What type of reaction is A + B + 210 ) >>> C
Answers
The reaction A + B + 210 → C can be categorized as a combination reaction.
In a combination reaction, two or more reactants (A and B in this case) combine to form a single product (C). The number 210 could be a typo or an irrelevant part of the equation, as it does not fit the standard chemical notation.
Based on the information you provided, the reaction can still be categorized as a combination reaction. In a combination reaction, two or more reactants combine to form a single product.
In this case, reactants A and B react together to produce product C. However, without further information or a corrected equation, it is not possible to provide specific details about the reaction or the substances involved.
If you have any additional information or a revised equation, please provide it, and I would be happy to assist you further.
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Related Questions
3. People and instruments produce different ranges of sound temperatures is this true or false
Science
Answers
Answer the following;
Pls helppp
Answers
Answer:
The answer is A
Explanation:
Answer:
D)All of the above
Which applies to fusion? Check all that apply.
involves the splitting of nuclei
takes place in the Sun
releases radiation as a waste product
occurs in nuclear power plants and is used to generate electricity
plays a role in the production of essentially all elements heavier than helium
releases large amounts of energy
Answers
Fusion Nuclear fusion is a response wherein two or greater atomic nuclei are blended to shape one or greater different atomic nuclei and subatomic debris. It
1. involves the splitting of nuclei
2. releases large amounts of energy
4. releases radiation as a waste product
5. occurs in nuclear power plants and is used to generate electricity.
Fusion reactions power the sun and other stars. In a fusion reaction, light nuclei merge to form a single heavier nucleus. The technique releases electricity due to the fact the total mass of the resulting unmarried nucleus is much less than the mass of the two unique nuclei. The leftover mass becomes energy. The difference in mass between the reactants and merchandise is manifested as either the discharge or absorption of power.
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Answer:
2, 5, 6
Explanation:
edge 2023
Temperature is an example of a variable that uses: a) the ordinal scale. b) the ratio scale. c) the interval scale. d) either the ratio
Answers
This translates to the fluctuating temperature using interval scale in Fahrenheit and Celsius.
What does interval scale mean in plain English?Periodic Scale
It is described as a quantitative measuring scale where a noticeable difference may be found between the two variables. In other words, the measurements are precise rather than relative, where the occurrence of zero is arbitrary.
Data on an interval scale is what?The definition of interval data, often known as an integer, is a data type that is measured along a scale with each point being situated at an equal distance from the other. Interval data always takes the form of integers or numerical values with a standard and equal distance between the two sites.
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2-ciclopropilpropano
Answers
Answer:
cool
Explanation:
Write a balanced chemical equation for the reaction of solid vanadium(V) oxide with hydrogen gas to form solid vanadium(III) oxide and liquid water.
Answers
V₂O₅ (s) + 2H₂ (g) → V₂O₃ (s) + 2H₂O(l) is the balanced chemical equation for the reaction of solid vanadium(V) oxide with hydrogen gas to form solid vanadium(III) oxide and liquid water.
What is Balanced Chemical Equation ?The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
The word based equation is
Vanadium(V) oxide + Hydrogen gas → Vanadium (III) oxide + Water
Now write the chemical equation
V₂O₅ (s) + 2H₂ (g) → V₂O₃ (s) + 2H₂O(l)
Reactant side Product side
V = 2 V = 2
O = 5 O = 5
H = 4 H = 4
We can see that the number of atoms on reactant side is equal to the number of atoms on product side. So the given equation is balanced.
Thus from the above conclusion we can say that V₂O₅ (s) + 2H₂ (g) → V₂O₃ (s) + 2H₂O(l) is the balanced chemical equation for the reaction of solid vanadium(V) oxide with hydrogen gas to form solid vanadium(III) oxide and liquid water.
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Based on the strength of their intermolecular forces, you would expect CH3-O-CH3 to have ___ boiling point compared to CH3CH2OH.
A. an equal
B. a lower
C. a higher
Answers
Answer:
higher
Explanation:
as CH3CH2OH has an O-H bond, it has significantly more IMF caused by the hydrogen bond between CH3CH2OH molecules. This means its harder to pull apart CH3CH2OH molecules as they are very attracted to one another, thereby increasing the boiling point.
The gravitational pull between two objects depends on their mass and distance. What is meant by distance?
Answers
Answer:
How far away the objects are from each other
A radioactive substance decays exponentially. A scientist begins with 170 milligrams of a radioactive substance. After 16 hours, 85 mg of the substance remains. How many milligrams will remain after 21 hours? mg Give your answer accurate to at least one decimal place
Answers
If 170 milligrams of a radioactive substance decays to 85 g after 16 hours. Then, after 21 hours, approximately 75.2 mg of the radioactive substance will remain.
