Answer:
A log burning to ashes is a chemical change.
Explanation:
Answer:
Chemical Change
Explanation:
Burning of wood is a chemical change as new substances which cannot be changed back [e.g. carbon dioxide] are formed. For example, if wood is burned in a fireplace, there is not wood anymore but ash.
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:
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 addedHow does the chemical formula relate to the number of particles that result from the dissociation of a molecule
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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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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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.
Indicate whether each structure is aromatic, nonaromatic, or antiaromatic. Assume planarity. Compound A is a 5 membered heterocyclic ring containing two pi bonds. On one of the pi bonds is a sulfur with one lone pair and a positive charge. There is an N H group between the two pi bonds with one lone pair. Compound B is a 6 membered ring with three pi bonds. One of the ring atoms is a protonated oxygen with one lone pair. Compound C is an 8 membered ring with a double bond between carbons 2 and 3, between carbons 5 and 6 and between carbons 7 and 8. There is a positive charge on carbon 4 and a negative charge and lone pair on carbon 1. Compound D is a 7 membered ring with three pi bonds. In between two alkenes is a boron attached to a hydrogen. Comppound E is a 7 membered ring with three pi bonds. In between two alkenes is a negative charge with a lone pair.
Answer:
this some next level stuff homie
Explanation:
imma just guess 7?
The thing being described is shaped like a ring with five parts. Inside the ring, there are two special types of bonds. This has a sulfur that is positively charged and has one unshared pair, also it has an NH group with one unshared pair.
What is the structureCompound B is a substance that is made up of two or more different elements or molecules.
The thing being described has a ring with six members and three pi bonds. It has an oxygen atom with an extra proton and one extra electron. This substance is not aromatic.
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Consider the reaction 2Al + 6HBr → 2AlBr3 + 3H2. If 12 moles of Al react with 12 moles of HBr, what is the limiting reactant?
Answer:
the limiting reactant is HBr
Explanation:
if you tried to make the products using 12 mol Al and 12 mol HBr, the HBr will run out first
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.
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]
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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19. 1 mole = _____________________ _________________ g
Answer:
344.1g
Explanation:
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
Excess silver(I) nitrate was added to a 8.500 g mixture containing some amount of barium chloride, and 7.123 g of silver chloride was obtained. The unbalanced equation is
AgNO3 + BaCl2(aq) --> Ba(NO3)2(aq) + AgCl(s).
What is the mass% of BaCl2 in the mixture?
Answer:
60.88%
Explanation:
The balanced equation of the reaction is given as;
2 AgNO3 (aq) + BaCl2 (aq) → 2 AgCl (s) + Ba(NO3)2 (aq)
Since AgNO3 is in excess, the limiting reactant is BaCl2. From the reaction;
1 mol of BaCl2 produces 2 mol of AgCl
Converting to masses;
Mass = Number of mol * Molar mass
BaCl2;
Mass = 1 * 208.23 g/mol = 208.23 g
AgCl;
Mass = 2 * 143.32 g/mol = 286.64 g
208.23 g BaCl2 produces 286.64 g of AgCl
x g BaCl2 produces 7.123 g of AgCl
Solving for x;
x = 7.123 * 208.23 / 286.64 = 5.1745 g
Mass percent = Mass / Total mass of Mixture * 100
Mass Percent = 5.1745 / 8.500 = 0.6088 * 100 = 60.88%
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:
A chemist determined by measurements that 0.0650 moles of gallium participated in a chemica reaction. Calculate the mass of gallium that participated in the chemica reaction
Answer:
4.53gm
Explanation: n=m/M
n is # of moles
m is mass
M is molar mass
so m=n*M when you rearrange the equation
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]
A sample of acid rain turned
universal indicator yellow. What would you expect its pH to be? Is it a strong or a weak acid?
Answer:
i think it would be 3 or lower
Answer:
A universal indicator shows a greenish-yellow colour for pH 6 and orange yellow for pH 5. This means that if the indicator turns yellow, the pH should be between these two values. Hence, it must have a pH value between 5 and 6. As the pH is closer to 7 (between 5 and 6), it is a weak acid.Explanation:
Trust the correct answer
If a solution containing 24.0 g of a substance reacts by first-order kinetics, how many grams remain after three half-lives?
Answer:
3.0g remain
Explanation:
The half-life is defined as the time required for a reactant to decrease its concentration in exactly the half of the initial amount of reactant. Having this in mind:
In one half-life, the mass will be:
24.0g / 2 = 12.0g
In two half-lifes:
12.0g / 2 = 6.0g
And in three half-lifes, the mass that remain is:
6.0g / 2 =
3.0g remain
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.
No pain no gain . which figure of speech is this
Answer:
No pain, no gain is a proverb that means in order to make progress or to be successful, one must suffer. This suffering may be in a physical or mental sense. The phrase no pain, no gain was popularized in the 1980s by the American actress, Jane Fonda.
1. What would be the molarity of the sodium ion in solution.
MnO4 - is a stronger oxidizing agent than ReO4 - . Both ions have charge-transfer (LMCT) bands; however, the charge-transfer band for ReO4 - is in the ultraviolet, whereas the corresponding band for MnO4 - is responsible for its intensely purple color. Are the relative positions of the charge-transfer absorptions consistent with the oxidizing abilities of these ions? Explain
Answer:
[tex]$MnO^-_4$[/tex] is a strong oxidizing agent.
