Answer:
CF₂
Explanation:
Let's assume we have 100 g of the gas. If that were the case we'd have
24 g of C76 g of FNow we convert both masses into moles, using their respective molar mass:
24 g C ÷ 12 g/mol = 2 mol C76 g F ÷ 19 g/mol = 4 mol FWe can express those results as C₂F₄.
To determine the empirical formula we reduce those coefficients to the lowest possible integers, leaving us with CF₂.
how hydrogen chloride gas is prepared on labrotary by conc sulpheric acid
Answer:
It is prepared small amounts of hydrogen cloride for uses in the lab.
It can be "generated in an HCl generator by dehydrating hydrochloric acid with either sulfuric acid or anhydrous calcium chloride."
Two common methods to generate an aldehyde is by oxidation of an alcohol and through ozonolysis.
a. True
b. False
Answer:
a. True.
Explanation:
Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.
weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C
take an example of ethanol:
[tex]{ \bf{CH _{3} CH_{2}OH \: \: \frac{Ag/O_{2} }{500 \degree C} > \: \:CH _{3} CHO}}[/tex]
[tex]{ \sf{CH _{3} CHO \: \: is \: ethanal}} [/tex]
By ozonolysis:
Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.
take an example of propanol:
if it undergoes ozonolysis, it gives ethanal and methanal.
Answer:
A. True
Explanation:
Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.
weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C
take an example of ethanol:
By ozonolysis:
Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.
take an example of propanol:
if it undergoes ozonolysis, it gives ethanal and methanal.
What volume of water is produced when 38.5 g of ethanol reacts with oxygen at 500°C at 1.75 atm?
CH3CH2OH(g) + 3 O2(g)→ 2 CO2(g) + 3 H2O(g)
Answer:
90.99 or 91.0
Explanation:
Using the balanced equation, you convert 38.5g of ethanol to moles of water. From there, you plug the values into the Ideal Gas Equation: PV=nRT.
Answer: The volume of oxygen gas is 91.4 L.
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of ethanol = 38.5 g
Molar mass of ethanol = 46 g/mol
Plugging values in equation 1:
[tex]\text{Moles of ethanol}=\frac{38.5g}{46g/mol}=0.840 mol[/tex]
The given chemical equation follows:
[tex]CH_3CH_2OH(g)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)[/tex]
By stoichiometry of the reaction:
If 1 mole of ethanol produces 3 moles of water
So, 0.840 moles of ethanol will produce = [tex]\frac{3}{1}\times 0.840=2.52mol[/tex] of water
The ideal gas equation is given as:
[tex]PV=nRT[/tex] .......(2)
where
P = pressure = 1.75 atm
V = volume of oxygen gas = ?
n = number of moles= 2.52 moles
R = Gas constant = 0.0821 L.atm/mol.K
T = temperature of the tank = [tex]500^oC=[500+273]K=773K[/tex]
Putting values in equation 2, we get:
[tex]1.75 atm\times V=2.52mol\times 0.0821L.atm/mol.K\times 773K\\\\V=\frac{2.52\times 0.0821\times 773}{1.75}=91.4L[/tex]
Hence, the volume of oxygen gas is 91.4 L.
Consider the reaction: A(aq) + 2B (aq) === C (aq). Initially 1.00 mol A and 1.80 mol B
were placed in a 5.00-liter container. The mole of B at equilibrium was determined to
be 1.00 mol. Calculate K value.
0.060
5.1
25
17
Ugh
Answer:
17
Explanation:
Step 1: Calculate the needed concentrations
[A]i = 1.00 mol/5.00 L = 0.200 M
[B]i = 1.80 mol/5.00 L = 0.360 M
[B]e = 1.00 mol/5.00 L = 0.200 M
Step 2: Make an ICE chart
A(aq) + 2 B(aq) ⇄ C(aq)
I 0.200 0.360 0
C -x -2x +x
E 0.200-x 0.360-2x x
Then,
[B]e = 0.360-2x = 0.200
x = 0.0800
The concentrations at equilibrium are:
[A]e = 0.200-0.0800 = 0.120 M
[B]e = 0.200 M
[C]e = 0.0800 M
Step 3: Calculate the concentration equilibrium constant (K)
K = [C] / [A] × [B]²
K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17
2. Calculate the wavelength of the emitted photon from hydrogen for the transition from ni = 3 to nf = 2. What part of the visible spectrum is this wavelength? Visible wavelengths are: Red 700 - 620 nm, Yellow 620 - 560 nm, Green 560 - 500 nm, Blue 500 - 440 nm, and Violet 440 - 400 nm.
