Answer:
The correct answer is 1.60 Liters.
Explanation:
The given reaction:
CO (g) + H₂(g) ⇔ CH₃OH (g)
Based on the given reaction, two moles of H₂ reacts with one mole of CO and produce one mole of CH₃OH.
It is mentioned that 3.20 L of H₂ is reacted, therefore, there is a need to convert it into moles.
As 22.4 L at standard temperature and pressure is equivalent to 1 mole.
Therefore, 1 L at STP will be, 1/22.4 mole
Now 3.20 L at STP will be,
= 1/22.4 × 3.20
= 0.1428 mole
And as mentioned in the reaction that 2 moles of H₂ gives 1 mole of CH₃OH, therefore, 1 mole of H₂ will give 1/2 mole of CH₃OH
Now, 0.1428 mole of H₂ will give,
= 0.1428/2 = 0.071 mole of CH₃OH
= 0.071 × 22.4 = 1.60 L
The volume, in liters, of CH₃OH gas formed is 1.60 L
From the question,
We are to determine the volume of CH₃OH formed
The given chemical equation for the reaction is
CO(g)+ H₂(g) → CH₃OH
The balanced chemical equation for the reaction is
CO(g)+ 2H₂(g) → CH₃OH
This means
1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH
Now, we will determine the number of moles of H₂ present in the 3.20 L H₂ at STP
1 mol of an ideal gas has a volume of 22.4 L at STP
Then,
x mole of the H₂ gas will have a volume of 3.20 L at STP
x = [tex]\frac{3.20 \times 1}{22.4}[/tex]
x = 0.142857 mole
∴ The number of mole of H₂ present is 0.142857 mole
Since
2 moles of H₂ reacts to produce 1 mole of CH₃OH
Then,
0.142857 mole of H₂ will react to produce [tex]\frac{0.142857}{2}[/tex] mole of CH₃OH
[tex]\frac{0.142857}{2} = 0.0714285[/tex]
∴ The number of moles of CH₃OH produced = 0.0714285 mole
Now, for the volume of CH₃OH formed
Since
1 mol of an ideal gas has a volume of 22.4 L at STP
Then,
0.0714285 mol of CH₃OH will have a volume of 22.4 × 0.0714285 at STP
22.4 × 0.0714285 = 1.5999984 L ≅ 1.60 L
Hence, the volume of CH₃OH gas formed is 1.60 L
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Name the following alkane molecule:
Br
CI
A. 6-bromo-2-chlorodecane
B. 2-chloro-6-bromodecane
C. 5-bromo-9-chlorodecane
Answer:
6- bromo- 2-chlorodecane
The name of molecule is 6- bromo- 2-chlorodecane.
What is 6- bromo- 2-chlorodecane?
It is organic molecule. There are 10 carbon present in this molecule. The bromine atom is attached with C-6 carbon where chlorine is attached with C-2 carbon.
What is molecule?A molecule is a collection of more than one atom bound together by chemical bonds; depending on the context, the phrase may or may not include ions that meet this requirement.
Therefore, the name of the given molecule will be 6- bromo- 2-chlorodecane
To know more about molecule and 6- bromo- 2-chlorodecane.
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A compound is found to contain 31.42 % sulfur, 31.35 % oxygen, and 37.23 % fluorine by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 102.1 g/mol. What is the molecular formula for this compound?
Explanation:
To obtain the empirical and molecular formula of this compound from the percent composition of the elements, we follow the steps below;
Step 1: Divide the percentage composition by the atomic mass
Sulphur = 31.42 / 32 = 0.9819
Oxygen = 31.35 / 16 = 1.9594
Flourine = 37.23 / 19 = 1.9595
Step 2: Divide by the lowest number
Sulphur = 0.9819 / 0.9819 = 1
Oxygen = 1.9594 / 0.9819 ≈ 2
Flourine = 1.9595 / 0.9819 ≈ 2
This means the ratio of the elements is 1 : 2: 2
The empirical formular (simplest formular of a compound) of the compound is;
SO₂F₂
To obtain the molecular formular (Actual formular of a compound);
(SO₂F₂)n = 102.1
Inserting the atomic masses and solving for n;
(102)n = 102.1
n ≈ 1
The molecular formular is; (SO₂F₂)₁ = SO₂F₂
Which of the following elements is in the same group as Sulfur (S)?
