He radioisotope radon-222 has a half-life of 3.8 days. how much of a 65-g sample of radon-222 would be left after approximately 15 days?
Copper(i) oxide, cu2o, is reduced to metallic copper by heating in a stream of hydrogen gas. what mass of water is produced when 10.00 g copper is formed?
Final answer:
The mass of water produced when 10.00 g of copper is formed by the reduction of Cu2O with hydrogen gas is 1.4131 g. This is calculated using stoichiometry and the molar mass of water.
Explanation:
The mass of water produced when 10.00 g of copper is formed from the reduction of copper(I) oxide (Cu2O) by hydrogen gas (H2) can be determined by using stoichiometry. The reaction is as follows:
Cu2O(s) + H2(g) → 2 Cu(s) + H2O(g)
First, calculate the moles of copper produced using its molar mass. Since the copper produced is 10.00 g, and the molar mass of copper is approximately 63.55 g/mol, the moles of copper formed are:
10.00 g Cu × (1 mol Cu / 63.55 g Cu) = 0.157 mol Cu
According to the reaction, 1 mole of Cu2O produces 2 moles of Cu, so we have:
0.157 mol Cu × (1 mol Cu2O / 2 mol Cu) = 0.0785 mol Cu2O
For each mole of Cu2O reduced, 1 mole of water (H2O) is produced:
0.0785 mol Cu2O × (1 mol H2O / 1 mol Cu2O) = 0.0785 mol H2O
Finally, to find the mass of water produced:
0.0785 mol H2O × (18.015 g H2O / 1 mol H2O) = 1.4131 g H2O
Therefore, the mass of water produced is 1.4131 g.
Write a balanced half-reaction for the oxidation of manganese ion mn 2 to permanganate ion mno−4 in basic aqueous solution. be sure to add physical state symbols where appropriate.
Answer : The balanced oxidation half reaction in basic medium will be :
[tex]Mn^{2+}(aq)+8OH^-(aq)\rightarrow MnO_4^-(aq)+4H_2O(l)+5e^-[/tex]
Explanation :
Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.
Rules for the balanced chemical equation in basic solution are :
First we have to write into the two half-reactions.Now balance the main atoms in the reaction.Now balance the hydrogen and oxygen atoms on both the sides of the reaction.If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion [tex](OH^-)[/tex] at that side where the less number of hydrogen are present.Now balance the charge.The balanced oxidation half reaction in basic medium will be :
[tex]Mn^{2+}(aq)+8OH^-(aq)\rightarrow MnO_4^-(aq)+4H_2O(l)+5e^-[/tex]
According to the valence bond theory the triple bond in ethyne consists of
AlCl3 + Na NaCl + Al Did Cl change oxidation number?
Answer : There is no change takes place in the oxidation number of Cl.
Explanation :
The given chemical reaction is,
[tex]AlCl_3+Na\rightarrow NaCl+Al[/tex]
This reaction is an unbalanced reaction because the chlorine atoms are not balanced.
In order to balance the chemical reaction, the coefficient 3 is put before the Na and NaCl.
The balanced chemical reaction will be,
[tex]AlCl_3+3Na\rightarrow 3NaCl+Al[/tex]
Now we have to calculate the oxidation number of all the elements.
In [tex]AlCl_3[/tex], the oxidation number of Al and Cl are, (+3) and (-1) respectively.
The oxidation number Na and Al are, zero (0)
In [tex]NaCl[/tex], the oxidation number of Na and Cl are, (+1) and (-1) respectively.
From this we conclude that the oxidation number of Cl changes from (-1) to (-1) that means remains same.
Therefore, there is no change takes place in the oxidation number of Cl.
In a lead-acid storage battery, pbo2 is reduced to
What is the emf of a cell consisting of a pb2+ / pb half-cell and a pt / h+ / h2 half-cell if [pb2+] = 0.83 m, [h+] = 0.064 m and ph2 = 1.0 atm ?
The emf of the electrochemical cell has been calculated to be 0.0532 V.
