Answer:
0.1 M
Explanation:
The overall balanced reaction equation for the process is;
IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)
Generally, we must note that;
1 mol of IO3^- require 6 moles of S2O3^2-
Thus;
n (iodate) = n(thiosulfate)/6
C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6
Concentration of iodate C(iodate)= 0.0100 M
Volume of iodate= V(iodate)= 26.34 ml
Concentration of thiosulphate= C(thiosulfate)= the unknown
Volume of thiosulphate=V(thiosulfate)= 15.51 ml
Hence;
C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)
0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M
1. Rank the following compounds in order of decreasing acid strength using periodic trends. Rank the acids from strongest to weakest.
A. HCl
B. H2S
C. HBr
D. BH3
2. Without consulting the table of acid-dissociation constants, match the following acids to the given Ka1 values.
1. H2S
2. H2SO3
3. H2SO4
A. Kal = 1.7 x 10^-7
B. Kal = 1.7 x 10^-2
C. Kal = very large
Answer:
ESCALAS MAYORES (D, E, G, A, B) Porfavor necesito ayuda,te lo agradecería muchísimo!!
Es urgente
Describe the reactions during the electrolysis of water in an electrolytic cell. Describe the reactions during the electrolysis of water in an electrolytic cell. Oxygen and hydrogen are both reduced. Oxygen is oxidized and hydrogen is reduced. Neither oxygen or hydrogen are oxidized or reduced. Oxygen and hydrogen are both oxidized. Oxygen is reduced and hydrogen is oxidized.
Answer:
Oxygen is oxidized and hydrogen is reduced
Explanation:
In the electrolysis of water, a pair of Platinum electrodes are immersed in water. Th water has a small quantity of either an acid, salt or base, in most cases H2S04, added to it, to aid ionization. This is because water on its own does not posses enough ions to undergo electrolysis. At the platinum anode, water is oxidized to oxygen gas and hydrogen ions. At the platinum cathode, water is reduced to hydrogen gas and hydroxide ions. The proportion of oxygen and hydrogen produced should be theoretically 1 : 2 respectively, but is not usually so, due to competing side reactions. The hydrogen by product is usually used as a fuel source, and it usually combine with the hydroxide ion to form water back again.
Explain the term isomers?
Answer:
Isomers are molecules that have the same molecular method, however have a unique association of the atoms in space. That excludes any extraordinary preparations which can be sincerely because of the molecule rotating as an entire, or rotating about precise bonds.
Ethane burns according to the following reaction: 2C2H6 + 7O2 → 4CO2 + 6H2O + 2.86 x 103 kJ How much heat can be generated when 185 grams of oxygen gas (MW = 32 g/mol) are consumed?
Answer:
question is not clear please send clear question
what bsic difference is between NMR and MS spectroscopic techniques?
Answer:
The Nuclear magnetic resonance is the process this technique does not use radiation.
The ms is an sensitive technology can be a massive number and small sample of the blood.
Explanation:
The Nuclear magnetic resonance we look at the both side of that coin.
The technique provides that fatty acid composition and various including amino acids.
These are contain the complementary these biomarkers, that are suitable for all kinds of studies. there are many types of research:-
(1) A powerful tool metabolic (2) A versatile tool research (3) Quick analysis (4) Low cost analysis.
The MS is an extremely sensitive technology using a very small number of the blood.
(1) Powerful techniques (2) Highly method (3) Large number of metabolites (4)Small sample volume
MS can be fine mapping metabolic pathways to sign analytical strategy.
A compound is found to contain 18.28 % phosphorus , 18.93 % sulfur , and 62.78 % chlorine by mass. To answer the question, enter the elements in the order presented above. QUESTION 1: The empirical formula for this compound is . QUESTION 2: The molar mass for this compound is 169.4 g/mol. The molecular formula for this compound is
Answer:
1. EF = PSCl₃; 2. MF = PSCl₃
Explanation:
1. Empirical formula
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our first job is to calculate the molar ratio of P:S:Cl.
