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Related Questions
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)
Answers
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.
Preparation the buffer solution: initial pH of buffer solution: ____ Titration of a weak acid with a strong base: initial pH of weak acid: ____ final pH of weak acid: ____ Amount of NaOH added: ____ Titration Curve for Weak Acid with a Strong Base (Paste curve here.)
Answers
Answer:
pH of buffer solution is 7.0
Initial pH of Weak acid is 3.27
Final pH of weak acid is 3.07
Amount of NaOH added is 1ml
Explanation:
Titration is a process in which acid and base are introduced together until a neutral solution is achieved whose pH value is near to buffer solution which is 7.0, the pH value for acid is below 7 while pH value for base is above 7.
What is Bose Einstein state of matter and their examples
Answers
Answer:
A BEC ( Bose - Einstein condensate ) is a state of matter of a dilute gas of bosons cooled to temperatures very close to absolute zero is called BEC.
Examples - Superconductors and superfluids are the two examples of BEC.
Explanation:
#19.
An unknown sample weighs 45.2 g and takes 58.2 kJ to vaporize. What is
its heat of vaporization?
Answers
Butanal, an aldehyde, can be made from 1-bromobutane, but it requires two reactions in sequence. What are the two reagents that can be used to make butanal from 1-bromobutane if reagent A is first reacted with 1-bromobutane and then the product of that reaction, is reacted with reagent B.
Answers
Answer:
See explanation and image attached
Explanation:
My aim is to convert 1-bromobutane to butanal. The first step is to react the 1-bromobutane substrate with water. This reaction occurs by SN2 mechanism to yield 1-butanol. Hence reagent A is water.
1-butanol is now reacted with an oxidizing agent such as acidified K2Cr2O7 (reagent B) to yield butanal. Note that primary alkanols are oxidized to alkanals.
These sequence of reactions are shown in the image attached.
If atom X had 3 valence electrons and atom Y had 7 valence electrons, the correct chemical formula for the ionic compound they would form is Choices: A) XY3 B) X3Y6 C) X2Y3 D) X3Y7
Answers
Answer:
A) XY3
In both of the atom, they want to have 8 valence electrons on the outer shell so they can become stable.
The electronic arrangement of an atom shows how electrons are distributed across the different energy levels. Which of the following elements is represented by the electron arrangement 2, 8, 18, 6?
А Сa
B. Mg
C. S
D Se
E. Ga
Answers
Answer:
The answer is D which is Selenium
The Nernst equation at 20oC is:
Eion= 58 millvolts/z. [log10 (ion)out/(ion)in]
Calculate the equilibrium potential for Cl- if the concentration of Cl- outside of the cell is 100 and the concentration inside of the cell is 10 mmol/liter.
a. 58 millivolts
b. +58 millivolts
c. -116 millivolts
d. 0
Answers
Answer:
a. -58 millivolts
Explanation:
The given Nernst equation is:
[tex]E_{ion} = 58 millivolts /z \Big[ log_{10} \Big( \dfrac{[ion]_{out}}{[ion]_{in}}\Big) \Big]}[/tex]
The equilibrium potential given by the Nernst equation can be determined by using the formula:
[tex]E_{Cl^-} = \dfrac{2.303*R*T}{ZF} \times log \dfrac{[Cl^-]_{out}} {[Cl^-]_{in}}[/tex]
where:
gas constant(R) = 8.314 J/K/mol
Temperature (T) = (20+273)K
= 298K
Faraday constant F = 96485 C/mol
Number of electron on Cl = -1
[tex]E_{Cl^-} = \dfrac{2.303*8.314*298} {(-1)*(96845)} \times log \dfrac{100} {10}[/tex]
[tex]E_{Cl^-} = - 0.05814 \ volts[/tex]
[tex]\mathsf{E_{Cl^-} = - 0.05814 \times 1000 \ milli volts}[/tex]
[tex]\mathsf{E_{Cl^-} \simeq - 58\ milli volts}[/tex]
An antacid tablet weighing 1.30g was fully neutralized at 42.00 mL(an excess amount) of 0.250MHCl. 10.00 mL of 0.100 M NaOH was then used to back titrate the excess HCl. How many moles of acid did the antacid neutralize
Answers
Answer:
0.0095 moles of acid were neutralized by the antiacid
Explanation:
The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:
Moles HCl added:
42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl
Moles NaOH to titrate the excess:
10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.
