An SDOF system is one whose motion is governed by a single, second-order differential equation. Only two variables, position and velocity are needed to describe the trajectory of the system. Many structures can be idealized as single degree-of-freedom systems.
The LDAP Data Interchange Format is a standard plain text data interchange format for representing LDAP directory content and update requests. LDIF conveys directory content as a set of records, one record for each object.
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:
Give the structure that corresponds to the following molecular formula and H1 NMR spectrum: The proton N M R shows a triplet at 0.9 p p m integrating to 11238 and a quartet at 1.4 integrating to 7473. THere is a singlet at 1.35 p p m integrating to 1275 that disappears after a D 2 O shake. The unknown compound has a molecular formula of C 7 H 16 O. The magenta numbers in the spectrum are the relative integrals in arbitrary units. Draw the structure.
Answer:
Kindly check the attached picture for the diagram of the chemical compound.
Explanation:
So, the following parameters were given from the question above;
=> A triplet at 0.9 ppm and a quartet at 1.4 ppm, a singlet at 1.35 ppm. Now, the unknown compound has a molecular formula of C7H16O.
For a triplet at 0.9ppm, there are nine (9) atoms of hydrogen, for the quartet at 1.4ppm there are six(6) atoms of hydrogen and for the singlet at 1.35 ppm, the number of hydrogen atoms is one(1). Hence, the total number of hydrogen atoms = 16.
Therefore, number of bondings = [(2 × number of carbon atoms) + 2 - number of hydrogen atoms present on the compound)/2 .
Thus number of bonds =[( 2× 7) + 2 - 16] ÷ 2 = 0.
Hence, there is no double bond or ring in the compound.
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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Given:
BO3 + 3Mg → 3MgO + B
1. How many moles of Mg are required if 2 moles of B are produced?
Answer:
6moles of Mg
Explanation:
The reaction expression is given as:
BO₃ + 3Mg → 3MgO + B
Given:
Number of moles of B produced = 2 moles
Unknown:
Number of moles of Mg required = ?
Solution:
From the balanced reaction expression:
1 mole of B is produced from 3 moles of Mg
2 moles of B will be produced from 2 x 3 = 6moles of Mg
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
Which rule states that cracks on glass tend to form at a certain angle on the opposite side of the polnt of Impact?
The
rule states that cracks tend to form at a (n)
angle on the opposite side from the point of Impact.
Answer:
The 3R rule states that cracks tend to form at a (n) right angle on the opposite side from the point of impact.
Explanation:
if LiCl and MgBr2 were to enter into a chemical reaction which member of MgBr2 will chlorine replace?
Answer:
It replaces Br in MgBr2.
Explanation:
Hello!
In this case, when carrying out double displacement reactions, we need to take into account that cations and anions are switched; thus, when LiCl and MgBr2 react, since Li and Mg are the cations, Br and Cl switch as shown below:
[tex]2LiCl+MgBr_2\rightarrow 2LiBr+MgCl_2[/tex]
Thus, chlorine replace Br in MgBr2 to form MgCl2.
Best regards!
For the reaction of ammonia (NH3) with oxygen (O2) to produce water and nitric oxide (NO), how many moles of water are produced when 2.2 moles of ammonia are reacted?
Answer:
3.3 moles of H₂O.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
4NH₃ + 5O₂ —> 6H₂O + 4NO
From the balanced equation above,
4 moles of NH₃ reacted to produce 6 moles of H₂O.
Finally, we shall determine the number of mole of H₂O produced by the reaction of 2.2 moles of NH₃. This can be obtained as follow :
From the balanced equation above,
4 moles of NH₃ reacted to produce 6 moles of H₂O.
Therefore, 2.2 moles of NH₃ will react to produce = (2.2 × 6)/4 = 3.3 moles of H₂O.
Thus, 3.3 moles of H₂O were obtained from the reaction.