Assume 100 g of the compound.
(a) Calculate the mass of each element.
Then we have 18.28 g P, 18.93 g S, and 67.28 g Cl.
(b) Calculate the moles of each element
[tex]\text{Moles of P} = \text{18.28 g C} \times \dfrac{\text{1 mol P}}{\text{30.97 g P}} = \text{0.5902 mol P}\\\\\text{Moles of S} = \text{18.93 g S} \times \dfrac{\text{1 mol S}}{\text{32.06 g S }} = \text{0.5905 mol S}\\\\\text{Moles of Cl} = \text{62.78 g Cl} \times \dfrac{\text{1 mol Cl}}{\text{35.45 g Cl }} = \text{1.771 mol Cl}[/tex]
(c) Calculate the molar ratio of the elements
Divide each number by the smallest number of moles
P:S:Cl = 0.5902:0.5905:1.898 = 1:1.000:3.000 ≈ 1:1:3
(d) Write the empirical formula
EF = PSCl₃
The empirical formula for this compound is PSCl₃.
2. Molecular formula
(a) Calculate the ratio of the molecular and empirical formula masses
n = (169.4 u)/(169.40 u) = 1.000 ≈ 1
(b) Calculate the molecular formula
MF = (EF)ₙ = (EF)₁ = PSCl₃
The molecular formula for this compound is PSCl₃.
The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast furnace. After cooling the pure liquid iron, Steve determines that he has produced 288g of iron ingots. How much of the excess reactant (in grams) is left over after the theoretical yield of liquid iron is produced?
Answer:
Amount of excess Carbon (ii) oxide left over = 23.75 g
Explanation:
Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂
Molar mass of Fe₂O₃ = 160 g/mol;
Molar mass of Carbon (ii) oxide = 28 g/mol
From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide
450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide
Therefore the excess reactant is carbon (ii) oxide.
Amount of excess Carbon (ii) oxide left over = 260 - 236.25
Amount of excess Carbon (ii) oxide left over = 23.75 g
Write the condensed formula from left to right, starting with (CH3)x where x is a number.
Complete question:
Write the condensed formula from left to right, starting with (CH3)x where x is a number.
See attached image for the structure formula of the compound
Answer:
(CH₃)₂CHC(CH₃)₃ named as 2,2,3-Trimethylbutane
Explanation:
If we number the longest chain of the carbon starting from the left, we will observe that there are four carbons in the straight chain as shown in the image.
Starting from first carbon from the left of the carbon chain, at carbon number number 2, there two alkyl group, that is two methyl (CH3 is two). Also at carbon number 3, there are three alkyl group, that is three methyl (CH3 is three).
The condensed formula will be written as;
(CH₃)₂CHC(CH₃)₃
This compound is named as 2,2,3-Trimethylbutane, an isomer of Heptane
Not all bonds are "created equal". From the following molecules, which one contains the most easily broken carbon to carbon bond? Group of answer choices H3C—CH3 F2C=CF2 H2C=CH2 HCCH
Answer:
H3C—CH3
Explanation:
The strength of a bond is indicated by the value of its bond dissociation energy. Simply put, the bond dissociation energy is the energy required to break the bond.
Carbon forms single, double and triple bonds with itself. As a matter of fact, carbon atoms can link to each other indefinitely. This is known as catenation and has been attributed to the low bond energy of the carbon-carbon single bond.
The bond energy of the carbon-carbon single bond is about 90KJmol-1 while that of carbon-carbon double bond is about 174KJmok-1. The carbon-carbon triple bond has the highest bond dissociation energy of about 230KJmol-1.
Hence, it is easier to break carbon-carbon single bonds than double and triple bonds respectively, hence the answer.
According to the forces of attraction, the molecule which can be easily broken is CH₃-CH₃ as it has a single bond with low dissociation energy as compared to double or triple bonds.