Answer:
PLEASE SHOW ME THE ELEMENTS OR I WOULD ENLIST ALL THE ELEMENTS.
Explanation:
Group 6A (or VIA) of the periodic table are the chalcogens: the nonmetals oxygen (O), sulfur (S), and selenium (Se), the metalloid tellurium (Te), and the metal polonium (Po)
A critical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis of adenosine triphosphate, ATP, to adenosine diphosphate, ADP, as described by the reaction
ATP(aq)+H2O(l)→ADP(aq)+HPO2−4(aq) for which ΔGrxn=30.5 kJ/mol at 37.0∘C and pH 7.0.
Required:
Calculate the value of ΔGrxn in a biological cell in which [ATP]=5.0 mM, [ADP]=0.60 mM, and [HPO2−4]=5.0 mM
Answer:
ΔG = -49.64 KJ/mol
Explanation:
The actual free energy change of the reaction under the given conditions, ΔG is given by the formula below;
ΔG = ΔG'° + RT ln([ADP][HPO₂⁴⁻] / [ATP])
where ΔG'° = -30.5 KJ/mol; R = 8.315 J/mol.K; T = 37°C = 310 K; [ADP] = 5.0 mM = 0.005 M; [HPO₂⁴⁻] = 0.60 mM = 0.0006 M; [ATP] = 5.0 mM = 0.005 M
ΔG = -30.5 KJ/mol + (8.315 J/mol.K)(310 K) ln {(0.005)(0.0006)/(0.005)}
ΔG = -30.5 KJ/mol + (2.58 KJ/mol * -7.4186)
ΔG = -30.5 KJ/mol - 19.14 KJ/mol
ΔG = -49.64 KJ/mol
The valves in the heart open and close to move blood in between heart chambers. O a) three directions O b) two directions O c) four directions d) one direction
Answer:
im not sure but i think its four directions
Hi,
The valves in the heart open and close to move blood in between heart chambers.
Answer:
d) one direction
A skydiver slows down from 65 m/s to 5 m/s by opening the parachute. If this
takes 0.75 seconds, what is the skydiver's acceleration?
A. 45 m/s2 up
B. 80 m/s2 up
C. 45 m/s2 down
D. 80 m/s2 down
Answer:
D. -80m/s^2
Explanation:
V = u + at
5 = 65 + a (0.75)
0.75a = -60
a = -60/0.75
a = -80m/s^2
Therefore, is decelerating at 80m/s^2
Answer:
[tex]\boxed {\boxed {\sf D. \ 80 \ m/s^2 \ down}}[/tex]
Explanation:
We are asked to find the acceleration of a skydiver. Acceleration is the change in velocity over the change in time, so the formula for calculating acceleration is:
[tex]a= \frac{v_f-v_i}{t}[/tex]
The skydiver was initially traveling 65 meters per second, then he slowed down to a final velocity of 5 meters per second. He slowed down in 0.75 seconds.
[tex]\bullet \ v_f = 5 \ m/s \\\bullet \ v_i= 65 \ m/s \\\bullet \ t= 0.75 \ s[/tex]
Substitute the values into the formula.
[tex]a= \frac{ 5 \ m/s - 65 \ m/s}{0.75 \ s}[/tex]
Solve the numerator.
[tex]a= \frac{-60 \ m/s}{0.75 \ s}[/tex]
Divide.