A sample of PCl5 weighting 2.69 gram was placed in 1.00 Litter container and completely vaporized at 250C. The pressure observed at that temperature was 1.00 atm. The possibility exists that some of the PCl5 dissociated according to PCl5 (g) ! PCl3 (g) Cl2 (g) . What must be the partial pressures of PCl5 PCl3 and Cl2 under these experimental conditions
Answer:
Partial pressures:
PCl₅ = 0.558 atm
PCl₃ = 0.22 atm
Cl₂ = 0.22 atm
Explanation:
From the given information:
The number of moles of PCl₅ associated with the evaporation is:
[tex]n_{PCl_5}= \dfrac {weight \ of \ PCl_5} {M.Wt. \ of \ PCl_5}[/tex]
[tex]n_{PCl_5}= \dfrac {2.69 \ g} {208.5 \ g/mol}[/tex]
[tex]n_{PCl_5}= 0.013 \ mol[/tex]
Temperature of the gas = 250° C = (250 + 273.15) K
= 523.15 K
Using the Ideal gas equation to determine the pressure exerted by the completely vaporized PCl₅
PV = nRT
[tex]P = \dfrac{nRT}{V}[/tex]
[tex]P = \dfrac{0.0013 \ mol \times 0.082 \ Latm^0 K^{-1} . mol ^{-1} \times 523.15 \ K}{1.0 \ L}[/tex]
P = 0.558 atm
Thus, at 250° C, decomposition of PCl₅ occurs.
In the container, PCl₅ decomposes to PCl₃ and Cl₂.
i.e.
[tex]PCl_{5(g)} \to PCl_{3(g)}+ Cl_{2(g)}[/tex]
Using Dalton's Law:
[tex]P_{total } =P_1 + P_2+P_3 +...[/tex]
[tex]P_1 = P_{Total} \times X_1[/tex]
where;
X = mole fraction
Then, the total no. of moles in the container is:
[tex]n = \dfrac{PV} {RT}[/tex]
[tex]n = \dfrac{1\ atm \times 1.0\ L}{0.0821 \ L \ atm \ K^{-1}.mol \times 523.15\ K}[/tex]
n = 0.023 mol
Now, the container contains a total amount of 0.023 mol where initially 0.013 mol are that of PCl₅ and remaining 0.005 mol of PCl₃ and 0.005 mol of Cl₂.
Thus, the partial pressure of PCl₃ is:
[tex]P__{PCL_3} }= P_{total} \times \dfrac{no. \ of \ moles \ of PCl_5}{total \ no. \ of \ moles}[/tex]
[tex]P__{PCL_3}} = 1 \ atm \times \dfrac{0.005}{0.023}[/tex]
[tex]P__{PCL_3}} = 0.22 \ atm[/tex]
Thus, since the no of moles of PCl₃ and Cl₂ are the same, then the partial pressure for Cl₂ is = 0.22 atm
A tank of nitrogen has a volume of 14.0 L and a pressure of 1 atm. Find the volume of the nitrogen when its pressure is changed to 0.8 atm while the temperature is held constant .
Answer:
17.5 L
Explanation:
Step 1: Given data
Initial volume of the tank of nitrogen (V₁): 14.0 LInitial pressure of nitrogen (P₁): 1 atmFinal volume of the tank of nitrogen (V₂): ?Final pressure of nitrogen (P₂): 0.8 atmStep 2: Calculate the final volume of the nitrogen
We have a gas that undergoes a change at a constant temperature. If we assume an ideal behavior, we can calculate the final volume of the nitrogen using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 1 atm × 14.0 L/0.8 atm = 17.5 L
What is the maximum mass of ammonia (NH3) that can be produced from the synthesis reaction of 781 g of nitrogen and 656 g hydrogen?
