Answer:
0.0258 mol Answer .......
Tengo que resolver con procedimiento
1. Un automóvil viaja a una velocidad de 100 Km/h durante dos horas. Calcular la distancia recorrida.
2. Un automóvil viaja a una velocidad de 68 Km/h durante tres horas. Calcular la distancia recorrida.
3. Un auto recorre 154 Km en dos horas. ¿Cuál fue su velocidad?
4. Un auto recorre 1500 Km, si lo hace a una velocidad de 70 Km/h, ¿cuánto tardo en hacerlo?.
Answer:
Distance = 200 km
Distance = 204 km
Speed = 77 km/h
Time = 21.42 h
Explanation:
Given:
A.
Speed = 100 km/h , Time = 2 h
Find:
Distance
B.
Speed = 68 km/h , Time = 3 h
Find:
Distance
C.
Distance = 154 km , Time = 2 h
Find:
Speed
D.
Distance = 1500 km speed = 70 km/h
Find:
Time
Computation:
Speed = distance / time
A.
Distance = 100 x 2
Distance = 200 km
B.
Distance = 68 x 3
Distance = 204 km
C.
Speed = 154 / 2
Speed = 77 km/h
D.
Time = 1500 / 70
Time = 21.42 h
10 points Part F
Identify at least two limitations of the simulation. How do these limitations influence the
simulation's ability to model the synthesis of ammonia?
Answer and Explanation:
When using simulations, it is important that the researcher knows some limitations. This is because the simulations do not behave exactly like what they represent, in addition to not presenting very important details for analysis and without approximations of what really happens with what they are representing. When modeling the synthesis of Ammonia, for example, these limitations can cause inaccuracies that will compromise the entire analysis.
When using simulations, it is important that the researcher knows some limitations. This is because the simulations do not behave exactly like what they represent, in addition to not presenting very important details for analysis and without approximations of what really happens with what they are representing. When modeling the synthesis of Ammonia, for example, these limitations can cause inaccuracies that will compromise the entire analysis.
How might the shoot systems and the digestive system in a human be similar? Help please
Answer: The both Deal with digesting and getting rid of waste
Explanation:
Digestive system deals with getting rid of waste in your body and breaking down food.
The digestive system in a human is similar to the shoot system in a plant because both systems are involved in the transport of food materials and waste products in the organism.
The plant is composed of the root system and shoot system. The root system is beneath the ground while the shoot system is above the ground.
In humans, the digestive system carries food materials and waste from the alimentary canal to the small intestine. In plants, the shoot system also performs the same function of carrying food materials and waste products.
Hence, In the shoot system in plants performs a similar function to the digestive system in humans.
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Krypton occupies 10.8 L at a pressure of 0.38 atm. Find the volume, in liters, of the krypton when the pressure is increased to 1.6 atm at the same temperature.
Answer:
New volume V2 = 2.565 L
Explanation:
Given:
Old volume V1 = 10.8 L
Old pressure P1 = 0.38 atm
New presuure P2 = 1.6 atm
Find:
New volume V2
Computation:
P1V1 = P2V2
So,
10.8 x 0.38 = V2 x 1.6
New volume V2 = 2.565 L
The pain reliever morphine contains 17.900 g C, 1.680 g H, 4.225 g O, and 1.228 g N. Determine the empirical Formula.
Answer: The empirical formula is [tex]C_{17}H_{19}O_3N[/tex]
Explanation:
Mass of C= 17.900 g
Mass of H = 1.680 g
Mass of O = 4.225 g
Mass of N = 1.228 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{17.990g}{12g/mole}=1.5moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.680g}{1g/mole}=1.680moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{4.225g}{16g/mole}=0.264moles[/tex]
Moles of N =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{1.228g}{14g/mole}=0.087moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{1.5}{0.087}=17[/tex]
For H = [tex]\frac{1.680}{0.087}=19[/tex]
For O =[tex]\frac{0.264}{0.087}=3[/tex]
For N = [tex]\frac{0.087}{0.087}=1[/tex]
The ratio of C : H: O: N = 17: 19: 3: 1
Hence the empirical formula is [tex]C_{17}H_{19}O_3N[/tex]
Why do you think the reaction between tetraphenylcyclopentadienone and dimethyl acetylenedicarboxylate can be conducted at relatively lower temperature compared to the reaction between tetraphenylcyclopentadienone and diphenylacetylene?
Answer:
See explanation
Explanation:
The reaction between tetraphenylcyclopentadienone and dimethyl acetylenedicarboxylate as well as the reaction of tetraphenylcyclopentadienone and diphenylacetylene are Diels Alder reactions. The former is performed in presence of a solvent while the former is performed neat.
