Answer:
Solution temperature.
Explanation:
Hello there!
In this case, considering this question about chemical kinetics, it will be possible for us to analyze two perspectives:
1. Molecular: here, we infer that the solution temperature will provide more energy to the molecules in order to undergo more effective crashes which will make more products and therefore, increase the rate constant.
2. Mathematical: by means of the Arrhenius equation, it will be possible to tell that the increase in the temperature of the system, the negative of the exponent present in such equation will increase and therefore turn the rate constant bigger.
In such way, we infer the answer is solution temperature.
Regards!
Could someone please help me out???
Answer:
Time is 2.2 seconds.
Explanation:
Time:
[tex]{ \boxed{ \bf{time = \frac{distance}{speed} }}}[/tex]
Substitute into the formula:
speed = 715 km/h = 198.61 m/s
[tex]{ \tt{time = \frac{435}{198.61} }} \\ { \tt{time = 2.2 \: seconds}}[/tex]
The reversible reaction 2H2 CO <------> CH3OH heat is carried out by mixing carbon monoxide and hydrogen gases is a closed vessel under high pressure with a suitable catalyst . After equilibrium is established at high temperature and pressure, all three substances are present. If the pressure on the system is lower, with the temperature kept constant, what will be the result
Answer:
The amount of CH3OH present in the mixture would decrease
Explanation:
According to Le Cha-telier's principle, when a reaction is in equilibrium and one of the constraints that influence the rate of reactions is applied, the equilibrium would shift so as to neutralize the effects of the constraint.
In this case, looking at the equation of the reaction:
2H2 + CO <------> CH3OH + heat
the total number of moles on the reactant's (left hand) side is 3 (2+1) while on the product's (right hand) side, it is 1. If the pressure on the system is increased, more CH3OH (and less of H2 and CO) will be produced because its side has the lower number of moles out of the two sides.
If the pressure on the system is otherwise lowered, more of H2 and CO would be produced while the amount of CH3OH present would gradually decrease.
The standard free energy that is required for the sodium-potassium ATPase to pump two K ions into the cell and three Na ions is 43.8 kJ/mol but the standard free energy change of hydrolysis of ATP is only -32 kJ/mol. This apparent imbalance of free energy can be accounted for because ________.
Answer:
Explanation:
This apparent disparity of the free energy can be taken into account because:
the free energy produced by the hydrolysis of one ATP is adequate enough under psychological circumstances.
The Na-K ATPase aids the pumping of Na+ ions out of the cell and K+ ions into the cell. These actions occurring against their potential(concentration) gradients, which may be produced by hydrolyzing one ATP molecule.
What is the trend in electropositivity in group 1 elements?
Answer:
Electro positivity increases down the group
If 50.0 g of sulfuric acid and 40.0 grams of barium chloride are mixed, how many grams of sulfuric acid and how many grams of barium chloride remain after the double replacement reaction is complete?
After the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.
First, we will write the balanced equation for the reaction
H₂SO₄ + BaCl₂ → BaSO₄ + 2HCl
This means 1 mole of BaCl₂ is needed to react completely with 1 mole of H₂SO₄ to give 1 mole of BaSO₄ and 2 moles of HCl
From the question, 50.0g of sulfuric acid is mixed with 40.0 grams of barium chloride. To determine the quantity of each substance remaining after the complete reaction, we will first determine the number of moles present in each of the reactant.
For H₂SO₄
mass = 50.0g
Molar mass = 98.079 g/mol
From the formula
Number of moles = Mass / Molar mass
∴ Number of moles of H₂SO₄ = 50.0g / 98.079 g/mol
Number of moles of H₂SO₄ = 0.5098 mol
For BaCl₂
mass = 40.0 g
Molar mass = 208.23 g/mol
∴ Number of moles of BaCl₂ = 40.0g / 208.23 g/mol
Number of moles of BaCl₂ = 0.1921 mol
Since the number of moles of H₂SO₄ is more than that of BaCl₂, then H₂SO₄ is the excess reagent and BaCl₂ is the limiting reagent (that is, it will be used up completely during the reaction)
From the equation, 1 mole of H₂SO₄ is needed to completely react with 1 mole of BaCl₂
∴ 0.1921 mol of H₂SO₄ will be needed to completely react with 0.1921 mol of BaCl₂.