HINT: You will need to write a balanced chemical equation and perform 2 calculations before you can determine the answers.
water is a unique material in that the density of the solid is lower than the density of the liquid (which is why ice forms at the top of a pond and why ice floats in our drinks). if the density for ice at 0C is .917g/mL and the density for water at 0C is .999 g/mL, what is the calculated free space (as %) for each of these materials. you will need to estimate the volume of water as the sum of 2 H atoms and 1 O atom with radii 37 and 66 pm respectively. note that you will also have to assume a quantity of water to perform this exercise
Answer:
% Free space in water = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%
% Free space in ice = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%
Explanation:
As given ,
Density for ice at 0⁰C = 0.917 g/ml
Density for water at 0⁰C = 0.999 g/ml
Radii of H atoms = 37 pm
Radii of O atoms = 66 pm
Now,
Consider 1 ml of water = 1 cm²
As , we know that mass of water in 1 cm² = 0.999 g
Moles of water = [tex]\frac{0.999}{18} = 0.056[/tex]
Volume of H₂O = 1.624×[tex]10^{-31}[/tex] m²
Now,
Volume occupied by water = 0.056×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²
= 5.48×[tex]10^{-9}[/tex] m²
⇒Volume occupied by water = 5.48×[tex]10^{-9}[/tex] m²
Now,
Free space = 1×[tex]10^{-6}[/tex] - 5.48×[tex]10^{-9}[/tex] = 9.95×[tex]10^{-7}[/tex] m²
% Free space = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%
Now,
Consider 1 ml of ice = 1 cm²
S.I unit of ice = 1×[tex]10^{-6}[/tex] m²
As , we know that mass of water in 1×[tex]10^{-6}[/tex] m² = 0.917 g
Moles of ice = [tex]\frac{0.917}{18} = 0.012[/tex]
Volume of H₂O = 6.022×[tex]10^{23}[/tex] ×0.012
Volume of ice unit = [tex]\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}[/tex]
Now,
Volume occupied by water = 0.012×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²
= 1.17×[tex]10^{-9}[/tex] m²
⇒Volume occupied by water = 1.17×[tex]10^{-9}[/tex] m²
Now,
Free space = 1×[tex]10^{-6}[/tex] - 1.17×[tex]10^{-9}[/tex] = 9.98×[tex]10^{-7}[/tex] m²
% Free space = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%
Which option is the basic unit of water, a compound?
A. A water atom
B. A hydrogen atom
C. A hydrogen molecule
D. A water molecule
Answer:
D. A water molecule
Explanation:
identify which element is oxidized and which element is reduced.
PLZ HELPP...
How many moles are in 25g of NaCI?
Nacl = (. 23+35. 5)
= 58.5g
.
1 mol of Nacl = 58.5g
X mol Of Nacl = 25g
X Mol of Na Cl =25 ÷ 58.5
X mol of Nacl = 0.4 mol
Ammonium phosphate is an important ingredient in many solid fertilizers. it can be made by reacting aqueous phosphoric acid with liquid ammonia. calculate the moles of ammonium phosphate produced by the reaction of 0.085 mol of ammonia. be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Answer:
0.028 mole of ammonium phosphate, (NH₄)₃PO₄.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
H₃PO₄ + 3NH₃ —> (NH₄)₃PO₄
From the balanced equation above,
3 moles of NH₃ reacted to produce 1 mole of (NH₄)₃PO₄.
Finally, we shall determine the number of mole of (NH₄)₃PO₄ produced by the reaction of 0.085 mole of ammonia, NH₃. This can be obtained as follow:
From the balanced equation above,
3 moles of NH₃ reacted to produce 1 mole of (NH₄)₃PO₄.
Therefore, 0.085 mole of NH₃ will react to produce = (0.085 × 1)/3 = 0.028 mole of (NH₄)₃PO₄.
Thus, 0.028 mole of ammonium phosphate, (NH₄)₃PO₄ were obtained from the reaction.
Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that liters per second of dinitrogen are consumed when the reaction is run at and the dinitrogen is supplied at . Calculate the rate at which ammonia is being produced. Give your answer in kilograms per second. Be sure your answer has the correct number of significant digits.
The question is incomplete. Here is the complete question.
In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous ammonia. This reaction is now the first step taken to make most of the world's fertilizer. Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that 505. liters per second of dinitrogen are consumed when the reaction is run at 172.°C and 0.88 atm. Calculate the rate at which ammonia is being produced. Give your answer in kilogram per second. Be sure your answer has the correct number of significant digits.