Forces of attraction is a force by which atoms in a molecule combine. it is basically an attractive force in nature. It can act between an ion and an atom as well.It varies for different states of matter that is solids, liquids and gases.
The forces of attraction are maximum in solids as the molecules present in solid are tightly held while it is minimum in gases as the molecules are far apart . The forces of attraction in liquids is intermediate of solids and gases.
The physical properties such as melting point, boiling point, density are all dependent on forces of attraction which exists in the substances.Single bonds have least dissociation energy while triple bonds have the maximum dissociation energy.
Thus,the molecule which can be easily broken is CH₃-CH₃.
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Carbon monoxide gas reacts with hydrogen gas to form methanol: CO (g_ + 2H2 (g) → CH3OH (g) A 1.50L reaction vessel, initially at 305 K, contains carbon monoxide gas at a partial pressure of 232 mmHg and hydrogen gas at a partial pressure of 397 mmHg. Identify the limiting reactant and determine the theoretical yield of methanol in grams.
:Answer : The limiting reactant is and the theoretical yield of methanol is, 0.96 grams.
Explanation :
First we have to calculate the moles of and .
where,
= pressure of CO gas = 232 mmHg = 0.305 atm (1 atm = 760 mmHg)
V = volume of gas = 1.65 L
T = temperature of gas = 305 K
= number of moles of CO gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
and,
where,
= pressure of gas = 374 mmHg = 0.492 atm (1 atm = 760 mmHg)
V = volume of gas = 1.65 L
T = temperature of gas = 305 K
= number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of
So, 0.0601 moles of react with moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 2 mole of react to give 1 mole of
So, 0.0601 moles of react with moles of
Now we have to calculate the mass of
Therefore, the theoretical yield of methanol is, 0.96 grams.
The theoretical yield of methanol is 0.496 g of methanol.
The reaction equation is CO (g) + 2H2 (g) → CH3OH (g).
From the partial pressures of each reactant, we can obtain the number of moles of reactants.
For CO;
P = 232 mmHg or 0.305 atm
V = 1.5 L
T = 305 K
n = ?
R = 0.082 atmL-1mol-1K-1
PV = nRT
n = PV/RT
n = 0.305 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K
n = 0.018 moles
For hydrogen;
P = 397 mmHg or 0.522 atm
V = 1.5 L
T = 305 K
n = ?
R = 0.082 atmL-1mol-1K-1
PV = nRT
n = PV/RT
n = 0.522 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K
n = 0.031 moles
From the reaction equation;
1 mole of CO reacted with 2 moles of H2
0.018 moles of CO will react with 0.018 moles × 2 moles/1 mole
= 0.036 moles of H2
We can see that there is not enough H2 to react with CO hence H2 is the limiting reactant.
2 moles of H2 yields 1 mole of methanol
0.031 moles of H2 yields 0.031 moles × 1 moles/2 mole
= 0.0155 moles of methanol
Mass of methanol produced = 0.0155 moles of methanol × 32 g/mol
= 0.496 g of methanol
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A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an elevation where the pressure is 0.71 atm, then the new volume is 1.8 L. What is the temperature (in °C) of the air at this elevation?
Answer: 44.37 degrees C
Explanation:
Use combined gas law: (P1)(V1)/T1=(P2)(V2)/T2
For most gas laws, you must convert to Kelvin:
K=deg C+273
K=25+273=298 K
Plug and chug:
(1.0 atm)(1.2 L)/(298 K)=(0.71 atm)(1.8 L)/(x)
Solve for x and get 317.37 K
Subtract 273 from this to convert to degrees Celsius. You will get 44.37 degrees Celsius.
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The reaction A + 2B → products was found to follow the rate law: rate = k[A] 2[B]. Predict by what factor the rate of reaction will increase when the concentration of A is doubled, the concentration of B is tripled, and the temperature remains constant
Answer:
By a factor of 12
Explanation:
For the reaction;
A + 2B → products
The rate law is;
rate = k[A]²[B]
As you can see, the rate is proportional to the square of the concentration of A and the of the concentration of B .
