Answer:
4.7 atm
Explanation:
From the question given above, the following data were obtained:
Total pressure (P) = 15.3 atm
Pressure of Helium, P(He) = 3.2 atm
Pressure of Oxygen, P(O) = 7.4 atm
Pressure of Argon, P(Ar) =?
The pressure of Argon, P(Ar) in the tank can be obtained as illustrated below:
P = P(He) + P(O) + P(Ar)
15.3 = 3.2 + 7.4 + P(Ar)
15.3 = 10.6 + P(Ar)
Collect like terms
15.3 – 10.6 = P(Ar)
4.7 = P(Ar)
P(Ar) = 4.7 atm
Therefore, the pressure of Argon, P(Ar) in the tank is 4.7 atm
pls help Which statement is true about the environment of urban areas?
Urban areas have higher temperatures.
Urban areas have few problems with soil erosion.
Urban areas have more rain infiltration into the soil.
Urban areas have less habitat fragmentation.
The statement that is correct about Urban areas is that they have higher temperatures.
Urban Area is a term to refer to cities. Urban areas are characterized by having a developed infrastructure (wide roads, vehicular bridges, wide platforms, tall buildings, residential areas, industrial areas, among others).
Recent studies affirm that urban areas are warmer than surrounding areas; This phenomenon is because the materials with are built buildings, roads, houses, and others, concentrate the sun's rays, increasing the temperature of cities. In addition, the lack of trees deepens this phenomenon, because the trees contribute to cooling by physicochemical processes such as evapotranspiration.
According to the above, it is possible to affirm that urban areas are hotter than their surrounding area because they lack vegetation, and the materials with which it is built contribute to the increase in temperature.
On the other hand, urban areas are characterized by habitat fragmentation, more problems with soil erosion, and less rain infiltration into the soil.
Learn more in: https://brainly.com/question/23587978
Answer:
A) higher temps
Explanation:
What is the name of Na
What type of reaction is illustrated?
C3H8 +502 + 3C02 + 4H2O
А
B
decomposition
combustion
single
replacement
Answer:
Combustion
Explanation:
In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) H2O (g) CO2 (g) H2 (g) In an experiment, 0.35 mol of CO and 0.40 mol of H2O were placed in a 1.00-L reaction vessel. At equilibrium, there were 0.19 mol of CO remaining. Keq at the temperature of the experiment is __________. A) 5.47 B) 1.0 C) 1.78 D) 0.75 E) 0.56
Answer: [tex]K_{eq}[/tex] at the temperature of the experiment is 0.56.
Explanation:
Moles of [tex]CO[/tex] = 0.35 mole
Moles of [tex]H_2O[/tex] = 0.40 mole
Volume of solution = 1.00 L
Initial concentration of [tex]CO[/tex] = [tex]\frac{0.35mol}{1.00L}=0.35M[/tex]
Initial concentration of [tex]H_2O[/tex] = [tex]\frac{0.40mol}{1.00L}=0.40M[/tex]
Equilibrium concentration of [tex]CO[/tex] = [tex]\frac{0.19mol}{1.00L}=0.19M[/tex]
The given balanced equilibrium reaction is,
[tex]CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)[/tex]
Initial conc. 0.35 M 0.40 M 0 M 0M
At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M
Given: (0.35-x) = 0.19
x= 0.16 M
The expression for equilibrium constant for this reaction will be,
[tex]K_{eq}=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}[/tex]
Now put all the given values in this expression, we get :
[tex]K_{eq}=\frac{0.16\times 0.16}{(0.35-0.16)\times (0.40-0.16)}[/tex]
[tex]K_{eq}=\frac{0.16\times 0.16}{(0.19)\times (0.24)}=0.56[/tex]
Thus [tex]K_{eq}[/tex] at the temperature of the experiment is 0.56.
Help plz:)))I’ll mark u Brainliest
Answer:
2.475 mol of O2 formed.
Explanation:
Given 1.65 moles of KClO3 as the target amount in the reactant, used the coefficient of the balanced chemical reaction involved to determine the number of moles of O2 molecules formed.
x mole of O2 = 1.65 mol KClO3 x [(3 mol O2)/ (2 mol KClO3)] = 2.475 mol of O2
x mole of O2 formed = 2.475 mol of O2
The lattice-like structure of a metal consists of negative metal ions in a "sea" of electrons.