[tex]a= -80 \ m/s^2[/tex]
The acceleration of the skydiver is -80 meters per second squared or 80 meters per second squared down. The skydiver is slowing down or decelerating, so the acceleration is negative or down.
Calculate the molarity of bromide ions in 250. mL of a solution containing 25.9 g NaBr and 0.155 moles of HBr.
Answer:
[tex]1.628 M[/tex]
Explanation:
From the question we were given 0.155 moles of HBr, but Br and H are in ratio 1:1, then there are 0.155 moles of Br- ions.
We were also told that the solution contain NaBr, of 25.9 g. Then it must be converted to moles.
molar mass of NaBr =(22.99g + 79.90 )
= 102.89 g per mol.
the moles of NaBr can be calculated as 25.9 / 102.89
=0.252 moles
But Na and Br are in a ratio 1:1 , then there are 0.252 moles of Br-.
Then to get two Br- mol , we will add the first and second mol of Br- together
= 0.155 + 0.252
=0.407 moles.
The given solution has volume of 250 mL, but we know that there are 1000 ml in a liter, then if we convert to L for unit consistency we have
= 250/1000
= 0.25 L
molarity=0.407 moles/0.25 L
= 1.628 M.
Therefore, Br ion molarity is 1.628 M.
The molarity of the Br ions in the 250 ml solution has been 1.628 M.
Moles can be defined as the mass per unit molecular mass. Moles can be expressed as:
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Moles of NaBr = [tex]\rm \dfrac{25.9}{102.89}[/tex]
Moles of NaBr = 0.252 mol
Moles of HBr = 0.155 mol.
Since both the compounds have 1:1 ratio of atom: Br, the Br produced has been equal to the concentration of the compound.
Br from NaBr = 0.252 mol
Br from HBr = 0.155 mol.
Total Br ions = 0.407 mol.
Molarity can be expressed as:
Molarity = [tex]\rm moles\;\times\;\dfrac{1000}{Volume\;(ml)}[/tex]
Molarity of Br ions = 0.407 × [tex]\rm \dfrac{1000}{250\;ml}[/tex]
Molarity of Br ions = 1.628 M.
The molarity of the Br ions in the 250 ml solution has been 1.628 M.
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Sean plated an unknown metal onto his silver ring which initially weighed 1.4 g. He constructs an electrolytic cell using his ring as one of the electrodes. After running the cell, 0.022 moles of the unknown metal was plated onto his ring and the mass of the ring increased to 3.137 g. What is the atomic weight of the unknown metal in g/mol
Answer:
79 g/mol
Explanation:
Mass of unknown metal deposited = 3.137 g - 1.4 g = 1.737 g
Number of moles of metal deposited = 0.022 moles
Since;
Number of moles = reacting mass/molar mass
Molar mass = reacting mass/number of moles
Molar mass = 1.737 g/0.022 moles
Molar mass= 79 g/mol
3. Write the following isotope in hyphenated form (e.g., "carbon-14”): Kr
a. Krypton-109
b. Krypton -37
c. Krypton -36
d. Krypton -73
Answer:
Krypton -73
Explanation:
There are 33 known isotopes of krypton (36Kr) with atomic mass numbers from 69 through 101.
Good luck!
Answer:
D. Krypton-73
Explanation:
An isotope of an element has the same atomic number and the same number of protons but a different number of neutrons and a different atomic weight. Krypton is the 36th element on the periodic table. It has 36 protons and 48 neutrons. Krypton-73 is one of 33 known isotopes of Krypton and is the only one that actually exists from the list of choices.
Hope that helps.
An unknown gas diffuses 5 times slower than that of H2.The moleculer mass of unknown gas is??
Answer:
50.
Explanation:
We can write Graham's Law of Diffusion as:
(Rate 1)^2 = Molecular Mass 2
-------------- -------------------------
(Rate 2)^2 Molecular Mass 1
So using the Given Information:
1^2 / (1/5)^2 = Molecular Mass of unknown gas / 2, so:
25 = M/2
M = 50.
