What type(s) of intermolecular forces are expected between ch3conhch3 molecules?
The intermolecular forces between CH3CONHCH3 molecules are dipole-dipole attractions and hydrogen bonding.
Explanation:The type of intermolecular forces that are expected between CH3CONHCH3 molecules are dipole-dipole attractions and hydrogen bonding.
CH3CONHCH3, also known as acetamide, has a polar carbonyl group (C=O) and an amide group (C-NH). The oxygen atom in the carbonyl group is more electronegative than the carbon atom, resulting in a partial negative charge on the oxygen and a partial positive charge on the carbon. This dipole moment allows for dipole-dipole attractions between acetamide molecules.
In addition, the hydrogen atoms in the amide group can form hydrogen bonds with the lone pairs of electrons on the nitrogen atom of neighboring acetamide molecules. Hydrogen bonding is a stronger type of dipole-dipole attraction and is responsible for the high boiling point and other properties of acetamide.
How many grams of cah2 are needed to generate 143 l of h2 gas if the pressure of h2 is 827 torr at 22 ∘c? g.com?
What is the [H+] if the pH of a solution is 7.90
Answer:
1.26 x 10^-8
Explanation
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What is the molarity of a solution which contains 58.4 g of sodium chloride dissolved in of solution?
The molarity of a solution is calculated by the formula: Molarity = moles of solute divided by volume of solution in liters. For the mentioned solution, we have roughly 1 mole of sodium chloride. The molarity can be obtained by dividing this number by the volume of the solution in liters.
Explanation:The molarity of a solution is given by the formula: Molarity (M) = moles of solute / volume of solution in liters.
Via the provided information, we know that one mole of sodium chloride (NaCl) weighs 58.44 g, and we have 58.4 g NaCl in our solution, roughly equivalent to 1 mole. The volume of the solution unfortunately is not given in the question. If this missing value can be retrieved, the molarity can be calculated by dividing the number of moles (which is 1 in this case) by the volume of the solution in liters.
Assuming a 1 L solution for simplicity, the molarity will be: 1 mol NaCl / 1 L solution = 1 M NaCl solution.
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Which of the following is not a group found in a protein?
Carboxyl Group
L Group
R Group
Amino Group
Answer:
L Group
Explanation:
Proteins are made up of amino acids, these amino acids determine the protein's functional groups, in the image I annexed you can observe that amino acids have an amino group (NH2), an acid group(COOH) and a R group, a lateral chain that varies depending on the amino acid, in the protein these amino acids are linked through a peptide bond leaving one side of the chain with an acid group and the other side with an amide chain, keeping the R groups as well.
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Marshall determines that a gas has a gage pressure of 276 kPa what's the absolute pressure of this gas
And object has a mass of 120 KG’s on the moon what is the force of gravity acting on the object of the moon
The ka of hypochlorous acid (hclo) is 3.0 x 10-8 at . what is the % ionization of hypochlorous acid in a 0.015 m aqueous solution of hclo at ?
Answer is: the % ionization of hypochlorous acid is 0.14.
Balanced chemical reaction (dissociation) of an aqueous solution of hypochlorous acid:
HClO(aq) ⇄ H⁺(aq) + ClO⁻(aq).
Ka = [H⁺] · [ClO⁻] / [HClO].
[H⁺] is equilibrium concentration of hydrogen cations or protons.
[ClO⁻] is equilibrium concentration of hypochlorite anions.
[HClO] is equilibrium concentration of hypochlorous acid.
Ka is the acid dissociation constant.
Ka(HClO) = 3.0·10⁻⁸.
c(HClO) = 0.015 M.
Ka(HClO) = α² · c(HClO).
α = √(3.0·10⁻⁸ ÷ 0.015).
α = 0.0014 · 100% = 0.14%.
The percent ionization of the solution is 0.14%.
The equation of the reaction is;HClO(aq) ⇄ H^+(aq) + ClO^-(aq)
I 0.015 0 0
C -x +x +x
E 0.015 - x x x
Now the Ka of the solution is obtained from;Ka = [ H^+] [ClO^-]/[HClO]
3.0 x 10^-8= x^2/ 0.015 - x
3.0 x 10^-8(0.015 - x) = x^2
4.5 x 10^-10 - 3.0 x 10^-8x = x^2
x^2 + 3.0 x 10^-8x - 4.5 x 10^-10 = 0
x = 0.000021 M
Percent ionization = 0.000021 M/0.015 M × 100/1Percent ionization = 0.14%
Hence, the percent ionization of the solution is 0.14%.