Let's say initially, [A] = x, [B] = y
The rate law in this case is equal to;
rate1 = k. x².y
Now you double the concentration of A and triple the concentration of B.
[A] = 2x, [B] = 3y
The new rate law is given as;
rate2 = k . (2x)². (3y)
rate2 = k . 4x² . 3y
rate2 = 12 k . x² . y
Comparing rate 2 and rate 1, the ratio is given as; rate 2/ rate 1 = 12
Therefore the rate has increased by a factor of 12.
most vegetables substantially diminish in quality in as little as days
Answer:
As little as 2 days
Hope this is correct
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Modern atomic theory states that atoms are neutral. How is this neutrality achieved in atoms? (2 points)
BeH2 has no lone pairs of electrons. What's the structure of this molecule?
Answer:
linear
Explanation:
nothing bends off
In this experiment, you will analyze your sample by TLC by spotting pure benzophenone and your product and eluting with 5:1 hexanes/ethyl acetate. Choose the statement that BEST describes what you should observe for a successful experiment.
A) The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value.
B) The benzophenone in the standard and reaction samples should travel the least and will have the same Rf value. No unreacted benzhydrol should be observed at a higher Rf value.
C) The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. Unreacted benzhydrol should be observed at a lower Rf value.
D) The benzophenone in the standard and reaction samples should travel the farthest and will have the different Rf values. No unreacted benzhydrol should be observed at a lower Rf value.
Answer:
The correct answer is statement A.
Explanation:
In a media comprising 20 percent ethyl acetate/hexane, as the benzophenone is non-polar, so it will travel farther with high Rf value. On the other hand, as benzohydrol is a polar molecule, therefore, it should be at lower Rf value and will not rise in the given media.
For an experiment to be successful, there should not be any unreacted benzohydrol to be left in the experimental system. The benzophenone in the reaction as well as the standard samples should exhibit similar Rf value.
The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value to analyze by TLC.
What is TLC ?In synthetic chemistry, thin-layer chromatography (TLC) is a method that is frequently used to identify compounds, assess their purity, and monitor the progress of a reaction. Additionally, it enables the solvent system for a specific separation problem to be optimized.
The foundation of TCL is the adsorption-based separation theory. The separation depends on how sensitive different chemicals are to the stationary and mobile phases.
TLC, or thin layer chromatography, is a technique for isolating the components of mixtures before analysis. TLC can be used to identify compounds, ascertain their identities, and ascertain the purity of a compound.
Thus, option A is correct.
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Rank the compounds in each set in order of increasing acid strength.
(a) CH3CH2COOH CH3CHBrCOOH CH3CBr2COOH
(b) CH3CH2CH2CHBrCOOH CH3CH2CHBrCH2COOH CH3CHBrCH2CH2COOH
Answer:
See explanation
Explanation:
For this question, we have to remember the effect of an atom with high electronegativity as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an inductive effect. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be weaker and the compound will be more acid (because is easier to produce the hydronium ion [tex]H^+[/tex]).
With this in mind, for A in the last compound, we have 2 Br atoms near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have more acidity. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.
In B, the difference between the molecules is the position of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a higher inductive effect and more acidity.
See figure 1
I hope it helps!