The reaction of tetraphenylcyclopentadienone and dimethyl acetylenedicarboxylate leads to the formation of a more resonance-stabilized aromatic ring(lower energy product) compared to the reaction of tetraphenylcyclopentadienone and diphenylacetylene.
Hence, the reaction between tetraphenylcyclopentadienone and dimethyl acetylenedicarboxylate can be conducted at relatively lower temperature compared to the reaction between tetraphenylcyclopentadienone and diphenylacetylene.
What is the difference between an introduced species and an invasive species?
Answer and Explanation:
The introduced species is, literally, a species that was introduced into an environment by human action. In other words, an introduced species is one that is not native to a region, does not occur naturally, but has been taken by humans to that region.
An invasive species, on the other hand, is one that was introduced naturally in an environment, but multiplied in a harmful way, causing a strong imbalance in the region.
How does the chemical formula relate to the number of particles that result from the dissociation of a molecule
In a technique used for surface analysis called auger electron spectroscopy (AES), electrons are accelerated toward a metal surface. These electrons cause the emissions of secondary electrons (called auger electrons) from the metal surface. The kinetic energy of the auger electrons depends on the composition of the surface. The presence of oxygen atoms on the surface results in auger electrons with a kinetic energy of approximately 520 eV .
Required:
What is the de Broglie wavelength of one of these electrons?
Answer:
the de Broglie wavelength of one of the electrons is 5.3787 × 10⁻¹¹ m
Explanation:
Given the data in the question;
we know that; ( electron volt ) 1 eV = 1.602 × 10⁻¹⁹ J
so given that Kinetic Energy KE = 520 eV
KE = 520 × 1.602 × 10⁻¹⁹ = 8.3304 × 10⁻¹⁷ J
we know that;
Kinetic Energy = 1/2 × mv²
where m is the electron mass
and mass of electron is 9.109 × 10⁻³¹ kilograms
so
8.3304 × 10⁻¹⁷ = 1/2 × 9.109 × 10⁻¹⁷× v²
v = √[(8.3304 × 10⁻¹⁷) / (1/2 × 9.109 × 10⁻³¹)]
v = 1.3524 × 10⁷ m/s
speed of the electron v = 1.3524 × 10⁷ m/s
now, de Broglie equation;
wavelength λ = h / mv
where h is Planck constant ( 6.626 × 10⁻³⁴ m² kg / s )
so we substitute
wavelength λ = 6.626 × 10⁻³⁴ / ( 9.109 × 10⁻³¹ × 1.3524 × 10⁷)
wavelength λ = 5.3787 × 10⁻¹¹ m
Therefore, the de Broglie wavelength of one of the electrons is 5.3787 × 10⁻¹¹ m
How much water needs to be added to 100 mL of a 20%w solution of copper (II) sulfate to prepare a 0.01-M solution?
Answer:
12430mL of water must be added
Explanation:
To solve this question we need to convert the 20%w of CuSO₄ to molarity. Then, using the diulution factor we can find the amount of water required:
20g CuSO₄ / 100mL * (1mol / 159.609g CuSO₄) = 0.1253 moles / 100mL = 0.1253mol / 0.1L =
1.25M is the concentration of CuSO₄. To dilute this concentration to 0.01M, the dilution factor must be of:
1.25M / 0.01M = 125 times must be diluted the solution.
As the volume of the concentrated solution is 100mL, the total volume of the solution to have a 0.01M solution must be of:
100mL * 125 times = 12530mL is the final volume of the solution. That means the amount of water added must be of:
12530mL - 100mL =
12430mL of water must be addedCalculate the volume of concentrated HCl (12 M) needed to convert the sodium benzoate back to benzoic acid.
The question is incomplete, the complete question is;
Calculate the volume of concentrated HCL (12 M) needed to convert 1.5 g of sodium benzoate back to benzoic acid.
Answer:
8.33 * 10^-4 L
Explanation:
Equation of the reaction;
C6H5COONa + HCl -------->>>> C6H5COOH + NaCl
Number of moles of sodium benzoate = mass/molar mass
molar mass of sodium benzoate =144.11 g/mol
mass of sodium benzoate = 1.5 g
Number of moles of sodium benzoate = 1.5g/144.11 g/mol
Number of moles of sodium benzoate = 0.01 moles
Since the reaction is 1:1, 0.01 moles of HCl reacted.
but;
n = CV
n = number of moles
C = concentration
V = volume
V = n/C
V = 0.01 moles/12 M
V = 8.33 * 10^-4 L
The proteins of all living things require
geometry of amino acids.
As the building blocks of proteins , amino acids are linked to almost every life process, but they also have key roles as precursor compounds in many physiological processes.