O True
O False
Answer:
True
Explanation:
does the moon really change shape? why
what is the major organic product obtained from the following sequence of reactions 1. naoch2ch3 ch3ch2oh 2. ph br 1. naoch2ch3 ch3ch2oh 2. br 1. lialh4 et2o 2. h3o
Select the correct answer.
Using this activity chart, which reaction will happen when a piece of copper is placed in a lead nitrate solution?
A.
2Cu + 3Pb(NO3)2 3Cu(NO3)2 + 2Pb
B.
No reaction occurs.
C.
2Cu + 3Pb(NO3)2 2Cu(NO3)2 + 3Pb
D.
3Cu + 3Pb(NO3)2 3Cu(NO3)2 + 3Pb
E.
The answer cannot be determined from the information given.
Answer:
B, No reaction will occur
Explanation:
Copper as compared to lead is less reactive. This is the reason when lead is added to copper nitrate solution, it replaces the copper and itself combines with nitrate to form lead nitrate aqueous solution
Lead + Copper(II) nitrate → Copper + Lead (II) nitrate
The same is not the case when the reaction is revered i.e Cu is added to Pb NO3 solution.
Hence, option B is correct
how to make 100 ml of 0.001 mM solution with 0.0405mM solution?
Answer:
Measure 2.47 mL of the stock solution (i.e 0.0405 mM) and dilute it to the 100 mL mark with water
Explanation:
To make 100 mL of 0.001 mM solution from 0.0405mM solution, we need to determine the volume of 0.0405mM solution needed. This can be obtained as follow:
Molarity of stock (M₁) = 0.0405 mM
Volume of diluted (V₂) = 100 mL
Molarity of diluted solution (M₂) = 0.001 mM
Volume of stock solution needed (V₁) =?
M₁V₁ = M₂V₂
0.0405 × V₁ = 0.001 × 100
0.0405 × V₁ = 0.1
Divide both side by 0.0405
V₁ = 0.1 / 0.0405
V₁ = 2.47 mL
Therefore, to make 100 mL of 0.001 mM solution from 0.0405mM solution, measure 2.47 mL of the stock solution (i.e 0.0405 mM) and dilute it to the 100 mL mark with water.
Calculate the Ka of a 0.35M weak acid with a pH of 4.2.
Answer:
Ka = 1.14x10⁻⁸
Explanation:
First we calculate [H⁺] from the pH:
pH = -log[H⁺][H⁺] = [tex]10^{-pH}[/tex][H⁺] = 6.31x10⁻⁵ MFor a monoprotic weak acid, the molar concentration of H⁺ of a solution can be expressed as:
[H⁺] = √(C*Ka)Where C is the molar concentration of the weak acid solution.
6.31x10⁻⁵ M = [tex]\sqrt{0.35M*Ka}[/tex]1.14x10⁻⁸ = KaA sample of air was collected on a day when the total atmosphere
pressure was 592 mmHg. The sample contained only oxygen and
nitrogen gas. If the oxygen in the sample had a pressure of 261
mmHg, how much pressure did the nitrogen have?
A. 853 mm Hg
B. 0.206 mm Hg
C. 4.76 mm Hg
D. 331 mm Hg
E. Other________
(Please show me how you did it)
Answer:
D. 331 mm Hg
Explanation:
We can solve this problem by keeping in mind the law of partial pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of its components.
In other words:
P oxygen + P nitrogen = Total PWe input the given data:
261 mmHg + P nitrogen = 592 mmHgAnd calculate the pressure of the nitrogen:
P nitrogen = 331 mmHgUse the scenario to answer the following question. A group of students working in a chemistry lab are planning a procedure to neutralize 10.0mL of 5.0 M hydrochloric acid (strong acid) with 5.0 M potassium hydroxide (strong base). In their procedure they plan on adding an equal volume of the base to the acid. What would be the expected outcome of carrying out this step as planned?
a- The resulting pH will be less than 7 because potassium hydroxide is less concentrated than the hydrochloric acid.
b- The resulting pH will not be able to be determined because the concentrations of the acid and base are not the same.
c- The resulting pH will be greater than 7 because potassium hydroxide is less concentrated than the hydrochloric acid.
d- The resulting pH will be equal to 7 because a strong base will neutralize a strong acid.