Therefore, after the reaction is complete, 0 mole (i.e 0 grams) of BaCl₂ will remain and (0.5098 mole - 0.1921 mole) of H₂SO₄ will remain.
Number of moles H₂SO₄ that will remain = 0.5098 mole - 0.1921 mole = 0.3177 moles
Now, we will convert this to grams
From the formula
Mass = Number of moles × Molar mass
Mass of H₂SO₄ that will remain = 0.3177 moles × 98.079 g/mol
Mass of H₂SO₄ that will remain = 31.1597 g
Mass of H₂SO₄ that will remain ≅ 31.16 g
Hence, after the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.
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Is Water and kerosine a mixture
Answer:
No.Kerosene oil and water do not mix with each other and form two separate layers.
Answer:
No
Explanation:
They cannot be mixed together they will form upper and lower layer
3 why does soldium produce blue colour when dissolve in ammonia?
Answer:
Because it is dissolving and has pigment.
Explanation:
Answer:
The solvated electron is responsible for a great deal of radiation chemistry. Alkali metals dissolve in liquid ammonia giving deep blue solutions which conduct electricity . The blue colour of the solution is due to ammoniated electrons which absorb energy in the visible region of light.
The density of an aqueous solution containing 25.0 percent of ethanol (C2H5OH) by mass is 0.950 g/mL. (a) Calculate the molality of this solution. m (b) Calculate its molarity. M (c) What volume of the solution would contain 0.275 mole of ethanol
Answer:
a. 7.24m
b. 5.15M
c. 53.4mL of the solution would contain this amount of ethanol.
Explanation:
Molality, m, is defined as the moles of solute (ethanol, in this case) per kg of solvent.
Molarity, M, are the moles of solute per kg of solvent
To solve this question we need to find the moles of solute in 100g of solution and the volume using its density as follows:
a. Moles ethanol -Molar mass: 46.07g/mol-:
25g ethanol * (1mol/46.07g) = 0.54265 moles ethanol
kg solvent:
100g solution - 25g solute = 75g solvent * (1kg / 1000g) = 0.075kg
Molality:
0.54265 moles ethanol / 0.075kg = 7.24m
b. Liters solution:
100g solution * (1mL / 0.950g) = 105.3mL * (1L / 1000mL) = 0.1053L
Molarity:
0.54265 moles ethanol / 0.1053L = 5.15M
c. 0.275 moles ethanol * (1L / 5.15moles Ethanol) = 0.0534L =
53.4mL of the solution would contain this amount of ethanol
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 6.90kg of water at 34.7 degrees C . During the reaction 57.1kJ of heat flows out of the bath and into the flask.
Calculate the new temperature of the water bath. You can assume the specific heat capacity of water under these conditions is 4.18J.g^(-1).K^(-1) . Round your answer to significant digits.
Answer:
[tex]T_2= 36.7 \textdegree C[/tex]
Explanation:
Mass of Water [tex]m_w=6.90kg[/tex]
Temperature [tex]T=34.7 degrees[/tex]
Heat Flow [tex]H=57.1kJ[/tex]
Specific heat capacity of water [tex]\mu= 4.18J.g^(-1).K^(-1)[/tex]
Generally the equation for Final Temperature is mathematically given by
[tex]M*\mu *T_1 + Q = M*\mu *T_2[/tex]
[tex]T_2=\frac{M*\mu *T_1 + Q }{M*\mu}[/tex]
Therefore
[tex]T_2=\frac{6.90*4.18*34.7 + 57.1}{6.90*4.18}[/tex]
[tex]T_2= 36.7 \textdegree C[/tex]
Congratulations! You are now the head biologist at the local "Cells and Bells" research lab! It has come to other cell biologists' attention recently that some cells are too small to contain all of the organelles inside of them. They decide that it's best to get rid of an organelle, but they're not sure which one. In the first process of this decision, they need to know "which organelle is the most important?"
