Answer:
1. 0.90 are the initial moles of X and Y
2. 0.60 moles are the moles of Y and Z after the reaction
3. 0.90 moles of X and 0.30 moles of Y
4. 3X + 1Y → 2Z
Explanation:
1. For the reaction, initial moles of X and Y are:
500mL = 0.500L × (1.8 moles / L) = 0.90 are the initial moles of X and Y
2. After the reaction. The total volume is 500mL + 500mL = 1L
Moles Y and Z = 1L × (0.60 moles / 1L) = 0.60 moles are the moles of Y and Z after the reaction
3. As there is no moles of X after the reaction, all X reacts, that is 0.90 moles of X. And moles of Y that reacts are 0.90 mol - 0.60mol = 0.30 moles of Y
4. That means 3 moles of X reacts per mole of Y 0.90/0.30 = 3. Also, 2 moles of Z are produced per mole of Y 0.60/0.30 = 2.
That means balanced equation is:
aX + bY → cZ
3X + 1Y → 2ZWhich functional group does the molecule below have?
A. Ether
B. Ester
C. Hydroxyl
D. Amino
Answer:
Hydroxyl
Explanation:
A hydroxyl group is a functional group that attaches to some molecules containing an oxygen and hydrogen atom, bonded together. Also spelled hydroxy, this functional group provides important functions to both alcohols and carboxylic acids.
The functional groups are the part of the organic chemistry that confers the characteristic feature of a molecule. The molecule has a hydroxyl group in its structure. Thus, option C is correct.
What are hydroxyl functional groups?Hydroxyl functional groups are the atoms or molecules that provide a distinctive property to a compound. It has a chemical formula of -OH that has oxygen covalently bonded to the hydrogen atom.
The hydroxyl group is called the alcohol group that is seen in methanol, ethanol, propanol, etc. The presence of hydrogen allows the compound to form a water bond with other molecules and makes them soluble and polar.
Therefore, option C. the molecule has a hydroxyl or alcoholic functional group attached to its carbon atom.
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A student mixes wants to prepare 24.1 mmol of benzamide from benzoyl chloride and NH4OH. If the student uses excess 15 M NH4OH, how many mL of Benzoyl chloride must be used
Answer:
2.81mL
Explanation:
Based on the reaction:
C₆H₃COCl + 2NH₃ → C₆H₅CONH₂ + NH₄Cl
Benzoyl chloride + ammonia → Benzamide
1 mole of benzoyl chloride in excess of ammonia produce 1 mole of Benzamide.
Thus, assuming a theoretical yield, to produce 24.1mmoles of benzamide you require 24.1mmoles of benzoyl chloride.
As molar mass of benzoyl chloride is 141g/mol, mg you require are:
mg Benzoyl chloride: 24.1mmol × (141mg / 1mmol) = 3398.1mg = 3.3981g of benzoyl chloride.
to convert this mass to mL, you require density of Benzoyl chloride (1.21g/mL). Thus, mL you need are:
3.3981g × (1mL / 1.21g) =
2.81mLWhat is a major product of the reaction in the box?
Answer:
Molecule C
Explanation:
In this case, on the first reaction, we will have the production of a Grignard reagent. This molecule will react with [tex]D_2O[/tex] and a deuterium atom will be transferrred to the benzene ring. Then at the top of the molecule, we will have an acetal structure. This acetal can be broken by the action of the acid [tex]DCl[/tex], In the mechanism at the end, we will obtain a carbonyl group bonded to a hydrogen atom. Therefore we will have in the final product the aldehyde group. See figure 1 to further explanations.
I hope it helps!
The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. A(g) + 2B(g) ↔ AB2(g) Kc = 59 AB2(g) + B(g) ↔ AB3(g) Kc = ? A(g) + 3B(g) ↔ AB3(g) Kc = 478
Answer:
The correct answer is 8.10
Explanation:
Given:
A(g) + 2B(g) ↔ AB₂(g) Kc = 59 ---- Eq. 1
A(g) + 3B(g) ↔ AB₃(g) Kc = 478 ----- Eq. 2
We have to rearrange the chemical equations in order to obtain:
AB₂(g) + B(g) ↔ AB₃(g) Kc = ?
