust.
Explanation:
hope this helps you :)
Answer:
thrust
Explanation:
The mathematics of combining quantum theory with wave motion of atomic particles is known as _____.
Combining quantum theory with wave motion of atomic particles is: Wave Mechanics
Which of the following would be useful for converting g/mol to g/L?
A. Mass percent
B. Avogadro's number
C. Molarity
D. Molar mass
Answer:
Molarity
Explanation:
The conversion of g/mol to g/L molarity can be used. Thus, option C is correct.
The g/mol has been the amount of solute present in a mole. The g/mol has been the molecular weight of the compound.
The g/L has been the mass of solute present in a L of solution. The g/L has the unit for density.
Molarity has been the moles of solute present in the liter of solution. It has been given as mol/L.
The product of g/mol and g/L gives the value of mol/L. Thus, to convert g/mol to g/L molarity can be used. Thus, option C is correct.
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A compound consisting of atoms of small atomic mass is more likely to require what
Answer:
a lower temperature to liquefy
Explanation:
Look up and record the boiling point of acetic acid, and explain why only some of it evaporates from the reaction mixture.
Answer:
Heating the mixture to a temperature above the boiling point of acetic acid, but below 100°C (the boiling point of water). The vapours from the acetic acid rise, and go into a tube. They are then condensed within the tube, and run off into a separate storage area. Because water can exist as a gas at pretty much any temperature above 0°C, it will result in an impure mixture, but repeatedly doing this will get the acetic acid to the desired purity.
a soluation of acetone in water has a molarity of 2.422M and a density of 0.970 g/mL. Calculate the mole fraction
Answer:
[tex]x_{acetone}=7.970x10^{-3}[/tex]
Explanation:
Hello,
In this case, for the given molarity, we can assume a volume of 1 L of solution, to obtain the following moles of acetone:
[tex]n=0.422mol/L*1L=0.422mol[/tex]
Then, with the density of solution, we can compute the mass of the solution for the selected 1-L volume basis:
[tex]m_{solution}=1L*\frac{1000mL}{1L}*\frac{0.970g}{1mL}=970g[/tex]
After that, we compute the mass of water in the solution, considering the mass of acetone (molar mass = 58.08 g/mol):
[tex]m_{H_2O}=970g-0.422molAcetone*\frac{58.08g\ Acetone}{1mol\ Acetone} =945.49gH_2O[/tex]
Next, the moles of water:
[tex]n_{H_2O}=945.49g*\frac{1molH_2O}{18gH_2O} =52.53molH_2O[/tex]
Finally, the mole fraction:
[tex]x_{acetone}=\frac{n_{acetone}}{n_{acetone}+n_{H_2O}}=\frac{0.422mol}{0.422mol+52.53mol}\\ \\x_{acetone}=7.970x10^{-3}[/tex]
Regards.
What is the [OH-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution
Answer: The [tex][OH^-][/tex] of a solution is [tex]10^{-12}[/tex] M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n = moles of solute
[tex]V_s[/tex] = volume of solution in ml
moles of [tex]HCl[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.0912g}{36.5g/mol}=0.0025mol[/tex]
Now put all the given values in the formula of molality, we get
[tex]Molarity=\frac{0.0025\times 1000}{250}=0.01[/tex]
pH or pOH is the measure of acidity or alkalinity of a solution.
[tex]HCl\rightarrow H^++Cl^{-}[/tex]
According to stoichiometry,
1 mole of [tex]HCl[/tex] gives 1 mole of [tex]H^+[/tex]
Thus [tex]0.01[/tex] moles of [tex]HCl[/tex] gives =[tex]\frac{1}{1}\times 0.01=0.01[/tex] moles of [tex]H^+[/tex]
Putting in the values:
[tex][H^+][OH^-]=10^{-14}[/tex]
[tex][0.01][OH^-]=10^{-14}[/tex]
[tex][OH^-]=10^{-12}[/tex]
Thus the [tex][OH^-][/tex] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is [tex]10^{-12}[/tex] M
The [OH-] of a solution is [tex]10^{12}[/tex] M.
What is Molarity?Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
M = n/ V..................(1)
where,
n = moles of solute
V = volume of solution in ml
Calculation for number of moles:
Moles of HCl = 0.0912 g/ 36.5 g/mol = 0.0025 mol
On substituting the values in equation 1:
M = n/ V
M= 0.0025*1000 / 250
M=0.01 M
pH or pOH is the measure of acidity or alkalinity of a solution.
