Answer:
It is important to separate mixture into pure or relatively pure substances when performing a chemical analysis SO AS TO KNOW THE PROPERTIES COMING FROM EACH PART MIXTURE WHICH MAY INTERFERE WITH THE SEPARATION.
Explanation:
In chemistry, Mixture is the combination of two or more substances which are not combine chemically.
Mixture contain different substances with different physical and chemical properties.
It is important to purify the substances in a mixture so as to identify what properties are coming from each mixture and also some part of the mixture can interfere with the properties of other mixture present for skewing analysis.
A 0.1-L unbuffered solution needs the pH adjusted from 3.5 to 1. How many microliters of a 6 molar HCl solution need to be added to adjust the pH
Answer:
1661μL of a 6M HCl you need to add
Explanation:
pH is defined as -log[H⁺] ([H⁺] =10^{-pH}), the initial and final concentrations of [H⁺] you need are:
Initial [H⁺] = 10^{-3.5} = 3.16x10⁻⁴M H⁺
Final [H⁺] = 10^{-1} = 0.1M H⁺
In moles, knowing volume of the solution is 0.1L:
Initial [H⁺] = 0.1L ₓ (3.16x10⁻⁴mol H⁺ / L) = 3.16x10⁻⁵moles H⁺
Final [H⁺] = 0.1L ₓ (0.1mol H⁺ / L) = 0.01 moles H⁺.
That means, moles of H⁺ you need to add to the solution is:
0.01mol - 3.16x10⁻⁵moles = 9.9684x10⁻³ moles of H⁺.
A solution of HCl dissociates in H⁺ and Cl⁻ ions, that means moles of HCl added are equal to moles of H⁺. As you need to add 9.9684x10⁻³ moles of H⁺ = 9.9684x10⁻³ moles of HCl:
9.9684x10⁻³ moles of HCl ₓ (1L / 6mol) = 1.6614x10⁻³L
In μL:
1.661x10⁻³L × (1x10⁶μL / 1L) =
1661μL of a 6M HCl you need to addHow many moles of hydrogen gas will be produced when 12 g of Mg will react completely with excess of an acid according to the following reaction? 2 Mg + 2 HCI — 2 MgCl2 + H,
Answer:
0.49 mol
Explanation:
Step 1: Write the balanced equation
Mg + 2 HCI ⇒ MgCl₂ + H₂
Step 2: Calculate the moles corresponding to 12 g of Mg
The molar mass of Mg is 24.31 g/mol.
[tex]12g \times \frac{1mol}{24.31g} = 0.49mol[/tex]
Step 3: Calculate the moles of H₂ produced by 0.49 moles of Mg
The molar ratio of Mg to H₂ is 1:1. The moles of H₂ produced are 1/1 × 0.49 mol = 0.49 mol.
Of the following species, ________ will have bond angles of 120°. A) PH3 B) ClF3 C) NCl3 D) BCl3 E) All of these will have bond angles of 120°.
Answer:
D. BCl₃
Explanation:
BCl₃ molecular geometry is trigonal planar and it has a bond angle of 120°.
Hope that helps.
According to the molecular geometry, BCl₃ has trigonal planar geometry and a bond angle of 120°.
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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Of the following, only ________ has sp2 hybridization of the central atom. Of the following, only ________ has sp2 hybridization of the central atom. ICl3 PBr3 SiH2Br2 HCN BF3
Answer:
BF3
Explanation:
Hybridization can be defined as the mixing of two or more atomic pure orbitals. ( s, p , and d) to produce two or more hybrid atomic orbitals that are similar and identical in shape and energy e.g sp,sp²,sp³ ,sp³d, sp³d². Usually , the central atom of a covalent molecules or ion undergoes hybridization.
in BF3; Boron is the central atom. Here, A 2s electron is excited from the ground state of boron ( 1s²2s²2p¹) to one empty orbitals of 2p.