Answer:
d- The resulting pH will be equal to 7 because a strong base will neutralize a strong acid.
Explanation:
The reaction between potassium hydroxide and hydrochloric acid of equal volume and equal concentration yields a solution of pH 7 at equivalence point. We must note that KOH is a strong base while HCl is a strong acid. This fact influences the pH of the system at equivalence point.
Owing to the fact that the acid is exactly neutralized by the base; at the equivalence point of such titration, it is expected that hydrogen ions(H+) and hydroxide ions (OH-) must have reacted to form water, this leads to a final pH of 7.
How does carbonic acid work to maintain blood pH? (Select all that apply.) Check All That Apply When blood is too basic, carbonic acid can ionize to bicarbonate and H+ ions, adding H+ ions to the blood. When blood is too basic, carbonic acid can ionize to bicarbonate and H+ ions, adding H+ ions to the blood. When blood becomes too acidic, bicarbonate combines with extra H+ ions to form carbonic acid, removing H+ ions from the blood. When blood becomes too acidic, bicarbonate combines with extra H+ ions to form carbonic acid, removing H+ ions from the blood. Carbonic acid is added to the blood until the pH reaches 7.4. Carbonic acid is added to the blood until the pH reaches 7.4. Carbonic acid always lowers the blood pH to 7.4. Carbonic acid always lowers the blood pH to 7.4. Carbonic acid can raise or lower the pH of blood. Carbonic acid can raise or lower the pH of blood.
Answer:
When blood is too basic, carbonic acid can ionize to bicarbonate and H+ ions, adding H+ ions to the blood.
When blood becomes too acidic, bicarbonate combines with extra H+ ions to form carbonic acid, removing H+ ions from the blood.
Carbonic acid can raise or lower the pH of blood.
Explanation:
A buffer is a solution that resists changes to its pH when small quantities of acids or bases are added to it. The human blood serves as a buffer as it contains a buffer of carbonic acid (H2CO3) and bicarbonate anion (HCO3-) which serves to maintain blood pH between 7.35 and 7.45. Other buffering systems in blood exist such as the Hydrogen ion and oxygen gas which affects oxygen binding to haemoglobin, however the carbonic-acid-bicarbonate buffer is the most important buffer for maintaining acid-base balance in the blood.
A buffer solution is made up of an acid and its conjugate base or a base and its conjugate acid. For carbonic acid-bicarbonate buffer, carbonic acid serves as the acid while bicarbonate serves as the base. When a little quantity of a base as hydroxide ions is added to a buffer, the acid reacts with it and remove it from the solution. On the other hand, when a little quantity of an acid as hydrogen ions are added to a buffer, the conjugate base reacts with it and remove it from the solution, thus keeping the pH of the solution fairly constant.
In the carbonic acid-bicarbonate buffer:
When blood is too basic, carbonic acid can ionize to bicarbonate and H+ ions, adding H+ ions to the blood.
When blood becomes too acidic, bicarbonate combines with extra H+ ions to form carbonic acid, removing H+ ions from the blood.
Thus, carbonic acid can raise or lower the pH of blood.
Carbonic acid work to maintain blood pH as follows:
When blood is too basic, carbonic acid can ionize to bicarbonate and H+ ions, adding H+ ions to the blood.When blood becomes too acidic, bicarbonate combines with extra H+ ions to form carbonic acid, removing H+ ions from the blood.Carbonic acid can raise or lower the pH of blood.WHAT IS BUFFER SOLUTION:A buffer is a solution that resists changes to its pH when small quantities of acids or bases are added to it. A buffer is made up of an acid and its conjugate base or a base and its conjugate acid. Carbonic acid is an example of buffer that contains an acid with it's conjugate base.This means that, carbonic acid works to maintain blood pH as follows:
When blood is too basic, carbonic acid can ionize to bicarbonate and H+ ions, adding H+ ions to the blood.When blood becomes too acidic, bicarbonate combines with extra H+ ions to form carbonic acid, removing H+ ions from the blood.Carbonic acid can raise or lower the pH of blood.Learn more about buffers at: https://brainly.com/question/24188850
Which chemical equation is balanced as written?
