Answer:
2200 L
Explanation:
Ideal gas law:
PV = nRT,
where P is absolute pressure,
V is volume,
n is number of moles,
R is universal gas constant,
and T is absolute temperature.
The initial number of moles is:
(110 atm) (80 L) = n (0.0821 L atm / K / mol) (30 + 273.15) K
n = 353.58 mol
After some gas is removed, the number of moles remaining is:
(80 atm) (80 L) = n (0.0821 L atm / K / mol) (30 + 273.15) K
n = 257.15 mol
The amount of gas removed is therefore:
n = 353.58 mol − 257.15 mol
n = 92.43 mol
At normal conditions, the volume of this gas is:
PV = nRT
(1 atm) V = (92.43 mol) (0.0821 L atm / K / mol) (273.15 K)
V = 2162.5 L
Rounded, the volume is approximately 2200 liters.
Which of the following is a salt that will form from the combination of a strong base with a weak acid?
Select the correct answer below:
A. NaHCO3
B. H2O
C. CH3CO2H
D. NH4Cl
Answer:
A. NaHCO₃
Explanation:
NaHCO₃ ⇒ NaOH + H₂CO₃
NaOH is a strong base and H₂CO₃ is a weak acid. Therefore, NaHCO₃ is a salt of a strong base-weak acid reaction. The salt is basic because carbonic acid (H₂CO₃) is a weak acid so it remains undissociated. So, there is a presence of additional OH⁻ ions that makes the solution basic.
Hope that helps.
"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"
Answer:
245.66g of NH₄Cl is the mass we need to add to obtain the desire pH
Explanation:
The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical
First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.
pKa NH₃/NH₄⁺pKb = - log Kb
pKb = -log 1.8x10⁻⁵ = 4.74
pKa = 14 - pKb
pKa = 14 - 4.74
pKa = 9.26
Moles NH₃2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃
H-H equation:pH = pKa + log [NH₃] / [NH₄Cl]
8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]
-1.06 = log [0.400 moles] / [NH₄Cl]
0.0087 = [0.400 moles] / [NH₄Cl]
[NH₄Cl] = 0.400 moles / 0.0087
[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):
4.59 moles NH₄Cl ₓ (53.491g / mol) =
245.66g of NH₄Cl is the mass we need to add to obtain the desire pH
Heterocyclic aromatic compounds undergo electrophilic aromatic substitution in a similar fashion to that undergone by benzene with the formation of a resonance-stabilized intermediate. Draw all of the resonance contributors expected when the above compound undergoes bromination
Answer:
See explanation
Explanation:
When we talk about electrophilic substitution, we are talking about a substitution reaction in which the attacking agent is an electrophile. The electrophile attacks an electron rich area of a compound during the reaction.
The five membered furan ring is aromatic just as benzene. This aromatic structure is maintained during electrophilic substitution reaction. The attack of the electrophile generates a resonance stabilized intermediate whose canonical structures have been shown in the image attached.
Convert 59800 kilograms to pounds
Answer:
131836.43 pounds
Explanation:
one kilogram is 2.20462 pounds. multiply 2.20462 by 59800
Answer: 131836.43
Formula: Multiply the mass value by 2.205
59800×2.205=131836.43
Half-cells were made from a nickel rod dipping in nickel sulfate solution and a copper rod dipping in copper sulfate solution. The cells were combined to construct a voltaic electrochemical cell. Sketch the cell and label anode and cathode with charges, electrode material and electrolyte solutions, half-reactions and overall reaction, give direction of electron flow and movement of ions.
Answer:
Check the Attachment.
Half-reactions:
Anode: (OXIDATION) Ni --> Ni2+ + 2e-
Cathode: (REDUCTION) Cu2+ +2e- --> Cu
Overall reaction: Ni + Cu2+ --> Ni2+ + Cu
Explanation:
Overall, reaction is basically Anode + Cathode, where electrons on both sides cancel out (if not, you need to multiply the equation in a way you can cancel them out).
Hope this helps.
How many grams is 5.8 moles of hydrochloric acid (HCI)?