AB₂(g) is a reactant, so we have to use the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant is the same: Kc= 478. The following is the sum of rearranged chemical equations, and the compounds in bold and italic are canceled:
AB₂(g) ↔ A(g) + 2B(g) Kc₁= 1/59
A(g) + 3B(g) ↔ AB₃(g) Kc₂= 478
-----------------------------------------
AB₂(g) + B(g) ↔ AB₃(g)
If we add reactions at equilibrium, the equilibrium constants Kc are mutiplied as follows:
Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10
The value of the missing equilibrium constant is 8.10.
The value of the missing equilibrium constant is 8.10
Chemical Equations:Since
A(g) + 2B(g) ↔ AB₂(g) Kc = 59 ---- Eq. 1
A(g) + 3B(g) ↔ AB₃(g) Kc = 478 ----- Eq. 2
Now we have to rearrange the chemical equations in order to obtain:
AB₂(g) + B(g) ↔ AB₃(g) Kc = ?
Here AB₂(g) represents a reactant, so we have to applied the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant should be the same: Kc= 478.
The following is the sum of rearranged chemical equations, and the compounds in bold and italic should be canceled:
AB₂(g) ↔ A(g) + 2B(g) Kc₁= 1/59
A(g) + 3B(g) ↔ AB₃(g) Kc₂= 478
-----------------------------------------
AB₂(g) + B(g) ↔ AB₃(g)
In the case when we add reactions at equilibrium, the equilibrium constants Kc are multiplied as follows:
Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10
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most vegetables substantially diminish in quality in as little as days
Answer:
As little as 2 days
Hope this is correct
HAVE A GOOD DAY!
If a radioactive isotope of thorium (atomic number 90, mass number 232) emits 6 alpha particles and 4 beta particles during the course of radioactive decay, what is the mass number of the stable daughter product?
Answer:
The mass number of the stable daughter product is 208
Explanation:
First thing's first, we have to write out the equation of the reaction. This is given as;
²³²₉₀Th → 6 ⁴₂α + 4 ⁰₋₁ β + X
In order to obtain the identity of X, we have to obtain it's mass numbers and atomic number.
There is conservation of matter so we expect the mass number to remain the same in both the reactant and products.
Mass Number
Reactant = 232
Product = (6* 4 = 24) + (4 * 0 = 0) + x = 24 + x
since reactant = product
232 = 24 + x
x = 232 - 24 = 208
Atomic Number
Reactant = 90
Product = (6* 2 = 12) + (4 * -1 = -4) + x = 8 + x
since reactant = product
90 = 8 + x
x = 90 - 8 = 82
What products result from mixing aqueous solutions of Cr(NO3)2(aq) and NaOH(aq)? Question 10 options: Cr(OH)2(s), Na+(aq), and NO3−(aq) Cr(OH)2(s) and NaNO3(s) Cr2(OH)2(aq) and NaNO3(aq) Cr(OH)2(aq) and NaNO3(s) Cr(OH)2(s), N2(g), and H2O(l)
Answer:
Cr(OH)2(s), Na+(aq), and NO3−(aq)
Explanation:
Let is consider the molecular equation;
2NaOH(aq) + Cr(NO3)2(aq) -----> 2NaNO3(aq) + Cr(OH)2(s)
This is a double displacement or double replacement reaction. The reacting species exchange their partners. We can see here that both the sodium ion and chromium II ion both exchanged partners and picked up each others partners in the product.
Sodium ions and nitrate ions now remain in the solution while chromium II hydroxide which is insoluble is precipitated out of the solution as a solid hence the answer.
One of the nuclides in spent nuclear fuel is U-235, an alpha emitter with a half-life of 703 million years. How long will it take for an amount of U-235 to reach 24.0% of its initial amount? Express your answer in years to three significant figures. View Available Hint(s)
Answer:
1447584654 years or 1.44 billion years
Explanation:
From the formula;
0.693/t1/2=2.303/t log No/N
Where;
t1/2 = half life of the U-235 nuclides
No= amount of U-235 initially present
N= amount of U-235 present after a time t
t= time taken for N amount of U-235 to remain
Since N= 0.24No
Substituting values
0.693/703×10^6 = 2.303/t log No/0.24No
9.8578×10^-10 = 2.303/t log 1/0.24
9.8578×10^-10 = 1.427/t
t= 1.427/9.8578×10^-10
t= 1447584654 years
A student accidentally let some of the vapor escape the beaker. As a result of this error, will the mass of naphthalene you record be too high, too low, or unaffected? Why?
