Answer:
the fraction of collisions with energy greater than Eact
Explanation:
The activation energy of a reaction stands as a sort of energy barrier between reactants and products. It is only reactants that possesses energy greater than the activation energy that can be converted from reactants to products.
When the temperature of the system is increased, more particles acquire energy greater than the activation energy. Hence, the fraction of collisions with energy greater than the activation energy increases so the rate of reaction increases likewise.
Suppose a 250.mL flask is filled with 1.7mol of H2 and 0.90mol of I2. The following reaction becomes possible:
+H2gI2g 2HIg
The equilibrium constant K for this reaction is 5.51 at the temperature of the flask.
Calculate the equilibrium molarity of I2. Round your answer to two decimal places.
Explanation:
here's the answer. I just plug the expression into my calculator and find the intercept to avoid the quadratic formula
A saturated solution of potassium iodide contains, in each 100 mL, 100 g of potassium iodide. The solubility of potassium iodide is 1 g in 0.7 mL of water. Calculate the specific gravity of the saturated solution
Answer:
Specific gravity of the saturated solution is 2
Explanation:
The specific gravity is defined as the ratio between density of a solution (In this case, saturated solution of potassium iodide, KI) and the density of water. Assuming density of water is 1:
Specific gravity = Density
The density is the ratio between the mass of the solution and its volume.
In 100mL of water, the mass of KI that can be dissolved is:
100mL * (1g KI / 0.7mL) = 143g of KI
That means all the 100g of KI are dissolved (Mass solute)
As the volume of water is 100mL, the mass is 100g (Mass solvent)
The mass of the solution is 100g + 100g = 200g
In a volume of 100mL, the density of the solution is:
200g / 100mL = 2g/mL.
The specific gravity has no units, that means specific gravity of the saturated solution is 2
why does D2O
have a higher boiling point than H2O
Answer:
deuterium is heavier isotope of hydrogen need
high weight of atom will lead to higher boiling point
A 1 liter solution contains 0.370 M hypochlorous acid and 0.493 M sodium hypochlorite. Addition of 0.092 moles of barium hydroxide will: (Assume that the volume does not change upon the addition of barium hydroxide.)
In the original solution you have the mixture of a weak acid (Hypochlorous acid) and its conjugate base (Sodium hypochlorite). That is a buffer.
The barium hydroxide will react with hypochlorous acid. If this reaction cause the complete reaction of hypochlorous acid, the buffer break its capacity and the pH change in several units. In this case:
The addition of barium hydroxide will raise the pH slightly because the buffer still working.
The initial moles of those species are:
Hypochlorous acid:
[tex]1L * \frac{0.370mol}{1L} = 0.370 moles[/tex]
Sodium hypochlorite:
[tex]1L * \frac{0.493mol}{1L} = 0.493 moles[/tex]
Now, a strong acid as barium hydroxide (Ba(OH)₂) reacts with a weak acid as hypochlorous acid (HClO) as follows:
Ba(OH)₂ + 2HClO → Ba(ClO)₂ + 2H₂O
For a complete reaction of 0.092 moles of barium hydroxide are required:
[tex]0.092 moles Ba(OH)_2*\frac{2mol HClO}{1molBa(OH)_2} = 0.184 moles HClO[/tex]
As there are 0.370 moles, the moles of HClO after the reaction are:
0.370 moles - 0.184 moles = 0.186 moles of HClO will remain
As you still have hypochlorite and hypochlorous acid you still have a buffer.
Thus, the pH will raise slightly because the amount of acid is decreasing and slightly because the buffer can keep the pH.
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Given the following reaction:
CO (g) + 2 H2(g) <==> CH3OH (g)
In an experiment, 0.42 mol of CO and 0.42 mol of H2 were placed in a 1.00-L reaction vessel. At equilibrium, there were 0.29 mol of CO remaining. Keq at the temperature of the experiment is ________.
A) 2.80
B) 0.357
C) 14.5
D) 17.5
E) none of the above
Answer:
Option D. 17.5
Explanation:
Equiibrium is: CO + 2H₂ ⇄ CH₃OH
1 mol of CO is in equibrium with 2 moles of hydrogen in order to make, methanol.
