Answer:
Explanation:
a )
Average reaction rate between 0 and 1500s
Time duration = 1500 s
moles reacted = .184 - .016 = .168 moles
Moles reacted per second = .168 / 1500
= 112 x 10⁻⁶ moles /s
b )
Average reaction rate between 200 and 1200s
Time duration = 1000 s
moles reacted = .129 - .019 = .11 moles
Moles reacted per second = .11 / 1000
= 110 x 10⁻⁶ moles /s
c )
the instantaneous rate of the reaction at t=800s
We shall assume that between 500 s and 1200 s , rate of reaction is uniform
rate between 500 and 1200
Time duration = 700 s
moles consumed = .069 - .019 = .05 moles
Rate of reaction = .05 / 700
= 71 .4 x 10⁻⁶ moles / s
This will also be instantaneous rate of reaction at t = 800 s .
How much heat must be used to raise the
temperature of 180. g water from 19° C to 96°C?
The specific heat of water is 4.18 J/gºC.
Answer:
Q = 57934.8 J
Explanation:
Given data:
Mass of water = 180.0 g
Initial temperature = 19°C
Fina temperature = 96°C
Specific heat capacity of water = 4.18 J/g.°C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 96°C - 19°C
ΔT = 77°C
Q = 180.0 g×4.18 J/g.°C×77°C
Q = 57934.8 J
You can use solid, wedged, and dashed lines to better represent the 3D structure of molecules. The solid lines represent bonds in the plane of the paper, wedged lines represent a bond coming out of the plane, and the dashed line represents a bond going back behind the plane of the paper. Draw a representation of methane that better depicts its 3D structure.
Answer:
See explanation and image attached
Explanation:
Often times, there is a need for a three dimensional representation of a molecule on paper. These three dimensional representations give us an idea of what the molecule really looks like if we were to be looking at it physically.
In order to make a three dimensional representation, we use wedged and dashed bonds. The wedged bonds are coming out of the plane of the paper towards you while the dashed bonds are going into the plane of the paper away from you.
In the image attached, you will find the three dimensional representation of the methane molecule.
How can you model the cycling of matter in the Earth system?
Answer:
The cycling of matter is important to many Earth processes and to the survival of organisms the existing matter must cycle continuously for this planet to support life Water, carbon, nitrogen, phosphorus, and even rocks move through cycles If these materials did not cycle, Earth could not support life.
Explanation:
Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.
What is Earth system?Rocks, as well as water, carbon, nitrogen, and phosphorus, go through cycles. The planet Earth could not support life if these materials did not cycle.
Subsystems exist within the Earth system. These subsystems include the exosphere, atmosphere, hydrosphere, lithosphere and geosphere, also referred to as the lithosphere, and the living environment (biosphere).
These systems are powered by energy that comes from both the Sun and the interior of the Earth. Through processes known as biogeochemical cycles, nutrients and elements also move through these systems along with energy.
Therefore, Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.
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What is an ecosystem? ?
Answer:
An ecosystem is a community or group of living organisms that live in and interact with each other in a specific environment.
Explanation:
That's the most straight forward explanation I could come up with
The total thermal energy of a system depends jointly on the temperature, total number of in the system _______________, and the _______________of the material. *
1. Molecules/State
2. Molecules/Heat
3. Matter/Heat
4. Atoms/State
Answer:
Molecules/State
Explanation:
We know from the kinetic molecular theory that matter is made up of molecules. These molecules are in constant motion at various velocities and energies.
The total thermal energy refers to the total kinetic energies of these particles. It depends on the temperature of the system, the energies of the molecules present in the material and the arrangement of the object's molecules (states of matter).
7. Use the concepts of relative abundance and relative weight to explain why carbon has an atomic mass of 12.011 amu when there are three isotopes of carbon weighing 12 amu, 13 amu and 14 amu. Why is the atomic mass not 13?