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Lakes that have been acidified by acid rain can be neutralized by limiting the addition of limestone how much limestone is required to completely neutralize a 4.3 billion liter lake with a ph of 5.5
The amount of limestone required to neutralize the lake is approximately [tex]\( 2.653 \times 10^4 \, \text{kg} \).[/tex]
To determine how much limestone ([tex]CaCO_3[/tex]) is required to neutralize a lake acidified by sulfuric acid ([tex]H_2SO_4[/tex]), we need to follow these steps:
1. Calculate the total mass of [tex]H_2SO_4[/tex] in the lake.
2. Determine the moles of [tex]H_2SO_4[/tex] present.
3. Use stoichiometry to find the moles of [tex]CaCO}_3[/tex] required to neutralize the [tex]H_2SO_4[/tex]
4. Convert the moles of [tex]CaCO}_3[/tex] to mass in kilograms.
Step 1: Calculate the Total Mass of [tex]H_2SO_4[/tex] in the Lake
The lake volume is [tex]\( 5.2 \times 10^9 \) liters.[/tex]
The concentration of [tex]H_2SO_4[/tex] is [tex]\( 5.0 \times 10^{-3} \) g/L.[/tex]
Total mass of [tex]H_2SO_4[/tex]
[tex]\[ \text{Total mass of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} \][/tex]
[tex]\[ \text{Total mass of H}_2\text{SO}_4 = 5.0 \times 10^{-3} \, \text{g/L} \times 5.2 \times 10^9 \, \text{L} \][/tex]
[tex]\[ \text{Total mass of H}_2\text{SO}_4 = 2.6 \times 10^7 \, \text{g} \][/tex]
Step 2: Determine the Moles of [tex]H_2SO_4[/tex]
Molar mass of [tex]H_2SO_4[/tex]
[tex]\[ \text{H}_2\text{SO}_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \, \text{g/mol} \][/tex]
Moles of [tex]H_2SO_4[/tex]
[tex]\[ \text{Moles of H}_2\text{SO}_4 = \frac{\text{Total mass of H}_2\text{SO}_4}{\text{Molar mass of H}_2\text{SO}_4} \][/tex]
[tex]\[ \text{Moles of H}_2\text{SO}_4 = \frac{2.6 \times 10^7 \, \text{g}}{98 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of H}_2\text{SO}_4 = 2.653 \times 10^5 \, \text{mol} \][/tex]
Step 3: Use Stoichiometry to Find the Moles of [tex]CaCO}_3[/tex] Required
The neutralization reaction between [tex]H_2SO_4[/tex] and [tex]CaCO}_3[/tex] is:
[tex]\[ \text{H}_2\text{SO}_4 + \text{CaCO}_3 \rightarrow \text{CaSO}_4 + \text{CO}_2 + \text{H}_2\text{O} \][/tex]
From the balanced equation, 1 mole of [tex]H_2SO_4[/tex] reacts with 1 mole of [tex]CaCO}_3[/tex]
Therefore, moles of [tex]CaCO}_3[/tex] required:
[tex]\[ \text{Moles of CaCO}_3 = \text{Moles of H}_2\text{SO}_4 \][/tex]
[tex]\[ \text{Moles of CaCO}_3 = 2.653 \times 10^5 \, \text{mol} \][/tex]
Step 4: Convert the Moles of [tex]CaCO}_3[/tex] to Mass in Kilograms
Molar mass of [tex]CaCO}_3[/tex]
[tex]\[ \text{CaCO}_3 = 40 + 12 + 3 \times 16 = 100 \, \text{g/mol} \][/tex]
Mass of [tex]CaCO}_3[/tex]
[tex]\[ \text{Mass of CaCO}_3 = \text{Moles of CaCO}_3 \times \text{Molar mass of CaCO}_3 \][/tex]
[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^5 \, \text{mol} \times 100 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^7 \, \text{g} \][/tex]
Convert grams to kilograms:
[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^7 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} \][/tex]
[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^4 \, \text{kg} \][/tex]
The complete Question is
Lakes that have been acidified by acid rain can be neutralized by the addition of limestone (CaCO3). How much limestone in kg would be required to completely neutralize a 5.2 x 10^9 L lake containing 5.0 x 10-3 g of H2SO4 per liter?
