Answer:
Fluorine, Chlorine, Bromine, or Iodine
Explanation:
These all have an ALMOST full valence shell. And they need one more electron so they'd react with hydrogen
Answer:
its chlorine
Explanation:
just trust me do i look like i would lie too you ;-)
btw i just took the test :-)
Write the equilibrium constant expression for the experiment you will be studying this week. 2. If the equilibrium values of Fe3+, SCN- and FeSCN2+ are 9.5 x 10-4 M, 3.6 x 10-4 M and 5.7 x 10-5 M respectively, what is the value of Kc? 3. Write the general form of the dilution equation. 4. A solution is prepared by adding 18 mL of 0.200 M Fe(NO3)3 and 2 mL of 0.0020 M KSCN. Calculate the initial concentrations of Fe3+ and SCNin the solution.
Answer:
Kc = 166.7
[Fe³⁺] = 0.18 M
[SCN⁻] = 2×10⁻⁴ M
Explanation:
In the reaction of Fe³⁺ and SCN⁻, it is formed a complex:
Fe³⁺ + SCN⁻ ⇄ FeSCN²⁺ Kc
Let's make the expression for Kc → [FeSCN²⁺] / [Fe³⁺] . [SCN⁻]
5.7×10⁻⁵ / 9.5×10⁻⁴. 3.6×10⁻⁴ = 166.7
We determine the mmoles, we add from each reactant:
18 ml . 0.2M = 3.6 mmoles of Fe³⁺
2 ml . 0.002M = 4×10⁻³ mmoles of SCN⁻
General form of the dilution equation is:
Concentrated [C] . Concentrated Volume = Diluted [C] . Diluted Volume
Total volume = 20mL
[Fe³⁺]: 3.6 mmoles /20mL = 0.18 M
[SCN⁻]: 4×10⁻³ mmoles /20mL = 2×10⁻⁴ M
The value should be 1.67 x 10^2
The initial concentration should be 0.18 M and 2.0 x 10^(-4) M
Calculation of the value and initial concentration:The value is
= 5.7 x 10^(-5)/(9.5 x 10^(-4) x 3.6 x 10^(-4))
= 167
= 1.67 x 10^2
we know that
Initial moles = volume x concentration
So,
= 18/1000 x 0.200
= 0.0036 mol
Now
Initial moles = volume x concentration
= 2/1000 x 0.0020
= 4.0 x 10^(-6) mol
So,
Total volume should be
= 18 + 2
= 20 mL
= 0.02 L
Now
Initial concentration
= moles /total volume
= 0.0036/0.02
= 0.18 M
Now
Initial concentration
= moles /total volume
= 4.0 x 10^(-6)/0.02
= 2.0 x 10^(-4) M
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Prepare 10.00 mL of 0.010 M NaOH by diluting the NaOH solution used in Trial 1. (Your procedure should clearly show your calculations and the glassware used to perform the dilution.)
Answer:
= 0.2 mL.
Explanation:
Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:
[tex]C_{1} V_{1} = C_{2} V_{2}[/tex]
where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.
Let C1 = 0.5M, V2 = ?
C2 = 0.010M; V2 = 10mL
⇒Volume of stock solution to be diluted, V2
= [tex]\frac{10}{0.5}[/tex] × 0.010
= 0.2 mL.
Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.
I hope this was a detailed explanation given the missing details of "Trial 1" in the question.
plez hurry Which is an important safety precaution that should be taken during a tornado? Stay away from doors and windows. Move to high ground to avoid flood waters. Try to avoid the storm by driving or running. Stay outside to avoid being trapped in a building.
Answer: stay away from doors and windows.
Explanation:
to aviod geting hit by glass
Answer:
Stay away from doors and windows.
Explanation:
Always stay in the center of the room during a tornado storm. Avoid windows, doors, and corners. If you’re near a window, the glass can shatter and hurt you.
Which of the following elements is in the same family as fluorine?
a. silicon
b. antimony
O c. iodine
O d. arsenic
e. None of these.