The 2s orbital is then mixed with two orbitals of 2p to form three sp² hybrid orbitals tat are trigonally arranged in the plane in order to minimize repulsion . Each of the three hybrid orbitals overlaps with p-orbital of fluorine atom to form three bonds of equal strength and with bond angles of 120⁰.
Energy
B ⇵ ║ ⇅ ║ ↑ -----------> *B ⇵ ║ ↑ ║ ↑ ║ ↑
1s 2s 2p 1s 2s 2p
Ground state Excited State
The above shows an illustrative example of how electrons move from the ground state to the excited state.
Calculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation of the solute. Express your answer in atmospheres to three significant figures. Pvap = atm
The question is incomplete, the solute was not given.
Let the solute be K₂CrO₄ and the solvent be water
Complete Question should be like this:
The density of a 0.438 M solution of potassium chromate (K₂CrO₄) at 298 K is 1.063 g/mL.
Calculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation.
Pvap = ________atm
Answer:
Pvap (of water above the solution) = 0.0306 atm
Dissolution of the solute
K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻
Explanation:
Given
volume of solution = 1 Litre = 1000 mL of the solution
density of the solution = 1.063 g/mL
concentration of the solution= 0.438M
temperature of the solution= 298 K
vapour pressure of pure water = 0.0313atm
Recall: density = mass/volume
∴mass of solution = volume x density
m = 1000 x 1.063 = 1063 g
To calculate the moles of K₂CrO₄ = volume x concentration
= 1 x 0.438 = 0.438 mol
Mass of K₂CrO₄ = moles x molar mass = 0.438 x 194.19 = 85.055 g
Mass of water = mass of solution - mass of K₂CrO₄
= 1063 - 85.055 = 977.945 g
moles of water = mass/molar mass
∴ moles of water = 977.945/18.02 = 54.27 mol
Dissolution of the solute
K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻
(dissolution is the process by which solute(K₂CrO₄) is passed into solvent(H₂O) to form a solution
moles of ions = 3 x moles of K₂CrO₄
= 3 x 0.438 = 1.314 mol
Vapor pressure of solution = mole fraction of water x vapor pressure of water
= 54.27/(54.27 + 1.314) x 0.0313 = 0.0306 atm
Chemistry question. Image attached.
Answer:
Pb(NO3)2(aq) + 2NaCl(aq) —> PbCl2(s) + 2NaNO3(aq)
Explanation:
When lead (II) nitrate, Pb(NO3)2 undergoes a double displacement reaction with aqueous sodium chloride, NaCl, the following products are obtained:
Pb(NO3)2(aq) + NaCl(aq) —>
Aqueous Pb(NO3)2 will dissociate in solution as follow:
Pb(NO3)2(aq) —> Pb2+(aq) + 2NO3-(aq)
On the other hand, aqueous NaCl will dissociate as follow:
NaCl(aq) —> Na+(aq) + Cl-(aq)
The double displacement reaction will take place as follow:
Pb(NO3)2(aq) + NaCl(aq) —>
Pb2+(aq) + 2NO3-(aq) + Na+(aq) + Cl-(aq) —> Pb2+(aq) Cl-(aq) + Na+(aq) 2NO3-(aq)
Pb(NO3)2(aq) + NaCl(aq) —> PbCl2(s) + NaNO3(aq)
We simply balance the equation by putting 2 in front of NaCl and 2 in front of NaNO3 as shown below:
Pb(NO3)2(aq) + 2NaCl(aq) —> PbCl2(s) + 2NaNO3(aq)
What is the density of a gas with a molar mass of 28.26g/mol at 3.510atm and 237k?
Answer:
density = 5.1g/m³
Explanation:
use ideal gas equation
Pv=nRT
28.26 x 3.51 = m/v x 0.08206 x 237
m/v = (28.26 x3.51)/(237 x 0.08206) = 5.1g/m³
0.1 dm³ of argon contains x atoms .How many atoms are there in 0.2 dm³ of hydrogen gas under the same conditions?
