Answer:
Option C, an increase of reaction rate with increase in temperature
Explanation:
Colligative properties are as follows
a) Decrease of vapor pressure
b) Increase of boiling point
c) Reduction of freezing point
d) Increase of osmotic pressure
There is no impact on reaction rate and hence it is not a colligative property.
Thus, option c is the right choice
An increase of reaction rate with increase in temperature isn't an example
of colligative properties of solutions
Colligative properties of solutions depend on the ratio of the number of
solutes to that of the solvent(concentration) and not on the nature of the
substances involved.
Examples of colligative properties include vapor pressure lowering, boiling
point elevation, freezing point depression, and osmotic pressure. Increase of
reaction rate with increase in temperature is therefore not an example of
colligative properties of solutions.
Read more on https://brainly.com/question/24260365
Plz help plz help me will mark Brainly
Answer:
Greenhouse gases absorb some of the energy and trap it in the lower atmosphere. Less heat radiates into space, and Earth is warmer. Many greenhouse gases occur naturally. Carbon dioxide, methane, water vapor, and nitrous oxide are naturally present in Earth's atmosphere. Since some of the extra energy from a warmer atmosphere radiates back down to the surface, Earth's surface temperature rises.
Hope this helped! :)
What type of reaction is illustrated?
C3H8 +502 + 3C02 + 4H2O
А
B
decomposition
combustion
single
replacement
Answer:
Combustion
Explanation:
One of the governing laws of the universe is the thermodynamic law of entropy. Which choice best describes entropy?
A. Particles and energy in a system become more and more disordered.
B. Energy is consumed in a chemical reaction.
C. Heat is evenly distributed throughout the universe.
D. More bonds are broken in a chemical reaction than bonds are formed.
A.
Particles and energy in a system become more and more disordered.
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:
can someone please do the balanced equation for potassium fluoride?
Answer:
K + F2 = KF
Explanation:
Thats the balancee equation for potassium fluoride. The equation is 2K + F2 → 2KF and the balanced equation is K + F2 = KF
explain the process of finishing
Answer:
[Your answer would be in this document of mine below] Thanks :D
Explanation:
The tertiary structure of a protein is a complex arrangement formed as the polypeptide chain folds and twists.
a. True
b. False
Answer:
True
Explanation:
The tertiary structure of a protein is a complex arrangement formed as the polypeptide chain folds and twists.
This folding & twisting of polypeptide chain leading to its complex structure, is true about tertiary structure of protein. It occurs due to different interactions between side chains of amino acids.
A 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 mmHgIn 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.
For many purposes we can treat butane as an ideal gas at temperatures above its boiling point of . Suppose the pressure on a sample of butane gas at is cut in half. Is it possible to change the temperature of the butane at the same time such that the volume of the gas doesn't change? yes no If you answered yes, calculate the new temperature of the gas. Round your answer to the nearest °C.
Answer:
A. Yes
B. The new temperature of the gas is -116 °C
Note: The question is incomplete. The complete question is given below :
For many purposes we can treat butane C H10) as an ideal gas at temperatures above its boiling point of - 1. °C. Suppose the pressure on a 500 mL sample of butane gas at 41.0°C is cut in half. Iyes Is it possible to change the temperature of the butane at the same time such that the volume of the gas doesn't change? yes no If you answered yes, calculate the new temperature of the gas. Round your answer to the nearest °C.
Explanation:
A. According to the pressure law of gases,for a fixed mass of gas the pressure of a gas is directly proportional to its Kelvin temperature once the volume is kept constant. This means that a change in temperature can bring about a change in pressurein a gas at constant volume.
B. From the pressure law of gasese: P1/T1 = P2/T2
Where initial pressure = P1, final pressure = P2
Initial temperature = T1, final temperature = T2
For the butane gas;
P1 = P
P2 = P/2
T1 = 41°C = (273 + 41 ) K = 314 K
T2 = ?
From the equation, T2 = T1 × P2 / P1
T2 = 314 × P/2 /P
T2 = 157 K
T2 = (157 - 273) °C = -116 °C
Therefore, the new temperature of the gas is -116 °C
What is the name of Na
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
How many liters of phosphine are produced when 34 L of hydrogen reacts with an excess of phosphorus under STP?
Answer:
22.67 L of PH₃
Explanation:
The balanced equation is:
[tex]P_4 (s) + 6H_2(g) \to 4PH_3(g)[/tex]
From the equation:
[tex]34 L \times \dfrac{1 \ mol \ of H_2 }{22.4 \ L \ H_2} \times \dfrac{4 \ mol \ of \ PH_3}{6 \ mol \ H_2} \times \dfrac{22.4 \ L \ PH_3}{1 \ mol \ PH_3}[/tex]
= 22.67 L of PH₃
Qué uso le darías en la vida diaria ayuda
Answer:
ako po kailangan ko po nga ayuda
Explanation:
thanks for the points ❤️
Calculate 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!
Students are asked to design an experiment with cookies to demonstrate their understanding of the scientific method. One group has decided to determine the amount of time it will take for a cookie to "dissolve" in a glass of milk.
Which variable would most likely NOT affect the time it takes for the cookie to dissolve?
a) the size of the cup of milk
b) the type of cookie
c) the number of students in the group
d) the temperature of the milk
e) the type of milk
I'm gonna guess E on this one, but I think you should choose either E or A
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!
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.
Use 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 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
The ionization potential of Be atom is more than expected because it has -
(a) half filled valence p orbitals
(b) fully filled valence s orbitals
(c) both a and b
Answer:
(b) fully filled valence s orbitals
Explanation:
Electron configuration of Be: 1s22s2
2s2 is fully filled
bakit mahalaga ang ilaw trapiko?
grade 3
Explanation:
#IHOPEIHELPYOU PA BRAINLY NA LANG POAn increase in temperature will cause an increase in diffusion rate because Multiple Choice hot molecules get smaller and therefore diffuse faster. the frequency and force of molecular collisions decreases. hot water changes to a solid, so diffusion is easier. hot water is thinner than cold water. the frequency and force of molecular collisions increases.
Answer:
hot water is thinner than cold water.
Explanation:
The rate of diffusion of a substance has a lot to do with the temperature of a body.
Let us take water for instance, cold water has a greater density than hot water. As a result of this, the molecules in cold water are slower when in motion and more sluggish.
On the other hand, the molecules of hot water are quite faster since hot water has a lower density(thinner than cold water).
Thus, an increase in temperature will cause an increase in diffusion rate because hot water is thinner than cold water.
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]
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
choose me physics subject
write a story of your life when you were hurted by someone whom you trusted blindly...
Answer:
Sis I think it happened with me but I am not able to remember if u want u can share if it happened with u
Which equation represents a transmutation?
Answer:
A
Explanation:
edge21
The lattice-like structure of a metal consists of negative metal ions in a "sea" of electrons.
O True
O False
Answer:
True
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 kg