Answer:
c iodine
Explanation:
fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride
Stearic acid (C18H36O2) is a fatty acid, a molecule with a long hydrocarbon chain and an organic acid group (COOH) at the end. It is used to make cosmetics, ointments, soaps, and candles and is found in animal tissue as part of many saturated fats. In fact, when you eat meat, you are ingesting some fats containing stearic acid. ( of C18H36O2 = –948 kJ/mol, CO2=-393.5kJ/mol, H2O=-241.826kJ/mol).
Calculate the heat (q) released in kcal when 2.831 g of stearic acid is burned completely.
The molar mass of stearic acid is 18*AC+36*AH+2*AO=18*12+36*1+16*2=284g/mol.
n=m/M=2.831/284=0.01 moles
C18H36O2+27O2-->18CO2+18H2O
we have 18*0.01=0.18 moles of CO2
18*0.01=0.18 moles of H2O
0.01*948=9.48kJ from stearic acid
0.18*393.5=70.83kJ from CO2
0.18*241.826=43.52kJ from H2O
9.48+70.83+43.52=123.83kJ
123.83*4.184=518.10kcal
Identify the true statements regarding alpha-1,6 linkages in glycogen.
a. New alpha- 1,6 linkages can only form if the branch has a free reducing end
b. The number of sites for enzyme action on a glycogen molecule is increased through alpha- 1,6 linkages
c. At least four glucose residues separate alpha-1,6 linkages
d. These linkages can only be formed on a chain that is exactly 11 residues long.
e. The reaction that forms alpha-1,6 linkages is catalyzed by a branching enzyme.
Answer:
a. New alpha- 1,6 linkages can only form if the branch has a free reducing end
b. The number of sites for enzyme action on a glycogen molecule is increased through alpha- 1,6 linkages
c. At least four glucose residues separate alpha-1,6 linkages
e. The reaction that forms alpha-1,6 linkages is catalyzed by a branching enzyme.
Explanation:
Glycogen i is the main storage polysaccharide in animals. It a homoplymer of (alpha-1-->4)-linked subunits of glucose molecules, with alpha-1--->6)-linked branches.
The alpha-1,6 branches are formed by the glycogen-branching enzyme which catalyzes the transfer of about 7 glucose residues from the non-reducing end of a glycogen branch having at least 11 residues to the C-6 hydroxyl group of a glucose residue which lies inside the same glycogen chain or another glycogen chain, thereby forming a new branch. This ensures that there are at least four glucose residues separating alpha-1,6 linkages.
The effect of branching is that it makes the glycogen molecule more soluble and also increases the number of non-reducing ends, thereby increasing the number of sites for the action of the enzymes glycogen phosphorylase and glycogen synthase.
Applied Exercises (40 points) Answer the following questions in complete sentences. 1) In molecules with the same number of electron groups but different molecular geometries, discuss what happens to the bond angle? 2) What happens to the bond angle as you increase the number of bonding groups? 3) In 5 electron group molecules, what is the difference between axial and equatorial positions? Which groups are removed as lone pairs are added? 4) What is the difference between tetrahedral bent and
Answer:
See explanation
Explanation:
In molecules with the same number of electron groups but different molecular geometries, the bond angles differ markedly owing to the presence of lone pairs on the central atom. Recall that lone pairs of electrons take up more space around the central atom and causes more repulsion thus squeezing the bond angle and making it less than the value expected on the basis of the Valence Shell Electron Pair Repulsion Theory.
As the number of bonding groups increases, the bond angle increases since the repulsion due to lone pairs of electrons is being progressively removed by increase in the number of bonding groups.
For 5 electron group molecules, the axial groups are oriented at a bond angle of 90° while the equatorial groups are oriented at a bond angle of 120°. In the presence of lone pairs, the equatorial bonds are removed because the equatorial bonds often have a greater bond length than the axial bonds.
In the tetrahedral geometry, four groups are bonded to the central atom while in a bent molecular geometry, only two groups are bonded to the central atom with two lone pairs present in the molecule.