Answer:
The correct answer is "654.54 nm".
Explanation:
According to the question,
⇒ [tex]\frac{1}{\lambda}= Rh(\frac{1}{n1^2} -\frac{1}{n2^2} )[/tex]
By substituting the values, we get
[tex]=1.1\times 10^7(\frac{1}{4} -\frac{1}{9} )[/tex]
[tex]=1.1\times 10^7(\frac{9-4}{36} )[/tex]
[tex]=1.1\times 10^7(\frac{5}{36} )[/tex]
[tex]=654.54\ nm[/tex]
Thus the above is the right solution.
Which is the electronic configuration for oxygen?
how many CH4 molecules are in 14.8 g of CH4
Answer:
[tex]\boxed {\boxed {\sf 5.56 \times 10^{23} \ molecules \ CH_4}}[/tex]
Explanation:
We are asked to find how many molecules of methane are in 14.8 grams of the substance.
1. Convert Grams to MolesFirst, we convert grams to moles. We use the molar mass, or the mass of 1 mole of a substance. These values are equivalent to the atomic masses found on the Periodic Table, however the units are grams per mole instead of atomic mass units.
We are given the compound methane, or CH₄. Look up the molar mass of the individual elements (carbon and hydrogen).
C: 12.011 g/mol H: 1.008 g/molCheck the formula for subscripts. Hydrogen (H) has a subscript of 4, so there are 4 moles of hydrogen in 1 mole of methane. We must multiply hydrogen's molar mass by 4, then add carbon's molar mass.
H₄: 1.008 * 4 = 4.032 g/mol CH₄: 12.011 + 4.032 = 16.043 g/molNow we use dimensional analysis to convert. To do this, we set up a ratio using the molar mass.
[tex]\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}[/tex]
Since we are converting 14.8 grams of methane to moles, we multiply by this value.
[tex]14.8 \ g \ CH_4 *\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}[/tex]
Flip the ratio so the units of grams of methane cancel.
[tex]14.8 \ g \ CH_4 *\frac{ 1 \ mol \ CH_4} {16.043 \ g \ CH_4 }[/tex]
[tex]14.8 *\frac{ 1 \ mol \ CH_4} {16.043}[/tex]
[tex]\frac {14.8}{16.043} \ mol \ CH_4= 0.9225207256 \ mol \ CH_4[/tex]
2. Moles to MoleculesNext, we convert moles to molecules. We use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are moles of methane. Set up another ratio using Avogadro's Number.
[tex]\frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]
Multiply by the number of moles we calculated.
[tex]0.9225207256\ mol \ CH_4 * \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]
The units of moles of methane cancel.
[tex]0.9225207256* \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 }[/tex]
[tex]5.55541981 \times 10^{23} \ molecules \ CH_4[/tex]
3. RoundThe original measurement of grams (14.8) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 5 in the hundredth place up to a 6.
[tex]5.56 \times 10^{23} \ molecules \ CH_4[/tex]
14.8 grams of methane is equal to approximately 5.56 × 10²³ molecules of methane.
Write chemical equations for the reactions that occur when solutions of the following substances are mixed:
a. HNO₂ (nitrous acid) and C₂H₇NO (aq) ethanolamine, a base.
b. H₃O+ and F-
a) HNO₂ + C₂H₇NO → N₂ + C₂H₆O + H₂O
b) H₃O⁺ + F⁻ → HF + H₂O
[tex]\large\color{lime}\boxed{\colorbox{black}{Answer : - }}[/tex]
a) HNO₂ + C₂H₇NO → N₂ + C₂H₆O + H₂O
b) H₃O⁺ + F⁻ → HF + H₂O
A wavelength of 489.2 nm is observed in a hydrogen spectrum for a transition that ends in the nf level of the Balmer series. What was ni for the initial level of the electron
Answer:
[tex]n_1=4[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda=489.2 nm =>4.86*10^{-7}[/tex]
nf level= Balmer series
nf level= 2
Generally the equation for Wavelength is mathematically given by
[tex]\frac{1}{\lambda}=R[\frac{1}{nf^2}-\frac{1}{n_1^2}][/tex]
Where
[tex]R=Rydberg Constant[/tex]
[tex]R=1.097*10^7[/tex]
Therefore
[tex]\frac{1}{4.86*10^{-7}}=1.097*10^7[\frac{1}{2^2}-\frac{1}{n_1^2}][/tex]
[tex]n_1=4.0021[/tex]
[tex]n_1=4[/tex]
H2+O=???????????????????