Answer to the nearest 0.01 g.
Answer:
211.47 grams
Explanation:
We need to set up a dimensional analysis to solve this problem by converting from moles to grams.
First, find the molar mass of HCl. Since the molar mass of H (hydrogen) is 1.01 g/mol and the molar mass of Cl (chlorine) is 35.45 g/mol, then the molar mass of HCl is:
1.01 + 35.45 = 36.46 g/mol
We have 5.8 moles of HCl, so multiply by its molar mass:
(5.8 mol) * (36.46 g/mol) = 211.468 ≈ 211.47 g
The answer is thus 211.47 grams.
~ an aesthetics over
Answer:
[tex]\large\boxed{211.47}\\[/tex] grams
Explanation:
First, you need to gather the atomic masses of the elements involved in the compound - hydrogen and chlorine. Referencing a modern periodic table will give you this information.
Hydrogen has an atomic weight of 1.00784 and Chlorine has an atomic mass of 35.453.Add those two values together - 1.00784 + 35.453 = 36.46084Multiply this value by 5.8 (one mole is equivalent to the atomic mass of the compound) - 5.8 x 36.46084 = 211.472872Round to the nearest 0.01 gram - 211.47[tex]\large\boxed{211.47}[/tex] is the final answer.
Which of the following combinations will result in a reaction that is spontaneous at all temperatures?
Negative enthalpy change and negative entropy change
Negative enthalpy change and positive entropy change
Positive enthalpy change and negative entropy change
Positive enthalpy change and positive entropy change
PLS EXPLAIN WHAT EACH MEANS AND THE VARIABLES AND THE EXPLANATION BEHIND IT
Answer:
[tex]\huge\boxed{Option \ 2}[/tex]
Explanation:
A reaction is spontaneous at all temperatures by the following combinations:
=> A negative enthalpy change ( [tex]\triangle H < 0[/tex] )
=> A positive entropy change ( [tex]\triangle S > 0[/tex] )
See the attached file for more better understanding!
from Gibbs Equation, [tex] \Delta G = \Delta H - T\Delta S [/tex]
reaction is spontaneous if $\Delta G$ is negative.
so, first option is not valid at high temperature, ($-h+ts$)
second, is always a spontaneous reaction, ($-h-ts$)
third, is never spontaneous ($+h+ts$)
4th is similar to second, spontaneous at higher temperatures ($+h-ts$)
The reaction of butadiene gas (C4H6) with itself produces C8H12 gas as follows: The reaction is second order with a rate constant equal to 5.76 × 10−2 L/mol/min under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration in molarity remaining after 10.0 min? Report your answer to 3 decimal places.
Answer:
[tex]C_{C_4H_6}=0.179M[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]2C_4H_6\rightarrow C_8H_{12}[/tex]
And the rate law is:
[tex]\frac{dC_{C_4H_6}}{dt}=kC_{C_4H_6}^2[/tex]
Which integrated is:
[tex]\frac{1}{C_{C_4H_6}} =\frac{1}{C_{C_4H_6}^0}+kt[/tex]
In such a way, the concentration after 10.0 min is:
[tex]\frac{1}{C_{C_4H_6}} =\frac{1}{0.200M}}+5.76x10^{-2}\frac{L}{mol*min}*10.0min\\ \\\frac{1}{C_{C_4H_6}}=5.58\frac{L}{mol} \\\\C_{C_4H_6}=\frac{1}{5.58\frac{L}{mol} } \\\\C_{C_4H_6}=0.179M[/tex]
Regards.
What is the ph of 0.36M HNO3 ?
Answer:
0.44
Explanation:
We know that the pH of any acid solution is given by the negative logarithm of its hydrogen ion concentration. Hence, if I can obtain the hydrogen ion concentration of any acid, I can obtain its pH.
For the acid, HNO3, [H^+] = [NO3^-]= 0.36 M
pH= -log [H^+]
pH= - log[0.36]
pH= 0.44
When we react a weak acid with a strong base of equal amounts and concentration, the component of the reaction that will have the greatest effect on the pH of the solution is:______.
a. the acid.
b. the base.
c. the conjugate acid.
d. the conjugate base.