Drag each image to the correct location on the model. Each image can be used more than once. Apply the rules and principles of electron configuration to draw the orbital diagram of aluminum. Use the periodic table to help you.
Answer:
The answer to your question is given below.
Explanation:
Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:
Al (13) —› 1s² 2s²2p⁶ 3s²3p¹
The orbital diagram is shown on the attached photo.
Answer: screen shot
Explanation:
If powdered platinum metal is used to speed up the following reaction: Cl2(g) 3F2(g) --> 2ClF3(g), what would you classify the platinum as
Answer:
Catalyst
Explanation:
For the reaction:
[tex]Cl_2_(_g_)~+~3F_2_(_g_)->2ClF_3_(_g_)[/tex]
We have a main observation: When platinum is added the reaction goes faster. With this in mind, we have to remember the kinetic equilibrium theory. In figure 1, we have an energy diagram. In which we have an specific energy for the reagents and the products. When the reaction takes place, the reaction has to must go through an energy peak. This energy peak is called "activation energy". When platinum is added the activation energy decreases and the reaction can go faster. Therefore, platinum is a "catalyst", a substance with the ability to reduce the activation energy.
I hope it helps!
Automotive antifreeze is typically a 50:50 mixture (by volume) of water and ethylene glycol. Discuss why this solution is useful for protecting automobile engines from both summer and winter temperature extremes.
Answer:
A 50:50 mixture of ethylene glycol and water is effective both summer and winter extremes in temperature because of high boiling point of 106°C and low freezing point of about -37°C. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.
Explanation:
Ethylene glycol or antifreeze is an organic compound which is used in automobile engines as a coolant and also as an anti-freezing agent, however it does not conduct heat effectively as water due to its lower heat capacity. It has a freezing point of -12.9°C and boiling point of 197.3°C.
Water is also used as a coolant in automobile engine but it has a limited range due to its boiling point of 100°C. It is also not a good anti-freezing agent due to it high freezing point of 0.°C
However, when ethylene glycol is mixed with water in a ratio of 50:50, the property of the mixture is enhanced to both serve as a coolant and as an antifreeze. The boiling point is elevated to about 106°C while its freezing point is lowered to about -37°C.
This temperature range is effective for both summer and winter temperatures. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.
Using appropriate chemical reactions for illustration, show how calcium present as the dissolved bicarbonate salt in water is easier to remove than other forms of hardness such dissolved Calcium chloride
Answer:
Explanation:
Calcium bicarbonate dissolved in hard water can easily be removed by heating the hard water . On heating , it decomposes to give calcium carbonate which is insoluble and therefore can be filtered out .
Ca( HCO₃)₂ = CaCO₃ + CO₂ + H₂O.
In this way hardness of water is removed .
A diode has IS = 10−17 A and n = 1.05. (a) What is the diode voltage if the diode current is 70 μA? (b) What is the diode current for VD = 0.1 mV?
Answer:
(a) The diode voltage, [tex]V_D =[/tex] 0.776 V
(b) The diode current, [tex]I_D =[/tex] 3.81 x 10⁻²⁰ A
Explanation:
Given;
saturation current in diode, [tex]I_s[/tex] = 10⁻¹⁷ A
nonideality factor, n = 1.05
(a) the diode voltage
Given diode current, [tex]I_D[/tex] = 70 μA = 7 x 10⁻⁶ A
Diode voltage is calculated as;
[tex]V_D = nV_Tln(1+ \frac{I_D}{I_S} )[/tex]
Where;
[tex]V_T[/tex] is thermal voltage at 25°C = 0.025
[tex]V_D = 1.05 * 0.025 ln(1+ \frac{70*10^{-6}}{1*10^{-17}})\\\\V_D = 0.02625ln(1+ 7*10^{12})\\\\V_D = 0.776 \ V[/tex]
b) the diode current for VD = 0.1 mV
[tex]V_D = nV_Tln(1 +\frac{I_D}{I_S} )\\\\ln(1 +\frac{I_D}{I_S} ) = \frac{V_D}{nV_T} \\\\ln(1 +\frac{I_D}{I_S} ) = \frac{0.1*10^{-3}}{1.05*0.025} \\\\ln(1 +\frac{I_D}{I_S} ) = 0.00381\\\\1 +\frac{I_D}{I_S} = e^{0.00381}\\\\1+ \frac{I_D}{I_S}= 1.00381\\\\ \frac{I_D}{I_S}=1.00381 - 1\\\\ \frac{I_D}{I_S}= 0.00381\\\\I_D = 0.00381(I_S)\\\\I_D = 0.00381(10^{-17})\\\\I_D = 3.81*10^{-20} \ A[/tex]
The SNArreaction requires 1 mmol methyl 4-fluoro-3-nitrobenzoateand 2 mmol of 4-chlorobenzyl amine. Calculate the mass (mg) of both reagents.(2
Answer: Mass of methyl 4-fluoro-3-nitrobenzoate = 199 mg;
Mass of 4-chlorobenzylamine = 282 mg
Explanation: The mass and mol of a molecule is related by its molar mass, which is given in g/mol.