Answer:
too low
Explanation:
If our aim is to recover the naphthalene and measure its mass after separation, then we must not allow any vapour to escape.
Naphthalene is a sublime substance, it can be separated by sublimation. It changes directly from solid to gas. This vapour must be kept securely so that none of it escapes. If part of the naphthalene vapour happens to escape accidentally, then the measured mass of naphthalene will be too low compared to the mass of naphthalene originally present in the mixture.
An acetic acid buffer solution is required to have a pH of 5.27. You have a solution that contains 0.010 mol of Acetic acid. What molarity of sodium acetate will you need to add to the solution
Answer:
Molarity of sodium acetate you will need to add is 0.0324M
Explanation:
Assuming volume of the buffer is 1L.
The pH of a buffer can be determined using Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
Where pKa is pKa of the weak acid, [A⁻] molar concentration of conjugate base and [HA] molar concentration of weak acid
Replacing for the acetic buffer (pKa = 4.76):
pH = 4.76 + log [Sodium Acetate] / [Acetic Acid]
As you have 0.010 moles of acetic acid in 1L:
[Acetic Acid] = 0.010mol / 1L = 0.010M
And you require a pH of 5.27:
5.27 = 4.76 + log [Sodium Acetate] / [0.010M]
0.51 = log [Sodium Acetate] / [0.010M]
10^0.51 = [Sodium Acetate] / [0.010M]
3.236 = [Sodium Acetate] / [0.010M]
3.236 [0.010M] = [Sodium Acetate]
0.0324M = [Sodium Acetate]
Molarity of sodium acetate you will need to add is 0.0324M
Why is it important that the primary standard chemical be non-hygroscopic and pure? Why is it important to dry the primary standard to a constant weight?
Answer:
It is extremely important for the primary standard chemical to be non – hygroscopic and pure and to also have a constant weight because you don't want any moisture or any impurities to alter the stoichiometric point in the reaction
It is important that the primary standard chemical be non-hygroscopic and pure to calculate the exact calculation of the reaction.
What is non hygroscopic chemicals?Non hygroscopic chemicals are those compounds which will not absorb water or mositure from the outside.
If we take any substance which are hygroscopic in nature and during the chemical reaction if they absorb water content or moisture then the mass of that substance will alter and changes all the calculation of the reaction.
So, to maintain the stability of calculation we use non hygroscopic materials.
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Determine which complex of the electron transport chain (respiratory chain) each phrase describes. (Coenzyme Q is also called ubiquinone or ubiquinol, depending on whether it is in oxidized or reduced form.)
Complex I:
Complex II:
Complex III:
Complex IV:
Here are the choices that need to be put in the correct complex:
1) NADH-ubiquinone(NADH-coenzyme Q oxidoreductase)
2) Coenzyme Q-cytochrome c oxidoreductase
3) Electron transfer from succinate to ubiquinone (coenzyme Q)
4) Electron transfer from cytochrome c to O2
5) Succinate-coenzyme Q Oxidoreductase (succinate dehydrogenase)
6) Cytochrome c oxidase
7) Electron transfer from ubiquinol (QH2) to cytochrom c
8) Electron transfer from NADH to ubiquinone (coenzyme Q)
Answer:
Complex I: (1) NADH-ubiquinone(NADH-coenzyme Q oxidoreductase), (8) Electron transfer from NADH to ubiquinone (coenzyme Q)
Complex II: (3) Electron transfer from succinate to ubiquinone (coenzyme Q) (5) Succinate-coenzyme Q Oxidoreductase (succinate dehydrogenase)
Complex III: (2) Coenzyme Q-cytochrome c oxidoreductase, (7) Electron transfer from ubiquinol (QH2) to cytochrome c
Complex IV: (4) Electron transfer from cytochrome c to O2, (6) Cytochrome c oxidase
Explanation:
The electron transport chain (ETC) in the mitochondria provides a pathway by which electrons are transferred from NADH and FADH₂ through a series of membrane-bound carriers to molecular oxygen reducing it to water.
The electron transport chain electron carriers are organized into four complexes, Complexes I - IV.
Complex I : It is also called NADH:ubiquinone reductase. It transfers electrons from NADH to ubiquinone (also known as coenzyme Q)
Complex II : It is also called succinate dehydrogenase. It functions to tranfer electrons from succinate to FAD and then to ubiquinone.