what mass of calcium nitrate can be prepared by the reaction of 18.9 grams of nitric acid with 7.4 grams of calcium hydroxide
Answer:
16.4 grams of calcium nitrate can be prepared by the reaction of 18.9 grams of nitric acid with 7.4 grams of calcium hydroxide
Explanation:
The balanced reaction is:
Ca(OH)₂ + 2 HNO₃ → Ca(NO₃)₂ + 2 H₂O
First, you must determine the limiting reagent. The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction):
Ca(OH)₂: 1 moleHNO₃: 2 molesCa(NO₃)₂: 1 mole H₂O : 2 molesBeing:
Ca: 40 g/moleO: 16 g/moleH: 1 g/moleN: 14 g/moleThen, the molar mass of the compounds participating in the reaction is:
Ca(OH)₂: 40 g/mole + 2*(16 g/mole + 1 g/mole)= 74 g/moleHNO₃: 1 g/mole + 14 g/mole + 3*16 g/mole= 63 g/moleCa(NO₃)₂: 40 g/mole + 2*(14 g/mole + 3*16 g/mole)= 164 g/moleH₂O : 2*1 g/mole + 16 g/mole= 18 g/moleThen, by stoichiometry of the reaction the following amounts of reagents and products participate:
Ca(OH)₂: 1 mole* 74 g/mole= 74 gHNO₃: 2 moles* 63 g/mole= 126 gCa(NO₃)₂: 1 mole* 164 g/mole= 164 gH₂O : 2 moles* 18 g/mole= 36 gThen apply the following rule of three: if 126 grams of nitric acid reacts with 74 grams of calcium hydroxide, 18.9 grams of nitric acid with how much mass of calcium hydroxide does it react?
[tex]mass of calcium hydroxide=\frac{18.9 grams of nitric acid*74 grams of calcium hydroxide}{126 grams of nitric acid}[/tex]
mass of calcium hydroxide= 11.1 grams
But 11.1 grams of calcium hydroxide are not available, 7.4 grams are available. Since you have less mass than you need to react with 18.9 grams of nitric acid, calcium hydroxide will be the limiting reagent.
Then, it is possible to determine the amount of mass of calcium nitrate produced by another rule of three: if 164 grams of calcium nitrate are formed by stoichiometry from 74 grams of calcium hydroxide, how much mass of calcium nitrate will form from 7.4 grams of calcium hydroxide?
[tex]mass of calcium nitrate=\frac{7.4 grams of calcium hydroxide*164 grams of calcium nitrate}{74 grams of calcium hydroxide}[/tex]
mass of calcium nitrate= 16.4 grams
16.4 grams of calcium nitrate can be prepared by the reaction of 18.9 grams of nitric acid with 7.4 grams of calcium hydroxide
When a 2.75g sample of liquid octane (C8H18) is burned in a bomb calorimeter, the temperature of the calorimeter rises from 22.0 °C to 41.5 °C. The heat capacity of the calorimeter, measured in a separate experiment, is 6.18 kJ/°C. Determine the ΔE for octane combustion in units of kJ/mol octane.
Answer:
THE HEAT OF COMBUSTION IS 4995.69 kJ/mol OF OCTANE.
Explanation:
Heat capacity = 6.18 kJ/C
Temperature change = 41.5 C - 22.0 C = 19.5 C
Heat required to raise the temperature by 19.5 °C is:
Heat = heat capacity * temperature change
Heat = 6.18 kJ/ C * 19.5 C
heat = 120.51 kJ of heat
120.51 kJ of heat is required to raise the temperature of 2.75 g sample of a liquid octane.
Molar mass of octane = ( 12* 8 + 1 * 18) = 114 g/mol
So therefore, the heat of the reaction per mole of octane will be:
120.51 kJ of heat is required for 2.75 g of octane
x J of heat will be required for 114 g of octane
x J = 120.51kJ * 114 / 2.75
x = 4995.69 kJ of heat per mole.
In conclusion, the heat of the combustion reaction in kJ / mole of octane is 4995.69 kJ/mol
This substituent deactivates the benzene ring towards electrophilic substitution but directs the incoming group chiefly to the ortho and para positions.
A) -F
B) -OCH2CH3
C) -CF3
D) -NHCOCH3
E) -NO2
Answer:
F
Explanation:
Halogens may interact with the benzene ring via inductive or resonance effects. Halogens deactivate the benzene ring by inductive effect rather than by resonance effects.
The lone pairs of electrons present on the halogen atoms may be donated to the ring by resonance, but an opposite effect, the inductive pull (-I inductive effect) of the halogen atoms on electrons away from the benzene ring due to the high electro negativity of the halogens leads to a deactivation of the ring towards electrophilic substitution.