The molar mass of methyl 4-fluoro-3-nitrobenzoate, which has molecular formula: [tex]C_{8}H_{6}FNO_{4}[/tex] is:
[tex]C_{8}H_{6}FNO_{4}[/tex] = 12.8 + 6.1 + 19 + 14 + 16.4 = 199 g/mol
Since it is asking in mg: MM = 199.10³mg/mol
For 4-chlorobenzylamine, with molecular formula [tex]C_{7}H_{8}ClN[/tex]:
[tex]C_{7}H_{8}ClN[/tex] = 12.7 + 8.1 + 35 + 14 = 141 g/mol
In mg: MM = 141.10³mg/mol
The reaction requires 1 mmol of [tex]C_{8}H_{6}FNO_{4}[/tex] , then its mass is:
m = 1.10⁻³ mol * 199.10³mg/mol = 199 mg
For [tex]C_{7}H_{8}ClN[/tex], it requires 2mmol:
m = 2.10⁻³ mol * 141.10³ mg/mol = 282 mg
For the SNAr reaction, it is necessary 199 mg of methyl 4-fluoro-3-nitrobenzoate and 282 mg of 4-chlorobenzylamine
The mass of 1 mmol methyl 4-fluoro-3-nitrobenzoate is 199.14 mg and the mass of 2 mmol of 4-chlorobenzyl amine is 283.2 mg
To determine the masses of both reagents,
First we will determine their molar masses
For methyl 4-fluoro-3-nitrobenzoate (C₈H₆FNO₄)Molar mass = 199.14 g/mol
Now, for the mass of 1 mmol of methyl 4-fluoro-3-nitrobenzoate
Using the formula
Mass = Number of moles × Molar mass
Mass = 1 mmol × 199.14 g/mol
Mass = 199.14 mg
For 4-chlorobenzyl amine (C₇H₈ClN)Molar mass = 141.6 g/mol
Now, for the mass of 2 mmol of 4-chlorobenzyl amine
Mass = 2 mmol × 141.6 g/mol
Mass = 283.2 mg
Hence, the mass of 1 mmol methyl 4-fluoro-3-nitrobenzoate is 199.14 mg and the mass of 2 mmol of 4-chlorobenzyl amine is 283.2 mg
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Which of these are elimination reactions? Check all that apply.
CH3OH + CH3COOH → CH3CO2CH3 + H20
C3H7OH → C3H6 + H20
H9C2Br + NaOH → C2H4 + NaBr + H20
Answer:
C3H7OH → C3H6 + H20
Explanation:
If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.
Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.
Answer:
B and C. Just finished my lesson on Edge.
mechanism of 1-iodobutane reacts with pyridine
Answer:
It is an example of elimination reaction through the E2 mechanism.
Explanation:
The reaction between 1-iodobutane and pyridine is an example of an E2 (bimolecular elimination) elimination reaction.
Pyridine acts predominantly as a base and the given alkyl halide is a primary alkyl halide. Both of these two factors facilitate the E2 mechanism.