A. )×/2
B.)2x
C.)4x
Answer:
B.) 2x
Explanation:
Hello,
In this case, we can apply the following rule of three, knowing that 0.1 dm³ equals x and 0.2 dm³ is the unknown:
[tex]0.1dm^3\longrightarrow x\\0.2dm^3 \longrightarrow ?[/tex]
Thus, solving for the unknown we find:
[tex]?=\frac{0.2dm^3*x}{0.1dm^3} \\\\?=2*x[/tex]
Therefore, the answer is B.) 2x.
Best regards.
If the particles of matter that make up a substance are relatively far apart and can move freely, the substance is in what state?
gaseous
liquid
solid
Answer:
Gaseous
Explanation:
Gasses can move freely and do not form the shape of their containers
Liquids are more free than solids, but they conform to the shape of their container
Solids are not free
Ethylene glycol, C2H6O2, is infinitely miscible (soluble) in water. It is a nonelectrolyte that is used as antifreeze. What is the lowest possible melting point for engine coolant that is 26.1% ethylene glycol
Answer: The lowest possible melting point for engine coolant that is 26.1% ethylene glycol is [tex]-7.83^0C[/tex]
Explanation:
Depression in freezing point or melting point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=T_f^0-T_f=(0-T_f)^0C[/tex] = Depression in freezing point
i= vant hoff factor = ( 1 for non electrolyte)
[tex]K_f[/tex] = freezing point constant for water= [tex]1.86^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Given : 26.1 g of ethylene glycol in 100 g of water
Weight of solvent (water)= 100 g = 0.1 kg (1kg=1000g)
Molar mass of solute (ethylene glycol) = 62 g/mol
Mass of solute (ethylene glycol) added = 26.1 g
[tex](0-T_f)^0C=1\times 1.86\times \frac{26.1g}{62g/mol\times 0.1kg}[/tex]
[tex](0-T_f)^0C=7.83[/tex]
[tex]T_f=-7.83^0C[/tex]
Thus the lowest possible melting point for engine coolant that is 26.1% ethylene glycol is [tex]-7.83^0C[/tex]
Choose the substance with the highest surface tension.
a. CH3CH2OH
b. CH3CH2Cl
c. CH3CH2CH3
d. CH2Br2
e. HOCH2CH2OH
Answer:
Hello
you're answer should be E.HOCH2CH2OH
hope this answer is correct
The substance with the highest surface tension - e. HOCH2CH2OH.
Surface tension is the elastic tendency of a fluid, caused by the attraction of particles in the surface which makes it acquire the least surface area.
it depends on intermolecular forces like hydrogen bonding, Vanderwall forces, ion-dipole, and ion-induced dipole interactions.The higher the intermolecular forces higher will be the surface tension.Among all these intermolecular forces, hydrogen bonding is the strongest.So from given options (a) and (e) have -OH functional group and forms H-bonding.Among these two options, option (e) has 2-OH groups so (e) will have more extensive H-bonding.Thus, The substance with the highest surface tension - e. HOCH2CH2OH.
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A 5.00 gram sample of an oxide of lead PbxOy contains 4.33 g of lead. Determine simplest formula for the compund
Answer: The empirical formula is [tex]PbO_2[/tex]
Explanation:
Mass of Pb = 4.33 g
Mass of O = (5.00-4.33) g = 0.67 g
Step 1 : convert given masses into moles
Moles of Pb =[tex]\frac{\text{ given mass of Pb}}{\text{ molar mass of Pb}}= \frac{4.33g}{207g/mole}=0.021moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.67g}{16g/mole}=0.042moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Pb = [tex]\frac{0.021}{0.021}=1[/tex]
For O = [tex]\frac{0.042}{0.021}=2[/tex]
The ratio of Pb O= 1: 2
Hence the empirical formula is [tex]PbO_2[/tex]
According to Le Châtelier's principle, how would a change in pressure affect a
gaseous system in equilibrium?
Answer:
Le Châtelier's principle states that when a chemical system at equilibrium is distributed by a change in conditions, the equilibrium position will shift in a direction that tends to counteract the change.
Therefore, when there is a change in pressure, the equilibrium will counteract the change by reducing/increasing the pressure through adjusting the no. of moles of gas.