Hence inductive electron withdrawal by the halogen atom predominates over electron donation by resonance effect and the benzene ring g is deactivated towards electrophilic substitution at the ortho and para positions.
1. In the simple cubic unit cell, the centers of ____________ identical particles define the ____________ of a cube. The particles do touch along the cube's ____________ but do not touch along the cube's ____________ or through the center. There is/are ____________ particle per unit cell and the coordination number is ____________ .
2. In the body-centered cubic unit cell, the centers of ____________ identical particles define the ____________ of the cube plus ____________ particle at the ____________ of ____________ . The particles do not touch along the cube's ____________ or faces but do touch along the cube's ____________ . There is/are ____________ particles per unit cell and the coordination number is ____________ .
3. In the face-centered cubic cell, the centers of ____________ identical particles define the ____________ of the cube plus ____________ particle in the ____________ of ____________ . The particles on the ____________ do not touch each other but do touch those on the ____________ . There is/are ____________ particles per unit cell and the coordination number is ____________ .
Answer:please see below for answers in the spaces given.
Explanation:
There are three types of cubic-unit cells of a cubic system which include Simple cubic unit cell, body-centered cubic unit cell and face-centered cubic-unit cell and Thier characteristics are completed below.
1) In the simple cubic unit cell, the centers of _______eight _____ identical particles define the _________corners___ of a cube. The particles do touch along the cube's _______edges_____ but do not touch along the cube's ____diagonal_______ or through the center. There is/are _______one_____ particle per unit cell and the coordination number is
__six______ .
2. In the body-centered cubic unit cell, the centers of _______eight _____ identical particles define the _______corners_____ of the cube plus ______one______ particle at the _______center_____ of ______the cube______ . The particles do not touch along the cube's _______edges_____ or faces but do touch along the cube's ____diagonal________ . There is/are _____two_______ particles per unit cell and the coordination number is _____eight_______ .
3. In the face-centered cubic cell, the centers of ______eight______ identical particles define the _______corner____ of the cube plus ________one____ particle in the _____center_______ of ______each face______ . The particles on the _____corners_______ do not touch each other but do touch those on the ______faces____ . There is/are ________four___ particles per unit cell and the coordination number is _____twelve_______ .
For carbon: What is the effective nuclear charge? In which orbitals do the valence electrons reside? For silicon: What is the effective nuclear charge? In which orbitals do the valence electrons reside?
Answer:
For carbon, the effective nuclear charge is 3.25 and the valence electrons will reside in the orbitals 2s^2 and 2p^2
For silicon, the effective nuclear charge is 4.15 and its valance electrons will reside in the orbitals 3s^2 and 3p^2
Explanation:
Carbon
The effective nuclear charge of carbon is 3.25
To get the orbitals in which it’s valence electron reside, let’s write the electronic configuration
The atomic number of carbon is 6
So the configuration will be;
1s^2 2s^2 2p^2
So the valence electrons will reside in the orbitals 2s^2 and 2p^2
For silicon;
It’s effective nuclear charge is +4.15
The electronic configuration of silicon with atomic number 14 is;
1s^2 2s^2 2p^6 3s^2 3p^2
So the valence electrons will reside in the orbitals 3s^2 and 3p^2
During which part of the scientific method would error bars be used?
A. Conclusion
B. Analysis
C. Hypothesis
D. Research
please helppppppppppp
Answer:
The correct answer is B. Analysis
Explanation:
Error bars are part of the statistical analysis in the scientific method. Once the scientist have collected the data, he or she proceed to the data analysis. A very common way of comparing the data variability is to use error bars in ghaphical representations. From these bars, it can be estimated the error of a determination and experimental groups are compared.
The error bars would be used during B. Analysis.
What is an Error Bar?A blunders bar is a line through a factor on a graph, parallel to one of the axes, which represents the uncertainty or variant of the corresponding coordinate of the point. In IB Biology, the error bars most often represent the same old deviation of an information set.