Answer:
d. the conjugate base.
Explanation:
The general reaction of a weak acid, HA, with a strong base YOH, is:
HA + YOH → A⁻ + H₂O + Y⁻
Where A⁻ is the conjugate base of the weak acid and Y⁻ usually is a strong electrolyte.
That means after he complete reaction you don't have weak acid nor strong base, just conjugate base that will be in equilibrium with water, thus (Strong electrolyte doesn't change pH:
A⁻ + H₂O ⇄ HA + OH⁻
As the equilibrium is producing OH⁻, the pH of the solution is being affected for the conjugate base
Right option:
d. the conjugate base.please help guys the question is
give reasons
a. we have to separate the mixture
b. All impure substances are not harmful.
c. A mixture of iron fillings and sand can be separated by using a magnet
d. A sentences "shake before well use" is written on the bottle of the medicine.
Answer:
(a )people separate mixtures in order to ger a specific substance that they need.
Calculate the pH of a 0.20 M NH3/0.20 M NH4Cl buffer after the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.
Answer:
Explanation:
For pH of a buffer solution , the formula is
pH = pKa + log [ Base ] / [ conjugate acid ]
= pKa + log [ NH₃ ] / [ NH₄⁺ ]
Ka = Kw / Kb
Kb for NH₄OH = 1.8 x 10⁻⁵
Ka = 10⁻¹⁴ / 1.8 x 10⁻⁵
= 5.6 x 10⁻¹⁰
pH = - log ( 5.6 x 10⁻¹⁰ ) + log 0.2 / 0.2
= 10 - log 5.6
= 9.25
Effect of addition of HCl
H⁺ of HCl will react with NH₃ to produce NH₄⁺
25 mL of .1 HCl = 2.5 mM of HCl
25 mL of .1 NH₄⁺ = 2.5 mM of NH₄⁺
65 mL of .2 M NH₃ = 13 mM of NH₃
65 mL of .2 M NH₄⁺ = 13 mM of NH₄⁺
NH₃ + H⁺ = NH₄⁺
NH₄⁺ formed = 2.5 + 13 mM
15.5 mM of NH₄⁺
NH₃ = 13 mM
Concentration of NH₃ = 13 / 90
Concentration of NH₄⁺ = 15.5 / 90
pH of final buffer mixture
= 9.6 + log 13 / 15.5
= 9.25 - .076
= 9.174
The pH value is mathematically given as
pH= -6.332.
What is the pH of a 0.20 M NH3/0.20 M NH4Cl buffer?Question Parameters:
the pH of a 0.20 M NH3/0.20 M NH4Cl buffer
the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.
Generally, the equation for the Chemical Reaction is mathematically given as
HCl + NH3 --> NH4^+ + Cl^-
Therefore
pH= pka + log(13/14).
pH= -6.3 + log 0.93.
pH= -6.3+ (-0.032).
pH= -6.332.
Read more about Chemical Reaction
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A student is using a coffee-cup calorimeter to determine the enthalpy change of the endothermic reaction of two aqueous solutions. After both solutions are added to the cup, the student neglects to put the lid on the cup. This would cause the magnitude of the calculated ΔH° value to be: the answer is: too small, since the solution will absorb heat from the room. But why? Wouldn't depend on if the reaction releases or absorbs heat. Wouldn't it be too large because heat escapes the cup? I'm so confused
Answer:
Explanation:
In all calorimetric experiment , the calorimeter must be isolated from the surrounding . Otherwise the heat change in the experiment can not be determined with precision .
The reaction is endothermic . Hence, there is lowering of temperature due to absorption of heat in the reaction equal to ΔH°. The value of ΔH° can be calculated by measuring fall in the temperature of the content . The fall in the temperature will be less when heat is allowed to come from the surrounding . Less fall of temperature will result in less ΔH° to be calculated .
