Answer:
6 electrons are in the third level if the atom.
Which statement is true about the products of two reactants that combine chemically? A. They are different than the reactants. B. Some products are the same as the reactants. C. One product is the same as a reactant but the other is different. D. Products will spontaneously degrade back to reactants.
A kind of sedimentary rock formed from plant material over a long period of time is __________.
atp is a modified nucleotide used for cellular for energy that contains what sugar
ATP is a modified nucleotide used for cellular energy that contains the ribose sugar.
Further explanation
Ribose
It is a sort of sugar that is created by the body. It is utilized as a medicine. Ribose is utilized by mouth to diminish chest torment and improve heart work in individuals with coronary course illness.
ATP (Adenosine triphosphate)
It is an instant energy source. It is called as the energy currency of the cell. It is used as a coenzyme in many processes. ATP is synthesized during Krebs cycle and glycolysis. Mitochondria and chloroplast are involved in the production of ATP. ATP is involved in the active transport of molecules, in the processes like muscle contraction, protein synthesis and Calvin cycle. We synthesize 60 Kg of ATP per day. ATP is a large molecule but its energy lies in the terminal phosphate bond. ATP provides energy to cell by breaking its phosphate bond. By breaking one phosphate group ATP is converted to ADP molecule but it can be regenerated. ATP is also consumed by plants in the process of photosynthesis. Animals use ATP in the breakdown of carbohydrate and lipids. In fact, it is the principal molecule used in biochemical reactions.
Structure of ATP
ATP is composed of adenine which is a purine base, a ribose sugar and three phosphoric acid molecules. It also has ester linkage and high energy phosphate bond.
Phosphate bond is highly reactive bond but when it reacts with nucleoside, adenosine it becomes less reactive.
Answer details
Subject: Chemistry
Level: College
Keywords
Ribose Adenosine triphosphate Structure of ATPLearn more to evaluate
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The phosphate functional group in the non cyclic adenosine monophosphate molecule contains "acidic hydrogens".Explain what this phrase means.
Is the world really going to end Saturday
While in Europe, if you drive 119 km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per liter and your car's gas mileage is 22.0 mi/gal ? Assume that 1euro=1.26dollars.
The weekly cost of gas for driving 119 km per day in Europe, considering the car's gas mileage of 22 mi/gal and gas cost of 1.10 euros per liter, is approximately $123.45 when converted from euros.
To calculate the weekly cost of gas while driving in Europe, we need to figure out the total number of kilometers driven in a week, convert this distance to miles, and then determine how much gas would be used, taking into account the car's gas mileage, and finally convert the cost to dollars.
Firstly, the student will drive 119 km per day. Over a week, this amounts to:
119 km/day × 7 days/week = 833 km/week.
Next, we convert kilometers to miles using the conversion factor 1 km ≈ 0.621371 miles:
833 km × 0.621371 mi/km ≈ 517.781 mi/week.
Given the car's gas mileage of 22.0 mi/gal, the amount of gas needed for the week is:
517.781 mi / 22.0 mi/gal = 23.535 gal/week.
Now we convert gallons to liters since gas in Europe is sold by the liter. There are approximately 3.78541 liters in one gallon:
23.535 gal × 3.78541 liters/gal ≈ 89.071 liters/week.
With the cost of gas at 1.10 euros per liter, this totals:
89.071 liters × 1.10 euros/liter = 97.978 euros/week.
Finally, we convert euros to dollars:
97.978 euros × 1.26 dollars/euro = $123.452.
Therefore, the cost of gas for one week is approximately $123.45.
During an ige-mediated hypersensitivity reaction, which leukocyte is activated?
"how many atoms are in 169 g of calcium"
First calculate the number of moles. Molar mass of Calcium = 40.08 g/mol
Number of moles = 169 g / (40.08 g / mol) = 4.217 moles
Using Avogadros number, we can get the number of atoms:
Number of atoms = 4.217 moles * (6.022 x 10^23 atoms / moles) = 2.54 x 10^24 atoms
Which shows an electron being ejected from the atom? explain your reasoning?