When would error bars be used?Blunders bars can be used to examine visual quantities if various other situations preserve. This can decide whether or not differences are statistically sizable. Mistake bars can also propose the goodness of match of a given characteristic, i.e., how well the function describes the facts.
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If 50 ml of 1.00 M of H2SO4 and 50 ml of 2.0 M KOH are mixed what is the concentration of the resulting solutes?
Answer:
0.5 M
Explanation:
First, let us look at the balanced equation of the reaction.
[tex]H_2SO_4 + 2KOH --> K_2SO_4 + 2H_2O[/tex]
The solute formed is [tex]K_2SO_4[/tex].
Recall that: mole = molarity x volume
Hence,
50 ml, 1.00 M H2SO4 = 0.05 x 1 = 0.05 mole
50 ml, 2.0 M KOH = 0.05 x 2 = 0.1 mole
From the equation
1 mole of H2SO4 reacts with 2 moles of KOH to give 1 mole of K2SO4.
Hence,
0.05 mole H2SO4 reacting with 0.1 mole KOH will give 0.05 mole [tex]K_2SO_4[/tex].
Also recall that: concentration = mole/volume
Total volume of resulting solution = 50 ml + 50 ml = 100 ml or 0.1 liter
Concentration of [tex]K_2SO_4[/tex] = mole of [tex]K_2SO_4[/tex]/volume of resulting solution
= 0.05/0.1 = 0.5 M
The concentration of the resulting solute = 0.5 M
Limiting Reagent
1.) A student chose the wrong result of the two calculations of BaSO4, namely, the higher value. What would you expect to happen to the value of the % yield? Explain.
2.) In the process of filtration, what do you think has happened to the excess reagent which has not reacted? Where does it go, and do you think you could recover it, if needed? Explain.
Answer:
a) the percentage yield will exceed 100%
b) the excess reactant is filtered along with the barium sulphate precipitate. It is possible to recover the excess reactant by carefully washing the precipitate with water.
Explanation:
In the precipitation of barium sulphate, the ions in the reactants exchange partners in the product leading to an insoluble product.
In every reaction, there is a limiting reactant whose amount determines the amount of product that can be obtained. The reactant in excess remains in the system even after the reaction is completed and may be recovered alongside the product which leads to a percentage yield above 100%.
If the excess reactant is soluble in water, it can be recovered from the precipitate if needed by washing the precipitate with water.
Which equation best represents the net ionic equation for the reaction that occurs when aqueous solutions of potassium phosphate and iron(II) nitrate are mixed? Question 8 options: 3Fe2+(aq) + 2PO43–(aq) → Fe3(PO4)2(s) 2K+(aq) + Fe(NO3)2(aq) → 2KNO3(aq) + Fe2+(aq) 3Fe2+(aq) + 2PO43–(aq) → Fe3(PO4)2(aq) 2K3PO4(aq) + 3Fe2+(aq) → Fe3(PO4)2(s) + (K+)6(aq) 2K3PO4(aq) + 3Fe(NO3)2(aq) → Fe3(PO4)2(s) + 6KNO3(aq)
Answer:
2PO₄³⁻ + 3Fe²⁺ → Fe₃(PO₄)₂(s)
Explanation:
In a net ionic equation you list only the ions that are participating in the reaction.