The emf has been the potential of the cell in the reaction with the change in the electrons in the reaction. The emf of the cell has been given by the Nernst equation as;
[tex]emf=E^\circ _{cell}-\dfrac{0.059}{n}\;log\;\dfrac{1}{\rm concentration} [/tex]
Computation for the emf of the cellThe given cell has the number of electrons transfer, [tex]n=2[/tex]
The concentration of [tex]\rm Pb^2^+=0.83\;M[/tex]
The concentration of [tex]\rm H^+=0.064\;M[/tex]
The cell potential of the reaction has been, [tex]E^\circ =-0.126\;\text V[/tex]
Substituting the values for the emf of the cell:
[tex]emf=-0.126\;-\dfrac{0.059}{2}\;\times\;log\;\dfrac{1}{[0.83]\;\times\;[0.064]^2}\\ emf=0.0532\;\text V [/tex]
The emf of the electrochemical cell has been calculated to be 0.0532 V.
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Which of the following properties increases down the periodic table? A. Number of valence electrons B. Electronegativity C. Atomic radius D. Ionization energy
The property from the provided options that increases down the periodic table is the atomic radius, due to the addition of electron shells as we move down groups.
Explanation:In the periodic table, as we move down a group, there are certain properties that increase, and others that decrease. Among the options provided: the number of valence electrons, electronegativity, ionization energy, and atomic radius, the property that increases down the periodic table is the atomic radius.
This is because as we go down each group in the periodic table, there's an additional electron shell added to the atoms. So, even though the positive charge in the nucleus also increases, it's largely screened or shielded by the inner-shell electrons from interacting with the outer-shell or valence electrons. The result is that the atomic size or radius increases.
On the other hand, electronegativity and ionization energy generally decrease down a group because the increasing atomic radius means the outer electrons are less tightly bound to the nucleus.
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The theoretical yield for the reaction above was 2.78 grams silver (Ag). The experimental yield was 2.55 grams. Calculate the percent yield
To calculate the percent yield of silver, divide the experimental yield (2.55 g) by the theoretical yield (2.78 g) and multiply by 100, resulting in a percent yield of 91.73%.
The concept of percent yield is a key aspect of laboratory work in chemistry that compares what was actually obtained from a reaction to what could be obtained based on stoichiometry. The percent yield is calculated using a simple formula: (actual yield/theoretical yield) imes 100. In the student's experiment where the theoretical yield was 2.78 grams of silver and the experimental yield was 2.55 grams, we calculate the percent yield by dividing 2.55 by 2.78 and then multiplying by 100.
Therefore, the calculation is:
(2.55 g / 2.78 g) imes 100 = 91.73%
So, the percent yield for the reaction is 91.73%.
When water reaches its boiling point and turns into water vapor what happens to the molecules structure
A weak acid is a dilute acid that is not very powerful
a. True
b. False
Which tool is used to hold workpieces tightly so that both of your hands can be free to work on them? A. Snap-ring pliers B. Needlenose pliers C. Vise D. Bench
How much energy is required to ionize hydrogen when it is in the n = 4 state? express your answer to three significant figures?
The energy required to ionize a hydrogen atom in the n = 4 state is 0.85 eV.
Explanation:The energy required to ionize a hydrogen atom when it is in the n = 4 state can be calculated using the energy levels of the hydrogen atom. The energy for any state n is given by En = -13.6 eV / n2, where 13.6 eV is the ionization energy of the hydrogen atom from its ground state. To find the ionization energy from n = 4, we substitute 4 for n.
E4 = -13.6 eV / 42
E4 = -13.6 eV / 16 = -0.85 eV
The negative sign indicates that this is the energy the electron has relative to the energy of a free electron at rest (which is defined as 0 eV). To ionize the atom, we need to provide enough energy to bring the electron's energy up to 0 eV. Therefore, the ionization energy required is the absolute value of E4.
Ionization Energy = |E4| = 0.85 eV
The electron dot structure for CI is
A. Cl (w/ a dot above the C)
B. Cl (w/ a dot above, to the left of, and under the C, as well as 2 dots to the right of the 1)
C. Cl (w/ 2 dots above and to the left of the C, 1 underneath, and 2 dots to the right of the 1)
D. Cl (w/ 2 dots above, to the left of, and under the C, as well as 2 dots to the right of the 1)
The electron dot structure for Cl is ... (P.S. The l in Cl is a lowercase "L", for Chlorine, in case anyone else may have been confused.)