Complex III : It is also called ubiquinone:cytochrome c oxidoreductase. It functions to transfer electrons from ubiquinol (reduced ubiquinone) to cytochrome c.
Complex IV : It is also called cytochrome oxidase. It functions to transfer electrons from cytochrome c to molecular oxygen reducing it to water.
The electron transporter chain is a series of enzymatic reactions to produce and store energy for the organism’s correct functioning. Complex I: 1 and 8. Complex II: 3 and 5. Complex III: 2 and 7. Complex IV: 4 and 6.
---------------------------------
Electron transporter chain
The electron transporter chain is located in the internal mitochondrial membrane. It constitutes a series of enzymatic reactions to release and save energy for the organism’s correct functioning.
Along the chain, there are four proteinic complexes in the membrane, I, II, III, and IV, that contain the electrons transporters and the enzymes necessary to catalyze the electrons' transference from one complex to the other.
Different redox reactions occur to pass electrons along the chain.
Released energy creates a proton concentration gradient used to synthesize ATP.
1) NADH provides electrons to the first complex, Complex I (NADH-
ubiquinone or NADH-coenzyme Q oxidoreductase).
From there, electrons go to the coenzyme Q (Ubiquinone) that carries them to complex II and III. Meanwhile, complex I pomp four protons to the intermembrane space.
2) Complex II (succinate-dehydrogenase) receives electrons from CoQ and also receives electrons from FADH2. Electrons are sent from complex II to ubiquinone Q that carries these electrons to complex III.
3) Complex III (Cytochrome C-reductase) receives electrons from ubiquinone Q and pomps protons to the intermembrane space.
Electrons are transferred to Cytochrome c.
Electrons travel from cytochrome c to complex IV.
4) Complex IV (Cytochrome C-oxidase) is the last complex that pomps protons to the intermembrane space. It takes electrons from cytochrome C and sends them to oxygen.
5) Electrons are sent to O₂ molecules, which also receive protons in the matrix to create water molecules.
Four electrons are needed to produce two water molecules from one O₂ molecule.
The proton gradient is used to produce ATP molecules.
Now, we can join the complexes with the phrases.
Complex I:
1) NADH-ubiquinone (NADH-coenzyme Q oxidoreductase)
8) Electron transfer from NADH to ubiquinone (coenzyme Q)
Complex II:
3) Electron transfer from succinate to ubiquinone (coenzyme Q)
5) Succinate-coenzyme Q Oxidoreductase (succinate dehydrogenase)
Complex III:
2) Coenzyme Q - cytochrome c oxidoreductase
7) Electron transfer from ubiquinol (QH₂) to cytochrom c
Complex IV:
6) Cytochrome C oxidase
4) Electron transfer from cytochrome c to O₂
-----------------------------------
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Consider the following reaction (X = Cl or Br) which statement s is are correct?
CH3CH2CH3 + X2---------CH3CHCH3--X + CH3CH2CH2--X
i. statistically the 1 halopropane should be the major isomer
ii. the 2 halopropane to 1 halopropane ratio is largest when X = Br
iii. the 2 halopropane to 1 halopropane ratio is largest when X = Cl
A. only Il
B. only Ill
C. I and II
D. I and III
A.S OLOS kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkll
Choose the most appropriate indicator for the titration of a weak acid with NaOH, where the expected equivalence point of the titration is at pH 8.8.
a. methyl orange, pH range 3.2-4.4
b. methyl red, pH range 4.8 6.0
c. bromothymol blue, pH range 6.0-7.6
d. phenolphthalein, pH range 8.2-10.0
e. alizarin yellow R. pH range 10.1-12.0
Answer:
D phenolphthalein,pH range 8.2-10.0
14. A piece of titanium at 100.0°C was dropped into 50.0 g of water at 20.0°C. The final temperature of the system was 22.6°C. What is the mass of the titanium? (Specific Heat of titanium = 0.54 J/g°C)
Answer:
[tex]m_{Ti}=13.0g[/tex]
Explanation:
Hello,
In this case, based on the given, we can infer that as titanium is hot and water cold, it cools down whereas the water is heated up, therefore, in terms of heat, we have that the heat lost by the titanium is gained by the water:
[tex]-Q_{Ti}=Q_{H_2O}[/tex]
That in terms of mass, specific heat and temperatures is:
[tex]-m_{Ti}Cp_{Ti}(T_2-T_{Ti})=m_{H_2O}Cp_{H_2O}(T_2-T_{H_2O})[/tex]
In such a way, for computing the mass of titanium, considering the heat capacity of water 4.18 J/g°C, we have:
[tex]m_{Ti}=\frac{m_{H_2O}Cp_{H_2O}(T_2-T_{H_2O})}{-Cp_{Ti}(T_2-T_{Ti})} \\\\m_{Ti}=\frac{50.0g*4.18\frac{J}{g\°C}(22.6-20.0)\°C}{-0.54\frac{J}{g\°C}*(22.6-100.0)\°C} \\\\m_{Ti}=13.0g[/tex]
Regards.