If you ix 20.0 ml of a 3.00m sugar solution with 30.0 ml of a 5.69 m sugar solution, you will end up with a sugar solution of?
If a person mixes 20.0 ml of a 3.00m sugar solution with 30.0 ml of a 5.69m sugar solution, you will end up with a sugar solution of:
4.614m solutionAccording to the given question, we can see that there is a mixture of 20 milliliters of 3 moles of sugar solution with 30 milliliters of 5.69 moles of another sugar solution, then we need to find the total sugar solution from the mixtures.
As a result of this, we need to first convert the milliliters to moles.
20 mL/1000 mL x 3 moles = 0.06 moles of sugar
30 mL/ 1000 mL x 5.69 moles = 0.1707 moles of sugar
Please note that we are dividing by 1000 mL to convert to liters.
Now, we add up the two values, 0.06 + 0.1707 = 0.2307 moles
Next, we add the total millimeters we have so far which would be
20 mL + 30 mL = 50 mL
Finally, we would convert the 50 mL to liters before we get our final answer.
1000 mL / 50 mL = 20 L
Now, we multiply 20 L by 0.2307 moles = 4.614 M solution
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Which symbol can be used to indicate a chemical reaction that takes place in the presence of a catalyst?
A: 350° C
B: Pt
C: 0.5 atm
D: the triangle
What are the products obtained in the electrolysis of molten nai?
The electrolysis of molten sodium iodide (NaI) yields sodium metal and iodine gas. The process involves the migration of ions to respective electrodes, gaining or losing electrons, thus getting reduced or oxidized respectively.
Explanation:The products obtained from the electrolysis of molten sodium iodide (NaI) are sodium (Na) and iodine (I2). During electrolysis, sodium ions are reduced at the cathode to form sodium metal, shown by the half-equation 2Na+ + 2e- → 2Na. Meanwhile, iodide ions get oxidized at the anode to produce iodine gas, as depicted by 2I- → I2 + 2e-.
In a nutshell, the process involves electrolysis of molten sodium iodide using a Downs cell. Positively charged sodium ions migrate to the negatively charged cathode and gain electrons, reducing them to sodium metal. Conversely, the negatively charged iodide ions migrate to the positively charged anode and lose electrons, getting oxidized to iodine gas.
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The electrolysis of molten sodium iodide (NaI) produces sodium (Na) and iodine (I2). In the process, sodium ions migrate to the cathode and are reduced to sodium metal, while iodide ions migrate to the anode and are oxidized to iodine gas.
Explanation:The process of electrolyzing molten sodium iodide, or NaI, is similar both in concept and execution to the electrolysis of molten sodium chloride, which is a better-known and more commonly discussed process. Electrolysis of molten sodium iodide will produce sodium (Na) and iodine (I) as products. Here's a description of how it works:
In the set-up, you would have a Downs cell. The Downs cell contains molten sodium iodide. The passage of a direct current through the cell causes the sodium ions to migrate to the negatively charged cathode and pick up electrons, reducing the ions to sodium metal. Iodide ions migrate to the positively charged anode, lose electrons, and undergo oxidation to iodine gas.
The overall cell reaction would be: 2NaI -> 2Na + I2
This is a simple explanation of the process and should suffice for your understanding. Do keep in mind that actual industrial processes could have additional steps and complexities not addressed here.
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Suppose a student titrates a 10.00-ml aliquot of saturated ca(oh)2 solution to the equivalence point with 16.08 ml of 0.0199 m hcl. what was the initial [oh − ]?
Which of these do not obey the octet rule? select all that apply. select all that apply. clo clo− clo2− clo3− clo4−?
[tex]\boxed{{\text{ClO,ClO}}_{\text{2}}^ - {\text{,ClO}}_{\text{3}}^ - {\text{,ClO}}_{\text{4}}^ - }[/tex] does not follow the octet rule.
Further Explanation:
Octet rule: states that for the stability of any element it must have a valence shell of eight electrons (octet means a group of eight). This rule is given by Kossel and Lewis.
Atoms of same or different elements with an incomplete electronic configuration that is having electrons less than 8 are unstable and they combine together to form stable molecules with a complete octet. They can combine by sharing of electrons or loss or gain of electrons.
In [tex]{\mathbf{ClO}}[/tex], Chlorine (Cl) has 7 electrons in its outer shell and oxygen has 6 electrons in its outer shell. So chlorine and oxygen share 2 electrons each to complete their octet and attain stability. Chlorine atom in the compound has ten electrons in its outer shell so it doesn’t follow the octet rule.