PLS ANSWER FAST
Answer:
THERE ANSWER IS C
Explanation:
How many valance electrons does He need to get to 8
Answer:
Any element in group 18 has eight valence electrons (except for helium, which has a total of just two electrons
HELP ME PLEASE I AM STRUGGLING
maybe umm think so pick think so
During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
What is the amount of the leftover reactant?
0.74 grams of methane
0.89 grams of methane
1.22 grams of oxygen
1.45 grams of oxygen
Answer:d
Explanation:
In order to use the Ideal Gas Constant of 0.0821, what units must be used for volume, pressure, amount, and temperature?
Answer:
The value becomes 0.0821 L-atm/K⁻¹-mol⁻¹.
Explanation:
The value of gas constant is 0.0821.
We know that, the ideal gas law is as follows :
[tex]PV=nRT\\\\R=\dfrac{PV}{nT}[/tex]
Where
P is pressure
V is volume
T is temperature
In order to use the Ideal Gas Constant of 0.0821, the units are follows :
Volume = Litre (L)
Pressure = atm
Amount = mol⁻¹
Temperature = K⁻¹
So, the value becomes 0.0821 L-atm/K⁻¹-mol⁻¹.
If you have a solution that is 15 percent by mass of KCl in benzene, what is the new boiling point?
Answer:
https://www.chegg.com/homework-help/questions-and-answers/1-10-points-solution-15-percent-mass-kcl-benzene-new-boiling-point--901-c-b-921-c-c-821-c--q63751186
Explanation: Thats your answer
All of the following are advantages of nuclear power EXCEPT: uranium mines cause less environmental damage than coal mines because less uranium is needed to generate power. nuclear power plants generate no nitrogen oxides and sulfur dioxide. uranium generates far greater amounts of energy than coal by weight or volume. the power-generating process is emission-free. nuclear wastes can be safely disposed of.
Answer:
nuclear wastes can be safely disposed of
Explanation:
In comparing fossil fuel power plants with nuclear power plants, it is obvious that fossil fuel power plants lead to a large volume of emission of oxides of carbon and sulphur.
Also, a larger volume of coal needs to be burnt to generate energy compared to a minute amount of uranium fuel that can sustain a nuclear power plant for a long period of time thereby reducing the environmental damage associated with mining of the fuel.
However, the problem of nuclear waste disposal have remained a thorn in the flesh. It is often difficult to safely dispose of spent uranium fuel. This is a major disadvantage of the use of nuclear power.
Can someone please help me with this question
Explanation:
a) 3Pb(NO3)2 + Al2(SO4)3 ---> 3PbSO4 + 2Al(NO3)3
Double displacement
b) 2Cl2 + 3O2 ---> 2Cl2O3
Synthesis
c) 2Fe2O3 + 3C ---> 4Fe + 3CO2
single displacement
For the following reaction conditions select the correct statement regarding the reaction below.
OH Cl OTs OTs N TsCl W X Y Z N
a. W is the correct product.
b. X is the correct product.
c. Y is the correct product.
d. Z is the correct product since pyridine acts as a nucleophile.
e. Both X and Y are both formed in this reaction as a racemic mixture
Answer:
B
Explanation:
The appropriate diagram of the question is shown in the first image attached below.
From the diagram, we see the reaction of Cyclopentanol taking place under Tscl pyridine. We are to show the reaction mechanism and determine from the options, which appropriate product fits in.
So, from the reaction, the hydroxyl substituent reacts with Tscl where cl is being lost. This process is followed by an attack of N substituent on the pyridine with the Hydrogen atom and cleaves off for the structure to form a stable structure. The stereochemistry of the compound remains unchanged and it maintains its stick formula.
Thus, X is the appropriate and the correct product.
A cylinder of gas with a pressure of 8.0 atm is heated from 300. K to 600. K. What will the new pressure of the gas be?
Answer:
[tex]P_2=16atm[/tex]
Explanation:
Hello there!