Nombre de la siguiente estructura de compuesto orgánico
Answer:
Nombre de la siguiente estructura de compuesto orgánico
Explanation:
An SN2 reaction is a type of _____________ in which the nucleophile attacks the electrophile.
Answer:
A bimolecular nucleophilic substitution (SN2) reaction is a type of nucleophilic substitution whereby a lone pair of electrons on a nucleophile attacks an electron deficient electrophilic center and bonds to it, resulting in the expulsion of a leaving group.
Identify a homogeneous catalyst. Identify a homogeneous catalyst. Pd in H2 gas N2 and H2 catalyzed by Fe SO2 over vanadium (V) oxide Pt with methane H2SO4 with concentrated HCl
Answer:
H2SO4 with concentrated HCl
What is the concentration of A after 50.7 minutes for the second order reaction A → Products when the initial concentration of A is 0.250 M? (k = 0.117 M⁻¹min⁻¹)
Answer:
0.101 M
Explanation:
Step 1: Given data
Initial concentration of A ([A]₀): 0.250 MFinal concentration of A ([A]): ?Time (t): 50.7 minRate constant (k): 0.117 M⁻¹.min⁻¹Step 2: Calculate [A]
For a second-order reaction, we can calculate [A] using the following expression.
1/[A] = 1/[A]₀ + k × t
1/[A] = 1/0.250 M + 0.117 M⁻¹.min⁻¹ × 50.7 min
[A] = 0.101 M
The two common properties of all solids are fixed _____ and _____.
Answer:
shape
volume
Hope this helps! （づ￣3￣）づ╭❤～
Explanation:
clacium hydroxide is slightly soluable in water about 1 gram will dissolve in 1 liter what are the spectator ions in the reaction ions in the reaction of such a dilute solution of calcium hydroxide with hydrochloric acid
Answer:
Ca²⁺ and Cl⁻
Explanation:
In a chemical reaction, spectator ions are ions that are not involved in the reaction, that means are the same before and after the reaction.
In water, calcium hydroxide, Ca(OH)₂ is dissociated in Ca²⁺ and OH⁻. Also, hydrochloric acid, HCl, dissociates in H⁺ and Cl⁻. The reaction is:
Ca²⁺ + 2OH⁻ + 2H⁺ + 2Cl⁻ → 2H₂O + Ca²⁺ + 2Cl⁻
The ions that react are H⁺ and OH⁻ (Acid and base producing water)
And the ions that are not reacting, spectator ions, are:
Ca²⁺ and Cl⁻For the reaction CO2(g) + H2(g)CO(g) + H20(g)
∆H°=41.2 kJ and ∆S°=42.1 J/K
The standard free energy change for the reaction of 1.96 moles of Co2(g) at 289 K, 1 atm would be_________KJ.
This reaction is (reactant, product)___________ favored under standard conditions at 289 K.
Assume that ∆H° and ∆S° are independent of temperature.
Answer:
The ΔG° is 29 kJ and the reaction is favored towards reactant.
Explanation:
Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,
ΔG° = ΔH°rxn - TΔS°rxn
= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K
= 41.2 kJ - 12.2 kJ
= 29 kJ
As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.
What are the correct half reactions for the following reaction: Cu2+ + Mg -> Cu + Mg2+
Answer:
Cu2 + 2Mg-> 2Cu+ Mg2
Explanation:
Balance the equation and make sure both the reactant and the products are the same
Hope it will be helpful
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
What is a balanced equation?A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total oxidation numbers is the same for both the reactants and the products.
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.
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What is the initial temperature (°C) of a system that has the pressure decreased by 10 times while the volume increased by 5 times with a final temperature of -123°C?
