Answers
As shown below, propanamide contains two hydrogen atoms directly attached to Nitrogen atom. In this case the hydrogen atoms become partially positive and can form Hydrogen Bond Interactions with the Nitrogen atoms of neighbor propanamide molecule.
Due to the involvement of Hydrogen bonding which is strongest among intermolecular interactions increases the boiling point of propanamide which are absent in N,N−dimethylformamide.
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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Related Questions
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?
Answers
find the moles of O2 used = mass/molar mass
= 40 g/32g/mol = 1.25 moles
write the reacting equation
4Na+ O2 = 2Na2O
by use of mole ratio between Na to O2 which is 4 :1
the moles of Na = 1.25 x 4 = 5 moles
mass of Na = mass x molar mass
= 5 moles x 23 g /mol= 115 moles
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.
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?
Answers
20 mL / 1000 mL x 3 M = .06 moles sugar
30 mL / 1000 mL x 5.69 M = .1707 moles sugar
Total = .2307 moles
Find how many mL of solution you have. 20 mL + 30 mL = 50 mL
1000 mL / 50 mL x .2307 moles = 4.614 M solution
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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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?
Answers
R₀ = 4.5 x 10⁻⁶ Ci (3.70 x 10¹⁰ decays/s / 1 Ci)
= 166500 decays / s
The number of nuclei is:
N₀ = (166500 decays/s) / (5730 yr * 3.154 x 10⁷s /1yr) = 9.2 x 10⁻⁷
The mass of a ₆C¹⁴ source is:
m = N₀m₀ = (9.2 x 10⁻⁷) * (2.34 x 10⁻²⁶) = 2.16 x 10⁻³² kg
What type(s) of intermolecular forces are expected between ch3conhch3 molecules?
Answers
________________
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.
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 − ]?
Answers
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O
Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O
Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L
∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
= 0.0199 X 16.08 X 10^(-3)
= 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion = [tex] \frac{number of moles of [OH-]}{volume of solution (L)} [/tex] = [tex] \frac{3.1999 X 10^(^-^4^)}{10 X 10^(^-^3^)} [/tex] = 0.03199 M
What is the molarity of a solution which contains 58.4 g of sodium chloride dissolved in of solution?
Answers
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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The voltage generated by the zinc concentration cell described by the line notation zn(s) ∣∣ zn2+(aq,0.100 m) ∥∥ zn2+(aq,? m) ∣∣ zn(s) is 14.0 mv at 25 °c. calculate the concentration of the zn2+(aq) ion at the cathode.
Answers
Answer: The concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.295 M
Explanation:
The half reactions for the cell is:
Oxidation half reaction (anode): [tex]Zn(s)\rightarrow Zn^{2+}(0.100M,aq.)+2e^-[/tex]
Reduction half reaction (cathode): [tex]Zn^{2+}(?M,aq.)+2e^-\rightarrow Zn(s)[/tex]
In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]_{anode}}{[Zn^{2+}]_{cathode}}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E_{cell}[/tex] = 14.0 mV = 0.014 V (Conversion factor: 1 V = 1000 mV)
[tex][Zn^{2+}]_{anode}[/tex] = 0.100 M
[tex][Zn^{2+}]_{cathode}[/tex] = ? M
Putting values in above equation, we get:
[tex]0.014=0-\frac{0.0592}{2}\log \frac{0.100M}{[Zn^{2+}]_{cathode}}[/tex]
[tex][Zn^{2+}]_{cathode}=0.295M[/tex]
Hence, the concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.295 M
The concentration of Zn²⁺ ion at cathode is 0.295 M
Redox half equationsThe redox half equations are those of oxidation and reduction
Oxidation half reaction: Zn(s) ---> Zn²⁺ + 2 e⁻ [Zn⁺] = 0.100 M
reduction half reaction: Zn²⁺ (aq) + 2 e⁻ ----> Zn (s) [Zn⁺] = y
The Ecell = 14.0mV = 0.014 V
Using the Nernst equation:
Ecell = E°cell - 0.0592/n * log([Zn⁺]anode/[Zn⁺]cathode)since it is a concentration cell, E°cell = 0where n is number moles; n = 2
0.014 = 0 - 0.0592/2 * log (0.1/y)
y = 0.295 M
Therefore, the concentration of Zn²⁺ ion at cathode is 0.295 M
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What is the [H+] if the pH of a solution is 7.90
Answers
pH is calculated using the hydrogen ion concentration
pH = -log[H⁺]
if we know the pH we can calculate the H⁺ concentration
[H⁺] = antilog(-pH)
[H⁺] = 1.26 x 10⁻⁸ M
therefore [H⁺] is 1.26 x 10⁻⁸ M
Answer:
1.26 x 10^-8
Explanation
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Marshall determines that a gas has a gage pressure of 276 kPa what's the absolute pressure of this gas
Answers
Atmospheric pressure is equivalent to 100 kPa.