In [tex]{\mathbf{Cl}}{{\mathbf{O}}^ - }[/tex], Chlorine(Cl) has 7 electrons in its outer shell and oxygen has a unit negative charge on it [tex]\left( {{{\text{O}}^ - }} \right)[/tex] so, it has 7 electrons also in its outer shell. So chlorine and oxygen share one electron each to complete their octet and attain stability. All the atoms have eight electrons in their outer shell so they follow the octet rule.
In [tex]{\mathbf{ClO}}_2^ -[/tex], Chlorine (Cl) has 7 electrons in its outer shell and there is 2 oxygen attached to Cl. One has a unit negative charge on it [tex]\left( {{{\text{O}}^ - }} \right)[/tex] and the other is neutral. So, chlorine and neutral oxygen share 2 electrons each to get a stable octet and form a double bond. Similarly, chlorine shares 1 electron with oxygen ion with a negative charge on it forming a single bond with it. A chlorine atom has ten electrons in its outer shell so the compound doesn’t follow the octet rule.
In [tex]{\mathbf{ClO}}_3^ -[/tex], Chlorine (Cl) has 7 electrons in its outer shell and there is 3 oxygen attached to Cl. One has a unit negative charge on it [tex]\left( {{{\text{O}}^ - }} \right)[/tex] and the other two are neutral. So, chlorine and both neutral oxygen share 2 electrons to get a stable octet and form a double bond. Similarly, chlorine shares 1 electron with oxygen ion with a negative charge on it, forming a single bond with it. Here, Cl has ten electrons in its outer shell, so it doesn’t follow the octet rule.
In [tex]{\mathbf{ClO}}_4^ -[/tex] , Chlorine (Cl) has 7 electrons in its outer shell and there is 4 oxygen attached to Cl. One has a unit negative charge [tex]\left( {{{\text{O}}^ - }} \right)[/tex]on it and the other three are neutral. So, chlorine and both neutral oxygen share 2 electrons to get a stable octet and form a double bond. Similarly, chlorine shares 1 electron with oxygen ion with a negative charge on it, forming a single bond with it.
Here, Cl has fourteen electrons in its outer shell, so it doesn’t follow the octet rule.
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Answer details:
Grade: Secondary School
Subject: Chemistry
Chapter: Chemical Bonding
Keywords: octet rule, stability, oxygen, chlorine, covalent, sharing, ClO2-, ClO3-, ClO4-, ClO-and ClO.
The species that do not obey the octet rule are; ClO, ClO2^-, ClO3^-, ClO4^-.
The octet rule states that atoms must have eight electrons in their outermost shell in order to attain stability. Hence, stable molecules, ions and atoms are expected to contain atoms that obey the octet rule.
However, in some chemical species, atoms of elements do not obey the octet rule. For instance, in ClO, chlorine has seven valence electrons and oxygen has six electrons. The octet rule is clearly violated in this case.
Again, chlorine is able to expand its octet (contain more than eight valence electrons). This is seen in the ions; ClO2^- and ClO3^- in which chlorine contains ten valence electrons and ClO4^- in which chlorine contains fourteen valence electrons.
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Give the oxidation state of the metal species in each complex. ru(cn)(co)4 -
The oxidation state of the Ru metal in the complex [tex]{\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}[/tex] is [tex]\boxed{{\text{zero}}}[/tex] .
Further explanation:
Oxidation number:
The oxidation number is used to represent the formal charge on an atom. It also shows that gain or loss of electrons by the atom. Oxidation number can be a positive or negative number but cannot be fractional.
The rules to identify the oxidation state:
(1) The oxidation state of an atom in the elemental form is zero.
(2) The total charge on the species is equal to the sum of the oxidation state of individual atoms.
(3) The oxidation state of halides is -1. For example, fluorine, chlorine, bromine, iodine have -1 oxidation state.
(4)Hydrogen has a +1 oxidation state.
(5) Oxygen has an oxidation state -2.
(6) In a coordination compound, neutral ligands have zero oxidation state and negative ligands such as CN have -1 oxidation.
The given compound is [tex]{\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}[/tex]
Here, CN is a negative ligand thus oxidation state is -1 and CO is a neutral ligand thus it has 0 oxidation state. Also, the complex has -1 negative charge.