Note: At constant temperature and volume the pressure of a gas is directly proportional to the number of moles of gas.
For example, when there's an increase in total pressure, the equilibrium position will shift to the side with a smaller no. of moles of gas so as to reduce the pressure.
Answer:
The equilibrium would shift to reduce the pressure change
Explanation:
10. Show as many ways as you can think of: a) to make benzamide from benzene; b) to make cyclohexyl methyl ketone from bromocyclohexane;
Answer:
See explanation
Explanation:
a) Benzamide from benzene
For this synthesis, we have to start with the Friedel-Crafts reaction to produce Toluene. Then with a strong oxidant, we can produce benzoic acid. In the next step, we can use an esterification reaction to produce the methyl benzoate. Finally, we can use an acyl substitution reaction using ammonia to produce the benzamide.
b) From bromocyclohexane to cyclohexyl methyl ketone
In this case, we can start with a Grignard reaction. The first step is to produce the Grignard reagent with using magnesium. Then if we add acetaldehyde we can form an alcohol, 1-cyclohexylethan-1-ol. Finally, we can reduce the alcohol to produce cyclohexyl methyl ketone.
See figure 1
I hope it helps!
What is the name of the compound Br8P4? A. Bromine tetraphosphate B. Octaboron tetrapotassium C. Boron tetraphosphide D. Octabromine tetraphosphide
Answer:
Option D.
Explanation:
This compound has in its formula:
- Eight bromines
- Four phosphorous
8 → octa prefix
4 → tetra prefix
Right answer is Octabromine tetraphosphide
It can not be option A, tetraphosphate is P4O7
It can not be B and C, we do not have B (boron)
Answer:
Octabromine tetraphosphide
Explanation:
Answer via Educere/ Founder's Education
which element will have higher electronegativity
Given the information below, which is more favorable energetically, the oxidation of succinate to fumarate by NAD+ or by FAD? Fumarate + 2H+ + 2e- → Succinate E°´ = 0.031 V NAD+ + 2H+ + 2e- → NADH + H+ E°´ = -0.320 FAD + 2H+ + 2e- → FADH2 E°´ = -0.219
Answer:
Oxidation by FAD
Explanation:
1. Oxidation by NAD⁺
Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V
NAD⁺ + 2H⁺ + 2e⁻ ⇌ NADH + H⁺; E°´ = -0.320 V
Succinate + NAD⁺ ⇌ Fumarate + NADH + H⁺; E°' = -0.351 V
2. Oxidation by FAD
Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V
FAD + 2H⁺ + 2e⁻ ⇌ FADH₂; E°´ = -0.219 V
Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V
Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.
FAD is the stronger oxidizing agent.
The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.
The pressure in an automobile tire is 2.0 atm at 27°C. At the end of a journey on a hot summer day the pressure has risen to 2.2 atm. What is the temperature of the air in the tire? a. 272.72 K b. 330 K c. 0.014 K d. 175 K
Hey there!
For this we can use the combined gas law:
[tex]\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}[/tex]
We are only working with pressure and temperature so we can remove volume.
[tex]\frac{P_{1} }{T_{1}} = \frac{P_{2} }{T_{2}}[/tex]
P₁ = 2 atm
T₁ = 27 C
P₂ = 2.2 atm
Plug these values in:
[tex]\frac{2atm}{27C} = \frac{2.2atm}{T_{2}}[/tex]
Solve for T₂.
[tex]2atm = \frac{2.2atm}{T_{2}}*27C[/tex]
[tex]2atm * T_{2}={2.2atm}*27C[/tex]
[tex]T_{2}={2.2atm}\div2atm*27C[/tex]
[tex]T_{2}=1.1*27C[/tex]
[tex]T_{2}=29.7C[/tex]
Convert this to kelvin and get 302.85 K, which is closest to B. 330 K.
Hope this helps!