Moles of acid that were neutralized:
0.0105 moles - 0.0010 moles =
0.0095 moles of acid were neutralized by the antiacidHydrogen is manufactured on an industrial scale by this sequence of reactions:
CH4(g) + H2O(g) ⇆ CO(g) + 3H2(g)
CO(g)+ H2O(g) ⇆ CO(g) + H2(g)
The net reaction is:
CH4(g) + H2O(g) ⇆ CO(g) + 4H2(g)
Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K1 and K2.
Answers
Answer:
[tex]K=K_1*K_2\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}[/tex]
Explanation:
Hello there!
In this case, for the given chemical reaction, it turns out firstly necessary to write the equilibrium expression for both reactions 1 and 2:
[tex]K_1=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \\\\K_2=\frac{[CO_2][H_2]}{[CO][H_2O]}[/tex]
Now, when we combine them to get the overall expression, we infer these two are multiplied to get:
[tex]K=K_1*K_2\\\\K=\frac{[CO][H_2]^3}{[CH_4][H_2O]} *\frac{[CO_2][H_2]}{[CO][H_2O]}\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}[/tex]
Regards!
When determining the amount of oxidant present by titration, you can use iodine/starch as an indicator. First, the oxidant, like hypochlorite, oxidizes Choose... When starch and iodine are both present, the solution is Choose... During the titration, a titrant like thiosulfate reduces the
Answers
The question is incomplete, the complete question is;
When determining the amount of an oxidant present by titration, you can use iodine and starch as an indicator.
First, the oxidant, like hypochlorite, oxidizes
Choose...
neutral iodine into iodide ion
iodide ion into neutral iodine
iodate polyatomic ion into iodide ion
When starch and iodine are both present, the solution is
Choose...
blue-black
brownish yellow
clear
During the titration, the titrant, like thiosulfate, reduces the
Choose...
iodide ion into iodate polyatomic ion
neutral iodine into iodide ion
iodide ion into neutral iodine
When the iodine has completely reacted at the endpoint of the titration, the solution should become
Choose...
clear
blue-black
brownish yellow
Answer:
1. iodide ion into neutral iodine
2. blue-black
3. neutral iodine into iodide ion
4. clear
Explanation:
Hypochlorite oxidizes the iodide ion to iodine molecule according to the reaction equation;
ClO-(aq) + 2H+(aq) + 2I-(aq) ---------> 6 I2(l) + Cl- (aq)+ H2O(l)
When iodine is added, the colour of the starch solution immediately changes to blue-black.
A reduction reaction occurs when the titrant, thiosulfate is added as follows;
I2 + 2S2O32- → 2I- + S4O62-
The solution at end point is found to become clear again.
What is the equilibrium expression for the reaction below?
Caco (s)
Cao(s) + CO (g)
A.
B.
[Cacoz]
[Cao]
[Caco.]
[Cao]+[CO,
[Cao][COL]
[Caco:]
C.
D. [co]
Answers
Answer:
D. [CO₂]
Explanation:
Let's consider the following equation at equilibrium.
CaCO₃(s) ⇄ CaO(s) + CO₂(g)
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients. It only includes gases and aqueous species.
Kc = [CO₂]
Arrange these elements according to electron affinity.
a. He
b. C
c. F
Answers
Answer:
[tex]F>C> H_e[/tex]
Explanation:
From the question we are given
a. He
b. C
c. F
Generally
The electron affinity of an atom or molecule is defined as the amount of energy released when an electron is attached to a neutral atom or molecule in the gaseous state to form a negative ion.