When potassium phosphate, K₃PO₄, reacts with iron (II) nitrate, Fe(NO₃)₂ producing iron (II) phosphate, Fe₃(PO₄)₂ that is an insoluble salt. The reaction is:
2K₃PO₄ + 3 Fe(NO₃)₂ → Fe₃(PO₄)₂(s) + 6NO₃⁻ + 6K⁺
The ionic equation is:
6K⁺ + 2PO₄³⁻ + 3Fe²⁺ + 6NO₃⁻→ Fe₃(PO₄)₂(s) + 6NO₃⁻ + 6K⁺
Subtracting the K⁺ and NO₃⁻ ions that are not participating in the reaction, the net ionic equation is:
2PO₄³⁻ + 3Fe²⁺ → Fe₃(PO₄)₂(s)A student is given an antacid tablet that weighs 5.8400 g. The tablet is crushed and 4.2800 g of the antacid is added to 200. mL of simulated stomach acid. It is allowed to react and then filtered. It is found that 25.00 mL of this partially neutralized stomach acid required 11.6 mL of a NaOH solution to titrate it to a methyl red end point. It takes 29.0 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid. How much of the original 200. mL of stomach acid (in mL) is neutralized by the 4.2800 g crushed sample of the tablet
Answer:
Explanation:
Given that:
mass of the antacid tablet = 5.8400 g
required mass of the antacid tablet = 4.2800 g was added to 200. mL of simulated stomach acid.
The amount of the original 200. mL of stomach acid (in mL) needed to neutralize the 4.2800 g crushed sample of the tablet can be calculated as:
= 11.6 mL of NaOH × 25.00 mL /29.0 mL NaOH
= 10.00 mL original stomach acid
Now; since it requires 11.6 mL of NaOH o neutralize 10.00 mL of original acid , then:
the antacid neutralized = 200 mL - 10.00 mL
the antacid neutralized = 190.00 mL
what is the differences between amorphous solid and crystalline solid
Answer:
Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points.
In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures.
of all the hydrogen nuclei in the ocean, 0.0156 how much deuterium could be obtained from 1.0 gal of ordinary tap water
Answer:
Poop Butt.
Explanation: Poop Butt.
When (R)-3-bromo-2,3-dimethylpentane is treated with sodium hydroxide, four different alkenes are formed. Draw all four products, and rank them in terms of stability.
Answer:
Most stable: 2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane : Least stable
Explanation:
Treatment of NaOH with (R)-3-bromo-2,3-dimethylpentane results in the elimination of HBr. Each H atoms present on each [tex]\beta[/tex]-carbon atoms can be eliminated result in the formation of four possible products: (1) 2,3-dimethylpent-2-ene, (2) (E)-3,4-dimethylpent-2-ene, (3) (Z)-3,4-dimethylpent-2-ene and (4) 2-methyl-3-methylenepentane.
The stability of these alkenes depends on the number of hyperconjugative H atoms present with respect to the double bond. In accordance with this, 2,3-dimethylpent-2-ene is the most stable alkene (11-hyperconjugative H atoms). Then, 3,4-dimethylpent-2-ene is the second most stable alkene (7-hyperconjugative H atoms). Among (E)-3,4-dimethylpent-2-ene and (Z)-3,4-dimethylpent-2-ene, (E)-3,4-dimethylpent-2-ene is more stable due to it's less sterically hindered structure. 2-methyl-3-methylenepentane is the least stable alkene (3-hyperconjugative H atoms).
So, decreasing order of stability of alkenes from most stable to least stable:
2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane
Three bromo As just a result of the nucleophilic substitution mechanism, 2,4 dimethylpentane generates a racemic mix containing both the r and s forms of the molecule.
In the first, sluggish phase, a two-degree cation with a positive charge is produced at carbon 3, and then it undergoes rearrangement to a 3-degree carbocation at carbon 2 either from the side. So because carbocation only has 6 electrons in its outermost shell, it is sp2 hybridized and thus planar in structure. In step two then occurs, as well as the attack of the -OH group, as the -OH (hydroxyl) group can strike from either side (top or bottom), leading to the formation of a racemic mixture of 2,4 dimethyl pentane-2-ol.This quantity of alpha-hydrogens in an alkyl group can be used to determine its stability. The greater the number of alpha-hydrogens inside an alkene, the overall larger the number of hyperconjugated structures, and thus the greater the stability. Due to the obvious symmetrical structure of a trans-isomer, whenever the amount of alpha-hydrogens is the same, the trans-alkene isomer is much more stable than that of the cis-alkene isomer.Please find the attached file.
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