Your job, as the head cell biologist, is to decide which organelle the cell cannot live without.
Write a research paper (intro, body, and conclusion) on which organelle is the most important and why.
Answer EIGHT questions.
1(a) Whai do you mean by generation of computer? Describe brieíly
5
about third and fourth generations of computer.
B
moga dele
Answer:
Generation in computer terminology is a change in technology a computer is/was being used. Initially, the generation term was used to distinguish between varying hardware technologies. Nowadays, generation includes both hardware and software, which together make up an entire computer system.
5 generation of computer
Main electronic component: based on artificial intelligence, uses the Ultra Large-Scale Integration (ULSI) technology and parallel processing method.
ULSI – millions of transistors on a single microchip
Parallel processing method – use two or more microprocessors to run tasks simultaneously.
Language – understand natural language (human language).
Power – consume less power and generate less heat.
Speed – remarkable improvement of speed, accuracy and reliability (in comparison with the fourth generation computers).
Size – portable and small in size, and have a huge storage capacity.
Input / output device – keyboard, monitor, mouse, trackpad (or touchpad), touchscreen, pen, speech input (recognise voice / speech), light scanner, printer, etc.
Example – desktops, laptops, tablets, smartphones, etc.
3 generation of computer
Main electronic component – integrated circuits (ICs)
Memory – large magnetic core, magnetic tape / disk
Programming language – high level language (FORTRAN, BASIC, Pascal, COBOL, C, etc.)
Size – smaller, cheaper, and more efficient than second generation computers (they were called minicomputers).
Speed – improvement of speed and reliability (in comparison with the second generation computers).
Input / output devices – magnetic tape, keyboard, monitor, printer, etc.
Examples – IBM 360, IBM 370, PDP-11, UNIVAC 1108, etc.
4 generations of computer
Main electronic component – very large-scale integration (VLSI) and microprocessor.
VLSI– thousands of transistors on a single microchip.
Memory – semiconductor memory (such as RAM, ROM, etc.)
RAM (random-access memory) – a type of data storage (memory element) used in computers that temporary stores of programs and data (volatile: its contents are lost when the computer is turned off).
ROM (read-only memory) – a type of data storage used in computers that permanently stores data and programs (non-volatile: its contents are retained even when the computer is turned off).
Programming language – high level language (Python, C#, Java, JavaScript, Rust, Kotlin, etc.).
Problem 7 (Diffusion due to viscosity) If the viscosity of a solution is quadrupled, the rms-average distance of a collection of diffusing molecules from their starting point would be _________ over the same amount of time.
Answer:
1/2 the distance
Explanation:
If the viscosity of a solution is quadrupled then the distance of collection of diffusing molecules would be half over the same amount of time. The viscosity of the molecules is dependent on density of the liquid. It is independent to the volume of the liquid.
An analytical chemist is titrating of a solution of hydrazoic acid with a solution of . The of hydrazoic acid is . Calculate the pH of the acid solution after the chemist has added of the solution to it.
Answer:
pH = 12.43
Explanation:
...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it.
To solve this question we need to know that hidrazoic acid reacts with KOH as follows:
HN3 + KOH → KN3 + H2O
Moles KOH:
0.5716L * (0.2900mol /L) =0.1658 moles of KOH
Moles HN3:
0.2127L * (0.6800mol/L) = 0.1446 moles HN3
As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:
0.1658 moles - 0.1446 moles =
0.0212 mol KOH
In 212.7mL + 571.6mL = 784.3mL = 0.7843L
The molarity of KOH = [OH-] is:
0.0212 mol KOH / 0.7843L = 0.027M = [OH-]
The pOH is defined as -log [OH-]
pOH = -log 0.027M
pOH = 1.57
pH = 14 - pOH
pH = 12.43
Help me with these please
Answer:
Help you with what hmm I do not know what you are talking about
A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.60 mL of a 0.400 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
A beaker with 120mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M. A student adds 6.60mL of a 0.300M HCl solution to the beaker. How much will the pH change?