What word or two-word phrase best describes the shape of the water ( H2O ) molecule?
Answer:
Water (H2O) is an inorganic chemical compound formed by two hydrogen (H) and one oxygen (O) atoms. 3 This molecule is essential in the life of living beings, serving as a medium for the metabolism of biomolecules, is found in nature in its three states and was key to its formation. It is necessary to distinguish between drinking water and pure water, since the first is a mixture that also contains salts in solution; this is why in the laboratory and in other areas distilled water is used.
Explanation:
I hope I've helped
According to the molecular geometry, the V-shape or bent structure best describes the shape of water molecule.
What is molecular geometry?Molecular geometry can be defined as a three -dimensional arrangement of atoms which constitute the molecule.It includes parameters like bond length,bond angle and torsional angles.
It influences many properties of molecules like reactivity,polarity color,magnetism .The molecular geometry can be determined by various spectroscopic methods and diffraction methods , some of which are infrared,microwave and Raman spectroscopy.
They provide information about geometry by taking into considerations the vibrational and rotational absorbance of a substance.Neutron and electron diffraction techniques provide information about the distance between nuclei and electron density.
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a. How many moles of copper equal 8.00 × 109 copper atoms?
b. How many moles of calcium equal 102.5 g calcium?
c. How many atoms of lead are present in 5.04 g lead?
d. How many aluminum atoms are present in 2.85 moles of aluminum?
e. What is the mass in grams of 1.08 × 103 moles of fluorine gas?
f. How many molecules of benzene (C6H6) are present in 0.584 g of benzene?
g. What is the mass of 5.09 × 109 atoms of hydrogen gas?
h. How many calcium atoms are present in 0.45 mol Ca3PO4?
Answer:
Explanation:
a ) one mole = 6.02 x 10²³ atoms
no of moles in given no of atoms
= 8 x 10⁹ / 6.02 x 10²³
= 1.329 x 10⁻¹⁴ moles .
b ) one mole of calcium = 40 gram
102 .5 g calcium
= 102 .5 / 40 moles
= 2.5625 moles
c )
no of moles in 5.04 g lead = 5.04 / 207
= 2.4347 x 10⁻² moles
= 2.4347 x 10⁻²x 6.02 x 10²³ no of atoms of lead
= 14.6568 x 10²¹ no of atoms .
d)
one mole = 6.02 x 10²³ atoms
2.85 mole = 17.157 x 10²³ atoms .
e )
moles of fluorine gas = 1.08 x 10³ / 6.02 x 10²³
= .1794 x 10⁻²⁰ moles
mass in grams = .1794 x 10⁻²⁰ x 38
= 6.8172 x 10⁻²⁰ grams
f )
no of moles in .584 g of benzene = .584 / 78
= 7.487 x 10⁻³ moles
no of molecules = 6.02 x 10²³ x 7.487 x 10⁻³
= 45.07 x 10²⁰ molecules .
g )
moles of atoms = 5.09 x 10⁹ / 6.02 x 10²³
= .8455 x 10⁻¹⁴ moles
mass in gram = .8455 x 10⁻¹⁴ x 1
= .8455 x 10⁻¹⁴ g
h )
.45 moles of Ca₃PO₄ = .45 x 6.02 x 10²³ molecules
= 2.709 x 10²³ molecules of Ca₃PO₄
no of atoms of Ca = 3 x 2.709 x 10²³
= 8.127 x 10²³ atoms of Ca .
what is the name of this molecule?
some students believe that teachers are full of hot air. If I inhale 3.5 liters of gas at a temperature of 19 degrees Celsius and it heats to a temperature of 58 degrees celsius in my lungs. what is the new volume of the gas?
Answer:
3.97 L
Explanation:
Data obtained from the question include the following:
Initial volume (V1) = 3.5 L
Initial temperature (T1) = 19 °C
Final temperature (T2) = 58 °C
Final volume (V2) =..?