Therefore
The order of electron Affinity is give as
[tex]F>C> H_e[/tex]
Elements arranged according to increasing electron affinity is He, C, F.
Electron affinity is a measure of the ability of an atom or molecule to attract and accept an electron. It quantifies the energy change that occurs when an electron is added to a neutral atom or ion in the gaseous state.
When an atom or ion gains an electron, it undergoes a process called electron attachment or electron capture. The electron affinity is the energy released or absorbed in this process. A higher electron affinity value indicates a stronger attraction for electrons.
The electron affinity is typically expressed in units of energy, such as electron volts (eV) or kilojoules per mole (kJ/mol). It can be either positive or negative. A positive electron affinity indicates that energy is released when an electron is added, while a negative electron affinity indicates that energy is required to add an electron.
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What is the oxidation state of rubidium (Rb)?
A. +1
B. -2
C. +2
D. -1
Answers
Answer:
The answer is A. +1
Explanation:
name a factor tht affects the value of electron affinity
Answers
Answer:
Atomic sizeNuclear chargesymmetry of the electronic configurationDetermine the number of atoms of O in 60.1 moles of Fe₂(SO₃)₃.
Answers
Answer:
3.310308*10^26
Explanation:
nO=9nFe2(SO3)3=9*60.1=540.9 moles
number of atoms: 540.9*6.02*10^23
What are fires classified by?
Answers
Answer:
A fire class is a system of categorising fire with regard to the type of material and fuel for combustion. Class letters are often assigned to the different types of fire, but these differ between territories. There are separate standards for the United States, Europe, and Australia
PLEASE HELP I NEED THIS ASAP
Select all that apply.
The spectrum of Star S is compared to a reference hydrogen spectrum. What can be concluded about Star S?
Star S has radial motion.
Star S has transverse motion.
Star S is moving toward Earth.
Star S is moving away from Earth.
Answers
Answer:
I say Star S has radial motion
Explanation:
I'm not sure if it right but let me know if you have any other questions
Two hydrogen atoms interact to form a hydrogen molecule.
a. True
b. False
Answers
Answer:
a. True
Explanation:
An atom can be defined as the smallest unit comprising of matter that forms all chemical elements. Thus, atoms are basically the building blocks of matters and as such determines or defines the structure of a chemical element.
Generally, atoms are typically made up of three distinct particles and these are protons, neutrons and electrons.
A chemical reaction can be defined as a chemical process which typically involves the transformation or rearrangement of the atomic, ionic or molecular structure of an element through the breakdown and formation of chemical bonds to produce a new compound or substance.
Hydrogen is the simplest chemical element that exists. The symbol for the chemical element hydrogen is "H" and it is a colourless, tasteless, odorless, and highly flammable gas.
It is a chemical element found in group (1) of the periodic table and as such it has one (1) electron in its outermost shell. Thus, hydrogen has an atomic number of one (1) and a single valence electron because it has only one (1) proton and one (1) electron in its nucleus.
In Stoichiometry, two hydrogen atoms interact to form a hydrogen molecule, with each atom having a single electron in the 1S orbital. These hydrogen atoms react with each other due to the presence of a balanced attractive and repulsive force, which produces a strong covalent bond (H-H) in the hydrogen molecule.
Energy is released in the form of heat when the electrons in the orbitals of the two (2) hydrogen atoms form a covalent bond (H-H); thus, it's an exothermic chemical reaction.
How many grams are of aluminum nitrate are contained in 0.150 moles of Al(NO3)3?
Answers
Answer:
31.95 g
Explanation:
find the molar mass of Al(NO3)3
(Al = 27, N = 14, O = 16)
molar mass of Al(NO3)3 = 27+ (14+16×3)×3
= 213 gmol^-
mass = 213 ×0.15
= 31.95g
What volume of each solution contains 0.14 mol of KCl? Express your answer using two significant figures.