Gage pressure is 276 kPa.
Then, we add both values.
N = 100 kPa + 276 kPa
N = 376 kPa
The absolute pressure of this gas is 376 kPa.
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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
Identify the lowest energy lewis structure for nitrogen oxide
Answers
Answer:
Among given Lewis structures, Structure-C has the lowest energy.
Explanation:
The greater the stability of any compound the lesser will be its energy. Among different Lewis structures of N₂O or any other compound, the one which comprises of less number of formal charges will be having lesser energy. I have selected structure-C but not Structure-B (as both have same number of formal charges) because in structure-C oxygen has a formal charge of -1. Being more electronegative oxygen tends to attract electrons toward itself. While, in structure-B one of the Nitrogen atom is having -1 charge which is not feasible and makes the structure energetically high.
Note:
Formula for calculating formal charge is as follow,
Formal Charge = # of Valence e⁻s - [e⁻s in lone pairs + 1/2 bonding e⁻s]
Ydrogen and fluorine combine according to the equation h2(g) + f2(g) → 2 hf(g) if 5.00 g of hydrogen gas are combined with 38.0 g of fluorine gas, the maximum mass of hydrogen fluoride that could be produced is
Answers
Molar mass (F2)=2*19.0=38.0 g/mol
Molar mass (HF)=1.0+19.0=20.0 g/mol
5.00 g H2 * 1mol H2 /2 g H2=2.50 mol H2
38.0 g F2*1mol F2/38.0 g F2=1.00 mol F2
H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 1 mol
From problem 2.50 mol 1 .00mol
We can see that excess of H2, and that F2 is a limiting reactant.
So, the amount of HF is limited by the amount of F2.
H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 2 mol
From problem 1.00 mol 2.00mol
2.00 mol HF can be formed.
2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed
A gas sample contains 4.00g of CH4 and 2.00g of He. What is the volume of the sample at STP?
Answers
It means that in order to calculate volume occupied both by CH₄ and He, we must calculate their moles.
Hence,
Moles are given as,
Moles = Mass / Mass
Moles of CH₄;
Mole-CH₄ = 4 g / 16 g.mol⁻¹
Mole-CH₄ = 0.25 mol
Moles of He;
Mole-He = 2 g / 4 g.mol⁻¹
Mole-He = 0.5 mol
Now adding moles of both gases,
= 0.25 + 0.5
= 0.75 mol
Therefore, when,
1 mole of gas occupy = 22.4 L of Volume
Then,
0.75 mol will occupy = X L of Volume
Solving for X,
X = (0.75 mol × 22.4 L) ÷ 1 mole
X = 16.8 L
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.