The electron dot structure for Cl is B. Cl (w/ a dot above, to the left of, and under the C, as well as 2 dots to the right of the 1).
Explanation:The electron dot structure for Cl is B. Cl (w/ a dot above, to the left of, and under the C, as well as 2 dots to the right of the 1). The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs) and one shared pair of electrons (written between the atoms). A dash (or line) is sometimes used to indicate a shared pair of electrons.
Methanol (ch3oh) can be made by the reaction of co with h2: co(g)+2h2(g)⇌ch3oh(g) to maximize the equilibrium yield of methanol, would you use a high or low temperature?
Is it possible for two yellow belied noombats to have a green bellied child?
Calculate the molarity of each solution. 28.33 g c6h12o6 in 1.28 l of solution
Explanation:
Molarity is the number of moles present in a liter of solution.
Mathematically, Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
And, no. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
Molar mass of [tex]C_{6}H_{12}O_{6}[/tex] is 180.15 g/mol. Therefore, number of moles present will be as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex] = [tex]\frac{28.33 g}{180.15 g/mol}[/tex]
= 0.157 mol
Hence, calculate the molarity as follows.
Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
= [tex]\frac{0.157 mol}{1.28 L}[/tex]
= 0.122 M
Thus, we can conclude that molarity of the solution is 0.122 M.
How many structural and geometrical isomers are there of chloropropene?
What is the mass occupied by 44.8 l of nitrogen (n2) at standard conditions?
Nonmetals gain electrons under certain conditions to attain a noble-gas electron configuration. how many electrons must be gained by the element c?
If a titration of hcl with naoh took 15.25ml of a 0.1250 m naoh solution, how many moles of naoh was used
Answer:
[tex]n_{HCl}=1.906x10^{-3}molHCl[/tex]
Explanation:
Hello,
Titration is widely used to determine the neutralized moles of either an acid or base. In this case, the idea is to titrate (neutralize) hydrochloric acid with sodium hydroxide based on the following reaction:
[tex]NaOH+HCl-->NaCl+H_2O[/tex]
Thus, one computes the neutralized moles of hydrochloric acid (equivalence of moles) as long as the mole ratio between the acid and the base is 1 to 1 and the moles of both of them must be equal for the neutralization to be successfully carried out as shown below:
[tex]n_{HCl}=n_{NaOH}\\n_{HCl}=0.1250mol/L*0.01525L\\n_{HCl}=1.906x10^{-3}molHCl[/tex]
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How many grams of glucose are needed to prepare 400. g of a 2.00% (m/m) glucose solution g?
To create 400 grams of a 2% mass/mass glucose solution, you need 8 grams of glucose.
Explanation:You're trying to find out how many grams of glucose are needed to prepare a 400 gram 2% mass/mass glucose solution. A 2% w/w glucose solution means that for every 100 grams of solution, 2 grams are glucose. Therefore, if you have 400 grams of solution, the amount of glucose required will be 2% of 400 grams.
To calculate this, you will multiply 400 grams by 0.02 (which is the decimal equivalent of 2%). So, 400 grams * 0.02 = 8 grams. Therefore, you need 8 grams of glucose to prepare 400 grams of a 2% w/w glucose solution.
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Vanillin, c8h8o3 (m = 152 g/mol), is the molecule responsible for the vanilla flavor in food. how many oxygen atoms are present in a 45.0 mg sample of vanillin?
In a 45.0 mg sample of vanillin, there are approximately 5.36 x 10^20 oxygen atoms. The calculation involves converting the mass of the sample to grams, calculating the number of moles, and using Avogadro's number to determine the number of oxygen atoms.
Explanation:The question is asking how many oxygen atoms are present in a 45.0 mg sample of vanillin, which is a molecule responsible for the vanilla flavor in food. The molecular formula of vanillin is C8H8O3 and its molar mass is 152 g/mol.
First, we have to convert the mass of the sample from milligrams (mg) to grams (g) because the molar mass is in grams. Therefore, 45.0 mg equals 0.045 g.