The pKa of acetic acid is 4.76.
Chemistry Buffer Calculations
1 Answer
Stefan V.
May 8, 2016
Δ
pH
=
0.29
Explanation:
!! LONG ANSWER !!
The idea here is that you need to use the Henderson-Hasselbalch equation to determine the ratio that exists between the concentration of the weak acid and of its conjugate base in the buffer solution.
Once you know that, you can use the total molarity of the acid and of the conjugate base to find the number of moles of these two chemical species present in the buffer.
So, the Henderson-Hasselbalch equation looks like this
∣
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
pH
=
p
K
a
+
log
(
[
conjugate base
]
[
weak acid
]
)
a
a
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
In your case, you have acetic acid,
CH
3
COOH
, as the weak acid and the acetate anion,
CH
3
COO
−
, as its conjugate base. The
p
K
a
of the acid is said to be equal to
4.76
, which means that you have
pH
=
4.76
+
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
The pH is equal to
5
, and so
5.00
=
4.76
+
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
=
0.24
This will be equivalent to
10
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
=
10
0.24
which will give you
[
CH
3
COO
−
]
[
CH
3
COOH
]
=
1.74
This means that your buffer contains
1.74
times more conjugate base than weak acid
[
CH
3
COO
−
]
=
1.74
×
[
CH
3
COOH
]
Now, because both chemical species share the same volume,
120 mL
, this can be rewritten as
n
C
H
3
C
O
O
−
120
⋅
10
−
3
L
=
1.74
×
n
C
H
3
C
O
O
H
120
⋅
10
−
3
L
which is
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
n
C
H
3
C
O
O
−
=
1.74
×
n
C
H
3
C
O
O
H
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
(
1
)
So, the buffer contains
1.74
times more moles of acetate anions that of acetic acid.
Now, the total molarity of the buffer is said to be equal to
0.1 M
. You thus have
[
CH
3
COOH
]
+
[
CH
3
COO
−
]
=
0.10 M
Once again, use the volume of the buffer to write
n
C
H
3
C
O
O
H
120
⋅
10
−
3
L
+
n
C
H
3
C
O
O
−
120
⋅
10
−
3
L
=
0.1
moles
L
This will be equivalent to
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
n
C
H
3
C
O
O
−
+
n
C
H
3
C
O
O
H
=
0.012
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
(
2
)
Use equations
(
1
)
and
(
2
)
to find how many moles of acetate ions you have in the buffer
1.74
⋅
n
C
H
3
C
O
O
H
+
n
C
H
3
C
O
O
H
=
0.012
n
C
H
3
C
O
O
H
=
0.012
1.74
+
1
=
0.004380 moles CH
3
COOH
This means that you have
n
C
H
3
C
O
O
−
=
1.74
⋅
0.004380 moles
n
C
H
3
C
O
O
−
=
0.007621 moles CH
3
COO
−
Now, hydrochloric acid,
HCl
, will react with the acetate anions to form acetic acid and chloride anions,
Cl
−
H
Cl
(
a
q
)
+
CH
3
COO
−
(
a
q
)
→
CH
3
COO
H
(
a
q
)
+
Cl
−
(
a
q
)
Notice that the reaction consumes hydrochloric acid and acetate ions in a
1
:
1
mole ratio, and produces acetic acid in a
1
:
1
mole ratio.