Answer:
L
Explanation:
1) _PA+ __02 → _P,03
19. 1 mole = _____________________ _________________ g
Answer:
344.1g
Explanation:
What is the half‑life of an isotope that decays to 25% of its original activity in 85.8 h
Answer:
42.9 h
Explanation:
From the question given above, the following data were obtained:
Original amount (N₀) = 100%
Amount remaining (N) = 25%
Time (t) = 85.8 h
Half-life (t½) =?
Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:
Original amount (N₀) = 100%
Amount remaining (N) = 25%
Number of half-lives (n) =?
N = 1/2ⁿ × N₀
25 = 1/2ⁿ × 100
Cross multiply
25 × 2ⁿ = 100
Divide both side by 25
2ⁿ = 100/25
2ⁿ = 4
Express 4 index form with 2 as the base
2ⁿ = 2²
n = 2
Thus, two half-lives has elapsed.
Finally, we shall determine the half-life of the of the isotope. This can be obtained as follow:
Time (t) = 85.8 h
Number of half-lives (n) = 2
Half-life (t½) =?
n = t / t½
2 = 85.8 / t½
Cross multiply
2 × t½ = 85.8
Divide both side by 2
t½ = 85.8 / 2
t½ = 42.9 h
Thus, the half-life of the of the isotope is 42.9 h
A student added several drops of a clear liquid to an unknown liquid in each of four test tubes. The results are shown above. These are all examples of a chemical reaction.
true or false
The following molecular equation represents the reaction that occurs when aqueous solutions of silver(I) nitrate and iron(III) chloride are combined
3AgNO3(aq) + FeCl3(aq) → 3AgCl(s) + Fe(NO3), (aq)
Write the balanced net ionic equation for the reaction.
Answer: [tex]3Ag^+(aq)+3Cl^-(aq)\rightarrow 3AgCl(s)[/tex]
Explanation:
Complete ionic equation : In complete ionic equation, all the substance that are strong electrolytes and are represented in the form of ions.
The complete ionic equation will be:
[tex]3Ag^+(aq)+3NO_3^-(aq)+Fe^{3+}(aq)+3Cl^-(aq)\rightarrow 3AgCl(s)+Fe^{3+}(aq)+3NO_3^-(aq)[/tex]
Net ionic equation : In the net ionic equations, we do not not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
In this equation, [tex]Fe^{3+}[/tex] and [tex]NO_3^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be:
[tex]3Ag^+(aq)+3Cl^-(aq)\rightarrow 3AgCl(s)[/tex]
1. What would be the molarity of the sodium ion in solution.
PLS HELP ME WITH THIS ILL GIVE BRAINLIEST
Answer:
1 qualiative physical
2 qualiative chemical
3 quantitative physcial
4 quantiative chemical
5qualiative physcial
6quatiative chemcial
7quanatiative physcial
8 qualiative chemical
9quatliate physcial
I have calculated 23 grams of water in a laboratory. What is the amount of molecules
present in this amount?
a. None of the above
b. 8.3x10^24 molecules
c. 6.9x10^23 molecules
d. 7.7x10^23 molecules
e. 5.6x10^25 molecules
Answer:
d. 7.7x10^23 molecules
Explanation:
Given the following data:
Mass of water (H2O) = 23g
To find the number of molecules;
First of all, we would determine the number of moles;
[tex] Number \; of \; moles = \frac {mass}{molar mass} [/tex]
Molar mass of water (H2O) = (1 * 2) + 16 = 18 g/mol
Substituting into the equation, we have
[tex] Number \; of \; moles = \frac {23}{18} [/tex]
Number of moles = 1.2778 moles
Now, to find the number of water molecules;
We know that Avogadro constant is equal to 6.02 * 10^23 mol¯¹
Number of water molecules = number of H2O moles * Avogadro constant
Substituting into the equation, we have;
Number of water molecules = 1.2778 × 6.02 * 10^23
Number of water molecules = 7.7 × 10^23 atoms.
Plz help it over science stuff
Answer:9.012
Explanation:
I used the periodic table, the mass is 9.012
Answer:
Pretty sure it would be 69(nice).
Explanation:
You add the number of protons and neutrons to find the mass
How many moles are there in 2.30 x 1024 atoms of silver?
Answer:
Step 1 should be convert atoms to moles (n). Step 2 should be convert moles (n) to mass (m).
Step 1
Use dimensional analysis to convert the number of atoms to moles.
1 mole atoms = 6.022 × 10²³ atoms
n(Ag) = 2.3 × 10²⁴ Ag atoms × (1 mol Ag/6.022 × 10²³ Ag atoms) = 3.8193 mol Ag
Step 2
Convert the moles of Ag to mass.
mass (m) = moles (n) × molar mass (M)
n(Ag) = 3.8193 mol Ag
M(Ag) = atomic weight on the periodic table in g/mol = 107.868 g Ag/mol Ag
m(Ag) = 3.8193 mol × 107.868 g/mol = 412 g Ag = 410 g Ag rounded to two significant figures
The mass of 2.3 × 10²⁴ Ag atoms is approximately 410 g.