[tex]HCl---- > H^++Cl^-[/tex]
According to stoichiometry,
1 mole of HCl gives 1 mole of [tex]H^+[/tex]
Thus, 0.01 moles of HCl gives = 1 / 1 *0.01 = 0.01 mole of [tex]H^+[/tex]
On adding the values:
[tex][H^+][OH^-]=10^{14}\\\\(0.01)[OH^-]=10^{-14}\\\\OH^-=10^{-12}[/tex]
Thus, the [OH-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is [tex]10^{-12}[/tex] M.
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Using bond energies, estimate the enthalpy of reaction for the following chemical reaction. CH4(g) + 4 F2(g) → CF4(g) + 4 HF(g) ΔHrxn = ?
Answer:
[tex]\mathbf{ \Delta H_{rxn} = -1936 \ kJ/mol}}[/tex]
Explanation:
The equation for the reaction is given as:
[tex]\mathbf{ CH_{4(g)} + 4 F_{2(g)} \to CF_{4(g)} + 4 HF_{(g) }}[/tex]
At standard conditions; the bond energies are as follows;
Bond Bond Energies (kJ/mol)
C-H 413
F-F 155
C-F 485
H-F 567
[tex]\mathbf{\Delta \ H_{rxn} = \sum \Delta H ( reactant) - \sum \Delta H (product)}[/tex]
[tex]\mathbf{\Delta \ H_{rxn} = \sum [\Delta H \ 4( C-H) + \Delta H \ 4(F-F) ]- \sum[ \Delta H \ 4( C-F)+\Delta H \ 4( H-F)] (product)}[/tex]
[tex]\mathbf{\Delta \ H_{rxn} = \sum{ \Delta \ H (4*413) + \Delta \ H (4*155) - \Delta \ H (4(485)) + \Delta H (4(567) }}[/tex]
[tex]\mathbf{ \Delta H_{rxn} = \sum ({ (1652 + 620) - (1940 + 2268)})}[/tex]
[tex]\mathbf{ \Delta H_{rxn} = \sum ({2272- 4208})}[/tex]
[tex]\mathbf{ \Delta H_{rxn} = -1936 \ kJ/mol}}[/tex]
Write the empirical formula
Answer:
[tex]Pb(CO_{3})_{2} \\Pb(NO_{3})_{4} \\FeCO_{3}\\Fe(NO_{3})_{2}[/tex]
Explanation:
[tex]Pb^{4+}(CO_{3}^{2-})_{2} --->Pb(CO_{3})_{2} \\Pb^{4+} (NO_{3}^{-})_{4} --->Pb(NO_{3})_{4} \\Fe^{2+} CO_{3}^{2-} --->FeCO_{3}\\Fe^{2+} (NO_{3}^{-})_{2}--->Fe(NO_{3})_{2}[/tex]
Covalent bonding occurs between two oxygen atoms to form O2. What's the partial charge on each oxygen atom in 02?
A. Both oxygen atoms have positive partial charge.
B. Both oxygen atoms have negative partial charge.
C. Both oxygen atoms have zero partial charge.
D. One oxygen atom has positive partial charge, and the other has negative partial charge.
Answer:
Both oxygen atoms have zero partial charge.
Explanation:
In covalent bonds, electrons are shared between the two bonding atoms. The shared electron pair of the covalent bond is positioned between the nuclei of the two bonding atoms.
A covalent bond may be formed between two or more elements of different electronegativities. When a difference in electro negativity exists between these atoms in a covalent bond, the molecule becomes polarized. The shared electron pair of the covalent bond becomes closer to the nucleus of the more electronegative atom (s). This more electronegative atom (s) becomes partially negative while the other atom becomes partially positive.
When the two bonding atoms are of equal electronegativities such that the electro negativity difference between the bonding atoms is zero, there is now no difference in electro negativity and no consequent partial charges.
Since the two oxygen atoms have the same electro negativity, there is no difference in electronegativity, hence there is no partial charge between the two oxygen atoms.
Answer:
Both oxygen atoms have zero partial charge.
Explanation:
The lock and key model and the induced fit model are two models of enzyme action explaining both the specificity and the catalytic activity of enzymes. Indicate whether each statement is part of the lock and key model, the induced fit model, or is common to both models.
a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex
c. Enzyme active site has a rigid structure complementary
d. Substrate binds to the enzyme through noncovalent interactions
Answer:
"The active site of the enzyme has a complementary rigid structure" belongs to the key and lock system
"The conformation of the enzyme changes when it binds to the substrate so that the active site conforms to the substrate." belongs to the induced fit system.