Answer:
27°C or 300K
Explanation
We were told that the pressureof the system decreased by 10 times implies that P2= P1/10
Where P2=final pressure
P1= initial pressure
Wew were also told that the volume of the system increased by 5 times this implies that V2= 5×V1
Where T2= final temperature =-123C= 273+(-123C)=150K
T1= initial temperature
But from gas law
PV=nRT
As n and R are constant
P1V1/T1 = P2V2/T2
T1= P1V1T2/P2V2
T1=2×T2
T1=2×150
T1=300K
=300-273
=27°C
the initial temperature (°C) of a system is 27°C
An unknown compound has the following chemical formula:
Co(OH),
where x stands for a whole number.
Measurements also show that a certain sample of the unknown compound contains 5.1 mol of oxygen and 2.59 mol of cobalt.
Write the complete chemical formula for the unknown compound.
since we are given the moles for Co and O, we'll divide both of those moles by the lowest mole quantity, which is, in this case, 2.59. After dividing, we see that the ratio of O to Co is 2:1. So, for every 1 Co atom, there has to be 2 O atoms. we can then insert the 2 in for OH to satisfy this ratio.
Describe the factors influencing chemical bonds and the
characteristics of an ionic and a covalent bond.
Explanation:
factors affecting chemical bonds
a. electronegativity difference
b. size
c. charges
characteristics of Ionic bond
a. high melting point
b. they are soluble in polar solvents
c. they don't conduct electricity in solid state
characteristics of covalent bonds
a. they are insoluble in polar solvents
b. they don't conduct electricity
c. low melting point
When 1604 J of heat energy is added to 48.9 g of hexane, C6H14, the temperature increases by 14.5 ∘C. Calculate the molar heat capacity of C6H14.
Answer:
THE MOLAR HEAT CAPACITY OF HEXANE IS 290.027 J/ C
Explanation:
1604 J of heat is added to 48.9 g of hexane
To calculate the molar heat capacity of hexane, it is important to note that the molar heat capacity of a substance is the measure of the amount of heat needed to raise 1 mole of a substance by 1 K.
Since 1604 J of heat = 48.9 g of hexane
Molar mass of hexane = 86 g/mol = 1 mole
then;
1604 J = 48.9 g
x = 86 g
x = 1604 * 86 / 48.9
x = 4205.4 J
Hence, 4205.4 J of heat will be added to 1 mole or 86 g of hexane to raise the temperature by 14.5 C.
In other words,
heat = molar heat capacity * temperature change
molar heat capacity = heat/ temperature change
Molar heat capacity = 4205.4 J / 14.5 C
Molar heat capacity = 290.027 J/C
The molar heat capacity of hexane is 290.027 J/ C
What is the molarity of 4 g of NaCl dissolved in 100mL of water?
Answer:
[tex]M=0.684M[/tex]
Explanation:
Hello,
In this case, considering that the solution is formed by NaCl as the solute and water as the solvent, we can compute the molarity as shown below:
[tex]M=\frac{mol_{solute}}{V_{solution}}[/tex]
Whereas the volume of the solution must be in liters. In such a way, since the addition of sodium chloride does not significantly changes the volume of the solution we can say it remains in 100 mL (0.100 L) and the moles of sodium chloride are computed by using its molar mass (58.45 g/mol):
[tex]mol_{solute}=4g*\frac{1mol}{58.45g} =0.0684mol[/tex]
Therefore, the molarity is:
[tex]M=\frac{0.0684mol}{0.100L} \\\\M=0.684\frac{mol}{L}=0.684M[/tex]
Regards.
Im really confused and select all that apply questions scare me.
Answer:
The 3rd one
Explanation:
What is the rate constant of a reaction if rate = 1 x 10-2 (mol/L)/s, [A] is 2 M,
[B] is 3 M, m = 2, and n = 1?