Hence in the given experiment , if the student neglects to put lid on the cup , the experiment will give less value of ΔH°.
a boy capable of swimming 2.1m/a in still water is swimming in a river with a 1.8 m/a current. At what angle must he swim in order to end up directly opposite his starting point?
Answer:
The boy must swim at an angle of 59°northwest to get to a position directly opposite his starting point.
Explanation:
To get to a point directly opposite his starting point, the boy must travel at an angle x, in a direction northwest of his starting point. The speed of the boy and the speed of the river current forms a right-angled triangle with an an opposite side of 1.8 m/a and a hypotenuse of 2.1 m/a having an angle x.
Sin x = opp/ hyp
Sin x = 1.8/2.1
x = sin⁻¹ (1.8/2.10
x = 58.99
x = 59°
Therefore, the boy must swim at an angle of 59° in the northwesterly direction to get to a position directly opposite his starting point.
An electrolysis cell has two electrodes. Which statement is correct? A. Reduction takes place at the anode, which is positively charged. B. Reduction takes place at the cathode, which is positively charged. C. Reduction takes place at the dynode, which is uncharged. D. Reduction takes place at the cathode, which is negatively charged. E. Reduction takes place at the anode, which is negatively charged.
Answer:
D. Reduction takes place at the cathode, which is negatively charged.
Explanation:
In an electrolytic cell there are two electrodes; the cathode and the anode. The anode is the positive electrode while the cathode is the negative electrode. Oxidation occurs at the anode while reduction occurs at the cathode.
At the anode, species give up electrons and become positively charged ions while at the cathode species accept electrons and become reduced.
Which of the following represents six molecules of water? 6HO 2 6H 2O 1 6H 2O H 6O
Answer:
6H20 represents six molecules of water
Answer:
6H20 represents six molecules of water
Explanation:
is a polyprotic acid. Write balanced chemical equations for the sequence of reactions that carbonic acid can undergo when it's dissolved in water.
Answer:
H₂CO₃ H₂O ⇄ HCO₃⁻ + H₃O⁺ Ka1
HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺ Ka2
CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb1
HCO₃⁻ + H₂O ⇄ H₂CO₃ + OH⁻ Kb2
Explanation:
Formula for carbonic acid is: H₂CO₃
It is a dyprotic acid, because it can release two protons. We can also mention that is a weak one. The equilibrums are:
H₂CO₃ H₂O ⇄ HCO₃⁻ + H₃O⁺ Ka1
HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺ Ka2
When the conjugate strong bases, carbonate and bicarbonate take a proton from water, the reactions are:
CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb1
HCO₃⁻ + H₂O ⇄ H₂CO₃ + OH⁻ Kb2
Notice, that bicarbonate anion can release or take a proton to/from water. This is called amphoteric,
What are the conjugate acid-base pairs in the following chemical reaction? HBr(aq)+ CH3COOH(aq) ⇌ CH3C(OH)2+(aq) + Br-(aq)
Answer:
HBr, CH3C(OH)2 and CH3COOH, Br-
Explanation:
The conjugate acid-base pairs acid reacts with base to form a conjugate acid and conjugate base.
Conjugate acid is formed when a bases receives a proton (H+) and a conjugate base is formed when an acid losses a proton (H+).
From the given equation:
HBr, CH3C(OH)2 and CH3COOH, Br- are conjugate acid-base pair, where HBr is an acid and CH3C(OH)2 is a conjugate acid while CH3COOH and Br- is the conjugate base.
Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: specific heat (ice) = 2.1 J/g·°C; specific heat (water) = 4.18 J/g·°C; ΔH fus = 6.0 kJ/mol.
Answer:
The amount of heat to absorb is 6,261 J
Explanation:
Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.
The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.
So:
Heat required to raise the temperature of ice from -20 °C to 0 °CBeing the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).