Give the oxidation state of the metal species in each complex. ru(cn)(co)4 -
Answers
[Ru (CN) (CO)₄]⁻
Where;
[ ] = Coordination Sphere
Ru = Central Metal Atom = Ruthenium
CN = Cyanide Ligand
CO = Carbonyl Ligand
The charge on Ru is calculated as follow,
Ru + (CN) + (CO)₄ = -1
Where;
-1 = overall charge on sphere
0 = Charge on neutral CO
-1 = Charge on CN
So, Putting values,
Ru + (-1) + (0)₄ = -1
Ru - 1 + 0 = -1
Ru - 1 = -1
Ru = -1 + 1
Ru = 0
Result:
Oxidation state of the metal species in each complex [Ru(CN)(CO)₄]⁻ is zero.
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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1. General statement that is not applied on metals: https://brainly.com/question/2474874
2. The neutral element represented by the excited state electronic configuration: https://brainly.com/question/9616334
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.
What is the effect of removing some so3 from a system initially at equilibrium?
Answers
2 SO₂(g) + O₂ (g) ⇄ 2 SO₃ (g)
According to Le Chatelier's Principle," when a system at equilibrium is subjected to any external stress (like changing temperature, pressure or concentration) the system will tend to adjust itself in such a way to minimize the effect of that stress.
In given reaction, when the equilibrium is disturbed by removing SO₃ from the system then the equilibrium will shift in the forward direction resulting in consumption of more reactants. Or if the Oxygen gas is in excess then the concentration of SO₂ will decrease more.
Answer:
Right
Left
Left
Right
Explanation:
For the equilibrium system described by this equation, what will happen if SO3 is removed?
The equilibrium shifts to the
✔ right
What will happen if NO is added?
The equilibrium shifts to the
✔ left
.For the equilibrium system described by this equation, what will happen if SO2 is removed?
The equilibrium shifts to the
✔ left
.
What will happen if NO2 is added?
The equilibrium shifts to the
✔ right
Magnesium hydroxide (Mg(OH)2): g/mol Iron(III) oxide (Fe2O3): g/mol
Answers
Answer:
Magnesium hydroxide (Mg(OH)2): 58.33 g/mol
Iron(III) oxide (Fe2O3): 159.70 g/mol
Explanation:
To find the molar masses of both magnesium hydroxide and iron(III) oxide, add each molar mass in each compound.
Solving:
[tex]\section*{Molar Mass of Magnesium Hydroxide (Mg(OH)_2):}\textbf{Identify Atomic Masses}\begin{itemize} \item Magnesium (Mg): 24.31 g/mol \item Oxygen (O): 16.00 g/mol (2 oxygen atoms) \item Hydrogen (H): 1.01 g/mol (2 hydrogen atoms)\end{itemize}[/tex]
[tex]\textbf{Calculate Molar Mass:}\[\text{Molar mass of Mg(OH)}_2 = \text{Mg} + 2 \times (\text{O} + \text{H})\]\[= 24.31 \, \text{g/mol} + 2 \times (16.00 \, \text{g/mol} + 1.01 \, \text{g/mol})\]\\\[= 24.31 \, \text{g/mol} + 2 \times 17.01 \, \text{g/mol}\]\[= 24.31 \, \text{g/mol} + 34.02 \, \text{g/mol} = \boxed{58.33 \, \text{g/mol}}\][/tex]
[tex]\hrulefill[/tex]
[tex]\section*{Molar Mass of Iron(III) Oxide (Fe_2\text{O}_3):}\textbf{Identify Atomic Masses}\begin{itemize} \item Iron (Fe): 55.85 g/mol (2 iron atoms) \item Oxygen (O): 16.00 g/mol (3 oxygen atoms)\end{itemize}[/tex]
[tex]\textbf{Calculate Molar Mass:}\[\text{Molar mass of Fe}_2\text{O}_3 = 2 \times \text{Fe} + 3 \times \text{O}\]\\\[= 2 \times 55.85 \, \text{g/mol} + 3 \times 16.00 \, \text{g/mol}\]\\\[= 111.70 \, \text{g/mol} + 48.00 \, \text{g/mol} = \boxed{159.70 \, \text{g/mol}}\][/tex]
[tex]\hrulefill[/tex]
Which of these do not obey the octet rule? select all that apply. select all that apply. clo clo− clo2− clo3− clo4−?