Answer:
[tex]0.10 \text{ L$^2$mol$^{-2}$s$^{-1}$}[/tex]
Explanation:
The general formula for a rate law is
[tex]\text{rate} = k\text{[A]}^m \text{[B]}^{n}[/tex]
With your numbers, the rate law becomes
1.2 mol·L⁻¹s⁻¹ = k(2 mol·L⁻¹)²(3 mol·L⁻¹)¹ = k × 4 mol²L⁻² × 3 mol·L⁻¹
= 12k mol³L⁻³
[tex]\\ k = \dfrac{\text{1.2 mol $\cdot$ L$^{-1}$s$^{-1}$} }{12\text{ mol$^{3}$L}^{-3}} = \mathbf{0.10} \textbf{ L$\mathbf{^2}$mol$^{\mathbf{-2}}$s$^{\mathbf{-1}}$}[/tex]
A 2.87g sample of carbon reacts with hydrogen to form 3.41g of car fuel. What is the empirical formula of the car fuel?
Answer:
The empirical formulae for the car fuel is C4H9
1. What happens to global temperature averages that start an ice age?
Answer:
Around 46° F (7.8° C)
Explanation:
"Scientists have predicted that the global average temperature during the ice age was around 46 degrees Farenheit (7.8 degrees Celsius.) However, the polar regions were far colder, around 25 degrees Fahrenheit (14 degrees Celsius) colder than the global average."
Source: http://www.weforum.org
Atoms are indivisible spheres. 1.plum pudding model 2.Dalton model 3.Bohr model
Answer: 2. Dalton Model
Explanation:
John Dalton proposed that atoms are indivisible spheres. Although his model of an atom was not entirely new to the scientific world since the ancient Greeks has made a similar statement in the past ( all matter are made up of small indivisible particle called atom).
As of when Dalton proposed his model of an atom, electrons and nucleus where yet to be discovered.
A compound is found to contain 11.21 % hydrogen and 88.79 % oxygen by mass. What is the empirical formula for this compound?
Answer:
H₂O
Explanation:
The empirical formular of the compound is obtained using the following steps;
Step 1: Divide the percentage composition by the atomic mass
Hydrogen = 11.21 / 1 = 11.21
Oxygen = 88.79 / 16 = 5.55
Step 2: Divide by the lowest number
Hydrogen = 11.21 / 5.55 = 2.02 ≈ 2
Oxygen = 5.55 / 5.55 = 1
This means the ratio of the elements is 2 : 1
The empirical formular (simplest formular of a compound) of the compound is;
H₂O
Answer:Empirical formula ======== H₂O
Explanation:The empirical formula of a compound shows the whole number ratio for each atom in a compound.
To find empirical formula. we follow the below steps
The total mass of the compound here is 100 grams, that is (11.21% of hydrogen + 88.79% of oxygen) we can then assume 11.21 grams of hydrogen and 88.79grams of oxygen
Hydrogen Oxygen
1.composition by mass 11.21 88.79
molecular weight 1.007g/mol 15.990g/mol
2.Divide composition by mass 11.21/1.007 88.79/15.99
by each molecular weight to get 11.13 5.553
no of moles
3 Divide by the least number of moles
to get atomic ratio 11.13/5.553 5.553/5.553
2.004 1.00
4.Convert to whole numbers 2 1
Empirical formula ======== H₂O
Indicate the peptides that would result from cleavage by the indicated reagent: a. Gly-Lys-Leu-Ala-Cys-Arg-Ala-Phe by trypsin b. Glu-Ala-Phe-Gly-Ala-Tyr by chymotrypsin
Answer:
a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe
b. Glu-Ala-Phe + Gly-Ala-Tyr
Explanation:
In this case, we have to remember which peptidic bonds can break each protease:
-) Trypsin
It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.
-) Chymotrypsin
It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.
With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the "Lis" and "Arg" (See figure 1).
In "peptide b", the peptidic bond that would be broken is the one in the "Phe" (See figure 2). The second amino acid that can be broken is tyrosine, but this amino acid is placed in the C terminal spot, therefore will not be involved in the hydrolysis.