In this case, m= 10 g, specific heat of the ice= 2.1 [tex]\frac{J}{g*C}[/tex] and ΔT=0 C - (-20 C)= 20 C
Replacing: Q= 10 g*2.1 [tex]\frac{J}{g*C}[/tex] *20 C and solving: Q=420 J
Heat required to convert 0 °C ice to 0 °C waterThe heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
In this case, being 1 mol of water= 18 grams: Q= 10 g*[tex]6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}[/tex]= 3.333 kJ= 3,333 J (being kJ=1,000 J)
Heat required to raise the temperature of water from 0 °C to 60 °CIn this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 [tex]\frac{J}{g*C}[/tex] and ΔT=60 C - (0 C)= 60 C
Replacing: Q= 10 g*4.18 [tex]\frac{J}{g*C}[/tex] *60 C and solving: Q=2,508 J
Finally, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
The amount of heat to absorb is 6,261 J
The amount of heat to absorb is 6,261 J.
Calculation for heat:Heat required to raise the temperature of ice from -20 °C to 0 °C.
The formula for specific heat is used to calculate the amount of heat
Q = c * m * ΔT
Where,
Q =heat exchanged by a body,
m= mass of the body
c= specific heat
ΔT= change in temperature
Given:
m= 10 g,
specific heat of the ice= 2.1
ΔT=0 C - (-20 C)= 20 C
On substituting the values:
Q= 10 g*2.1 *20 C
Q=420 J
Heat required to convert 0 °C ice to 0 °C water.
The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
Heat required to raise the temperature of water from 0 °C to 60 °C
m= 10 g,
Specific heat of the water= 4.18
ΔT=60 C - (0 C)= 60 C
On substituting:
Q= 10 g*4.18 *60 C
Q=2,508 J
Thus, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
The amount of heat to absorb is 6,261 J
Find more information about Specific heat here:
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Consider these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)|Cu(s) Ag+(aq)|Ag(s) -0.40 V -0.76 V ‑0.25 V +0.34 V +0.80 V Based on the data above, which species is the best reducing agent?
Answer:
The best reducing agent is Zn(s)
Explanation:
A reducing agent must to be able to reduce another compound, by oxidizing itself. Consequently, the oxidation potential must be high. The oxidation potential of a compound is the reduction potential of the same compound with the opposite charge. Given the reduction potentials, the best reducing agent will be the compound with the most negative reduction potential. Among the following reduction potentials:
Cd₂⁺(aq)|Cd(s) ⇒ -0.40 V
Zn²⁺(aq)|Zn(s) ⇒ -0.76 V
Ni²⁺(aq)|Ni(s) ⇒‑0.25 V
Cu²⁺(aq)|Cu(s) ⇒ +0.34 V
Ag⁺(aq)|Ag(s) ⇒ +0.80 V
The most negative is Zn²⁺(aq)|Zn(s) ⇒ -0.76 V
From this, the most reducing agent is Zn. Zn(s) is oxidized to Zn²⁺ ions with the highest oxidation potential (0.76 V).
3,3-dibromo-4-methylhex-1-yne
Explanation:
see the attachment. hope it will help you...Which of the following contains a nonpolar covalent bond?
O A. Co
B. NaCl
O C. 02
O D. HE
Answer:
The answer is o2
Explanation:
I took the test
A student sets up the following equation to convert a measurement. The (?) Stands for a number the student is going to calculate. Fill in the missing part of this equation. (0.030 cm^3) x ? =m^3
Answer:
\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}
Explanation:
0.030 cm³ × ? = x m³
You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.
For example, you know that centi means "× 10⁻²", so
1 cm = 10⁻² m
If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).
If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.
So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.
We choose the former because it has the desired units on top.
The "cm" is cubed, so we must cube the conversion factor.
The calculation becomes
[tex]\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 0.30 \times 10^{-6}\text{ m}^{3} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}\\\\\textbf{0.30 cm}^{\mathbf{3}} \times \left (\dfrac{\mathbf{10^{-2}}\textbf{ m}}{\textbf{1 cm}}\right )^{\mathbf{3}} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}[/tex]
Arrange the following in order of increasing boiling point: CH4, CH3CH3, CH3CH2Cl, CH3CH2OH. Rank from lowest to highest. To rank items as equivalent, overlap them.