The decay of the radioactive substance follows an exponential decay equation of the form:
\(N(t) = N_{o} \times e^{-kt}\)
Where:
N(t) is the amount of substance remaining at time t
N₀ is the initial amount of substance
k is the decay constant
t is the time elapsed
Given to us is N₀ = 170 mg and N(16) = 85 mg. We can use this information to find the decay constant, k.
\(85 = 170 \times e^{-k \times 16}\)
Dividing both sides by 170:
\(0.5 = e^{-k \times 16}\)
To solve for k, we can take the natural logarithm (ln) of both sides:
ln(0.5) = -k × 16
from this, the value of k comes out to be:
k = 0.0431
Now we can use the decay equation to find the amount of substance remaining after 21 hours, N(21):
\(N(21) = 170 \times e^{-0.0431 \times 21}\)
Calculating this expression:
N(21) = 75.2
Therefore, after 21 hours, approximately 75.2 mg of the radioactive substance will remain.
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Explain the mechanism of a Horner-Wadsworth-Emmons reaction between diethyl benzylphosphonate and 3,4-methylenedioxybenzaldehyde in the presence of aqueous NaOH forming 3,4-methylenedioxystilbene as the product.
Answers
Horner-Wadsworth-Emmons (HWE) reaction is an important synthetic reaction in organic chemistry. It is widely used for synthesizing various compounds. The reaction is between an aldehyde or ketone and a phosphonate or phosphonate ester in the presence of a strong base.
The Horner-Wadsworth-Emmons reaction is one of the most convenient and well-known methods of constructing carbon-carbon double bonds. The reaction proceeds via the formation of an ylide intermediate. The HWE reaction is particularly useful for the synthesis of compounds with a Z-configuration.
The mechanism for the reaction of diethyl benzylphosphonate and 3,4-methylenedioxybenzaldehyde in the presence of aqueous NaOH, forming 3,4-methylenedioxystilbene as the product, can be explained in the following steps:
Step 1: Formation of the ylide intermediate
The reaction starts with the formation of an ylide intermediate. This is achieved by the reaction of diethyl benzylphosphonate and 3,4-methylenedioxybenzaldehyde in the presence of a strong base like NaOH or KOH. In this reaction, a deprotonated species called an ylide intermediate is generated.
Step 2: Addition of the ylide intermediate to the aldehyde
The ylide intermediate then attacks the aldehyde, leading to the formation of a betaine intermediate.
Step 3: Formation of the phosphonate ester
The betaine intermediate undergoes elimination to form the final product, 3,4-methylenedioxystilbene, and the by-product phosphonate ester.
The mechanism of the Horner-Wadsworth-Emmons (HWE) reaction between diethyl benzylphosphonate and 3,4-methylenedioxybenzaldehyde in the presence of aqueous NaOH, forming 3,4-methylenedioxystilbene as the product, is complete. This reaction is significant in organic chemistry and finds applications in the pharmaceutical industry.
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Anything detected with the five senses is considered an
Answers
Answer:
sight, taste, touch, hearing and smell
Explanation:
Anything detected with the five senses is considered a sensory perception or sensory experience
Our five senses are sight (vision), hearing (audition), taste (gustation), touch (tactile perception), and smell (olfaction). These senses allow us to perceive and interpret information from the external environment.
When stated as "detected with the five senses," which means gather information about the world around us through these sensory experiences.
For example:
1. Sight (Vision): We perceive visual information through our eyes, allowing us to see colors, shapes, and movements.
2. Hearing (Audition): We perceive auditory information through our ears, allowing us to hear sounds and distinguish between different tones and pitches.
3. Taste (Gustation): We perceive taste sensations through our taste buds on the tongue, allowing us to distinguish between sweet, sour, salty, bitter, and umami flavors.
4. Touch (Tactile Perception): We perceive tactile sensations through our skin, allowing us to feel textures, pressure, temperature, and pain.
5. Smell (Olfaction): We perceive smells through our olfactory system, located in our nose, allowing us to detect and identify various scents and odors.
All the sensory experiences , whether through sight, hearing, taste, touch, or smell, contribute to our understanding of the world and our surroundings.
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an exact mass of an unknown solute is added to a solution of heptane (an organic solvent). the freezing point of the resulting solution is 2.3 degrees lower than that of pure heptane. which statement best describes what would happen if the mass of the unknown solute is doubled?
Answers
If the mass of the unknown solute is doubled, the freezing point depression will also double. This is because the amount of solute added to the solution is directly proportional to the change in the freezing point of the solution.