Here, both H and Cl are eliminated in a single step to produce 1-butene as the product of the reaction.
The reaction mechanism and the structure of the product are shown below.
The mechanism by which 1-iodobutane reacts with pyridine is by the E2 mechanism.
What is Bimolecular Elimination (E2 Mechanism)?
The E2 mechanism process (Bimolecular Elimination) is a one-step reaction mechanism whereby carbon-hydrogen (C-H) and carbon-halogen (C-X) bonds split to generate a double bond. (C = C πbond).
The following characteristics of the E2 reaction are:
It is a one-step elimination andHas only one transition stage.From the information given:
Pyridine functions primarily as a base, and the alkyl iodide in question is a primary alkyl halide that helps in the E2 mechanism.
In this case, both H and Cl are removed in a single step, yielding 1-butene as the byproduct of the reaction.
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How many moles of HNO3 will be produced 3 NO2+H2O=2HNO3+ NO
Answer:
2 moles of HNO3
Explanation:
The equation seems to be balanced correctly. The problem is we done know what you started with. We will assume it is 3 moles of NO2.
If that is the case then 2 moles of HNO3 will be produced.
Which of the following metals is paramagnetic?
A. Magnesium
B. Sodium
C. Beryllium
D. Calcium
Answer:
sodium
Explanation:
(Na) atom is paramagnetic and sodium is a na atom.
Be sure to answer all parts. Arrange the following substances in order of increasing strength of intermolecular forces. Click in the answer box to open the symbol palette.
a. H2
b. Ne
c. O2
d. NH3
Answer:
H2<Ne<O2<NH3
Explanation:
Intermolecular forces refer to the force of attraction between molecules of a substance in any given state of matter whether solid, liquid or gas. Molecules in a substance must be held together by intermolecular forces of attraction. The magnitude of these intermolecular forces of attraction depends on many factors.
For H2, He and O2, the intermolecular force present in these gases are London forces. As the relative molecular mass of individual gas molecules becomes greater, London forces increases significantly with molecular mass. This explains the sequence shown in the answer.
NH3 has the strongest intermolecular interaction because it contains hydrogen bonds since nitrogen is an electronegative element. This a greater intermolecular interaction than dispersion forces.
Intermolecular forces are those forces that bind the molecules of the substance and the polarity of molecules. These forces range from the strongest to the weakest in ion-dipole, hydrogen bonding, and dipole to dipole.
H2 being a noble gas has a weak dispersion or a weak dipole force. Ne has an intermolecular force being a noble gas it increases the molecular weight and thus has a modest increase of dipole bounding. The O2 has a strong dipole force than Ne and is stronger due to the two Oxygen molecules. The NH3 has the strongest dipole and intermolecular interaction force. The nitrogen atom strongly pulls the electrons.Hence the form the strongest to the weakest is NH3, 02, Ne and H2.
Learn more about the substances in order of increasing the strength of intermolecular forces.
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When the owners of some wells in Pallerla started using high-powered motors to
draw water from the wells, the owners of other wells noticed that their wells were
drying up. Discuss the possible solution to the problem solutions to the problem
Answer:
The possible solution is to balance the rate of water removal from the well to the rate of natural recharge of the well from its underground aquifer.
Explanation:
A well is an excavation in the earth, made with the aim of extracting water from the aquifers. The water from a well can be drawn up by the means of a pump, containers, such as buckets, or by hand. Aquifers can also be recharged through a well.
Well draw down occurs when water from the well is drained faster than it is naturally recharged from the aquifer. This can be as a result of over pumping, extended drought, among other factors. The use of the high-powered motor in this case, for pumping, might be the possible cause of the well drying up. The situation might have resulted from the pump drawing out water from the well at a rate tat exceeds the rate at which it is recharged naturally, causing the well water to start drying up. There's also a possibility that the well is pumped indiscriminately, possibly leading to wastage of water.