"The substrate binds to the enzyme at the active site, forming an enzyme-substrate complex" belongs to both, that is, the key and lock system and the induced fit system.
"The substrate binds to the enzyme through non-covalent interactions" can belong to both enzyme systems.
Explanation:
Enzymatic key and lock systems bear this name because the enzyme at its site of union with the substrate has an ideal shape so that its fit is perfect, similar to a headbreaker, so once they are joined they are not It can bind another substrate to the enzyme, since they are generally associated with strong chemical bonds.
The shape of the enzyme's active site is a negative of what the shape of the substrate would be.
On the other hand, in the mechanism or enzyme system of induced adjustment, the enzyme has an active site that is where it binds with the substrate and another site where another chemical component binds, which when this chemical component binds this enzyme changes its morphology and becomes "active" to bond with your substrate.
This happens a lot in the inactive enzymes that are usually activated in digestive processes since the fact that these enzymes are constantly active would be dangerous, therefore the body takes the induced enzyme system as a control mechanism, where a molecule or chemical compound induces change morphological of an enzyme by means of the allosteric union so that it joins its substrate and catalyzes or analyzes it, depending on the enzymatic character of the enzyme.
The density of methanol at 20 degree Celsius is 0.791 g/ml. What is the mass of a 23.8 ml sample of methanol?
Answer: 18.8 g
Explanation:
To find the mass of a 23.8 mL sample, all we have to do is set up a proportion.
[tex]\frac{0.791g}{mL} =\frac{x}{23.8mL}[/tex]
[tex]18.8358=x[/tex]
x=18.8 g
We round to 18.8 because of significant figures.
Taking into account the definition of density, the mass of a 23.8 mL sample of methanol is 18.83 g.
But first you must know that density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.
Then, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance. That is, it is the relationship between the weight (mass) of a substance and the volume it occupies.
Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:
[tex]density= \frac{mass}{volume}[/tex]
In this case:
density= 0.791 [tex]\frac{g}{mL}[/tex]mass= ?volume= 23.8 mLReplacing in the definition of density:
[tex]0.791 \frac{g}{mL} = \frac{mass}{23.8 mL}[/tex]
Solving:
mass= 0.791 [tex]\frac{g}{mL}[/tex]× 23.8 mL
mass= 18.83 g
In conclusion, the mass of a 23.8 mL sample of methanol is 18.83 g.
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Answer:
The correct answer is 1.17 moles
Explanation:
The ideal gas equation combines the pressure (P), volume (V), temperature (T) and number of moles of a gas (n):
PV= nRT
R is the gas constant (0.082 L.atm/K.mol)
In this case, we have:
V= 7.2 L
P= 4.0 atm
T= 27ºC + 273 = 300 K
We calculate the number of moles (n) of gas as follows:
n= (PV)/(RT)= (4.0 atm x 7.2 L)/(0.082 L.atm/K.mol x 300 K) = 1.17 mol
Cual es la diferencia entre agua pesada y agua ligera a) el agua pesada contiene mas minerales que el agua ligera b) el agua ligera es liquida mientras el agua pesada es solida c) el agua ligera es agua purificada y el agua pesada es agua contaminada d) el agua pesada contiene mas elementos estearato de sodio
Answer:
d) El agua pesada contiene mas elementos
Explanation:
La diferencia fundamental entre el agua pesada y el agua ligera es que la primera tiene una proporción mayor de deuterio que la segunda. El deuterio es un ión del hidrógeno que tiene un peso atómico mayor que el hidrógeno común y corriente. Por ende, la opción D ofrece la mejor aproximación.
Answer:
....................
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Explanation:ki
A 3.00-g sample of an alloy (containing only Pb and Sn) was dissolved in nitric acid (HNO3). Sulfuric acid was added to this solution, which precipitated 2.93 g of PbSO4. Assuming that all of the lead was precipitated, what is the percentage of Sn in the sample? (molar mass of PbSO4 = 303.3 g/mol)
Answer:
33.3% of Sn in the sample
Explanation:
The addition of SO₄⁻ ions produce the selective precipitation of Pb²⁺ to produce PbSO₄.
Moles of PbSO₄ (molar mass 303.26g/mol) in 2.93g are:
2.93g ₓ (1mol / 303.26) = 9.66x10⁻³ moles PbSO₄ = Moles Pb²⁺.