The uncertainties of the delta measurements and the uncertainty of the atomic weight derivedfrom the best measurement of isotopic abundances constrain the number of significant figures in theatomic-weight values of the upper and lower bounds. For carbon, the fifth digit after the decimal pointis uncertain because of the uncertainty value of 0.000 027. Therefore, the number of significant digitsin the atomic-weight value is reduced to four figures after the decimal point. The Commission may rec-ommend additional conservatism and reduce the number of significant figures further. For the lowerbound of carbon, 12.009 635 is truncated to 12.0096. For an upper bound, the trailing digit is increasedto ensure the atomic-weight interval encompasses the atomic-weight values of all normal materials. Inthe case of carbon, the upper bound is adjusted from 12.011 532 to 12.0116 to express four digits afterthe decimal point. The lower and upper bounds are evaluated so that the number of significant digits ineach is identical. If a value ends with a zero, it may need to be included in the value to express therequired number of digits. The following are examples of lower and upper atomic-weight bounds foroxygen that could be published by the Commission in its various tables.
---------------------------
Extracted from" Atomic weights of the elements 2009 (IUPAC Technical Report)"
Suppose a 500.mL flask is filled with 0.40mol of N2 and 1.0mol of NO. The following reaction becomes possible:
N2g+O2g ->2NOg
The equilibrium constant K for this reaction is 5.93 at the temperature of the flask. Calculate the equilibrium molarity of N2. Round your answer to two decimal places.
Answer:
[N₂] = 1.1M
Explanation:
Based on the chemical reaction:
N₂(g) + O₂(g) ⇄ 2 NO(g)
Equilibrium constant, K, is defined as:
K = 5.93 = [NO]² / [N₂] [O₂]
Where [] are equilibrium concentrations of each specie
As initial concentrations are:
N₂ = 0.40mol / 0.500L = 0.8M
NO = 1mol / 0.500L = 2M
The equilbrium concentrations are:
[NO] = 2M - 2X
[N₂] = 0.8M +X
[O₂] = X
Replacing:
5.93 = [2 - 2X]² / [0.8+X] [X]
5.93 = 4 - 8X + 4X² / 0.8X + X²
4.744X + 5.93X² = 4 - 8X + 4X²
1.93X² + 12.744X - 4 = 0
Solving for X:
X = -6.9M → False solution. There are no negative concentrations
X = 0.3M. Real solution.
[N₂] in equilibrium is:
[N₂] = 0.8M +0.3M
[N₂] = 1.1M
A series of dilute NaCl solutions are prepared starting with an initial stock solution of 0.100 M NaCl. Solution A is prepared by pipeting 10 mL of the stock solution into a 250-mL volumetric flask and diluting to volume. Solution B is prepared by pipeting 25 mL of solution A into a 100-mL volumetric flask and diluting to volume. Solution C is prepared by pipeting 20 mL of solution B into a 500-mL volumetric flask and diluting to volume. What is the molar concentration of NaCl in solutions A, B and C
Answer:
Solution A: 0.00400M
Solution B: 0.00400M
Solution C: 4.00x10⁻⁵M
Explanation:
Solution A is diluting the 0.100M NaCl from 10mL to 250mL. That is:
250mL / 10mL = 25 times.
That means molar concentration of sln A is:
0.100M / 25 = 0.00400M
Solution B is obtained diluting 25mL to 100mL:
100mL / 25mL = 4 times
0.00400M / 4 times = 0.00100M
And solution C is obtained diluting the solution C from 20mL to 500mL:
500mL / 20mL = 25 times
Solution C:
0.00100M / 25 times = 4.00x10⁻⁵M
The formula for serial dilution can be used to obtain the molarity of solution A, B , C.
For solution AM1V1 = M2V2
M2 = 0.100 M × 10 mL/250-mL
M2 = 0.004 M
For solution BM1V1 = M2V2
M2 = 0.004 M × 25 mL/100-mL
M2 = 0.001 M
For solution CM1V1 = M2V2
M2 = 0.001 M × 20 mL/500-mL
M2 = 0.00004 M
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what is the formula for H-H
Answer:
H-H equation is written as follows:
pH=pK + log
{HCO3-}(base)
{H2CO3}(acid)
Two volumes of nitric oxide react with one volume of oxygen gas to form two volumes of a reddish-brown gas. Deduce the formula of this gas and sketch particle representations of its molecules.