1.8 M KCl
Answers
Answer:
Solution given:
1 mole of KCl[tex]\rightarrow [/tex]22.4l
1 mole of KCl[tex]\rightarrow [/tex]74.55g
we have
0.14 mole of KCl[tex]\rightarrow [/tex]74.55*0.14=10.347g
74.55g of KCl[tex]\rightarrow [/tex]22.4l
10.347 g of KCl[tex]\rightarrow [/tex]22.4/74.55*10.347=3.11litre
volume of each solution contains 0.14 mol of KCl contain 3.11litre.
1 mol of any gas contains 22.4L of volume at STP
Moles of KCl=0.14Volume of KCl:-
0.14(22.4)3.14L……….is strong due to the ……………..between positive ions and negative delocalized electrons
Answers
Answer:
atom &bond
Explanation:
atom is strong due to the bond
Write chemical equations and corresponding equilibrium expressions for each of the two ionization steps of carbonic acid. Part A Write chemical equations for first ionization step of carbonic acid. Express your answer as a chemical equation. Identify all of the phases in your answer.
Answers
Solution :
[tex]H_2CO_3[/tex] is considered a diprotic acid.
Sp it can dissociate in solution by giving two protons.
Chemical equations for the first step of carbonic acid is :
First ionization
[tex]$H_2CO_3(aq) + H_0(1) \rightleftharpoons H_.O^+(aq) + HCO_3^-(aq)$[/tex]
Equilibrium constant expression is
[tex]$K_{a}_{1}=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}$[/tex]
Second ionization -
[tex]$HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CO_3^{2-}(aq)$[/tex]
Equilibrium constant expression is
[tex]$K_{a2}=\frac{[H_3O^+][CO_3^{2-}]}{[HCO_3^-]}$[/tex]
Tech A says that the PCM monitors the pre-cat and post-cat oxygen sensors to determine catalytic converter efficiency. Tech B says that a catalytic converter can be tested by graphing the oxygen sensor readings on a scan tool or lab scope and comparing them. Who is correct
Answers
Answer:
Both Tech A and Tech B.
Explanation:
Catalyst is an element used to start chemical reaction but is not used in the reaction. Catalysts material used in catalytic converter include Rhodium, Palladium and platinum. The pre cat and post cat oxygen sensors helps determine converter efficiency.
Answer:
Explanation:
B
Not following hazardous material safety policies and procedures can result in which of
the following? (Select all that apply.)
a. Serious illnesses
b. Injury
c. Death
d. HIPAA violation
Answers
Answer:
A, B, C and D
Explanation:
It can result for all of the choices mentioned.
Not following safety and procedure for handling hazardous material results in illness, death, injury, and HIPAA violation. Thus, all options are correct.
The hazardous material safety policy and measures are the standards set by HIPAA for the safety and precautionary measures that have been followed for reducing personal risk.
The hazardous materials have been chemicals, gases, flammables, and explosives. The inappropriate handling and not following the standard procedure results in illness, injury, death, and HIPAA violation. Thus, all the options are correct.
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Match the change to its definition.
Name of change Change
condenation gas to solid
freezing solid to liquid
melting gas to liquid
evaporation liquid to gas
sublimation solid to gas
deposition liquid to solid
Answers
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?
Answers
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.
Give four examples illustrating each of the following terms. a. homogeneous mixture b. heterogeneous mixture c. compound e. physical change d. element f. chemical change
Answers
1. homogenous: sugar solution
2. heterogeneous: sand solution
3. compound: water
4. physical change: ice melting
5. element: hydrogen
6. chemical change: burning fire
A solution is made by dissolving 5.84 grams of NaCl in enough distilled water to give a final volume of 1.00 L. What is the molarity of the solution
Group of answer choices
0.0250 M
0.400 M
0.100 M
1.00 M
Answers
Answer:
Explanation:
1. A solution is made by dissolving 5.84g of NaCl is enough distilled water to a give a final volume of 1.00L. What is the molarity of the solution? a. 0.100 M b. 1.00 M c. 0.0250 M d. 0.400 M 2. A 0.9% NaCl (w/w) solution in water is a. is made by mixing 0.9 moles of NaCl in a 100 moles of water b. made and has the same final volume as 0.9% solution in ethyl alcohol c. a solution that boils at or above 100°C d. All the above (don't choose this one) 3. In an exergonic process, the system a. gains energy b. loses energy c. either gains or loses energy d. no energy change at all
Answer:
[tex]\boxed {\boxed {\sf 0.100 \ M }}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter.