Use the molarity and volume of the hydrochloric acid solution to determine how many moles of strong acid you have
∣
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
c
=
n
solute
V
solution
⇒
n
solute
=
c
⋅
V
solution
a
a
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
In your case, this gets you
n
H
C
l
=
0.300 mol
L
−
1
⋅
volume in liters
6.60
⋅
10
−
3
L
n
H
C
l
=
0.001980 moles HCl
The hydrochloric acid will be completely consumed by the reaction, and the resulting solution will contain
n
H
C
l
=
0 moles
→
completely consumed
n
C
H
3
C
O
O
−
=
0.007621 moles
−
0.001980 moles
=
0.005641 moles CH
3
COO
−
n
C
H
3
C
O
O
H
=
0.004380 moles
+
0.001980 moles
=
0.006360 moles CH
3
COOH
The total volume of the solution will now be
V
total
=
120 mL
+
6.60 mL
=
126.6 mL
The concentrations of acetic acid and acetate ions will be
[
CH
3
COOH
]
=
0.006360 moles
126.6
⋅
10
−
3
L
=
0.05024 M
[
CH
3
COO
−
]
=
0.005641 moles
126.6
⋅
10
−
3
L
=
0.04456 M
Use the Henderson-Hasselbalch equation to find the new pH of the solution
pH
=
4.76
+
log
(
0.04456
M
0.05024
M
)
pH
=
4.71
Therefore, the pH of the solution decreased by
Δ
pH
=
|
4.71
−
5.00
|
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
0.29 units
a
a
∣
∣
−−−−−−−−−−−−−
Answer link
Related topic
Buffer Calculations
Questions
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What information does the first quantum number of an electron give?
A. The sublevel that the electron is in
B. The specific orbital the electron is in
C. The energy level the electron is in
D. The spin that the specific electron has
Answer:
с
Explanation:
the first quantum number of an electron gives the information about the energy level the electron is in
The information first quantum number of an electron give is the energy level the electron is in.
What are quantum numbers?Quantum numbers is a set of symbols which gives idea about the position of electron present inside an atom.
First quantum number is denoted by symbol 'n' which gives idea about the number of shell or energy level in which electron is present.The sublevel that the electron is in is the second quantum number denoted by symbol 'l'.The specific orbital in which electron is present is denoted by symbol 'm' and it is the third quantum number.The spin that the specific electron has is the forth quantum number and denoted by symbol 's'.Hence first quantum number of electron gives ideal about the energy level.
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The rate constant of an SN1 reaction depends on the nucleophile b. The rate constant of an SN2 reaction does not depend on the nucleophile c. SN1 reactions proceed via carbocation intermediates d. The SN2 mechanism does not involve an intermediate
Answer:
The rate constant of an SN1 reaction depends on the nucleophile
The rate constant of an SN2 reaction does not depend on the nucleophile
Explanation:
Let us recall that in an SN1 reaction, the rate determining step involves only the alkyl halide substrate and not the nucleophile. Hence;
Rate = k[RX]
Therefore;
k= Rate/[RX]
For an SN2 reaction, the rate determining step involves both the nucleophile and the alkyl halide substrate.
Hence;
Rate = k[Nu-] [RX]
k= Rate/[Nu-] [RX]
Note that;
[Nu-] = concentration of the nucleophile
[RX] =concentration of alkyl halide substrate
k= rate constant
We can see from the above derivations that;
1) The rate constant of an SN1 reaction does not depend on the nucleophile
2) The rate constant of an SN2 reaction depends on the nucleophile
7.7 cm
9.8 cm
0.00
0.162 m
Answer:
Volume = 1222.5cm³
Explanation:
If the question is about the volume of the rectangle:
The volume of a rectangle is obtained by the multiplication of its 3 dimensions: Length, width, height.
In the problem, the length of the rectangle is 0.162m = 16.2cm
The width is 7.7cm
And the height is 9.8cm
The volume is:
Volume = 16.2cm*7.7cm*9.8cm
Volume = 1222.5cm³