In this case, according to the given information, we can infer we need to use the Gay-Lussac's equation it order to understand this pressure-temperature relationship as shown below:
[tex]\frac{P_1}{V_1}=\frac{P_2}{V_2}[/tex]
Thus, we solve for P2 and plug in P1, T1 and T2, to obtain:
[tex]P_2=\frac{P_1T_2}{T_1}\\\\P_2=\frac{8atm*600K}{300K}\\\\P_2=16atm[/tex]
Regards!
Hydrogen iodide can decompose into hydrogen and iodine gases. 2HI(g) H2(g) I2(g) K for the reaction is 0.016. Of 0.148 atm of HI(g) is sealed in a flask, what is the pressure of each gas when equilibrium is established
Solution :
Given :
Hydrogen iodide decomposes to hydrogen and iodine gas
[tex]$2 HI \ \ \ \Leftrightarrow \ \ \ \ H_2 \ \ \\ + \ \ I_2 $[/tex]
I 0.148 0 0
C -2a +a +a
E 0.148-2a a a
We know
[tex]$k_p=\frac{P(H_2)P(I_2)}{P(HI)^2}$[/tex]
[tex]$0.016=\frac{a^2}{(0.148-2a)^2}$[/tex]
[tex]$0.016^{1/2}=\frac{a}{0.148-2a}$[/tex]
[tex]$0.12649=\frac{a}{0.148-2a}$[/tex]
0.0187 = 1.2529 a
a = 0.0149
Therefore
P(HI) = 0.148 - 2a
= 0.148 - 2(0.0149)
= 0.1182 atm
P([tex]$H_2$[/tex]) = a
= 0.0149 atm
P([tex]$I_2$[/tex]) = a
= 0.0149 atm
Find the % composition for each element in Zinc Chlorate
Answer:
chlorine ~ 30%
zinc ~ 28%
oxygen ~ 41%
Explanation:
If a particular ore contains 56.3 % % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg k g of phosphorus?
Answer:
34.44 kg
Explanation:
First we convert 1.00 kg of phosphorus (P) into moles, using its molar mass:
1.00 kg ÷ 32 kg/kmol = 0.03125 kmol PThen we convert 0.03125 kmoles of P into kmoles of Ca₃(PO₄)₂:
0.03125 kmol P * [tex]\frac{2kmolCa_3(PO_4)_2}{2kmolP}[/tex] = 0.0625 kmol Ca₃(PO₄)₂Now we calculate the mass of 0.0625 kmoles of Ca₃(PO₄)₂:
0.0625 kmol Ca₃(PO₄)₂ * 310.18 kg/kmol = 19.39 kgFinally we calculate the required mass of the ore, using the definition of content percentage:
% content = Mass of calcium phosphate / mass of ore * 100 %56.3 % = 19.39 kg / mass of ore * 100%Mass of Ore = 34.44 kgCalculate the freezing point of a solution made by dissolving 13 g potassium sulfide in 150 g H2O Kf for water = 1.86 C/m
Answer:
[tex]T_F=-4.4\°C[/tex]
Explanation:
Hello there!
In this case, according to the given information, it is possible for us to calculate the freezing point of an aqueous solution of potassium sulfide by using the following equation:
[tex]T_F=T_{solv}-i*m*Kf[/tex]
In such a way, we firstly calculate the molality of this solution according to:
[tex]m=\frac{\frac{13g}{110.262 g/mol} }{0.150kg} =0.79m[/tex]
Finally, since the Van't Hoff's factor for K2S is 3, the freezing point of the solution turns out to be:
[tex]T_F=0\°C-3*0.79m*1.86\°C/m\\\\T_F=-4.4\°C[/tex]
Regards!
Use the nutrition label to answer the following questions.
16.How much energy is contained in
the six-cookie serving size
recommended on the label?
17.How much energy in joules is provided by eating
six cookies? (1 cal = 4.184 J; ALSO: 1 Calorie = 1
kilocalorie)
Answer:
six cookies?(1 cal _4.184 j; ALSO: 1 Calorie -1 kilocalorie)
Can someone give me an example of balancing equations with a solution that is simple?
Explanation:
[tex]H _{2}O _{2(aq)} →H _{2}O _{(l)} + O _{2}(g) \\ solution : 2 \: and\: 2[/tex]