The solution to this problem is to give the well a time duration for it to recharge itself. Then, the rate of recharges should be calculated and determined by an hydrologist. When all these is done, a pump with a motor power that does not exceed the calculated recharge rate should be used in place of the high-powered motor. Also, water usage should be brought to the minimum level to prevent unnecessary pumping due to excessive, wasteful use of water.
How many grams of sodium phosphate are needed to have 1.67 moles of sodium ion?
Answer:
91.29 g of Na₃PO₄ are needed to produce 1.67 moles of sodium ions.
Explanation:
The formula for sodium phosphate is Na₃PO₄
Molar mass of sodium phosphate = 164 g/mol
The dissociation of one mole of sodium phosphate produces 3 moles of sodium ions;
Na₃PO₄ ------> 3Na⁺ + PO₄³
Number of moles of Na₃PO₄ that will produce 1.67 moles of Na⁺ = 1/3 * 1.67 = 0.556 moles of Na₃PO₄
Mass of 0.556 moles of Na₃PO₄ = 0.556 moles * 164 g/mol = 91.29 g
Therefore, 91.29 g of Na₃PO₄ are needed to produce 1.67 moles of sodium ions.
Please help asap! Giving brainliest.
What is the total number of electrons that can occupy the p sublevel? (3 points)
Select one:
a. 2 electrons
b. 6 electrons
c. 8 electrons
d. 10 electrons
Answer:
The answer is 6 because the p sublevel holds 3 orbitals and since each orbital can hold 2 electrons, the answer is 3 * 2 = 6.
Answer:
6 electrons
Explanation:
Each principal energy level above the first contains one s orbital and three p orbitals. A set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. So the answer is 6 electrons.
Write a balanced equation for the single-replacement oxidation-reduction reaction described, using the smallest possible integer coefficients. The reaction that takes place when chlorine gas combines with aqueous potassium bromide. (Use the lowest possible coefficients. Omit states of matter.)
Answer:
[tex]\rm Cl_2 + 2\; KBr \to Br_2 + 2\; KCl[/tex].
One chlorine molecule reacts with two formula units of (aqueous) potassium bromide to produce one bromine molecule and two formula units of (aqueous) potassium chloride.
Explanation:
Formula for each of the speciesStart by finding the formula for each of the compound.
Both chlorine [tex]\rm Cl[/tex] and bromine [tex]\rm Br[/tex] are group 17 elements (halogens.) Each On the other hand, potassium [tex]\rm K[/tex] is a group 1 element (alkaline metal.) EachTherefore, the ratio between [tex]\rm K[/tex] atoms and [tex]\rm Br[/tex] atoms in potassium bromide is supposed to be one-to-one. That corresponds to the empirical formula [tex]\rm KBr[/tex]. Similarly, the ratio between
The formula for chlorine gas is [tex]\rm Cl_2[/tex], while the formula for bromine gas is [tex]\rm Br_2[/tex].
Balanced equation for the reactionWrite down the equation using these chemical formulas.
[tex]\rm ?\; Cl_2 + ?\; KBr \to ?\;Br_2 + ?\; KCl[/tex].
Start by assuming that the coefficient of compound with the largest number of elements is one. In this particular equation, both [tex]\rm KBr[/tex] and [tex]\rm KCl[/tex] features two elements each.
Assume that the coefficient of [tex]\rm KCl[/tex] is one. Hence:
[tex]\rm ?\; Cl_2 + 1 \; KBr \to ?\;Br_2 + ?\; KCl[/tex].
Note that [tex]\rm KBr[/tex] is the only source of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms among the reactants of this reaction.
There would thus be one [tex]\rm K[/tex] atom and one [tex]\rm Br[/tex] atom on the reactant side of the equation.
Because atoms are conserved in a chemical equation, there should be the same number of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms on the product side of the equation.
In this reaction, [tex]\rm Br_2[/tex] is the only product with [tex]\rm Br[/tex] atoms.
One [tex]\rm Br[/tex] atom would correspond to [tex]0.5[/tex] units of [tex]\rm Br_2[/tex].