Zinc Mass If a 1.85 g mass of zinc produces 475 mL of gas and your balloon weighs 0.580 g and the room temperature is 21.5°C. Calculate the amount of zinc needed to produce enough gas to get your balloon airborne by adding 1 mL to the required balloon volume so that its density is less than that of the surrounding air. (Hint: Complete balloon volume calculation as you did in Question 2 above)
Express the concentration of a 0.0610 m aqueous solution of fluoride, f–, in mass percentage and in parts per million. assume the density of the solution is 1.00 g/ml.
The solution concentration of 0.0610 m aqueous fluoride is 0.1159% by mass and 1159 ppm. These values are calculated using the molarity and molar mass of fluoride and assuming a solution density of 1.00 g/mL.
Explanation:To express the concentration of a 0.0610 m aqueous solution of fluoride (F−) in mass percentage and parts per million (ppm), given a density of 1.00 g/mL, we need to do some calculations based on the definitions of these concentration units.
First, to find the mass percentage, we use the molarity of the solution (moles of solute per liter of solution) and the molar mass of fluoride (F−, approximately 19.00 g/mol). Since the solution has a density of 1 g/mL, we assume that 1 liter of the solution weighs 1000 g.
The mass of fluoride in 1 L of 0.0610 M solution is:
0.0610 moles/L × 19.00 g/mole = 1.159 g of F−
To express this as a percentage, we divide the mass of fluoride by the total mass of the solution and multiply by 100:
(1.159 g / 1000 g) × 100 = 0.1159%
Now, to convert this to parts per million, we use the fact that 1 ppm equals 1 mg of solute per kg of solution. Considering that 1.159 g is 1159 mg, we have:
1159 mg of F− per 1000 g of solution = 1159 ppm
Therefore, the concentration of the fluoride solution is 0.1159% by mass and 1159 ppm.
How many ammonium ions and how many sulfate ions are present in an 0.370 mol sample of (nh4)2so4?
In a 0.370 mol sample of [tex](NH_4)_2SO_4[/tex], there are 0.740 mol of [tex]NH_4^+[/tex] ions and 0.370 mol of [tex]SO_4^{2-}[/tex] ions.
To determine the number of ammonium ([tex]NH_4^+[/tex]) and sulfate ([tex]SO_4^{2-}[/tex]) ions in a 0.370 mol sample of [tex](NH_4)_2SO_4[/tex], we use the stoichiometry of the compound. Each unit of ammonium sulfate contains two ammonium ions and one sulfate ion.
We calculate the number of ions as follows:
For ammonium ions: 0.370 mol of [tex](NH_4)_2SO_4[/tex]x 2 mol of [tex]NH_4^+[/tex] per mol of [tex](NH_4)_2SO_4[/tex] = 0.740 mol of [tex]NH_4^+[/tex]For sulfate ions: Since there is one sulfate ion per formula unit, we have 0.370 mol SO42-.The sample therefore contains 0.740 mol of [tex]NH_4^+[/tex] ions and 0.370 mol of [tex]SO_4^2^-[/tex] ions.
In two or more complete sentences explain how to balance the chemical equation, KClO3 ⟶ KCl + O2 and include all steps
A 0.2 m solution of a weak acid ha dissociates such that 99.4% of the weak acid remains intact (i.e., remains as ha). what is the pka of the weak acid?
The pKa of a weak acid can be determined using the equation: pKa = -log10(Ka). In this case, we are given that 99.4% of the weak acid remains intact, which means that only 0.6% has dissociated. Therefore, we can calculate the equilibrium concentration of HA by multiplying the initial concentration of HA by 0.006.