Answer:
H₂O
Explanation:
Two molecules of Hydrogen and one molecules of Oxygen, when mixed, create H₂O, or water. There is no scientific name for H₂O due to it's common name. It is just refereed to as "water" or H₂O.
một chất hữu cơ có cấu tạo c2h2 cho khí br2 vào ta được hỗn hợp khí
Answer:
C2H2 + Br2 → C2H2Br2
Explanation:
A number is three times the difference between twenty and the number. What is the number?
Answer:
the number is 7
Explanation:
"Three times" means multiply by 3
"Difference" means subtract
"Sum" means add
3(x - 7) = 23 - (3x + 2)
3x - 21 = 23 - 3x - 2
3x - 21 = 21 - 3x
6x = 42
x = 7
A substance which is made up of the same kind
of atom is known as?
Answer:
Element
Element : A pure substance composed of the same type of atom throughout. Compound : A substance made of two or more elements that are chemically combined in fixed amounts.
Explanation:
Manganese-55 has _____neutrons.
55 Mn
25
A. 55
B. 30
C. 25
QUESTION:- Manganese-55 has _____neutrons.
OPTIONS :-
A. 55
B. 30
C. 25
ANSWER:- NUMBER OF NEUTRONS IS EQUAL TO THE DIFFERENCE BETWEEN THE MASS IF THE ATOM AND ATOMIC NUMBER
SO DIFFERENCE IS EQUAL TO :- 55-25 = 30 NEUTRONS.
SO THERE IS 30 NEUTRONS IN SINGLE ATOM OF THE MANGANESE-55 ATOM.
Answer:
the mass of an atom is the sum of proton and neutron which are both concentrated in nocleus of an atom. from the question the mass is given as 55 and the proton is 25.
What mass of water is formed in the reaction of 4.16g H with excess oxygen gas.
Answer:
Explanation:
Start with a balanced equation.
2H2 + O2 → 2H2O
Calculate mole H2 using the formula: n = m/M, where:
n = mole
m = mass (g)
M = molar mass (g/mol)
Calculate molar mass of H2.
M H2 = 2 × 1.008 g/mol = 2.016 g/mol
Calculate moles H2.
n H2 = 4.16 g H2/2.016 g/mol = 2.063 mol H2
Calculate moles H2O by multiplying moles H2 by the mole ratio between H2O and H2 from the balanced equation, so that moles H2 cancel.
2.063 mol H2 × (2 mol H2O/2 mol H2) = 2.063 mol H2O
The mass of water will be calculated by rearranging the n = m/M formula to isolate m;
m = n × M
Calculate the molar mass H2O.
M H2O = (2 × 1.008 g/mol) + (1 × 15.999 g/mol) = 18.015 g/mol
Calculate the mass H2O.
m = n × M = 2.063 mol H2O × 18.015 g/mol = 37.2 g H2O
4.16 g H2 with excess O2 will produce 37.2 g H2O.
Explain why you get a basic solution when you dissolve NaF in water.
Answer:
The fluoride ion is capable of reacting, to a small extent, with water, accepting a proton. The fluoride ion is acting as a weak Brønsted-Lowry base. The hydroxide ion that is produced as a result of the above reaction makes the solution slightly basic.
So
we get a basic solution when you dissolve NaF in water.
Na is a alkaline earth metal
Metallic compound dissolved in water acts as base
Also there is another reason
Na F is ioniC
Fluorine is known as having highest electron affinity in world
It can accept a line pair of OH-
So it's Bronsted Lowry base .
It can also acts as Arrhenius baseHence its basic
What is the biggest cause of change in Earth's systems?
A. Heat
B. Motion
C. Friction
D. Plate tectonics
Answer:
heat
Explanation:
because it's the cause of change
Answer:
heat
Explanation:
because it is a natural factor that causes the change in Earth's system
What is ethane?
A. A polymer
B. An alkyne
C. An alkane
D. An alkene
Answer:
D. An alkene
Explanation:
because Ethane is C2H4
Answer:
It's a alkANE. C.
Explanation:
The easiest way to memorize this is to look at the endings. Substances that end in -ANE are alkANEs. Substances that end in -ENE are alkENEs. Substances that end in -YNE are alkYNEs.
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
Write balanced equations for the reaction of each of the following carboxylic acids with NaOH. Part A formic acid Express your answer as a chemical equation. A chemical reaction does not occur for this question. Request Answer Part B 3-chloropropanoic acid Express your answer as a chemical equation. nothing A chemical reaction does not occur for this question.