Next, we shall convert celsius temperature to Kelvin temperature. This can be done as shown below:
Temperature (K) = temperature (°C) + 273
T (K) = T (°C) + 273
Initial temperature (T1) = 19 °C
Initial temperature (T1) = 19 °C + 273 = 292 K
Final temperature (T2) = 58 °C
Final temperature (T2) = 58 °C + 273 = 331 K
Finally, we shall determine the new volume of the gas by using Charles' law equation as shown below:
Initial volume (V1) = 3.5 L
Initial temperature (T1) = 292 K
Final temperature (T2) = 331 K
Final volume (V2) =..?
V1 /T1 = V2 /T2
3.5 /292 = V2 /331
Cross multiply
292 x V2 = 3.5 x 331
Divide both side by 292
V2 = (3.5 x 331) / 292
V2 = 3.97 L
Therefore, the new volume of the gas is 3.97 L.
Nitrogen has different oxidation states in the following compounds: nitrite ion, nitrous oxide, nitrate ion, ammonia, and nitrogen gas. Arrange these species in order of increasing nitrogen oxidation state. Select the correct answer below: A. ammonia, nitrogen gas, nitrite, nitrous oxide, nitrate B. nitrogen gas, ammonia, nitrous oxide, nitrite, nitrate C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate D. ammonia, nitrogen gas, nitrate, nitrite, nitrous oxide
Answer:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
Explanation:
To establish the oxidation number of nitrogen in each compound, we know that the sum of the oxidation numbers of the elements is equal to the charge of the species.
Nitrite ion (NO₂⁻)
1 × N + 2 × O = -1
1 × N + 2 × (-2) = -1
N = +3
Nitrous oxide (NO)
1 × N + 1 × O = 0
1 × N + 1 × (-2) = 0
N = +2
Nitrate ion (NO₃⁻)
1 × N + 3 × O = -1
1 × N + 3 × (-2) = -1
N = +5
Ammonia (NH₃)
1 × N + 3 × H = 0
1 × N + 3 × (+1) = 0
N = -3
Nitrogen gas (N₂)
2 × N = 0
N = 0
The order of increasing nitrogen oxidation state is:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
Calculate the molality of a solution prepared by dissolved 19.9 g of kcl in 750ml of water
First, we find the molar mass of KCl which is AK+ACl=39+35.5=74.5g/mole
now we find out the number of moles by dividing the given mass to the molar mass n=m/M=19.9/74.5=0.26 moles. The molarity of a solution is equal to the number of moles divided by the volume of the solution. =0.26moles/0.75liters=0.346M
13C NMR is a technique in which the total number of signals represents the number of unique carbon atoms in a molecule. Propose a structure that is consistent with the following data.
a. The IR includes peaks at 1603 and 1495 cm^-1
b. The 13c NMR has a total of 7 signals
c. The compound has one acidic proton.
Answer:
D. Poop Butt.
Explanation:
Based on the given data, we can propose a possible structure that fits the criteria: a. carbonyl group (C=O) and an aromatic ring b. there are seven unique carbon environments. c. Presence of a functional group like a carboxylic acid or phenol .
a. The IR peaks at 1603 [tex]cm^{-1}[/tex] and 1495[tex]cm^{-1}[/tex]suggest the presence of both a carbonyl group (C=O) and an aromatic ring.
b. The 13C NMR having a total of 7 signals indicates that there are seven unique carbon environments in the molecule.
c. Considering the presence of an acidic proton, it suggests the presence of a functional group like a carboxylic acid (COOH) or phenol ([tex]C_6H_5OH[/tex]).
Putting all this information together, a possible structure that fits the data could be benzoic acid ([tex]C_6H_5COOH[/tex]). It contains a benzene ring (giving 6 unique carbon environments), a carbonyl group (giving 1 unique carbon environment), and an acidic proton in the carboxylic acid group. This structure satisfies all the given data.
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Which of the following properties should carbon (C) have based on its position on
the periodic table?
A. Shiny
B. Dense
C. Malleable
D. Poor conductor
Answer:
D- poor conductor
Explanation:
metallic properties decrease as we go on the right of the periodic table. Carbon is a non metal hence it is dull and a poor conductor.
it has a low density and is ductile.