The gas evolved in the metal carbonate reaction with acid turns limewater milky The milky substance formed is
Answer:
The gas evolved because of reaction of acid with metal carbonate or metal hydrogen carbonate turns lime water milky. This shows that the gas is carbon dioxide gas. This happens because of formation of white precipitate of calcium carbonate.
Hope it helps you!
The milky substance formed is CO₂ gas.
What leads to the formation of white precipitate of calcium carbonate ?The gas evolved because of reaction of acid with metal carbonate or metal hydrogen carbonate turns lime water milky. This shows that the gas is carbon dioxide gas.
This happens because of formation of white precipitate of calcium carbonate.
Lime water turns milky when the gas liberated from an acidified carbonate solution is passed into it.
The liberated CO₂ reacts with lime water to give calcium bicarbonate as the precipitate.
Hence, the milky substance formed is CO₂ gas.
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A sample of a compound is made up of 57.53 g C, 3.45 g H, and 39.01 g F. Determine the empirical formula of this compound.
Answer:
C7H5F3
Explanation:
The following data were obtained from the question:
Mass of Carbon (C) = 57.53g
Mass of Hydrogen (H) = 3.45g
Mass of Fluorine (F) = 39.01g
The empirical formula of the compound can be obtained as follow:
C = 57.53g
H = 3.45g
F= 39.01g
Divide each by their molar mass
C = 57.53/12 = 4.79
H = 3.45/1 = 3.45
F = 39.01/19 = 2.05
Divide each by the smallest
C = 4.79/2.05 = 2.3
H = 3.45/2.05 = 1.7
F = 2.05/2.05 = 1
Multiply through by 3 to express in whole number
C = 2.3 x 3 = 7
H = 1.7 x 3 = 5
F = 1 x 3 = 3
Therefore, the empirical formula for the compound is C7H5F3
Calculate Ecell at 80 ºC for a voltaic cell based on the following redox reaction: H2(g, 1.25 atm) + 2AgCl(s) → 2Ag(s) + 2H+(aq, 0.10 M) + 2Cl–(aq, 1.5 M) The standard cell potential Eºcell = +0.18 V at this temperature.
Answer:
Ecell = +0.25V
Explanation:
the half-cell reactions for a voltanic cell
cathode(reduction): 2H⁺(aq) + 2e⁻ ------- H₂(g)
anode(oxidation): 2AgCl(s) ------- 2Ag⁺(aq) + 2Cl⁻ + 2e⁻
we have the standard cell potential E⁺cell = 0.18V at 80C respectively
Q = [H⁺]/[Cl⁻]
sub for [H+] = 0.10M and [Cl-] = 1.5M
Q= 0.1M/1.5M
Q = 0.067
Ecell = E⁺cell - [tex]\frac{0.059}{n}[/tex] logQ
= 0.18 - [tex]\frac{0.056}{1}[/tex] log 0.067
0.18- 0.059(-1.174)
Ecell = +0.25V
Phosphoric acid is a polyprotic acid, with p K values of 2.14, 6.86, and 12.38. Which ionic form predominates at pH 9.3
Answer:
HPO₄⁻² predominates at pH 9.3
Explanation:
These are the equilibriums of the phosphoric acid, a tryprotic acid where 3 protons (H⁺) are realesed.
H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺ pKa 2.14
H₂PO₄⁻ + H₂O ⇄ HPO₄⁻² + H₃O⁺ pKa 6.86
HPO₄⁻² + H₂O ⇄ PO₄⁻³ + H₃O⁺ pKa 12.38
The H₂PO₄⁻ works as amphoterous, it can be a base and acid, according to these equilibriums.
H₂PO₄⁻ + H₂O ⇄ HPO₄⁻² + H₃O⁺
H₂PO₄⁻ + H₂O ⇄ H₃PO₄ + OH⁻
pH 9.3 is located between 6.86 and 12.38 where we have this buffer system HPO₄⁻² / PO₄⁻³, where the HPO₄⁻² is another amphoterous:
HPO₄⁻ + H₂O ⇄ H₂PO₄⁻ + OH⁻
HPO₄⁻² + H₂O ⇄ PO₄⁻³ + H₃O⁺
The media from the two pKa, indicates the pH where the protonated form is in the same quantity as the unpronated form, so:
(6.86 + 12.38) /2 = 9.62
Above this pH, [PO₄⁻³] > [HPO₄⁻²].