As molar mass of Pb is 207.2g/mol, mass in 9.66x10⁻³ moles of Pb²⁺ is:
9.66x10⁻³ moles of Pb²⁺ ₓ (207.2g / mol) = 2.00g of Pb²⁺
As mass of the sample is 3.00g, mass of Sn²⁺ is 3.00g - 2.00g = 1.00g
And the percentage of Sn in the sample is:
1.00g / 3.00g ₓ 100 =
33.3% of Sn in the sampleCH3-CHCl-CH2-CH2-CH2CHCl-CH3 +concentrated KCN
Answer:
what is the question is it
calculate the amount of carbon dioxid gas in 1.505x10^23 molecules of the gas.
[L=6.02x10^23 mol^-1]
Explanation:
We need to find the amount of carbon dioxide gas in [tex]1.505\times 10^{23}[/tex] molecules of the gas. We know that, 1 mole weighs 44 gram of carbon dioxide which contains [tex]6.022\times 10^{23}[/tex] number of molecules. It means that, [tex]6.022\times 10^{23}[/tex] number of molecules present in 44 grams of carbon dioxide molecule. So,[tex]1.505\times 10^{23}[/tex] number of molecules present in :[tex]=\dfrac{1.0505\times 10^{23}}{6.022\times 10^{23}}\times 44\\\\=7.675\ \text{grams}[/tex]
Hence, 7.675 grams of carbon dioxide is present in [tex]1.505\times 10^{23}[/tex] molecules of the gas.A solid white substance A is heated strongly in the absence of air. It decomposes to form a new white substance B and a gas C. The gas has exactly the same properties as the product obtained when carbon is burned in an excess of oxygen. Based on these observations, can we determine whether solids A and B and the gas C are elements or compounds?
Answer:
A, B and C are compounds
Explanation:
First of all, I need to establish that when carbon is burnt in excess oxygen, carbon dioxide is obtained as shown by this equation; C(s) + O2(g) ----> CO2(g).
Looking at the presentation in the question, A was said to be heated strongly and it decomposed to B and C. Only a compound can decompose when heated. Elements can not decompose on heating. Secondly, compounds usually decompose to give the same compounds that combined to form them. Compounds hardly decompose into their constituent elements.
Again from the information provided, the compound A is a white solid. This is likely to be CaCO3. It decomposes to give another white solid. This may be CaO and the gas was identified as CO2.
Hence;
CaCO3(s)--------> CaO(s) + CO2(g)
A chemist observed an unknown Balmer Series decay through an emission of 410 nm. Using the experimental wavelength, determine the energy levels transition involved in the
emitted wavelength.
Answer:
Option D is correct.
n = 6 to n = 2
Explanation:
Like all waves emitted from the movement of electrons from one energy level to another, the wavelength (λ) is given by the equation involving Rydberg's constant
(1/λ) = Rₕ [(1/n₂²) - (1/n₁²)]
where Rₕ = 10973731.57 m⁻¹ = (1.0974 × 10⁷) m⁻¹
n₂ = principal quantum number corresponding to the final energy level of the electron = 2 (For Balmer Series)
n₁ = principal quantum number corresponding to the final energy level of the electron = ?
λ = 410 nm = (410 × 10⁻⁹) m
(1/λ) = (2.439 × 10⁶) m⁻¹
2.439 × 10⁶ = (1.0974 × 10⁷) [(1/2²) - (1/n₁²)]
0.25 - (1/n₁²) = (2.439 × 10⁶) ÷ (1.0974 × 10⁷) = 0.2222602562
(1/n₁²) = 0.25 - 0.2222602562 = 0.0277397438
n₁² = (1/0.0277397438) = 36.05
n₁ = 6
Hope this Helps!!!
What volume (mL) of 0.0855 M phosphoric acid can be neutralized with 119 mL of 0.315 M sodium hydroxide? A 1320 B 438 C 9.61 D 146 E 1470
Answer:
146 mL
Explanation:
Step 1: Write the balanced equation
H₃PO₄ + 3 NaOH ⇒ Na₃PO₄ + 3 H₂O
Step 2: Calculate the reacting moles of sodium hydroxide
119 mL of 0.315 M NaOH react.
[tex]0.119L \times \frac{0.315mol}{L} = 0.0375mol[/tex]
Step 3: Calculate the reacting moles of phosphoric acid
The molar ratio of H₃PO₄ to NaOH is 1:3. The reacting moles of H₃PO₄ are 1/3 × 0.0375 mol = 0.0125 mol.
Step 4: Calculate the reacting volume of phosphoric acid
0.0125 moles of H₃PO₄ are in a 0.0855 M solution.
[tex]0.0125 mol \times \frac{1L}{0.0855mol} \times \frac{1,000mL}{1L} =146 mL[/tex]