Explanation:The pKa of a weak acid can be determined using the equation: pKa = -log10(Ka), where Ka is the acid dissociation constant. In this case, we are given that 99.4% of the weak acid remains intact, which means that only 0.6% has dissociated. Therefore, we can calculate the equilibrium concentration of HA by multiplying the initial concentration of HA by 0.006.
Since the [HA] is equal to the [HA]₀ * 0.006, we can use this calculated equilibrium concentration along with the given Ka value to find the pKa.
For example, let's assume the initial concentration of HA is 0.2 M. The equilibrium concentration of HA would be 0.2 M * 0.006 = 0.0012 M. Using the equation pKa = -log10(Ka), we can find the pKa value.
Tin(ii) fluoride (snf2) is often added to toothpaste as an ingredient to prevent tooth decay. what is the mass of f in grams in 36.5 g of the compound?
The mass of fluorine in 36.5 g of SnF2 is approximately 8.85 g, calculated by finding the fraction of the mass due to fluorine in tin(II) fluoride based on its molar mass and multiplying it by the total mass of the compound.
Explanation:To calculate the mass of fluorine (F) in grams in 36.5 g of tin(II) fluoride (SnF2), we need to consider the molar mass of SnF2 and the proportion of fluorine within the compound. The molecular weight of SnF2 is approximately 156.7 g/mol (119.7 for Sn and 37.0 for the two fluorine atoms). To find the amount of F in SnF2, we can set up a ratio based on the molar masses:
Molar mass of SnF2: 156.7 g/molMolar mass of F in SnF2: (2 × 19.0) = 38.0 g/molNext, calculate the fraction of the mass that is due to fluorine:
(38.0 g/mol F) / (156.7 g/mol SnF2) = 0.2426 (fraction of F in SnF2)
Finally, multiply the fraction by the total mass of SnF2 to obtain the mass of F:
36.5 g × 0.2426 = 8.8549 g F
Therefore, the mass of fluorine in 36.5 g of SnF2 is approximately 8.85 g.
Aside from carbon and hydrogen, which are always present in organic molecules, what are some other substances they may contain?
Please answer ASAP! Will mark brainliest:
According to its periodic table entry, how many electrons does nickel have in its valence level?
a.) 2
b.) 8
c.) 16
d.) 28
Option A= 2
ExplanationAtomic Number of Nickel (Ni) = 28
Electronic configuration is 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d8
By definition, the highest principle quantum number is the valence shell. In the case of Nickel, n = 4 is the highest QN and contains 2 electrons (max allowed) in the s-orbital.
A piece of metal has a volume of 38 cm and a mass of 277 g calculate the density of the metal
The density of the metal would be equal to 277/38 = 7.28 g/cm^3
What is density?
The ratio of an object's matter content to its volume is compared using the concept of density. High density describes an entity that contains a lot of matter in a small space. The density of a substance demonstrates the substance's density in a certain region.
Mass per unit volume is the definition of density for a substance. In essence, density is an indicator of how closely together matter is arranged. It is one of an object's special physical characteristics. Greek scientist Archimedes was the first to understand the density principle.
Knowing the formula and the corresponding units will make calculating density simple. Density can be symbolized by the letter D.
Therefore, the density of the metal is 7.28 g/cm^3.
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how many reference points are needed to make a thermometer scale?
Which two of the following elements would you expect to be most similar: nitrogen, chlorine, barium, fluorine, and sulfur?
Answer:
Chlorine and Flourine
Explanation:
It is important to identify the groups in which the elements being to. The elements in each group have the same number of electrons in the outer orbital. Those outer electrons are also called valence electrons.
Since elements in a group have the same number of valence electrons, they behave similarly in chemistry.
Nitrogen - Group 5
Chlorine - Group 7
Barium - Group 2
Flourine - Group 7
Sulphur - Group 6
This means the most similar elements would be flourine and Chlorine because they are both group 7 elements.