[tex]molarity = \frac{moles \ of \ solute}{liters \ of \ solution}}[/tex]
The solution has 5.84 grams of sodium chloride or NaCl and a volume of 1.00 liters.
1. Moles of SoluteWe are given the mass of solute in grams, so we must convert to moles. This requires the molar mass, or the mass of 1 mole of a substance. These values are found on the Periodic Table as the atomic masses, but the units are grams per mole, not atomic mass units.
We have the compound sodium chloride, so look up the molar masses of the individual elements: sodium and chlorine.
Na: 22.9897693 g/mol Cl: 35.45 g/molThe chemical formula (NaCl) contains no subscripts, so there is 1 mole of each element in 1 mole of the compound. Add the 2 molar masses to find the compound's molar mass.
NaCl: 22.9897693 + 35.45 = 58.4397693 g/molThere are 58.4397693 grams of sodium chloride in 1 mole. We will use dimensional analysis and create a ratio using this information.
[tex]\frac {58.4397693 \ g\ \ NaCl} {1 \ mol \ NaCl}[/tex]
We are converting 5.84 grams to moles, so we multiply by that value.
[tex]5.84 \ g \ NaCl *\frac {58.4397693 \ g\ NaCl} {1 \ mol \ NaCl}[/tex]
Flip the ratio. It remains equivalent and the units of grams of sodium chloride cancel.
[tex]5.84 \ g \ NaCl *\frac {1 \ mol \ NaCl}{58.4397693 \ g\ NaCl}[/tex]
[tex]5.84 *\frac {1 \ mol \ NaCl}{58.4397693 }[/tex]
[tex]0.09993194823 \ mol \ NaCl[/tex]
2. MolarityWe can use the number of moles we just calculated to find the molarity. Remember there is 1 liter of solution.
[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]
[tex]molarity= \frac{ 0.09993194823 \ mol \ NaCl}{1 \ L}[/tex]
[tex]molarity= 0.09993194823 \ mol \ NaCl/L[/tex]
3. Units and Significant FiguresThe original measurements of mass and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 9 in the ten-thousandths place tells us to round the 9 to a 0, but then we must also the next 9 to a 0, and the 0 to a 1.
[tex]molarity \approx 0.100 \ mol \ NaCl/L[/tex]
1 mole per liter is 1 molar or M. We can convert the units.
[tex]molarity \approx 0.100 \ M \ NaCl[/tex]
The molarity of the solution is 0.100 M.
The half-life of radon-222 is 3.8 days. How many grams of radon-222 remain
after 15.2 days if the original amount was 6.00 g?
A. 0.750 g
B. 0.375 g
C. 1.20 g
D. 3.00 g
Answers
The mass of radon-222 that will remain after 15.2 days given that it was originally 6 g is 0.375 g (Option B)
What is half life?This is the time taken for half a substance to decay.
How to determine the number of half-lives that has elapsedWe'll begin our calculation by calculating the number of half-lives that has elapsed after 15.2 days. This is illustrated below:
Half-life (t½) = 3.8 daysTime (t) = 15.2 day Number of half-lives (n) =?n = t / t½
n = 15.2 / 3.8
n = 4
Thus, 4 half-lives has elapsed.
How to determine the amount remainingOriginal amount (N₀) = 6 gNumber of half-lives (n) = 4Amount remaining (N) = ?The amount of radon-222 remaining can be obtained as illustrated below:
N = N₀ / 2ⁿ
N = 6 / 2⁴
N = 6 / 16
N = 0.375 g
Thus, the amount of radon-222 remaining after 15.2 days is 0.375 g
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