In conclussion, at pH 9.3, [HPO₄⁻²] > [PO₄⁻³]
1.60 mL of a suspension of 320.0 mg/5.00 mL aluminum hydroxide is
added to 2.80 mL of hydrochloric acid. What is the molarity of the
hydrochloric acid?
Answer:
1.41 M
Explanation:
First we must use the information provided to determine the concentration of the aluminum hydroxide.
Mass of aluminum hydroxide= 320mg = 0.32 g
Molar mass of aluminum hydroxide= 78 g/mol
Volume of the solution= 5.00 ml
From;
m/M= CV
Where;
m= mass of aluminum hydroxide= 0.32 g
M= molar mass of aluminum hydroxide = 78 g/mol
C= concentration of aluminum hydroxide solution = the unknown
V= volume of aluminum hydroxide solution = 5.0 ml
0.32 g/78 g/mol = C × 5/1000
C = 4.1×10^-3/5×10^-3
C= 0.82 M
Reaction equation;
Al(OH)3(aq) + 3HCl(aq) -----> AlCl3(aq) + 3H2O(l)
Concentration of base CB= 0.82 M
Volume of base VB= 1.60 ml
Concentration of acid CA= the unknown
Volume of acid VA= 2.80 ml
Number of moles of acid NA = 3
Number of moles of base NB= 1
Using;
CA VA/CB VB = NA/NB
CAVANB = CBVBNA
CA= CB VB NA/VA NB
CA= 0.82 × 1.60 × 3/ 2.80 ×1
CA= 1.41 M
Therefore the concentration of HCl is 1.41 M
Begin by writing formulas for the reactants and products, including physical states. When necessary use coefficients to balance the equation.
A) When solid sodium hydride, NaH, is added to water, hydrogen gas is released and aqueous sodium hydroxide forms.
B) Aluminum metal reacts with chlorine gas to form solid aluminum chloride.
Answer:
NaH(s) + H₂O(aq) ⇒ NaOH (aq) + H₂ (g)
Al(s) + Cl₂(g) ⇒ AlCl₃(s)
Explanation:
NaH - sodium hydride
H₂O - water
NaOH - sodium hydroxide
Al - aluminium
Cl₂ - Chlorine
AlCl₃ - aluminium chloride
what is the chemical formula for deionized water?
Answer:
Formula: H2O Formula Weight: 18.02 CAS No.: 7732-18-5 Density: 1.000 g/mL at 3.98 °C(lit.)
Explanation:
If a mixture of a given percentage or ratio strength is diluted to twice its original quantity, its active ingredient will be contained in twice as many parts of the whole, and its strength, therefore, will be?
Answer:
if a mixture of a given percentage or ratio strength is diluted to twice its original quantity, its active ingredient will be contained in twice as many parts of the whole, and its strength therefore will be reduced by one-half
Write a balanced equation depicting the formation of one mole of NaBr(s) from its elements in their standard states.
Express your answer as a chemical equation. Identify all of the phases in your answer.
Write a balanced equation depicting the formation of one mole of SO3(g) from its elements in their standard states.
Express your answer as a chemical equation. Identify all of the phases in your answer.
For SO3(g) find the value of ΔH∘f. (Use Appendix C in the textbook.)
Express your answer using four significant figures.
For NaBr(s) find the value of ΔH∘f. (Use Appendix C in the textbook.)
Express your answer using four significant figures.
Write a balanced equation depicting the formation of one mole of Pb(NO3)2(s) from its elements in their standard states.
Express your answer as a chemical equation. Identify all of the phases in your answer.
For Pb(NO3)2(s) find the value of ΔH∘f. (Use Appendix C in the textbook.)
Express your answer using four significant figures.
Answer:
Check Explanation.