A sugar crystal contains approximately 2.2×1017 sucrose (c12h22o11) molecules. part a what is its mass in mg?
Final answer:
The mass of a sugar crystal with 2.2×1017 sucrose molecules is found by calculating the number of moles and multiplying by the molar mass of sucrose (342.297 g/mol), then converting to milligrams.
Explanation:
To find the mass of a sugar crystal with approximately 2.2×1017 sucrose (C12H22O11) molecules, we need to use the molar mass of sucrose. The molar mass of sucrose is 342.297 grams per mole. Since one mole of a compound contains Avogadro's number (approximately 6.022×1023) of molecules, we can use the following steps to calculate the mass:
Calculate moles of sucrose: moles = number of molecules ÷ Avogadro's number = 2.2×1017 molecules ÷ 6.022×1023 molecules/mole.Convert moles to grams using the molar mass: mass in grams = moles × molar mass of sucrose.Finally, convert the mass from grams to milligrams by multiplying by 1,000, since there are 1,000 milligrams in a gram.Performing these calculations will give us the mass of the crystal in milligrams (mg).
What would the % p, on the npk ratio, be reported as for a 100g sample containing 7.5g of phosphorous?
Ans: 7.50% of P
Given:
Mass of NPK sample = 100 g
Mass of P in sample = 7.5 g
To determine:
The % P in the given sample
Explanation:
The percent of a particular substance (say,X) in a given total amount (M) is generally expressed as:
% X = [Mass of X/Total mass]*100
In this case:
%P = [mass of P/mass of NPK sample]*100
= [7.5/100]*100 = 7.5%
Determine the mass of 2.5 cups of water if the density of water is 1.00 g/cm3 and 1 cup = 240 mL\
Hydrides are compounds made up of hydrogen and metal. Why ammonia has the highest boiling point among the hydrides of elements in Group 15?
Select one:
a. The bonds in ammonia are polar
b. The size of the nitrogen atom is the smallest
c. The ammonia molecules is the smallest among the hydrides of the elements in Group 15
d. Hydrogen atoms bonded to the nitrogen atom can form hydrogen bonds
(EMG) Which sequence contains elements listed from most reactive to least?
1. Click to view
Alkali metals, alkaline earth metals, noble gases
2.
ransition metals, alkali metals, alkaline earth metals
3.Click to view
Alkaline earth metals, alkali metals, halogens
4. Click to view
Transition metals, noble gases, halogen
Alkali metals are the most reactive, followed by alkaline earth metals. Noble gases are the least reactive. Thus, the correct sequence from most to least reactive is alkali metals, alkaline earth metals, noble gases. Option 1
Explanation:In a periodic table, the reactivity of elements tends to decrease from the left to the right. Therefore, the correct sequence from most reactive to least reactive elements would be option 1: Alkali metals, alkaline earth metals, and then noble gases.
This is because alkali metals (group 1) are the most reactive elements, followed by alkaline earth metals (group 2). Noble gases (group 18) are very stable and hence the least reactive. On the other hand, transition metals are generally less reactive than alkali and alkaline earth metals, while halogens are more reactive than noble gases but less reactive than alkali and alkaline earth metals.
Option 1
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A mixture of table salt and ice is used to chill the contents of hand-operated ice-cream makers. what is the molality of salt in a mixture of 2.00 lb of nacl and 12.00 lb of ice if exactly half of the ice melts? assume that all the nacl dissolves in the melted ice.
The molality of the table salt (NaCl) in the described mixture, where 2.00 lbs of NaCl is mixed with 12.00 lbs of ice (half of which melts), is approximately 5.71 mol/kg.
Explanation:The question asks for the calculation of the molality of salt (NaCl) in a mixture which involves a 2.00 lb sample of NaCl and 12.00 lb sample of ice that eventually melts to half its quantity. The steps for finding molality in such a scenario involve the conversion of weight measurements to mole measurements and utilize the fact that molality is calculated as moles of solute (NaCl) per kilogram of solvent (water).