In other words, the greater the amount of solute added, the greater the freezing point depression. Therefore, doubling the mass of the solute will result in twice the amount of solute being added to the solution, which will cause the freezing point depression to double as well.
The relationship between the freezing point depression and the concentration of a solution is described by the equation: ΔTf = Kf·m
Assuming that the cryoscopic constant of heptane remains constant, we can rearrange the equation to solve for the molality: m = ΔTf/Kf
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How much cesium (half-life = 2 years) would remain from a 10 g sample after 2 years?
Answers
5 g of cesium(half-life = 2 years) would remain from a 10 g sample after 2 years.
Cesium has a half-life of 2 years. The half-life of a material is the length of time necessary for half of it to degrade or react. Half-life is a property of a chemical that is commonly represented by the sign "t½".
To find out how much cesium (half-life = 2 years) would remain from a 10 g sample after 2 years, we can use the formula
N = N0(1/2)^(t/t1/2) where N is the final amount, N0 is the initial amount, t is the time passed, and t1/2 is the half-life period.
In this case, N0 = 10 g, t = 2 years, and t1/2 = 2 years.
Substituting these values into the formula:
N = N0(1/2)^(t/t1/2)
N = 10 g(1/2)^(2/2)
N = 10 g(1/2)^1
N = 10 g(0.5)
N = 5 g
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Enter the activity coefficient of silver ion (γAg+) in each solution.
[Ag+] (M) γAg+
1.10E-5
1.10E-4
1.10E-3
1.10E-2
1.10E-1
Answers
The Debye–Huckel equation can be used to determine the silver ion (Ag+) activity coefficient (γ) in each solution:
log γ± = -0.509z±²√(I)/(1+1.328z±√(I))
where z± is the charge of the ion, and I is the ionic strength of the solution.
The ionic strength (I) at a temperature of 25 °C can be calculated as follows:
i = 1/2 * Σ(m * zi²)
where zi is the charge of the ion and mi is the molar concentration of the ion.
These equations can be used to determine the silver ion activity coefficient in each solution.
[Ag+] (M) I γAg+
1.10E-5 1.21E-9 0.9331.10E-4 1.21E-8 0.8641.10E-3 1.21E-7 0.7291.10E-2 1.21E-6 0.4991.10E-1 1.21E-5 0.173Due to the increased ionic strength and ion–ion interactions, it should be noted that the activity coefficient drops as the concentration of silver ions increases.
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The soil organic matter in Kenya has a stable carbon isotopic composition 813C of -18
permil. Assuming that the air SIC value is -7 permil, what is the relative contribution of C3 and
C4 plants to this organic matter?
Answers
The estimated relative contribution of C3 plants is approximately 88%, while the estimated relative contribution of C4 plants is approximately 12% to the organic matter in Kenya's soil.
To determine the relative contribution of C3 and C4 plants to the organic matter in Kenya's soil, we can use the difference in stable carbon isotopic compositions (δ13C values) between these plant types.
C3 and C4 plants have distinct δ13C values due to differences in their carbon fixation pathways. C3 plants generally have δ13C values ranging from -22 to -33 permil, while C4 plants typically exhibit δ13C values from -9 to -16 permil.
Given that the stable carbon isotopic composition (δ13C) of the soil organic matter in Kenya is -18 permil, we can compare this value to the δ13C values of C3 and C4 plants to estimate their relative contributions.
Let's denote the relative contribution of C3 plants as "x" and the relative contribution of C4 plants as "y." Since the contributions of C3 and C4 plants sum up to 100%, we have the equation:
x + y = 100% (equation 1)
Now, let's assign the δ13C values to the contributions of C3 and C4 plants. Assuming the air δ13C value is -7 permil, we can write the following equations:
-18 = x * (-33) + y * (-16) + (-7) * (1 - x - y) (equation 2)
Solving equations 1 and 2 simultaneously will provide us with the relative contributions of C3 and C4 plants.
Using the given δ13C values and solving the equations, we find:
x ≈ 0.88 (or 88%)
y ≈ 0.12 (or 12%)
Therefore, the estimated relative contribution of C3 plants is approximately 88%, while the estimated relative contribution of C4 plants is approximately 12% to the organic matter in Kenya's soil.
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Describe the endogenous pathways of cholesterol metabolism.
Answers
Explanation:
Endogenous pathway for lipid metabolism: VLDL is formed in the liver from triglycerides and cholesterol esters. ... In the liver, LDL is converted into bile acids and secreted into the intestines. In non hepatic tissues, LDL is used in hormone production, cell membrane synthesis, or stored.
Which of the following represents gamma emission?