Answer: Rate = 0.41 kg/s
Explanation: The balanced Haber reaction is
[tex]N_{2}+3H_{2}\rightarrow2NH_{3}[/tex]
As all the components are gases, we can use Ideal Gas Law, which relates Pressure (P), Volume (V), Temperature (T) and Moles (n) in the following formula:
PV = nRT
where
R is gas constant and, in this case, is R = 0.082 L.atm.K⁻¹mol⁻¹
T is in Kelvin
Converting Celsius in Kelvin:
T = 273 + 172
T = 445 K
Calculating moles
[tex]n=\frac{PV}{RT}[/tex]
[tex]n=\frac{0.88(505)}{0.082(445)}[/tex]
n = 12.18 moles
According to the balanced equation, for 1 mol of dinitrogen gas consumed, 2 moles of ammonia is produced.
With 12.18 moles of dinitrogen, the reaction will result in
2(12.18) = 24.36 moles of ammonia
Molar mass of ammonia is M = 17.031 g/mol.
In 24.36 moles, there are
[tex]m=n.M[/tex]
m = 24.36.17.031
m = 414.87 grams
Since it's asking in kilograms: m = 0.41 kg.
In the beginning, it is said that dinitrogen gas is consumed at a rate of liters per second. So, the production rate of ammonia will be 0.41 kg/s.
0
Which is not one of Earth's layers?
A А
crust
B)
inner core
mantle
D
ocean
The ocean is not a part of Earth's layers.
Answer:
Ocean
Explanation:
In the reaction 2 HgO --> 2 Hg + O2
how many moles of O2 are produced when 5 moles of HgO are decomposed?
Answer:
2.5 moles of oxygen are produced.
Explanation:
Given data:
Number of moles of O₂ produced = ?
Number of moles of HgO decomposed = 5 mol
Solution:
Chemical equation:
2HgO → 2Hg + O₂
now we will compare the moles of HgO and O₂.
HgO : O₂
2 : 1
5 : 1/2×5 = 2.5
Thus, from 5 moles of HgO 2.5 moles of oxygen are produced.
A change of state is a(n)
process.
A. irreversible
B. reversible
Answer:
Changes of states are reversible, you can go from a solid to liquid and liquid to solid.Answer:
Reversible
Explanation:
Changes of state are physical changes in matter. Common changes of the state include melting, freezing, sublimation, deposition, condensation, and vaporization.
Which of the following elements is NOT a metal?
A) arsenic
B) indium
C) iridium
D) lithium
E) lutetium
Answer:
arsenic don't mind does it have to be longer
The statement, that describes the elements is not a metal is "arsenic.
"What is a metal?Metals are shiny, opaque elements that transfer heat and electricity well. They are malleable and easily lose electrons to produce positive ions (cations).
Arsenic is the third element in the periodic table's fifteenth column. It is a metalloid or semi-metal because it possesses chemical and physical properties that are transitional between a metal and a non-metal.
Hence the correct option is A.
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What volume (in L) of water vapor will be
produced from the reaction of 24.65 L of oxygen?
2C2H6(9) + 702(9) — 4CO2(g) + 6H20(9)
Enter
Answer:
21.13 L
Explanation:
Step 1: Write the balanced equation
2 C₂H₆(g) + 7 O₂(g) ⇒ 4 CO₂(g) + 6 H₂O(g)
Step 2: Determine the appropriate volume ratio
Since all the gases are in the same container at the same temperature and pressure, the volume ratio is equal to the molar ratio, because the volume depends on the number of moles. The volume ratio of O₂(g) to H₂O(g) is 7:6.
Step 3: Determine the volume of H₂O produced from 24.65 L of O₂
24.65 L O₂ × 6 L H₂O/7 L O₂ = 21.13 L H₂O
When of a certain molecular compound X are dissolved in of benzene , the freezing point of the solution is measured to be . Calculate the molar mass of X. If you need any additional information on benzene, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to the correct number of significant digits.
The question is incomplete. Here is the complete question.