Step 1: Convert the amount of NaCl and water (melted ice) from lb to grams. 1 lb is approximately 453.592 grams. So, 2.00 lb of NaCl is 907.185 grams of NaCl and 6.00 lb of water is 2721.55 grams (since half of the 12 lb of ice melts). This needs to be converted to kilogram giving us 2.72155 kg.
Step 2: Convert grams of NaCl to moles. The molar mass of NaCl is approximately 58.44 grams per mole. Therefore, 907.185 grams of NaCl is approximately 15.54 moles.
Step 3: Determine the molality using the formula molality = moles of solute (NaCl) / kg of solvent (melted ice). This gives us a molality of approximately 5.71 mol/kg.
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how many pounds does 1.00 quart of mercury weight? its density is 13.6g/mL
1.00 quart of mercury weight approximately 28.4 pounds.
To find the weight of 1.00 quart of mercury, given its density of 13.6 g/mL, follow these steps:
1. Convert quarts to milliliters:
1 quart = 946.35 mL2. Calculate the mass using density:
Mass = Density × VolumeMass = 13.6 g/mL × 946.35 mLMass ≈ 12,868.76 g3. Convert grams to pounds:
1 pound ≈ 453.592 gramsMass in pounds = 12,868.76 g / 453.592 g/poundMass ≈ 28.4 poundsWhen making a guess and retesting this information a theory or _____ may be formed which explains why something has occurred or what it may look like.
4. Which one of the following groups of chemical compounds is composed entirely of organic compounds? A. Ch3OCH3, Ca3(PO4)2, CO2, H2CO3 B. C2H2, CH4, CaCl2, CaCN2 C. C2H4O, CH2O, CaSO4, C3H5(OH)3 D. C6H6, C2H5OH, C6H5CH3, C3H5(NO3)3
Answer: The correct answer is Option D.
Explanation: Organic compounds are defined as the compounds which have hydrogen and carbon atoms in it. They are also known as hydrocarbons.
From the given options:
Option 1: [tex]Ca_3(PO_4)_2\text{ and }CO_2[/tex] are not organic compounds because these two compounds do not contain hydrogen element in it.
Option 2: [tex]CaCl_2\text{ and }CaCN_2[/tex] are not organic compounds because these two compounds do not contain hydrogen and carbon elements in it.
Option 3: [tex]CaSO_4[/tex] is not an organic compound because this compound do not contain hydrogen and carbon elements in it.
Option 4: All the compounds contain hydrogen and carbon elements in it and hence, all the compounds are organic compounds.
Therefore, the correct option is Option D.
Draw the structure(s) of all alkane(s) that have 3-6 carbons and does not have secondary or tertiary carbons.
Final answer:
Here are each alkane and heir structures:
1. Propane (C₃H₈):
CH₃ - CHH₂ - CH₃
2. Butane (C₄H₁₀):
CH₃ - CHH₂ - CHH₂ - CH₃
3. Pentane (C₅H₁₂):
CH₃ - CHH₂ - CHH₂ - CHH₂ - CH₃
4. Hexane (C₆H₁₄):
CH₃ - CHH₂ - CHH₂ - CHH₂ - CH₂ - CH₃
Explanation:
In all of these alkanes, there are no secondary or tertiary carbons, as they have a linear structure without any branching. These structures follow the general formula CnH₂ₙ₊₂, where n is the number of carbon atoms.
Remember that these are just a few examples of alkanes with 3-6 carbons. There can be multiple isomers and variations of these alkanes based on different arrangements of the carbon atoms.
An alkane is a saturated hydrocarbon composed of carbon and hydrogen atoms, with only single covalent bonds between carbon atoms.
To draw the structures of alkanes with 3-6 carbon atoms and no secondary or tertiary carbon atoms, we need to consider the straight-chain (normal) isomers.