The expression to calculate the oxidation state in [tex]{\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}[/tex] is,
[tex]\left[{\left({{\text{oxidation state of Ru}}}\right)+\left({{\text{oxidation state of CN}}}\right)+4\left({{\text{oxidation state of CO}}}\right)}\right]=-1[/tex]
…… (1)
Rearrange equation (1) for the oxidation state of Ru.
[tex]{\text{Oxidation state of Ru}}=\left[{-\left({{\text{oxidation state of CN}}}\right)-4\left({{\text{oxidation state of CO}}}\right)-1}\right][/tex]
…… (2)
Substitute -1for the oxidation [tex]{\text{state}}[/tex] of CN and 0 for the oxidation state of CO in equation (2).
[tex]\begin{aligned}{\text{Oxidation state of Ru}}&=\left[{-\left({-{\text{1}}}\right)-4\left({\text{0}}\right)-1}\right]\\&=0\\\end{aligned}[/tex]
The oxidation state of Ru is zero.
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Answer details:
Grade: Senior school
Subject: Chemistry
Chapter: Coordination complex
Keywords: Oxidation state, metal complex, ru(cn)(co)4-, formal charge, cynide, carbonyl, zero, hydrogen, oxygen, 0 and -1.
The reaction 4 a + c + h → d has the mechanism below. what is the rate law? 1) 2 a → b fast 2) 2 b → b2 fast 3) b2 + c → g slow 4) g + h → d fast
a. rate = k[a]2[c]
b. rate = k[a]4[h][c]
c. rate = k[a]2
d. rate = k[b]2[c]
e. rate = k[a]4[c]
Final answer:
To determine the rate law, the rate-determining step must be identified, which is the third step in the given reaction mechanism. The rate law correlates with the concentration of reactants in this slow step, resulting in a rate law of rate [tex]= k[A]^2[C][/tex] correct answer to the question is a. rate = [tex]k[A]^2[C].[/tex]
Explanation:
To determine the rate law for a reaction with a given mechanism, it is essential to identify the rate-determining step, as the rate law is based on this slowest step. In the provided mechanism, the slow step is the third one, where B2 reacts with C to form G. The overall reaction is 4 A + C + H → D, and the steps are:
2 A → B (fast)
2 B → B2 (fast)
B2 + C → G (slow)
G + H → D (fast)
Because step 3 is the slow, rate-determining step, the rate law will be based on the concentrations of the reactants in this step. Since B is formed from A in a fast step and B2 is formed from 2 B, the rate of formation of B2 is dependent on the concentration of A. However, as B2 forms immediately before the slow step, we look at the concentration of A instead of B2 when writing the rate law.
With the stoichiometry of 2 A forming B and then 2 B forming B2, we can note that the concentration of B (and hence B2) is proportional to [tex][A]^2[/tex]efore, for the rate-determining step, the rate law is:
[tex]rate = k[B2][C] → rate = k[A]^2[C][/tex]
So the correct rate law for the overall reaction is:
[tex]rate = k[A]^2[C][/tex]
[tex]rate = k[A]^2[C][/tex]
What element is found in all organic compounds?
a.carbon?
If 255 g of water has 10.0 g of NaCl dissolved into initially, how much NaCl must be added in order to raise the mass percent of NaCl by 10%?
Fill in the missing blank 2C4H6 + _______ → 8CO2 + 6H2O
What is the mass of a 4.50-μci 146c source? the half-life of 146c is 5730 yr.
express your answer to three significant figures and include the appropriate units?
How many grams of NAOH are contained within 0.785 moles of NAOH?
Answer: The mass of NaOH for given number of moles is 31.4 grams.
Explanation:
To calculate the mass for given number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We are given:
Molar mass of NaOH = 40 g/mol
Moles of NaOH = 0.785 moles
Putting values in above equation, we get:
[tex]0.785mol=\frac{\text{Mass of NaOH}}{40g/mol}\\\\\text{Mass of NaOH}=(0.785mol\times 40g/mol)=31.4g[/tex]
Hence, the mass of NaOH for given number of moles is 31.4 grams.
A gas sample contains 4.00g of CH4 and 2.00g of He. What is the volume of the sample at STP?
To calculate the volume of the gas sample at standard temperature and pressure (STP), we find the moles of each gas (CH4 and He) using their molar masses and then multiply by the molar volume (22.4 L/mol) to get their individual volumes at STP. Adding these volumes together gives us the total volume of the gas sample, which is 16.8 L.