Answer:
In order from lowest to highest:
Methane < Ethane < Chloroethene < Methanol
i.e: CH4 < CH3CH3 < CH3CH2OH < CH3CH2Cl
Explanation:
Compounds with stronger molecular fore have higher boiling points, thus making the molecules more difficult to pull apart. The presence of chains also increases the molecular dispersion. The dipole force of ethanol makes it have a very high boiling point.
I'm positive this explanation would suffice. Best of luck.
The order of increasing boiling points of the substances listed is; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH.
Intermolecular interactions occur between molecules. The boiling point and melting points of substances depends on the nature and magnitude of intermolecular interaction between the molecules of the substance.
The order of increasing boiling points of the substances listed is as follows; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH. CH3CH2OH has the highest boiling point due to intermolecular hydrogen bonds in the molecule. Though CH4 and CH3CH3 are both alkanes, CH3CH3 has a higher molecular mass and consequently greater dispersion forces and a higher boiling point.
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When 91.96g of Na reacts with 32.o g of O2 how many grams of NaO2 are produced
Answer:
123.96 g Na₂O
Explanation:
4 Na + O₂ ⇒ 2 Na₂O
You first need to find the limiting reagent. Convert the reactants to moles and see which produces the least amount of product using the mole ratios in the chemical equation.
(91.96 g Na)/(22.99 g/mol Na) = 4 mol Na
(4 mol Na) × (2 mol Na₂O/4 mol Na) = 2 mol Na₂O
(32.0 g O₂)/(32.0 g/mol) = 1 mol O₂
(1 mol O₂) × (2 mol Na₂O/1 mol O₂) = 2 mol Na₂O
Since they both produce the same amount of product, you don't need to pick a limiting reagent. Now, convert moles of Na₂O to grams.
(2 mol Na₂O) × (61.98 g/mol Na₂O) = 123.96 g Na₂O
Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation you submit is balanced.
_______ → Ba(ClO)2 + H2O(l)
Answer:
2HClO(aq) + Ba(OH)₂(aq) → Ba(ClO)₂(aq) + 2H₂O(l)
Explanation:
The reaction corresponds to a neutralization reaction between an acid and a base, as follows:
2HClO(aq) + Ba(OH)₂(aq) → Ba(ClO)₂(aq) + 2H₂O(l)
From the equation above we have that the acid HClO reacts with the base Ba(OH)₂ to obtain a salt Ba(ClO)₂ and water.
In the balanced reaction, we have that 2 moles of HClO react with 1 mol of Ba(OH)₂ to produce 1 mol of Ba(ClO)₂ and 2 moles of water.
I hope it helps you!
9
What might happen if acidic chemicals were emitted into
the air by factories? Choose the best answer.
A
The acid would destroy metallic elements in the air
B
The acid would be neutralized by bases within clouds
C
Acid rain might destroy ecosystems and farmland
D
Violent chemical reactions would take place within the
atmosphere
co search
O
BI
A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 198 picometers and its molar mass is 195.08 g/mol. Calculate the density of the metal in g/cm3.
Answer:
7.38 g/cm³ is the density of the metal
Explanation:
In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).
To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.
Volume of the unit cellVolume = a³
a = √8×r
(r = 198x10⁻¹²m)
a = 5.6x10⁻¹⁰ m
Volume = 1.756x10⁻²⁸ m³
1m = 100cm → 1m³ = (100cm)³:
1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =
1.756x10⁻²² cm³ → Volume of the unit cell in cm³Mass of the unit cell:There are 4 atoms of gold:
4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold
As 1 mole weighs 195.08g:
6.64x10⁻²⁴ moles of gold × (195.08g / mol) =
1.296x10⁻²¹g is the mass of the unit cellDensity of the metal:1.296x10⁻²¹g / 1.756x10⁻²² cm³ =
7.38 g/cm³ is the density of the metalThe density of the metal is 7.40 g/cm³
In cubic crystal system, face-centered cubic FFC is the name given to sort of atom arrangement observed in which structure is made up of atoms organized in a cube with a portion of an atom in each corner and six extra atoms in the center of each cube face.