Explanation:
Formation reactions are chemical reactions where one mole of a compound is produced from its constituent elements in their standard states.
NaBr (s)
The Standard formation reaction is
Na (s) + (1/2)Br₂ (g) → NaBr (s)
Using appendix C, the standard heat of formation of NaBr(s) is
ΔH∘f = -359.8 kJ/mol.
SO₃ (g)
The Standard formation reaction is
S (s) + (3/2) O₂ (g) → SO₃ (g)
Using appendix C, the standard heat of formation of SO₃(g) is
ΔH∘f = -395.2 kJ/mol.
Pb(NO₃)₂ (s)
The Standard formation reaction is
Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)
Using appendix C, the standard heat of formation of Pb(NO₃)₂(s) is
ΔH∘f = -451.9 kJ/mol.
Hope this Helps!!!
Answer:
Explanation:
Formation reactions are chemical reactions where one mole of a compound is produced from its constituent elements in their standard states.
How many grams of C are required to react with 64.6 g of Fe2O3 ?
Answer:
14.6 g
Explanation:
Fe₂O₃ + 3 C → 2 Fe + 3 CO₃
First, convert grams of Fe₂O₃ to moles. The molar mass is 159.69 g/mol.
(64.6 g)/(159.69 g/mol) = 0.4045 mol
Now, use stoichiometry to convert moles of Fe₂O₃ to moles of C. By looking at the chemical equation, you can see that for every 1 mole of Fe₂O₃, you need 3 moles of C. Use this relationship.
(0.4045 mol Fe₂O₃) × (3 mol C/1 mol Fe₂O₃) = 1.214 mol C
Now, convert moles of C to grams. The molar mass is 12.01 g/mol.
(1.214 mol C) × (12.01 g/mol) = 14.58 g
You need to use significant figures.
14.58 g ≈ 14.6 g
g If 1.00 mol of ethane gas and 5.00 mol of oxygen gas react, what is the limiting reactant and how many moles of water are produced from the reaction
Answer:
First, write a balanced equation of the reaction.
A hydrocarbon reacting with oxygen usually gives out carbon dioxide and water (combustion)
2C2H6 + 7O2 ---> 4CO2 + 6H2O
From the equation, we can see the mole ratio of ethane : oxygen is 2:7, meaning 2 moles of ethane reacts with 7 moles of oxygen.
If there's 1 mole of ethane gas, we need y moles of oxygen gas for complete reaction.
[tex]\frac{2}{7} =\frac{1}{y}[/tex]
y= 3.5 moles
Since 5 moles of oxygen is more than the required 3.5 moles, we can deduce oxygen gas is in excess, meaning ethane is limiting.
(You can get the same results too if you take y as ethane gas required).
The no. of moles of product produced ALWAYS depend on the no. of moles of the limiting reactant, since they are the ones which reacts completely.
So, from the equation, since the mole ratio of ethane : water = 2:6,
hence, (take the no. of moles of water produced as z),
[tex]\frac{2}{6} =\frac{1}{z} \\z= 3[/tex]
Therefore, 3 moles of water is produced.
Ethane is the limiting reagent in the given reaction. The number of moles of water produced from the reaction is 3 moles.
What is a limiting reagent?A limiting reagent can be defined as that reactant in the chemical reaction which is consumed first among the other reactants during the completion of a chemical reaction.
The limiting reagent is the reactant which decides the yield of the product when the quantity of the reactants is not taken in stoichiometry.
Given, chemical reaction of ethane and oxygen is:
[tex]C_2H_6 + \frac{7}{2} O_2 \longrightarrow 2CO_2 +3H_2O[/tex]
Given, the number of moles of ethane = 1 mol
The number of moles of oxygen = 5 mol
One more of ethane reacts with moles of oxygen = 3.5 mol
Therefore, ethane is the limiting reagent in this reaction as the oxygen is given in excess.
From the balanced equation, one mole of ethane produces moles of water equal to 3 moles.
Learn more about limiting reagent, here:
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Name the compound Ga S3
Answer:
Gallium(III) sulfide