Explanation:To calculate the volume of a gas sample at standard temperature and pressure (STP), we can use the molar volume concept where one mole of any gas occupies 22.4 liters at STP. Methane (CH4) has a molar mass of 16.00 g/mol, and helium (He) has a molar mass of 4.00 g/mol.
To find the moles of CH4, we divide 4.00 g by its molar mass, 16.00 g/mol, which gives us 0.250 moles. Similarly, for He, we divide 2.00 g by its molar mass, 4.00 g/mol, which gives us 0.500 moles. To find the total volume, we sum the volumes of CH4 and He after multiplying their mole quantities by the molar volume (22.4 L/mol).
Total volume of CH4 = 0.250 moles × 22.4 L/mol = 5.60 L
Total volume of He = 0.500 moles × 22.4 L/mol = 11.2 L
Therefore, the total volume of the gas sample at STP is 5.60 L + 11.2 L = 16.8 L.
A gas sample of argon, maintained at constant temperature, occupies a volume of 500. l at 4.00 atm. what is the new volume if the pressure were charged to 8.0 atm
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Suppose that 26.89 ml of vinegar solution requires 33.23 ml of the 0.09892 m sodium hydroxide solution to reach the endpoint. calculate the molar concentration of the vinegar solution.
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Nuclear fission and fusion both affect the nucleus of an atom. Choose all of the items below that are correct.
The final products of fission and fusion are elements that are different than the original
Fusion involves the splitting apart of large atoms into smaller ones
Fission occurs mostly with elements heavier than lead on the periodic table
Two hydrogen atoms can go through fission to form helium
Fusion and fission involve a loss or gain of electrons
The answer options about a nuclear reaction which are correct include the following:
A. Nuclear fission and fusion both affect the nucleus of an atom.
B. The final products of fission and fusion are elements that are different than the original.
D. Fission occurs mostly with elements heavier than lead on the periodic table.
A nuclear reaction refers to a reaction in which the nucleus of an atom is transformed or transmuted by either joining (fusion) or splitting (fission) the nucleus of another atom of a radioactive element. Also, a nuclear reaction is always accompanied by a release of energy.
Generally, the two (2) main types of nuclear reaction are:
Nuclear fusion: it involves the joining of two smaller nuclei of atoms to form a single massive or heavier nucleus with the release of energy. Nuclear fission: it involves the collision of a heavy atomic nucleus with a neutron, thereby causing a split and release of energy.From the above, we can deduce the following about a nuclear reaction:
I. The nucleus of an atom is affected by both nuclear fission and fusion.
II. The final products of both nuclear fission and fusion are chemical elements that are different than the original radioactive chemical elements.
III. Nuclear fission occurs mostly with chemical elements heavier than lead (Pb) on the periodic table because the nuclear force holding their atoms together is lesser than the electromagnetic force pushing the nucleus apart.
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Which statement about ionic bonds is true?
A. In ionic bonds atoms share electrons to achieve a stable outer shell.
B. Ionic bonds occur between non-metals.
C. Ionic bonds occur between two metals.
D. In ionic bonds one atom accepts electrons from another atom to achieve a stable outer shell.
Answer:
In ionic bonds one atom accepts electrons from another atom to achieve a stable outer shell.
Explanation:
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What mass of solute is contained in 25.4 ml of a 1.56 m potassium bromide solution?
Answer : The mass of solute is 4.72 grams.
Explanation :
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
[tex]\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
or,
[tex]\text{Molarity}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
In this question, the solute is potassium bromide.
Given:
Volume of solution = 25.4 mL
Molarity = 1.56 M
Molar mass of potassium bromide = 119 g/mole
Now put all the given values in this formula, we get:
[tex]1.56M=\frac{\text{Mass of solute}\times 1000}{119g/mole\times 25.4L}[/tex]
[tex]\text{Mass of solute}=4.72g[/tex]
Therefore, the mass of solute is 4.72 grams.
Na has an atomic mass of 23.0g and O has an atomic mass of 16.0g. How many grams of Na are needed to completely react with 40.0g of O2?
Final answer:
To completely react with 40.0g of O2, 115.0g of Na are required. This calculation uses stoichiometry and the balanced chemical equation 4Na + O2 → 2Na2O, considering the atomic masses of Na and O.