Initially we have 0.42 moles of CO and 0.42 moles of H₂
If 0.29 moles of CO remained, (0.42 - 0.29) = 0.13 moles have reacted.
So in the equilibrium we may have:
0.29 moles of CO, and (0.42 - 0.13 . 2) = 0.16 moles of H₂
Ratio is 1:2, if 0.13 moles of CO haved reacted, (0.13 . 2) moles have reacted of hydrogen
Finally 0.13 moles of methanol, are found after the equilibrium reach the end.
Let's make expression for KC: [Methanol] / [CO] . [Hydrogen]²
0.13 / (0.29 . 0.16²)
Kc = 17.5
How is the compound NH3 classified?
A. As a salt
B. As a base
C. As an acid
D. As ionic
Answer:
B
Explanation:
Ammonia is considered a base as it's pH is 11
Answer from Gauthmath
The compound NH3 (Ammonia) can be classified as a weak Base. Below you can learn more about Ammonia.
What is Ammonia (NH3)?Ammonia is a chemical compound which is derived from the combination of Nitrogen and Hydrogen. It is denoted by the chemical formula NH3.
Ammonia is a base and when it reacts with acids to gives out salts. Physically, It is a colorless gas with a distinct characteristic of a pungent smell.
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Why do we need Chemistry in Nursing?
Answer:
We need chemistry in nursing because it deals with various kinds of drugs and the reactions of these drugs on the human body as well as with each other.
The elements present in a group of periodic table have
Similar chemical properties) Give reason and a
suitable element?
Answer:
group 1 elements(hydrogen,sodium,etc)
Explanation:
bexause if noticed all the element in the same group have the same eletron in thr outer most shell for example the group 1 elements are said to have 1 outermost elect ron which make them react so the same
What is represented by a straight line on a graph?
o the sum of the independent and dependent variables
O only the independent variable
O only the dependent variable
o the relationship between independent and dependent variable
1 2
3
4
5
Answer:
the relationship between independent and dependent variable
Explanation:
A straight line or linear graph is one of the ways to represent a given data. It shows the relationship between two given set of data; one called the independent variable is plotted on the x-axis (horizontal) while the other called the dependent variable is plotted on the y-axis (vertical).
The straighter the line is, the stronger the relationship between the two variables and vice versa. Hence, the straight line in the graph represents the relationship between independent and dependent variable.
Arrange the following in order of increasing melting point: NaCl, H2O, CH4, C6H4(OH)2.
a. NaCl < H2O < CH4 < C6H4(OH)2
b. CH4 < H2O < NaCl < C6H4(OH)2
c. CH4 < H2O < C6H4(OH)2 < NaCl
d. CH4 < C6H4(OH)2 < H2O < NaCl
e. CH4 < NaCl < C6H4(OH)2 < H2O
Explanation:
one thing to know is that higher surface area = higher boiling point.
NaCl has the smallest surface area, so it's the first one.
H2O has less surface area than methane, so it's second.
Methane has more surface area than H20, so it's third.
The big molecule has the most surface area, so it's last
Temperature measures the average kinetic energy of particles of the substances. Melting point is directly proportional to surface area. Therefore, the correct option is option C.
What is temperature?Temperature is used to measure degree or intensity of heat of a particular substance. Temperature is measured by an instrument called thermometer.
Temperature can be measured in degree Celsius °c, Kelvin k or in Fahrenheit. Temperature is a physical quantity. Heat always flow from higher temperature source to lower temperature source.
We can convert these units of temperature into one another. The relationship between degree Celsius and Fahrenheit can be expressed as:
°C={5(°F-32)}÷9
Melting point is directly proportional to surface area. NaCl has the smallest surface area. Water has less surface area than methane. Methane has more surface area than H[tex]_2[/tex]O.
Therefore, the correct option is option C.