Next, we calculate the number of moles of vanillin in the sample using the equation:
Number of moles = Mass / Molar mass
Substituting the given values:
Number of moles = 0.045 g / 152 g/mol = 2.96 x 10^-4 moles.
Each molecule of vanillin contains 3 oxygen atoms. Therefore, in one mole of vanillin, there are 3 moles of oxygen atoms. From the concept of Avogadro's number, we know that one mole of any substance contains 6.02 x 10^23 entities (atoms, molecules, ions etc.).
Therefore, in 2.96 x 10^-4 moles of vanillin, we have (3 moles O atoms / mole of vanillin) x (2.96 x 10^-4 moles of vanillin) x (6.02 x 10^23 O atoms / mole of O atoms) = 5.36 x 10^20 oxygen atoms.
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Calculate the molarity of a solution that contains 0.175 mol of zncl2 n exactly 150 ml of solution.
Explanation:
Molarity is the number of moles present in liter of a solution.
Mathematically, Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
It is given that no. of moles present into the solution are 0.175 mol and volume is 150 ml.
As 1 ml = 0.001 L. So, 150 ml will be equal to 0.15 L.
Hence, calculate the molarity as follows.
Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
= [tex]\frac{0.175 mol}{0.15 L}[/tex]
= 1.16 M
Thus, we can conclude that molarity of the given solution is 1.16 M.
Question 1 a sample of 0.255 mole of gas has a volume of 748 ml at 28°c. calculate the pressure of this gas. (r= 0.0821 l ∙ atm / mol ∙ k) 0.784 atm 8.42 atm 0.00842 atm 7.84 × 10-4 atm none of the above
Answer : The pressure of the gas is, 8.42 atm
Explanation :
Using ideal gas equation,
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = ?
V = volume of the gas = 748 ml = 0.748 L
conversion used : (1 L = 1000 ml)
T = temperature of the gas = [tex]28^oC+273+28=301K[/tex]
n = number of moles of the gas = 0.255 mole
R = gas constant = 0.0821 L.atm/mole.K
Now put all the given values in the ideal gas equation, we get the pressure of the gas.
[tex]P\times (0.748L)=0.255mole\times (0.0821L.atm/mole.K)\times (301K)[/tex]
[tex]P=8.42atm[/tex]
Therefore, the pressure of the gas is, 8.42 atm
What is the ph of a 0.0055 m ha (weak acid) solution that is 8.2% ionized?
Determine the amount of heat(in Joules) needed to boil 5.25 grams of ice. (Assume standard conditions - the ice exists at zero degrees Celsius, melts at zero degrees Celsius, and boils at 100 degrees Celsius. Remember that you need to take into account three changes: melting ice, heating water, and vaporizing the water.)
The amount of heat needed to boil 5.25 grams of ice when we take into account three changes as melting, heating and vaporizing the water is 15797.93 J.
How do we calculate total heat?Total heat for the given condition will be calculated by the addition of the heat of fusion, heat of vaporization and specific heat of water.
In the question it is given that,
mass of ice = 5.25 grams
Change in temperature = 100 - 0 = 100 degree celsius
For the melting of ice:We know that heat of fusion of water = 334 J/g
Required heat for the melting of ice = 334 × 5.25 = 1753.5 J
For heating water:Amount of heat will be used by using the formula as,
Q = mcΔT, where
c = specific heat of water = 4.18 J/gK
Required heat for heating of water = 4.18 × 5.25 × 100 = 2195.18 J
For vaporization of water:We know that heat of vaporization of water = 2257 J/g
Required heat for vaporization of ice = 2257 × 5.25 = 11849.25 J
Total amount of heat involved = 1753.5 + 2195.18 + 11849.25 = 15797.93 J
Hence total amount of heat is 15797.93 J.
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Classify these solids as molecular, ionic, or atomic. co2 c cacl2 c6h12o6 kbr pbs
Explanation:
Molecular compounds : Molecular compounds are formed by the sharing of electrons with each elements.
Ionic compound : When compounds are formed by the transfer of electrons, they are called Ionic compounds.
Atomic species : Species comprised of only atom.