Explanation:
The question relates to how many grams of sodium (Na) are needed to react completely with 40.0 grams of oxygen (O2). This is a stoichiometry problem that involves understanding the molar mass of the reactants and the chemical reaction equation. The first step is to understand the reaction between sodium and oxygen, which typically forms sodium oxide (Na2O). The balanced chemical equation for this reaction is:
4Na + O2 → 2Na2O
From this equation, we can see that 4 moles of Na react with 1 mole of O2 to produce 2 moles of Na2O. Using the given atomic mass of O (16.0 g/mol) and knowing that oxygen is diatomic in its normal state (O2), the molar mass of O2 is 32.0 g/mol. Therefore, 40.0 g of O2 corresponds to 40.0 g / 32.0 g/mol = 1.25 moles of O2.
Since 4 moles of Na are required for every 1 mole of O2, the moles of Na needed is 4 * 1.25 moles = 5 moles of Na. Given the atomic mass of Na as 23.0 g/mol, the mass of Na required is 5 moles * 23.0 g/mol = 115.0 grams. Therefore, to completely react with 40.0 grams of O2, 115.0 grams of Na are needed.
What are the oxidation numbers of potassium, sulfur, and oxygen, respectively, in k2so4?
In the compound K2SO4, the oxidation numbers for potassium (K), sulfur (S) and oxygen (O) are +1, +6 and -2, respectively.
Explanation:In the compound K2SO4, the oxidation numbers for potassium (K), sulfur (S) and oxygen (O) can be determined based on established guidelines for assigning oxidation states.
In every stable (neutral) atom, the oxidation number is always zero. Therefore, for potassium, in its ionic form, it has an oxidation number of +1.
In general, the oxidation number of oxygen in its compounds is -2. The compound K2SO4 contains 4 oxygen atoms. Therefore, the total oxidation state contributed by oxygen is -8.
To ensure that the compound is electrically neutral, the total oxidation number should be zero. Hence, for sulfur, you would calculate its oxidation state as follows: total oxidation state of compound = (+1 x 2 for potassium) + oxidation state of sulfur + (-2 x 4 for oxygen) = 0. Solving this equation gives the oxidation state for sulfur as +6.
So, the oxidation numbers for potassium, sulfur and oxygen in K2SO4 are +1, +6 and -2, respectively.
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Select the single best answer. why is the boiling point of propanamide, ch3ch2conh2, considerably higher than the boiling point of n,n−dimethylformamide, hcon(ch3)2 (213°c vs. 153°c), even though both compounds are isomeric amides?
The higher boiling point of propanamide compared to N,N-dimethylformamide is due to the ability of propanamide to form stronger intermolecular hydrogen bonds, requiring more energy to break and hence, raising its boiling point.
Explanation:The boiling point of propanamide, CH3CH2CONH2, is considerably higher than that of N,N-dimethylformamide, HCON(CH3)2 because propanamide can form robust intermolecular hydrogen bonds within its molecules. These require a significant amount of energy to break, hence increasing its boiling point. However, in N,N-dimethylformamide, hydrogen bonding is inhibited because the hydrogen atoms are bonded to carbon rather than a more electronegative element like nitrogen or oxygen, which would contribute to stronger hydrogen bonding. Therefore, the boiling point of N,N-dimethylformamide is lower.
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In the nuclear transmutation represented by 168o(p, \alpha) 137n, the emitted particle is ________. in the nuclear transmutation represented by o(p, \alpha) n, the emitted particle is ________. a beta particle. a proton. an alpha particle. a neutron. a positron.
In the nuclear transmutation represented by 168o(p, \alpha) 137n, the emitted particle is alpha. in the nuclear transmutation represented by o(p, \alpha) n, the emitted particle is alpha.
In nuclear reactions, an alpha particle is a common emitted particle. An alpha particle is a relatively heavy particle compared to others, consisting of two protons and two neutrons. The emission of an alpha particle is a form of nuclear decay or transmutation, and it is frequently observed in certain nuclear reactions involving heavy elements.
The emission of an alpha particle is often associated with nuclear decay or transmutation processes. In these processes, the nucleus of an unstable atom transforms into a more stable one by releasing an alpha particle. This helps reduce the excess energy and stabilize the nucleus.
Alpha decay is more commonly observed in heavy elements, particularly those with a high atomic number. Elements with larger atomic nuclei tend to be less stable, and alpha decay is one way for them to reach a more stable configuration.