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Calculate the amount of water (in grams) that must be added to (a) 6.80 g of urea [(NH2)2CO] in the preparation of a 9.95 percent by mass solution: g (b) 29.3 g of MgBr2 in the preparation of a 1.70 percent mass solution: g
Explanation:
Amount of water required in each case:
(a)The mass% of the solution is:9.95
Mass of solute that is urea is 6.80 g
To determine the mass of solvent water use the formula:
[tex]mass percent=\frac{mass of solute}{mass of solution} x 100\\\\9.95=(6.80g/mass of solution )x100\\mass of solution =(6.80 /9.95)x100\\=68.3 g[/tex]
Hence the mass of solvent = mass of solution - the mass of solute
=68.3 g - 6.80g
=61.5 g
Hence, the answer is mass of solvent water required is 61.5 g.
(b) Given mass%=1.70
mass of solute MgBr2 = 29.3 g
The mass of solvent water required can be calculated as shown below:
[tex]mass percent=\frac{mass of solute}{mass of solution} x 100\\\\1.70=(29.3g/mass of solution )x100\\mass of solution =(29.3 g /1.70)x100\\=1720 g[/tex]
The mass of the solution is 1720 g.
Mass of solvent water = mass of solution - mass of solute
=1720 g - 29.3 g
=1690.7 g
Answer: The mass of water required is 1690.7 g.
All of the following are characteristics of metals except: Group of answer choices good conductors of heat malleable ductile often lustrous tend to gain electrons in chemical reactions
Answer:
Hence the correct option is the last option that is tends to gain electrons in chemical reactions to become anions.
Explanation:
Metals tend to donate electrons in chemical reactions to become cations.
Too many objects inside a laboratory fume hood can disrupt the airflow and possibly compromise you safety. Which of the following are considered best practices in the use of a laboratory fume hood?
a. Open the sash as much as possible
b. Work at least 25 cm inside the hood
c. Use fast, quick movements to limit your exposure
d. Place objects to one side—work on other side
e. Use a raised along the back of the hood
Best practices for fume hoods: work 25 cm inside, organize items to one side, use raised ledge; avoid open sash and quick movements.
Laboratory fume hoods must be used safely. Workers should operate at least 25 cm within the hood to preserve ventilation and avoid dangerous chemicals. Place things on one side of the hood to preserve ventilation and prevent clogging.
A raised ledge on the rear of the hood prevents things from falling in and impeding airflow. Avoid fully opening the sash to maintain ventilation and containment. Fast, rapid motions can interrupt airflow, so prevent them. These practises guarantee the fume hood contains harmful compounds, making the lab safer. Therefore, option (B), (D) and (E) are correct.
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Avogradro's number is the number of particles in one gram of carbon- 12 atom true or false?
Answer:
True
Explanation:
The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).
Given the reactants of the chemical reaction that will take place in Part D (construction of a lead concentration cell) prior to the assembly of the cell, determine the type of chemical reaction it is. Hint: Determine the products of the reaction.
Answer:
hi
Explanation:
Which of the following is a reduction half-reaction?
Solution : An oxidation reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between tow species an oxidaion reductin reaction is any chemical reaction in which the oxidation number of a molecule atom or ion changes by gaining or losing an electron
When 1 mole of CO(g) reacts with H2O(l) to form CO2(g) and H2(g) according to the following equation, 2.80 kJ of energy are absorbed. CO(g) + H2O(l)CO2(g) + H2(g) Is this reaction endothermic or exothermic? _________ What is the value of q? kJ
State what would be observed when the following pairs of reagents are mixed in a test tube.
C6H2COOH and Na2CO3(aq)
(ii) CH3CH2CH2OH and KMnO4 /H
(iii) CH3CH2OH and CH3COOH + conc. H2SO4 (iv) CH3CH = CHCH3 and Br2 /H2O
Answer:
(i). C6H2COOH and Na2CO3(aq)
observation: Bubbles of a colourless gas (carbon dioxide gas)
(ii) CH3CH2CH2OH and KMnO4 /H
observation: The orange solution turns green.
[This is because oxidation of propanol to propanoic acid occurs]
(iii) CH3CH2OH and CH3COOH + conc. H2SO4
observation: A sweet fruity smell is formed.