Answer: Poor conductor
Explanation:
Suppose of nickel(II) iodide is dissolved in of a aqueous solution of potassium carbonate. Calculate the final molarity of nickel(II) cation in the solution. You can assume the volume of the solution doesn't change when the nickel(II) iodide is dissolved in it. Round your answer to significant digits.
Answer:
0.619 M to 3 significant figures.
Explanation:
1 mole of [tex]NiI_{2}[/tex] - 312.5 g
? mole of [tex]NiI_{2}[/tex] - 2.9 g
= 2.9/312.5
= 0.0928 moles.
Concentration = no. of moles/vol in litres = [tex]\frac{0.0928}{0.150L}[/tex]
= 0.619 M
Treatment of 1 mole of dimethyl sulfate with 2 moles of sodium acetylide results in the formation of propyne as the major product.
A) Draw a reasonable mechanism accounting for the formation of the byproduct 2-butyne.
B) 2-Butyne is observed as a minor product of this reaction. Draw a mechanism accounting for the formation of this minor product and explain how your proposed mechanism is consistent with the observation that acetylene is present among the reaction products.
C) Predict the major and minor products that are expected if diethyl sulfate is used in place of dimethyl sulfate.
Answer:
(a) appended underneath is the inorganic ion shaped in the reaction and the mechanism of its formation
(b) 2-butyne framed as a minor product is appeared in the connection. It is shaped when the monosodium subordinate of dimethylsulphoxide gets a hydrogen from the propyne and reacts again with monosodium methylsulphoxide.
(c) The major product framed when diethylsulphoxide is utilized, would be butyne and minor product would be 3-hexyne.
Explanation:
attached below is diagram
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
Ammonia is oxidized with air to form nitric oxide in the first step of the production of nitric acid. Two principal gas-phase reactions occur:
Answer:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O
2NO(g) + O₂(g) → 2 NO₂
Explanation:
First of all, we need to consider the reaction for production of ammonia. In this reaction we have as reactants, nitrogen and hydroge.
3H₂ (g) + N₂(g) → 2NH₃ (g)
Afterwards, ammonia reacts to oxygen, to produce NO and H₂O
The equation for the process will be:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O
Then, we take the nitric oxide to make it react, to produce NO₂, in order to produce nitric acid, for the final reaction:
2NO(g) + O₂(g) → 2 NO₂
3NO₂(g) + H₂O(g) → 2 HNO₃ (g) + NO(g)
Hydrogen iodide decomposes according to the equation: 2HI(g) H 2(g) + I 2(g), K c = 0.0156 at 400ºC A 0.660 mol sample of HI was injected into a 2.00 L reaction vessel held at 400ºC. Calculate the concentration of HI at equilibrium.
Answer:
[HI] = 0.264M
Explanation:
Based on the equilibrium:
2HI(g) ⇄ H₂(g) + I₂(g)
It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:
Kc = 0.0156 = [H₂] [I₂] / [HI]²
As initial concentration of HI is 0.660mol / 2.00L = 0.330M, the equlibrium concentrations will be:
[HI] = 0.330M - 2X
[H₂] = X
[I₂] = X
Where X is reaction coefficient.
Replacing in Kc:
0.0156 = [X] [X] / [0.330M - 2X]²
0.0156 = X² / [0.1089 - 1.32X + 4X² ]
0.00169884 - 0.020592 X + 0.0624 X² = X²
0.00169884 - 0.020592 X - 0.9376 X² = 0
Solving for X:
X = - 0.055 → False solution, there is no negative concentrations
X = 0.0330 → Right solution.
Replacing in HI formula:
[HI] = 0.330M - 2×0.033M
[HI] = 0.264MDraw a Lewis structure for one important resonance form of HBrO4 (HOBrO3). Include all lone pair electrons in your structure. Do not include formal charges in your structure.
Answer:
The Lewis structure is attached with the answer -
Explanation:
Lewis structure or Lewis dot diagram are diagrams or representation of showing the bonding between different or same atoms of a molecule in any and also shows lone pairs of electrons that may exist in the molecule as dots.
HBrO₄ is bromine oxoacid which is also known as perbromic acid. It is a unstable inorganic compound.
The Lewis structure is attached in form of image with representation of lone pairs of electrons.
Decide which element probably has a density most and least similar to the density of lithium.
Answer:
Helium and potassium
Explanation:
The density of helium is 0.18 and potassium 0.86, while lithium is 0.53