Similarly, in this reaction, [tex]\rm KCl[/tex] is the only product with [tex]\rm K[/tex] atoms.
One [tex]\rm K[/tex] atom would correspond to one formula unit of [tex]\rm KCl[/tex].
Hence:
[tex]\displaystyle \rm ?\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].
Similarly, there should be exactly one [tex]\rm Cl[/tex] atom on either side of this equation. The coefficient of [tex]\rm Cl_2[/tex] should thus be [tex]0.5[/tex]. Hence:
[tex]\displaystyle \rm \frac{1}{2}\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].
That does not meet the requirements, because two of these coefficients are not integers. Multiply all these coefficients by two (the least common multiple- LCM- of these two denominators) to obtain:
[tex]\displaystyle \rm 1\; Cl_2 + 2 \; KBr \to 1\;Br_2 + 2\; KCl[/tex].
If the H+ concentration is 0.00001 M, what is the OH- concentration?
Answer:
1.00x10^-9
Explanation:
What is an ideal gas?
Answer:
a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.
Air contains nitrogen, oxygen, argon, and trace gases. Ifthe partial pressure of nitrogen is 592 mm Hg, oxygen is160 mm Hg, argon is 7 mm Hg, and trace gas is 1 mm Hg,what is the atmospheric pressure
Answer:
760 mmHg
Explanation:
Step 1: Given data
Partial pressure of nitrogen (pN₂): 592 mmHgPartial pressure of oxygen (pO₂): 160 mmHgPartial pressure of argon (pAr): 7 mmHgPartial pressure of the trace gas (pt): 1 mmHgStep 2: Calculate the atmospheric pressure
Since air is a gaseous mixture, the atmospheric pressure is equal to the sum of the gases that compose it.
P = pN₂ + pO₂ + pAr + pt = 592 mmHg + 160 mmHg + 7 mmHg + 1 mmHg = 760 mmHg
. Calculate ΔG for the following reaction at 25°C. Will the reaction occur (be spontaneous)? How do you know? NH3(g) + HCl(g) → NH4Cl(s) ΔH = -176.0 kJ ΔS = -284.8 J·K-1
Answer:
[tex]\triangle G = -911.296 \ kJ[/tex]
Explanation:
ΔG = ΔH-TΔS
Where ΔH = -176 kJ = -176000 J , T = 25°C + 273 = 298 K , ΔS = -284.8 JK⁻¹
=> [tex]\triangle G =-176000 - (298)(-284.8)[/tex]
=> [tex]\triangle G = -176000+84870.4[/tex]
=> [tex]\triangle G = -91129.6 \ J[/tex]
=> [tex]\triangle G = -911.296\ kJ[/tex]
Since the value is negative, the reaction is spontaneous under standard conditions at 298 K and the reactants have more free energy than the products.
2,4-Dimethylpent-2-ene undergoes an electrophilic addition reaction in the presence of HBr to form 2-bromo-2,4-dimethylpentane. Complete the mechanism of this addition and draw the intermediates formed as the reaction proceeds.
Draw all missing reactants and/or products in the appropriate boxes by placing atoms on the canvas and connecting them with bonds. Add charges where needed. Electron flow arrows should start on the electron(s) of an atom or a bond and should end on an atom, bond, or location where a new bond should be created.
Answer:
the answer is in the diagram
Explanation:
when 2,4-dimethylpent-2-ene undergo electrophilic addition reaction in the presence of HBr to form 2-bromo-2,4-dimethylpentane, it firstly lead to an intermediate carbocation
A carbocation can be describe as an organic molecule, which serves as an intermediate, that has a carbon atom bearing a positive charge and three bonds instead of four
Calculate the [H+] and pH of a 0.0040 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20×10−5. Use the method of successive approximations in your calculations or the quadratic formula.