[This is because an ester, diethylether is formed]
(iv) CH3CH = CHCH3 and Br2 /H2O
observation: a brown solution is formed.
How many grams of CO2 are formed if 44.7 g C5H12 is mixed with 108 g O2?
Explanation:
here's the answer to your question
Given the equation: 2C6H10(l) 17 O2(g) ---> 12 CO2(g) 10 H2O(g) MM( g/mol): 82 32 44 18 If 115 g of C6H10 reacts with 199 g of O2 and 49 g of H2O are formed, what is the percent yield of the reaction
Answer:
74%
Explanation:
Step 1: Write the balanced equation
2 C₆H₁₀(l) + 17 O₂(g) ⇒ 12 CO₂(g) + 10 H₂O(g)
Step 2: Determine the limiting reactant
The theoretical mass ratio (TMR) of C₆H₁₀ to O₂ is 164:544 = 0.301:1.
The experimental mass ratio (EMR) of C₆H₁₀ to O₂ is 115:199 = 0.578:1.
Since EMR > TMR, the limiting reactant is O₂.
Step 3: Calculate the theoretical yield of H₂O
The theoretical mass ratio of O₂ to H₂O 544:180.
199 g O₂ × 180 g H₂O/544 g O₂ = 65.8 g H₂O
Step 4: Calculate the percent yield of H₂O
%yield = (experimental yield/theoretical yield) × 100%
%yield = (49 g/65.8 g) × 100% = 74%
Answer:
Percentage yield of H₂O = 74.24%
Explanation:
The balanced equation for the reaction is given below:
2C₆H₁₀ + 17O₂ —> 12CO₂ + 10H₂O
Next, we shall determine the masses of C₆H₁₀ and O₂ that reacted and the mass of H₂O produced from the balanced equation. This is can be obtained as follow:
Molar mass of C₆H₁₀ = 82 g/mol
Mass of C₆H₁₀ from the balanced equation = 2 × 82 = 164 g
Molar mass of O₂ = 32 g/mol
Mass of O₂ from the balanced equation = 17 × 32 = 544 g
Molar mass of H₂O = 18 g/mol
Mass of H₂O from the balanced equation = 10 × 18 = 180 g
SUMMARY:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂ to produce 180 g of H₂O.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂.
Therefore, 115 g of C₆H₁₀ will react to produce = (115 × 544)/164 = 381 g of O₂.
From the calculations made above, we can see that a higher mass (i.e 381 g) of O₂ than what was given (i.e 199 g) is needed to react with 115 g of C₆H₁₀.
Therefore, O₂ is the limiting reactant and C₆H₁₀ is the excess reactant.
Next, we shall determine the theoretical yield of H₂O. This can be obtained by using the limiting reactant as shown below:
From the balanced equation above,
544 g of O₂ reacted to produce 180 g of H₂O.
Therefore, 199 g of O₂ will react to produce = (199 × 180)/544 = 66 g of H₂O.
Thus, the theoretical yield of H₂O is 66 g.
Finally, we shall determine the percentage yield. This can be obtained as follow:
Actual yield of H₂O = 49 g
Theoretical yield of H₂O = 66 g
Percentage yield of H₂O =?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield of H₂O = 49/66 × 100
Percentage yield of H₂O = 74.24%
Consider a galvanic (voltaic) cell that has the generic metals X and Y as electrodes. If X is more reactive than Y (that is, X more readily reacts to form a cation than Y does), classify the following descriptions by whether they apply to the X or Y electrode.
i. anode
ii. cathode
iii. electrons in the wire flow toward
iv. electrons in the wire flow away
v. cations from salt bridge flow toward
vi. anions from salt bridge flow toward
vii. gains mass
viii. loses mass
Answer:
X
anode
electrons in the wire flow away
anions from salt bridge flow toward
loses mass
Y
cathode
electrons in the wire flow toward
cations from salt bridge flow toward
gains mass
Explanation:
In a galvanic cell, oxidation occurs at the anode while reduction occurs at the cathode. The metal that is more reactive functions as the anode while the less reactive metal functions as the cathode.