Answer:
Explanation:
Nitric oxide is the gas NO, it reacts with oxygen as shown below;
2NO(g) + O2(g) -----> 2NO2(g)
Now the gas formed is the gas NO2 which is known to be reddish brown in colour.
A diagrammatic representation of this reaction is shown in the image attached to this answer.
Image credit: Chemlibretext
Which statement correctly describes ionic bonds? Multiple Choice An ionic bond only forms between two atoms of the same element. Ionic bonds usually form between electrically neutral, stable atoms. An ionic bond is the electrostatic force that holds ions together when they form bonds. All of the answer choices are correct.
Answer:
An ionic bond is the electrostatic force that holds ions together when they form bonds
Explanation:
An ionic bond is formed when a metal looses electron(s) to a non metal leading to the formation of a positive ion and a negative ion.
An ionic compound is actually an ion pair, the ions are held together by strong electrostatic forces.
This strong electrostatic force that holds the ion pair together in ionic compounds is what we commonly call the IONIC BOND.
Answer:
An ionic bond is the electrostatic force that holds ions together when they form bonds
Explanation:
I took this test and it was the correct answer :)
16. Using the average atomic masses given in the inside front cover of this book, calculate the indicated quantities.
d. the number of moles of cobalt represented by 5.99 x 1021 cobalt atoms e. the mass of 4.23 mol of cobalt
f. the number of cobalt atoms in 4.23 mol of cobalt
g. the number of cobalt atoms in 4.23 g of cobalt
Answer:
d. 9.95 × 10⁻³ mol
e. 249 g
f. 2.55 × 10²⁴ atoms
g. 4.32 × 10²² atoms
Explanation:
d. the number of moles of cobalt represented by 5.99 x 10²¹ cobalt atoms
We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.
5.99 x 10²¹ atoms × 1 mol/6.02 × 10²³ atoms = 9.95 × 10⁻³ mol
e. the mass of 4.23 mol of cobalt
The molar mass of cobalt is 58.93 g/mol.
4.23 mol × 58.93 g/mol = 249 g
f. the number of cobalt atoms in 4.23 mol of cobalt
We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.
4.23 mol × 6.02 × 10²³ atoms/1 mol = 2.55 × 10²⁴ atoms
g. the number of cobalt atoms in 4.23 g of cobalt
First, we will calculate the moles of cobalt using the molar mass of cobalt.
4.23 g × 1 mol/58.93 g = 0.0718 mol
Then, we will calculate the number of cobalt atoms using Avogadro's number.
0.0718 mol × 6.02 × 10²³ atoms/1 mol = 4.32 × 10²² atoms
Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of the visible spectrum. Suppose a particular cone cell absorbs light with a wavelength of 519.nm. Calculate the frequency of this light. Round your answer to 3 significant digits.
Answer:
5.78 × 10¹⁴ Hz
Explanation:
Step 1: Given and required data
Wavelength of this light (λ): 519. nmFrecquency of this light (ν): ?Speed of light (c): 3.00 × 10⁸ m/sStep 2: Convert "λ" to meters
We will use the conversion factor 1 m = 10⁹ nm.
519. nm × 1 m/10⁹ nm = 5.19 × 10⁻⁷ m
Step 3: Calculate the frecquency of this light
We will use the following expression.
c = λ × ν
ν = c/λ
ν = (3.00 × 10⁸ m/s)/5.19 × 10⁻⁷ m
ν = 5.78 × 10¹⁴ s⁻¹ = 5.78 × 10¹⁴ Hz
The gas carbon dioxide is a pure substance. Which of the following is true about carbon dioxide? (5 points)
Select one:
a. Carbon and oxygen are chemically bonded in it.
b. Carbon and oxygen retain their original identity in it.
c. It can be separated into carbon and oxygen using physical methods.
d. The proportion of carbon and oxygen is different in different samples of the gas.
Answer:
Carbon and oxygen are chemically bonded in it.