Answer:
[tex][H^+]=0.000285[/tex]
[tex]pH=3.55[/tex]
Explanation:
In this, we can with the ionization equation for the hydrazoic acid ([tex]HN_3[/tex]). So:
[tex]HN_3~<->~H^+~+~N_3^-[/tex]
Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:
[tex]Ka=\frac{[H^+][N_3^-]}{[HN_3]}[/tex]
For each mol of [tex]H^+[/tex] produced we will have 1 mol of [tex]N_3^-[/tex]. So, we can use "X" for the unknown values and replace in the Ka equation:
[tex]Ka=\frac{X*X}{[HN_3]}[/tex]
Additionally, we have to keep in mind that [tex]HN_3[/tex] is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:
[tex]Ka=\frac{X*X}{0.004-X}[/tex]
Finally, we can put the ka value and solve for "X":
[tex]2.2X10^-^5=\frac{X*X}{0.004-X}[/tex]
[tex]2.2X10^-^5=\frac{X^2}{0.004-X}[/tex]
[tex]X= 0.000285[/tex]
So, we have a concentration of 0.000285 for [tex]H^+[/tex]. With this in mind, we can calculate the pH value:
[tex]pH=-Log[H^+]=-Log[0.000285]=3.55[/tex]
I hope it helps!
The [H+] and pH of a 0.0040 M hydrazoic acid solution is 0.000296648 and 3.527759
pH based problem:What information do we have?
Hydrazoic acid solution = 0.0040 M
Ka of hydrazoic acid = 2.20 × 10⁻⁵
We know that weak acids
[H+] = √( Ka × C)
[H+] = √( 2.2 × 10⁻⁵ × 0.0040)
[H+] = 0.000296648
So,
pH = -log [H+]
pH = -log [0.000296648]
Using log calculator
pH = 3.527759
Find more information about 'pH'.
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Draw the Lewis structure of N₂O₄ and then choose the appropriate pair of hybridization states for the two central atoms. Your answer choice is independent of the orientation of your drawn structure.
Answer:
See explanation
Explanation:
In this case, we have to keep in mind the valence electrons for each atom:
N => 5 electrons
O => 6 electrons
If the formula is [tex]N_2O_4[/tex], we will have in total:
[tex](5*2)+(6*4)=34~electrons[/tex]
Additionally, we have to remember that each atom must have 8 electrons. So, for oxygens 5 and 3 we will have 3 lone pairs and 1 bond (in total 8 electrons. For oxygens, 6 and 4 we will have 2 lone pairs and 2 bonds (in total 8 electrons) and for nitrogens 1 and 2 we will have 4 bonds (in total 8 electrons).
To find the hybridization, we have to count the atoms and the lone pairs around the nitrogen. We have 3 atoms and zero lone pairs. If we take into account the following rules:
[tex]Sp^3~=~4[/tex]
[tex]Sp^2~=~3[/tex]
[tex]Sp~=~2[/tex]
With this in mind, the hybridization of nitrogen is [tex]Sp^2[/tex].
See figure 1
I hope it helps!
The central nitrogen atoms in N2O4 are both sp2 hybridized.
The Lewis structure shows the number of electron pairs that surround the atoms in a molecule as dots. It is quite easy to determine the number of valence electrons in a molecule simply by observing its Lewis dot structure.
The molecule N2O4 has 34 valence electrons as shown in its dot electron structure. The central nitrogen atoms in N2O4 are both sp2 hybridized as shown. The formal charges on each atom in N2O4 are also shown.
Learn more:
Predict the order of acid strengths in the following series of cationic
species: CH3CH2NH3
+, CH3CH=NH2
Answer:
CH3CH=NH2+>CH3CH2NH3+
Explanation:
A close examination of both structures will reveal that they are both amines hence they must have the polar N-H bond.
Electrons usually move towards the nitrogen atom and this makes both compounds acidic. We must also remember that some features of a compound may make it more acidic than another of close resemblance. Being more acidic may imply that the proton of the N-H is more easily lost.
CH3CH=NH2+ has an sp2 hybridized carbon atom in its structure which is known to be very electronegative due to increasing s character of the bond. It will withdraw electron density from the N-H bond thereby leading to a greater acidity of CH3CH=NH2+ in comparison to CH3CH2NH3+