Electrons leave the anode and travel via a wire to the cathode. At the anode cations give up electrons and enter into the solution.
At the cathode, cations pick up electrons and are deposited on the cathode leading to a gain in mass at the cathode.
Positive ions from the salt bridge flow towards the cathode while negative ions from the salt bridge flow towards the anode.
Balance the redox reaction Al(s) + MnO4^- (aq) --> MnO2 (s) + Al(OH)4^- (aq) in aqueous basic solution
Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
e. Which of the following is a mixture? i. Water ii. Hydrogen iii. Air iv. Iron
water is known as the mixture
Answer:
iv. Iron
water is not a mixture
hydrogen is the simplest element
air is pure
Could someone help me with this too? Much appreciated! I am a bit stuck.
Answer:
CO
Explanation:
1.Q= {n: 7 <n<31}, list the members of the set Q
Q={x:x[tex]\epsilon[/tex]n,7<n<31}
[tex]\\ \sf\longmapsto Q=\left\{8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30\right\}[/tex]
You can write it like this too
[tex]\\ \sf\longmapsto Q=\left\{8,9......30,31\right\}[/tex]
Calculate the mass of isoborneol in 2.5 mmol of isoborneol and the theoretical yield (in grams) of camphor from that amount of isoborneol
isoborneol = 154.25 g mol?1
Camphor, Molar mass = 152.23 g/mol
Answer:
[tex]m_{isoborneol }=0.39g\\\\m_{Camphor}=0.38g\\[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the reaction whereby isoborneol goes to camphor occurs in a 1:1 mole ratio, that is why the theoretical yield of the latter is also 2.5 mmol (0.0025 mol) but the masses can be calculated as follows:
[tex]m_{isoborneol }=0.0025mol*\frac{154.25g}{1mol} =0.39g\\\\m_{Camphor}=0.0025mol*\frac{152.23 g}{1mol} =0.38g\\[/tex]
Because of the fact this is a rearrangement reaction whereas the number of atoms is not significantly modified.
Regards!
Để xác định hàm lượng Cu trong hợp kim Cu-Zn người ta làm như sau: Hòa
tan hoàn toàn 2,068g mẫu hợp kim Cu-Zn trong lượng dư axit HNO3, thu được dung
dịch X. Đun đuổi axit dư, điều chỉnh tới pH 3 thu được 100mL dung dịch Y. Lấy
10mL dung dịch Y, thêm KI dư, rồi chuẩn độ dung dịch tạo thành bằng dung dịch
Na2S2O3 0,1M thì thấy hết 15,0 mL. Viết các phương trình phản ứng xảy ra. Tính
hàm lượng Cu trong mẫu hợp kim trên.
Low-density polyethylene is formed because _______ polymerization is very unpredictable and difficult to control.
dehydration-condensation
anionic-initiated
radical-initiated
esterification
Answer:
radical-initiated
Explanation:
Radical-initiated polymerization is unpredictable and difficult to control. The reaction proceeds indiscriminately and produces shortened chains, loops, and branches that create holes in the polymer. This reduces its mass to volume ratio.
PLEASE HELP!!
this is on USAtestprep
a)
b)
c)
d)
At elevated temperatures, hydrogen iodide may decompose to form hydrogen gas and iodine gas, as follows:
2HI(g) ⇌ H2 (g) + I2 (g)
In a particular experiment, the concentrations at equilibrium were measured to be [HI] = 0.85 mol/L, [I2] = 0.60 mol/L, and [H2] = 0.27 mol/L. What is Kc for the above reaction?
Explanation:
Since Kc is
[tex]k = \frac{(products)}{(reactants)} [/tex]
You can insert the Hydrogen and Iodine gas on top, and Hydrogen Iodide in the denominator.
Note: you can only include gases and aqueous species in an equilibrium expression, and all the species in this reaction are gaseous so you're good.
Inserting their molarity at equilibrium into their places, and you can solve. Don't forget to make the coefficient of HI turn into a power.