Explanation:
The other answer choices do not apply for compounds, but rather for mixtures instead.
definition of solubility
(science)
Answer:
th relative ability of a solute to devolve into a solvent
Determine where each type of cleaning solution should be discarded after use. Solvent used to rinse chemicals out of a beaker ______Acid solution used to clean a crucible _________Water used to rinse detergent out of a flask ________
Answer:
Acidic solution used to clean a crucible
Explanation:
This liquid dissolves alcoholic solvents such as crucible, that is why it was selected as the ideal for cleaning.
Although it would be ideal to know in detail which chemical compound is the one you want to clean so that the cleaning technique has better effectiveness.
WASTE CONTAINER refers to the solvent used to rinse chemicals out of a beaker, and it also refers to the acid solution used to clean a crucible. Water used to rinse the detergent out of a flask refer to the SINK.
The disposal containers in a lab can be used for recycling, disposal of trash, glassware disposal box, sharp box, etc.A waste container is a container usually used to dispose of waste in a laboratory, which may be made of plastic.Moreover, a laboratory sink can be used to wash tools and/or hands without the hazard of damaging the health or the sink.In conclusion, WASTE CONTAINER refers to the solvent used to rinse chemicals out of a beaker, and it also refers to the acid solution used to clean a crucible. Water used to rinse the detergent out of a flask refer to the SINK.
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An unknown object has a mass of 150 grams and a volume of 5 cm3. What is the density of this
object?
Answer:
1.2 cm
Explanation:
The reason how I got my answer is by dividing 150 by 125 and that gave me 1.2 cm.
is C5H10 ionic or covalent?
Why is observational evidence important in an experiment?
Answer:
Observational evidence is essential for investigating the way disease affects populations, the patterns and distribution of risk within them, and the emergence of trends in health and disease over time.
Answer:
It tests a prediction It supports the results. It asks a testable question It predicts what will happen
Explanation:
A container holds 100.0 mL of nitrogen at 21° C and a pressure of 736 mm Hg. What will be its volume if the temperature increases by 35° C?
Answer:
V₂ = 104.76 mL
Explanation:
Given data:
Initial volume = 100.0 mL
Initial temperature = 21°C (21 + 273.15 K = 294.15 K)
Final temperature = 35°C (35 + 273.15 K = 308.15 k)
Final volume = ?
Solution:
Charles Law:
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ =100.0 mL × 308.15 K / 294.15 K
V₂ = 30815 mL.K /294.15 K
V₂ = 104.76 mL
1. Each substance written to the right of the arrow in a chemical equation is a
(1 point)
O catalyst
O reactant
O precipitate
O product
Answer: product
Explanation:
Each substance written to the right of the arrow in a chemical equation is referred to as a product.
When writing a chemical equation, the substance that's written to the left of arrow in the equation is the reactants.
On the other hand which is the right side is the product.
An element has five isotopes. Calculate the atomic mass of this element using the information below. Show all your work. Using the periodic table, identify the element this is likely to be and explain your choice. (18 pts)
A) Isotope 1 – mass: 64 amu; percent abundance: 48.89%
B) Isotope 2 – mass: 66 amu; percent abundance: 27.81%
C) Isotope 3 – mass: 67 amu; percent abundance: 4.11%
D) Isotope 4 – mass: 68 amu; percent abundance: 18.57%
E) Isotope 5 – mass: 70 amu; percent abundance: 0.62%
Answer: Sol:-
Data provided in the question is :-
Atomic mass of isotope -1 = 64 amu
Atomic mass of isotope -2 = 66 amu
Atomic mass of isotope -3 = 67 amu
Atomic mass of isotope -4 = 68 amu
Atomic mass of isotope - 5 = 70 amu
Percentage abundace of isotope - 1 = 48.89 %
Percentage abundance of isotope -2 = 27.81 %
Percentage abundance of isotope - 3 = 4.11%
Percentage abundance of isotope-4 = 18.57%
Percentage abundance of isotope - 5 = 0.62 %
Formula used :-
Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]
Calculation :-
Put all the value in the formula :-
Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu
= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu
= {(6528.04) / 100} amu
= 65.2804 amu
Average atomic mass of an element is = 65.2804 amu
Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc
atomic mass of zinc = 65.38 \approx 65.2804 amu
How many cm 3 are in 0.014 in 3? (1 in = 2.54 cm)
Answer:
0.229 cm³.