Explanation:
The 5d orbitals of Re are higher in energy than 3 d orbitals of Mn. So an LMCT ligand to metal charge transfer excitation requires more energy of [tex]$ReO^-_4$[/tex].
Also, since the molecular orbitals are derived primarily from 3d orbitals of [tex]$MnO^-_4$[/tex] are lower in energy than the corresponding MO's of [tex]$ReO^-_4$[/tex], [tex]$MnO^-_4$[/tex] is better able to accept the electrons.
So it is a better oxidizing agent.
The ligand to metal charge transfer band in ReO4- occurs in the near UV region hence ReO4 - appears colorless.
The electron configuration of Re is Xe 4f14 5d5 6s2 and the electron configuration of Mn is [Ar] 3d5 4s2. We can see that Mn^7+ and Re^7+ have empty d orbitals.
The color of MnO4 - must result from ligand to metal charge transfer hence the purple color of MnO4 -. In the case of ReO4 -, the ligand to metal charge transfer occurs at a much higher energy owing to the fact that 5d orbitals are involved. This transition occurs in the near UV region hence ReO4 - appears colorless. The ligand to metal charge transfer in MnO4- involves lower energy 3d orbitals hence it occurs in the visible region of the spectrum.
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1) _PA+ __02 → _P,03
When a chemical reaction occurs, what happens to the atoms of the two substances?
Answer:
In a chemical reaction, only the atoms present in the reactants can end up in the products. No new atoms are created, and no atoms are destroyed. In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange and form new bonds to make the products.
Explanation:
Methyl isocyanate, shown as resonance structure 1, can also be represented by other resonance structures. Draw the next most important resonance contributor. Then add curved arrows to each structure to show delocalization of electron pairs to form the other structure.
Include lone pairs of electrons, formal charges, and hydrogen atoms. You can add condensed hydrogens using the More menu, selecting +H and clicking on the carbon as many times as needed.
Solution :
Structure I
The formal charge on both Carbon (C) atom is = 4 valance [tex]$e^-$[/tex] - bonds = 0
Formal charge (O) = 6 V.E - 2 bonds - 4 non bonding electrons = 0
Formal charge on (N) = 5 V.E - 3 bonds - 2 non bonding electrons = 0
F.C. on H = 1 V.E. - 1 bond = 0
Overall charge on the molecule = 0 charge
Structure II
Formal charge on both C atom = 4 valence [tex]$e^-$[/tex] - 4 bonds = 0
Formal charge (O) = 6 V.E. - 1 bonds - 6 non bonding electrons = -1 charge
Formal charge on (N) = 5 V.E. - 4bonds - 0 non bonding electrons = +1 charge
F.C on H = 1 V.E. - 1 bond = 0
Overall charge on the molecule = +1 -1
= 0 charge
looking at the pure subtances :water,oxygen,copper, can you tell by looking at them if it is pure or not ? explain your answer using example...
Answer:
Yes water is a pure, Water, H2O, is a pure substance, a compound made of hydrogen and oxygen. Although water is the most abundant substance on earth, it is rarely found naturally in its pure form. Most of the time, pure water has to be created. Pure water is called distilled water or deionized water.
Oxygen is a chemical element – a substance that contains only one type of atom. Its official chemical symbol is O, and its atomic number is 8, which means that an oxygen atom has eight protons in its nucleus. Oxygen is a gas at room temperature and has no colour, smell or taste
Usually by pure substances we mean either pure elements (all one type of atom), regular crystals (atoms arranged in a repeating pattern), or things made of only one type of molecule (a tightly bound structure of one or more types of atoms). So copper is a pure substance in any form (only copper atoms).
Enter the electron configuration for I+ using noble gas shorthand notation.
In the first box enter the noble gas (notice the brackets). In the following boxes enter the number that goes in front of the orbital followed by the superscript.
For example, the electron configuration for sulfur is: [Ne]3s2 3p4
so the first box would have Ne in it followed by 3, then 2, then 3 then 4.
If you do not need an orbital, just enter 0 (zero) in the boxes for the coefficient and superscript.
[ ] s f d p
Find the element in the periodic table and count over
to the right the number of negative charges on your anion.
This element has the same electron configuration as your anion.
Which noble gas precedes the element? Knowing the s, p, d, and f blocks
of elements in the periodic table, deduce the the electron configuration
of the element from the preceding noble gas. Remember that the
1p 1d, 2d, 1f, 2f, and 3f orbitals are forbidden energy levels (they do not exist).
Answer:
[Kr] 4d10 5s2 5p4
Explanation:
The Symbol I represents Iodine. It has atomic number of 53. The full electronic configuration is given as;
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p5
However the question requested for the configuration of I+.
I+ is a cation and it simply refers to an iodine atom that has lost a single electron. The electronic configuration of I+ is given as;
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p4
Using Noble gas shorthand representation, we have;
[Kr] 4d10 5s2 5p4
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:
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.