Co2 -- Molecular
C -- Atomic
CaCl2 -- Ionic
C6H12O6 --- Molecular
KBr --- Ionic
PbS ---- Ionic
The above compounds can be classified as;
CO₂- Molecular compound
C - Atomic compound
CaCl₂ - ionic compound
C₆H₁₂O₆ - Molecular compound
KBr - Ionic compound
PbS - Ionic compound
Further Explanation:A compound A compound is a substances that contains two or more different atoms that are bonded together.When the atoms are similar the substance is known as a molecule, therefore not all molecules are compounds.Types of compounds Ionic compounds Ionic compounds are compounds that contain ions. They are as a result of ionic bonding between a metal atom and a non-metal atom.During ionic bonding formation the metallic atom looses electrons while the non-metallic atom gains electrons.Ionic compounds contains both negatively charged ion(anion) and positively charged ion (cation).Examples of ionic compounds, NaCl, LiF, KCl, Cs2S, KBr, PbS etc.
Molecular compounds
Molecular compounds are covalent compounds that are formed when atoms of elements share electrons in a covalent bond to form molecules.Molecular covalent compounds are formed between non-metal atoms as a result of covalent bond between non-metal atoms.These compounds are electrically neutral.Examples of molecular compounds include; CO₂, H₂O, NO, SF₆, etc.Atomic covalent compounds
Atomic compounds are also an example of covalent compounds that result from sharing of electrons between non-metal atoms in a covalent bond.Usually atoms of the same element such as carbon are joined together by a covalent bond.Examples of such compounds are, diamond and graphite.Keywords: Compound, types of compounds, ionic compounds, molecular compounds, atomic compounds
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Subject: Chemistry
Topic: Structure and bonding
Sub-topic: Types of structures and compounds
which of the following represents a stable octet? A 1s2 2s2 2p6 3s2 3p6 4s2
B [He] 2s2 2p3
C 1s2 2s2 2p1
D [Ne] 3s2 3p6
If 4.27 g sucrose (c12h22o11) are dissolved in 15.2 g water, what is the boiling point of the resulting solution? kb for water = 0.512c/m.
Answer : The boiling point of a solution is, [tex]100.42^oC[/tex]
Explanation :
Formula used for Elevation in boiling point :
[tex]\Delta T_b=k_b\times m[/tex]
or,
[tex]T_b-T^o_b=\frac{1000\times k_b\times w_2}{w_1\times M_2}[/tex]
where,
[tex]T_b[/tex] = boiling point of solution = ?
[tex]T^o_b[/tex] = boiling point of pure water = [tex]100^oC[/tex]
[tex]k_b[/tex] = boiling point constant for water = [tex]0.512^oC/m[/tex]
m = molality
[tex]w_2[/tex] = mass of solute (sucrose) = 4.27 g
[tex]w_1[/tex] = mass of solvent (water) = 15.2 g
[tex]M_2[/tex] = molar mass of solute (sucrose) = 342.3 g/mole
Now put all the given values in the above formula, we get the boiling point of a solution.
[tex]T_b-100^oC=\frac{1000\times 0.512^oC/m\times 4.27g}{15.2g\times 342.3g/mole}[/tex]
[tex]T_b=100.42^oC[/tex]
Therefore, the boiling point of a solution is, [tex]100.42^oC[/tex]
The temperature of the solution is 100.42°C.
Given that;
ΔT = K m i
Where;
ΔT = boiling point elevation
K = boiling point constant for water
m = molality of the solution
i = Van't Hoff factor
Number of moles of solute = 4.27 g /342 g/mol = 0.0125 moles
Molality of the solution = 0.0125 moles/15.2 × 10^-3 Kg = 0.822 m
Since the boiling point of pure water = 100°C
Let the boiling point of pure water be Ta
Let the boiling point of the solution be Tb
ΔT = Tb - Ta = Tb - 100
Substituting values;
Tb - 100 = 0.512c/m × 0.822 m × 1
Note that the Van't Hoff factor (i) = 1 because the solute is molecular
Tb = [0.512°C/m × 0.822 m × 1] + 100°C
Tb = 100.42°C
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