When 2.10 g of a certain molecular compound X are dissolved in 65.0 g of benzene (C₆H₆), the freezing point of the solution is measured to be 3.5°C. Calculate the molar mass of X. If you need any additional information on benzene, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to 2 significant digits.
Answer: MM = 47.30 g/mol.
Explanation: There is a relationship between freezing point depression and molality. With this last one, is possible to calculate molar mass or molar weight of a compound.
Freezing Point Depression occurs when a solute is added to a solvent: the freezing point of the solvent decreases when a non-volatile solute is incremented.
Molality or molal concentration is a quantity of solute dissolved in a certain mass, in kg, of solvent. Its symbol is m and it's defined as
[tex]m=\frac{moles(solute)}{kg(solvent)}[/tex]
Freezing point depression and molal are related as the following:
[tex]\Delta T_{f}=K_{f}.m[/tex]
where
[tex]\Delta T_{f}[/tex] is freezing point depression of solution
[tex]K_{f}[/tex] is molal freezing point depression constant
m is molality
Now, to determine molar mass, first, find molality of the mixture:
[tex]\Delta T_{f}=K_{f}.m[/tex]
[tex]m=\frac{\Delta T_{f}}{K_{f}}[/tex]
For benzene, constant is 5.12°C/molal. Then
[tex]m=\frac{3.5}{5.12}[/tex]
m = 0.683 molal
Second, knowing the relationship between molal and moles of solute, determine the last one:
[tex]m=\frac{moles(solute)}{kg(solvent)}[/tex]
[tex]mol(solute)=m.kg(solvent)[/tex]
mol(solute) = 0.683(0.065)
mol(solute) = 0.044 mol
The definition for Molar mass is the mass in grams of 1 mol of substance:
[tex]n(moles)=\frac{m(g)}{MM(g/mol)}[/tex]
[tex]MM=\frac{m}{n}[/tex]
In the mixture, there are 0.044 moles of X, so its molecular mass is
[tex]MM=\frac{2.1}{0.044}[/tex]
MM = 47.30 g/mol
The molecular compound X has molecular mass of 47.30 g/mol.
Please help with whatever this is and go in number order
what state of matter travels in straight lines
Answer:
light
Explanation:
light is plasma, which is a state of matter
What is the mass of 1.75 moles of Ca(H2C302)2?
Explanation:
first we have to find molar mass of ca(H2c3o2)2
40+(1*2)2+(12*3)2+(16*2)2
40+4+72+64=180g/mole
m=n*Mm
m=1.75mole*180g/mole
m=315g
helpppp nowww plssss nowww!!
A student dissolves of aniline in of a solvent with a density of . The student notices that the volume of the solvent does not change when the aniline dissolves in it. Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits.
Answer:
Molarity: 0.21M
Molality: 0.20m
Explanation:
...dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL...
To solve this question, we need to find the moles of aniline in 3.9g using its molar mass. Then, we need to find the kg and Liters of solution in order to find molarity (Moles/L solution) and molality (Moles/kg of solvent):
Moles aniline:
Molar mass:
6C: 6* 12.01g/mol = 72.06g/mol
7H: 7*1.008g/mol = 7.056g/mol
N: 1*14.007g/mol = 14.007g/mol
72.06g/mol+7.056g/mol+14.007g/mol = 93.123g/mol
Moles of 3.9g: 3.9g * (1mol / 93.123g) = 0.04188moles
Liters solution:
200mL * (1L / 1000mL) = 0.200L
kg solvent:
200mL * (1.05g/mL) * (1kg/1000g) = 0.210L
Molarity:
0.04188mol / 0.200L = 0.21M
Molality:
0.04188mol / 0.210L =0.20m
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.
What is wrong with the following electron configuration?
Answer:
a. without completing 2p stuff...electrons cant go to 3s. the correct configuration is
1s2, 2s2, 2p6, 3s1
b. after 4s, 3d comes not 4d. the correct configuration is 1s2, 2s2, 2p6, 3s2, 4s2, 3d6
c. after 4s, it is 3d and then 4p the correct configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6.
Hope it helps you ｡◕‿◕｡