It is expressed by using the formula:
[tex]\mathbf{\rho = \dfrac{Z \times M}{N_A\times a^}}[/tex]
where;
[tex]\rho[/tex] = density of the metalZ = atoms coordination no = 4 (for FCC)Molar mass (M) = 195.8 g/molAvogadro's constant (NA) = 6.022 × 10²³ /mola = edge lengthFor face-centered cubic FFC;
The edge length [tex]\mathbf{a =2 \sqrt{2}\times r }[/tex]
[tex]\mathbf{a =2 \sqrt{2}\times 198 \ pm }[/tex]
[tex]\mathbf{a =560.0285 \ pm }[/tex]
a = 5.60 × 10⁻⁸ cm
Replacing it into the previous equation, we have:
[tex]\mathbf{\rho = \dfrac{4 \times 195.8}{6.022 \times 10^{23} \times( 5.60 \times 10^{-8} )^3}}[/tex]
[tex]\mathbf{\rho = 7.40\ g/cm^3 }[/tex]
Learn more about face-centered cubic arrangement here:
https://brainly.com/question/14786352?referrer=searchResults
Im really confused and select all that apply questions scare me.
Answer:
The 3rd one
Explanation:
Assume that you are provided with the following materials:
• Strips of metallic zinc, metallic copper, metallic iron
• 1M aqueous solutions of ZnSO4, CuSO4, FeSO4, and aqueous iodine(I2)
• Other required materials to create Voltaic cells such as beakers, porous containers, graphite rods, a voltmeter, and a few wires with alligator clips.
In this modified version of the lab, after thoroughly studying the lab hand out and watching the videos,identify 4 different combinations of Voltaic cells that are possible to be created with the above materials.For each cell created, include the following details.
A) Which electrode was the anode,and which was the Cathode?
B) The anode and cathode half reactions.
C) Balanced equation for each cell you propose to construct.
D) Calculated Eocelle Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed).
Answer:
See explanation
Explanation:
First voltaic cell;
Zn(s)|Zn^2+(aq)||Cu^2+(aq)|Cu(s)
Anode;
Zinc
Cathode;
Copper
Oxidation half equation;
Zn(s)------> Zn^2+(aq) + 2e
Reduction half equation;
Cu^2+(aq) +2e -----> Cu(s)
Overall; Zn(s) + Cu^2+(aq) -----> Zn^2+(aq) + Cu(s)
E°cell = 0.34 -(-0.76) =1.1 V
Second voltaic cell;
Zn(s)|Zn^2+(aq)||Fe^2+(aq)|Fe(s)
Anode;
Zinc
Cathode;
Iron
Oxidation half equation;
Zn(s)------> Zn^2+(aq) + 2e
Reduction half equation;
Fe^2+(aq) +2e -----> Fe(s)
Overall; Zn(s) + Fe^2+(aq) -----> Zn^2+(aq) + Fe(s)
E°cell = (-0.44) -(-0.76) = 0.32 V
Third voltaic cell;
Fe(s)|Fe^2+(aq)||Cu^2+(aq)|Cu(s)
Anode;
Iron
Cathode;
Copper
Oxidation half equation;
Fe(s)------> Fe^2+(aq) + 2e
Reduction half equation;
Cu^2+(aq) +2e -----> Cu(s)
Overall; Fe(s) + Cu^2+(aq) -----> Fe^2+(aq) + Cu(s)
E°cell = 0.34 -(-0.44) = 0.78 V
Fourth voltaic cell
Cu(s)|Cu^2+(aq)||I2(aq)|C(s)|I^-(aq)
Anode;
Copper
Cathode;
Graphite rod
Oxidation half equation;
Cu(s)------> Cu^2+(aq) + 2e
Reduction half equation;
I2(aq) +2e -----> 2I^-(aq)
Overall; Cu(s) + I2(aq) -----> Cu^2+(aq) + 2I^-(aq)
E°cell = 0.54 -0.34 = 0.20 V