Answer:
Part A
HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)
Part B
ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)
Explanation:
The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;
RCOOH + NaOH ----> RCOONa + H2O
We have to note the fact that the net ionic reaction still remains;
H^+(aq) + OH^-(aq) ---> H2O(l)
In both cases, the reaction can occur and they actually do occur as written.
Critique this statement: Electrons can exist in any position
outside of the nucleus.
Answer:
However, there has to be 2 electrons on the first shell, and 8 on the others.
Explanation:
Hope this helps :)
Choose the most correct answer: The endothermic (∆H > 0) reaction:
a) Cannot occur at all temperature.
b) Can occur with the positive ∆S at high temperature.
c) Can occur with the negative ∆S at low temperature.
d) Cannot occur with the positive ∆S at high temperature.
Answer:
B
Explanation:
In order to create spontaneity, an endothermic process has to occur along with positive entropy and high temperature
Determine the number of hydrogen atoms connected to each carbon atom: The bond-line structure of a compound has a SMILES string of CC1CCN(CC1N(C)C2=NC=NC3=C2C=CN3)C(=O)CC#N. All the carbon atoms of the compound are highlighted and labeled a through p.
Answer:
dsgsdfd
Explanation:
Read the following statement:
Energy cannot be created or destroyed.
Does the statement describe a scientific law? (3 points)
a
No, because it universally applies to all objects
b
No, because it is not true in all circumstances
c
Yes, because it universally applies to all objects
d
Yes, because it is not true in all circumstances
Answer:
C. yes, because it is universally applies to all objects
Forcus on the yellow highlighted texts, your help is appreciated.
[tex]{ \sf{ \red{no \: pranks}}}[/tex]
Answer:
Transition temperature is the temperature at which a substance changes from one state to another.
Allotropy is the existence of an element in many forms.
How many moles of Al are needed to react exactly with 10.00 moles of Fe2O3 according to the following
equation?
Fe2O3 + 2 Al → Al2O3 + 2Fe
A) 15.0 moles
1
B) 20.0 moles
C) 30.0 moles
D) 60.0 moles
E) 35.0 moles
Answer:
Answer is B) 20.0 moles
Explanation:
From the equation,
1 mole of Fe2O3 = 2 moles of Al
therefore 10.0 moles of Fe2O3 = 10×2
= 20.0 moles.
# of protons
# of neutrons
# of electrons
Atomic Number
Mass Number
18
17
35
17
37
6
8
6
6
15
Answer:
35
Explanation:
is the answer for your question
Lewis Structures are used to describe the covalent bonding in molecules and ions. Draw a Lewis structure for NO3- and answer the following questions based on your drawing.
1. For the central nitrogen atom:
The number of lone pairs = ________
The number of single bonds=_______
The number of double bonds= ______
2. The central nitrogen atom :
Answer:
The lewis structure for NO₃⁻ is shown in the attachment below
For the central nitrogen atom:
The number of lone pairs = 0
The number of single bonds = 2
The number of double bonds= 1
Explanation:
The lewis structure for NO₃⁻ is shown in the attachment below.
From the Lewis structure
For the central nitrogen atom:
The number of lone pairs = 0
The number of single bonds = 2
The number of double bonds= 1
Determine the number of moles of aluminum in 2.154 x 10-1 kg of Al. Group of answer choices 5816 mol 7.984 mol 6.02 X 1023 mol 4.801 mol 8.783
Answer:
Avogadro's number is 1 mol = 6.02 * 10^23 elements
It means that 1 mol of atoms is 6.02 * 10^23 atoms
1 mol of atoms = 6.02 * 10^23 atoms
From there, if you divide both sides by 1 mol of atoms, you get
1 = 6.02 * 10^23 atoms / 1 mol of atoms.
That means, that to pass from a number of moles of atoms to number of atoms you have to multipby by the conversion factor
6.02*10^23 atoms Al/ 1 mol Al
That is the second option of the list.
Explanation:
Each 5-ml teaspoon of Extra Strength Maalox Plus contains 450 mg of magnesium hydroxide and 500 mg of aluminum hydroxide. How many moles of hydronium ions H3O are neutralized by 1 teaspoon of antacid product?
Answer:
0.0347 moles of hydronium ions
Explanation:
The equation of the neutralization reaction between hydroxide and hydronium ions is given below:
H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)
From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.
The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:
Number of moles = mass / molar mass
Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol
Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol
Mass of magnesium hydroxide = 450 g = 0.45 g
Mass of aluminium hydroxide = 500 mg = 0.5 g
Moles of magnesium hydroxide = (0.45/58) moles
Moles of aluminium hydroxide = (0.5/78) moles
Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:
Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)
Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)
Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles
Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles
Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions
Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.