Explanation:
The following data were obtained from the question:
Volume (in in³) = 0.014 in³
Volume (in cm³) =?
1 in = 2.54 cm
Next, we shall determine a conversion scale to convert from in³ to cm³. This can be obtained as follow:
1 in = 2.54 cm
Therefore,
1 in³ = 2.54³ cm³
1 in³ = 16.387 cm³
Finally, we shall convert 0.014 in³ to cm³. This can be obtained as follow:
1 in³ = 16.387 cm³
Therefore,
0.014 in³ = 0.014 in³ × 16.387 cm³ / 1 in³
0.014 in³ = 0.229 cm³
Thus, 0.014 in³ is equivalent to 0.229 cm³.
A compound is made of 6.00 grams of oxygen, 7.00 grams of nitrogen, and 20.00grams of hydrogen. Find the percent composition of the compound.
A O-18.18%, N-21.21%, H-60.60%
B O-11.18%, N-22.21%, H-69.60%
C O-20%, N-30%, H-50%
D O-60.60%, N-21.21%, H-18.18%
The percent composition of the compound.
A O-18.18%, N-21.21%, H-60.60%
Further explanationGiven
6.00 grams of oxygen,
7.00 grams of nitrogen,
20.00 grams of hydrogen.
Required
The percent composition
Solution
Total mass :
= mass of O + mass of N + mass of H
= 6 + 7 + 20
= 33 g
% O = 6/33 x 100%= 18.18%
% N = 7/33 x 100%=21.21%
% H = 20/33 x 100% = 60.6 %
a sample of fully saturated clay weighs 1350 g in its natural state and 975 g after drying. what is the natural water content of the soil
Answer:
The soil has a percentage of water by mass of 27.8 %.
Explanation:
Keeping in mind that
Mass of Clay = Mass of water + Mass of Dry Soilwe can calculate the mass of water:
1350 g = Mass of Water + 975 gMass of Water = 375 gWe can then calculate the mass percentage of water in the soil:
375 / 1350 * 100% = 27.8 %Find the mass in grams of 1.38 moles of Sr
Answer:
116.78 grams.
Explanation:
1 mol of Strontium (Sr) = 87.62 grams
1.38 mol of Strontium = x
Cross Multiply
1 * x = 1.38 * 87.62
x = 116.78 grams
A state of matter where the particles that make up a substance start to break apart
Answer:
Liquid
Explanation:
Which of the following has the largest atomic radius?
Answer:
b
Explanation:
According to the Michaelis-Menten equation, when an enzyme is combined with a substrate of concentration s (in millimolars), the reaction rate (in micromolars/min) is
Answer:
The answer is "A"
Explanation:
Please find the complete question in the attachment file.
[tex]\to R(s)= \frac{As}{K+s}[/tex]
when the s in the approach, that is infinity R(s) tends
[tex]\to \frac{A}{\frac{K}{s}+1} \\\\ \to\frac{A}{0+1} \\\\ \to\frac{A}{1} \\\\ \to A[/tex]
A chemist is studying the rate of the Haber synthesis: N2 + 3H2 2NH3
Starting with a closed reactor containing 1.25 mol/L of N2 and 0.50 mol/L of H2, the chemist finds that the H2 concentration has fallen to 0.25 mol/L in 44 seconds.What is the N2 concentration after 44 seconds?
Answer:
1.17 M
Explanation:
Step 1: Write the balanced equation
N₂ + 3 H₂ ⇒ 2 NH₃
Step 2: Calculate the rate of disappearance of H₂
We will use the following expression.
rH₂ = - Δ[H₂]/t = - (0.25 M - 0.50 M)/44 s = 0.0057 M/s
Step 3: Calculate the rate of disappearance of N₂
The molar ratio of N₂ to H₂ is 1:3.
0.0057 mol H₂/L.s × 1 mol N₂/3 mol H₂ = 0.0019 mol N₂/L.s
Step 4: Calculate the final concentration of N₂
We will use the following expression.
[N₂] = [N₂]₀ - rN₂ × t
[N₂] = 1.25 mol/L - 0.0019 mol/L.s × 44 s
[N₂] = 1.17 M