Explanation:
The number of moles that are present is 2.3 × 10²⁴ atoms of silver are 3.8193 moles that can be determined using Avogadro's law.
The calculation for number of moles:One mole:
A mole is defined as 6.022 × 10²³ of some chemical unit, be it atoms, molecules, ions, or others.
1 mole = 6.022 × 10²³ atoms
n(Ag) = 2.3 × 10²⁴ Ag atoms × (1 mol Ag/6.022 × 10²³ Ag atoms)
n(Ag) = 3.8193 mol Ag
Thus, the number of moles is 3.8193 moles.
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I need help answering that middle question
Answer: 153
Explanation:
californium is a synthetic radioactive element found on the bottom row of the periodic table, along with other actinides.
It was first discovered in 1950 at the University of California Radiation Laboratory in Berkeley by bombarding curium with alpha particles (helium nuclei). Pure californium metal has a silvery-white appearance and is soft enough to be easily cut with razor blade.
In initial experiments with californium in its concentrated form began in 1958.
Since then, 20 isotopes of californium have been characterize, with mass numbers ranging from 237 to 256. One of these isotopes is californium-252, which was first isolated from a neutron-irradiated sample of plutonium-239. As an exclusively synthetically made element (it cannot be found in nature),
californium-252 is very rare an is produced via neutron bombardment in high flux isotope reactor or particle accelerators.
On this,page we walk through some of the properties and uses of his unique isotope and provide additional resources for your reference.
Hope this helps! :)
3
A student has a 5.00 gram sample of calcium chloride (CaCl2) solid. How many moles of calcium chloride are
contained in this sample?
Answer: 0.0450 moles of [tex]CaCl_2[/tex]
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
[tex]\text{Number of moles of} CaCl_2=\frac{5.00g}{110.98g/mol}=0.0450moles[/tex]
Thus there are 0.0450 moles of [tex]CaCl_2[/tex]
In determining the energy of activation, why was it prudent to run the slowest trial done at room temperature in the hot water bath and the fastest trial done at room temperature in the cold water bath?
Answer:In determining the energy of activation, why was it prudent to run the slowest trial done at room temperature in the hot water bath and the fastest trial done at room temperature in the cold water bath?
Explanation:
cuantas moléculas de aspartame están presente en 10.00 gramos de aspartame
Mole measure the number of elementary entities of a given substance that are present in a given sample. Therefore, 1.98×10²²molecules of aspartame are present in 10.00 grams of aspartame.
What is mole?The SI unit of amount of substance in chemistry is mole. The mole is used to measure the quantity or amount of substance. We know one mole of any element contains 6.022×10²³ atoms which is also called Avogadro number.
Mathematically,
mole =given mass ÷ molar mass
=10.00 ÷ 294.30
=0.033moles
number of molecules= Avogadro's number × number of moles
=6.022×10²³× 0.033
= 1.98×10²²molecules
Therefore, 1.98×10²²molecules of aspartame are present in 10.00 grams of aspartame.
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Does ionic or covalent form solid crystals?
Answer:
I hope it help
your welcome
Detection of iron ions Fe+3 and Fet+2
Answer:
The diffrence between Fe2+ and Fe3+ is the Fe2+has a pale green color and turns violet when water is added to it. While Fe3+ forms blood red when it reacts with thiocyanate ions.
Fe2+ has paramagnetic properties whereas
Fe3+ has diamagnetic properties.
Explanation:
I may not be correct.
Answer:
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How are algae and polyps related?
The polyps receive carbon dioxide from the algae.
The polyps receive shelter from the algae.
The polyps get most of their food from the algae.
The polyps embed themselves in the flesh of the algae.
Answer:
The polyps get most of their food from the algae.
Explanation:
You dissolve 0.64 g of potassium chloride (KCl) in 725 ml of water.
What is the molarity of the solution?
(From the periodic table: 1 mol K = 39.10 g; 1 mol Cl = 35.453 g)
Enter the value rounded to three decimal places with no units
Will give BRAINLEST
Answer:
0.012 M
Explanation:
Step 1: Given and required data
Mass of KCl (m): 0.64 gVolume of water (V): 725 mL (We will assume it is equal to the volume of the solution)Molar mass of KCl (M): 74.553 g/molStep 2: Calculate the moles (n) of solute (KCl)
We will use the following expression.
n = m/M = 0.64 g / (74.553 g/mol) = 8.6 × 10⁻³ mol
Step 3: Calculate the molar concentration (C) of the solution
The molarity is equal to the moles of solute divided by the liters of solution.
C = 8.6 × 10⁻³ mol/0.725 L = 0.012 M