Answers
[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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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
Answers
Nuclear fission and fusion both affect the nucleus of an atom.
The final products of fission and fusion are elements that are different than the original.
Fission occurs mostly with elements heavier than lead on the periodic table.
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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How many grams of NAOH are contained within 0.785 moles of NAOH?
Answers
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.
What is the final step in the scientific method
Answers
Hope this helps!
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.
Answers
= 3.287
Thus, number of millimoles of vinegar present in sample = 3.287
Now, Molarity of vinegar solution = [tex] \frac{\text{number of millimoles of vinegar}}{\text{volume of solution (ml)}} [/tex]
= [tex] \frac{3.287}{26.89} [/tex]
= 0.1222 mol/dm3
Molar concentration of vinegar present in sample solution is 0.1222 M
What element is found in all organic compounds?
a.carbon?
Answers
The presence of which magnetic feature best explains why a magnet can act at a distance on other magnets or on objects containing certain metals?
Answers
Answer:
Magnetic fields.
Explanation:
Hello,
The presence of a magnetic fields best explains why a permanent magnet can act on another magnetic objects when are separated by a certain distance. There exist magnetic field lines which are emanated from the north pole to the south pole whose path is curved. Now, take into account that the longer the distance between the objects, the lower the magnetic field or force.
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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.
Answers
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.
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
Answers
It states that at constant temperature, pressure is inversely proportional to volume of gas.
PV = k
where P - pressure, V - volume and k - constant
P1V1 = P2V2
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting these values in the equation
4.00 atm x 500 L = 8.0 atm x V
V = 250 L
new volume is 250 L
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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%?
Answers
m₁(NaCl) = 10.0 g.
m₁(solution) = 255 g + 10 g = 265 g.
ω₁ = 10 g / 265 g · 100%.
ω₁ = 3.77% ÷ 100% = 0.0377.
ω₂= 10% ÷ 100% = 0.1.
ω₂= m₁(NaCl) + m₂(NaCl) / m₁(solution) + m₂(NaCl).
0.1 = 10 g + m₂(NaCl) / 265 g + m₂(NaCl).
26.5 g + 0.1·m₂(NaCl) = 10 g + m₂(NaCl).
0.9·m₂(NaCl) = 16.5 g.
m₂(NaCl) = 18.33 g.
Carbon tetrachloride will have which shape?
A) bent
B) linear
C) tetrahedral
D) triangular
Answers
Fill in the missing blank 2C4H6 + _______ → 8CO2 + 6H2O
Answers
that is
= 2C4H6 + 11O2 = 8 CO2 + 6H2O
C4H6 reacted with oxygen (O2 )through the process of combustion to form carbon iv oxide (CO2) and water (H2O) . 11 infront of O2 is to make sure the molecules of O2 is balanced in both reactant side and the product side
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]
Answers
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]
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?
Answers
by use of ideal gas equation Pv=nRT where n is number of moles, calculate the moles of H2 produced
by making n the subject of the formula
n= PV/RT
p= 827 torr
R(gas constant)= 62.36 L.torr/mol.K
T= 273 +22=295 k
V=143 L
n =(827 torr x143 L)/ 62.36 L.torr/mol.k x295 k) =6.43 moles
write the reacting equation
that is caH2 +2H2O= Ca(OH)2 + 2H2
by use of mole ratio between CaH2 to H2 which is 1 :2 the moles of CaH2 = 6.43 x1/2=3.215 moles
mass of CaH2 is therefore= moles of CaH2 x molar mass of CaH2
=3.215 moles x 42 g/mol = 135.03 grams
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.
Answers
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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