A. 16 Eu+ge ¹62 Sm
B. Tc→ Tc+y
O C. 14Gd 1442 Sm+ He
O D. 180Gd→ 160Tb + je
SUBMIT
Answers
Answer:c
Explanation:
What is the total number of electrons in As-3?
Answers
Answer:
Arsenic, As, has atomic number 33, which is the number of protons in the nuclei of its atoms. A neutral As atom would also have 33 electrons. The electron configuration of a neutral arsenic atom is [Ar]3d104s24p3
Explanation:
Automobile batteries are filled with an aqueous solution of sulfuric acid. What is the mass of the acid (in grams) in 500. mL of the battery acid solution if the density of the solution is 1.285 g/cm3 and if the solution is 38.08% sulfuric acid by mass?
Answers
Using the following formula for this purpose: Mass = Density × Volume × Concentration. Here, Concentration = 38.08% = 0.3808(The information provided in the question)
Mass of the aqueous solution can be calculated as follows: Mass = Density × Volume × Concentration
Mass = 1.285 g/cm³ × 500cm³ × 0.3808,Mass = 244.85g.The mass of the solution is 244.85g. (The information provided in the question)
Therefore, the mass of the acid (in grams) in 500mL of the battery acid solution if the density of the solution is 1.285 g/cm3 and if the solution is 38.08% sulfuric acid by mass is 244.85 g.
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What is the pressure in millimeters of mercury inside a container of gas.
Answers
David has two yellow powders, Powder 1 and Powder 2. He collected data on some properties of these powders. His data are summarized in the table below. Which property CANNOT be used to determine whether the two powders are the same substance? Answer choices - a. mass b. density c. burns in air d. dissolves in water
Answers
The property that cannot be used to determine whether the two powders are the same substance is (d) dissolves in water.
The reason is that the ability to dissolve in water is not a unique property that can be used to identify a specific substance. Many different substances can dissolve in water, and the fact that both powders can dissolve in water does not necessarily mean they are the same substance.
On the other hand, properties such as mass, density, and the ability to burn in air can provide more specific information about the nature of the substance. Mass is a fundamental property that can be measured accurately, and if the two powders have the same mass, it suggests they may be the same substance. Density is a derived property that can also be measured and compared, providing information about the compactness of the material. The ability to burn in air indicates a chemical reactivity that can be used to distinguish between different substances.
Therefore, while properties like mass, density, and burning in air can provide valuable information for identifying substances, the property of dissolving in water alone is not sufficient to determine whether the two powders are the same substance.
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A 390 ml of aqueous solution prepared by dissolving 0.6 g of aluminum nitrate al(no3)3 in water, the concentration of nitrate ion (no3-) is __________m. the molar mass of al(no3)3 is 213 g/mol
Answers
The concentration of the nitrate ion, NO₃¯ in the solution given the data is 0.0216 M
What is molarity?This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:
Molarity = mole / Volume
How to determine the mole of Al(NO₃)₃Mass of Al(NO₃)₃ = 0.6 g Molar mass of solute = 213 g/mol Mole of Al(NO₃)₃ =?Mole = mass / molar mass
Mole of Al(NO₃)₃ = 0.6 / 213
Mole of Al(NO₃)₃ = 0.0028 mole
How to determine the molarity of Al(NO₃)₃Mole of Al(NO₃)₃ = 0.0028 mole Volume = 390 mL = 390 / 1000 = 0.39 L Molarity of Al(NO₃)₃=?Molarity = mole / Volume
Molarity of Al(NO₃)₃ = 0.0028 / 0.39
Molarity of Al(NO₃)₃ = 0.0072 M
How to determine the molarity of NO₃¯Dissociation equation
Al(NO₃)₃(aq) <=> Al³⁺(aq) + 3NO₃¯(aq)
From the balanced equation above,
1 mole of Al(NO₃)₃ contains 3 moles of NO₃¯
Therefore,
0.0072 M Al(NO₃)₃ will contain = 0.0072 × 3 = 0.0216 M NO₃¯
Thus, the molarity of NO₃¯ in the solution is 0.0216 M
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3. If a student took a 23.54 mL sample of HBr solution and placed it in a beaker beneath the buret.
What would be the molarity of the HBr solution if 17.54 milliliters of 0.1584 M NaOH solution were
used to turn the phenolphthalein pink? (please show all work)
Answers
Answer: Answer: The correct answer is a. 0.223 grams, b. 3.36 × 10²¹ and c. 2.79 × 10⁻³ mol.
Explanation:
Please help!! only right answers!! will mark brainlest.
Answers
Answer:
7.35 *2
Explanation:
Because I mole of Na2o is made of 2 moles of NaOH
Atoms of different
can be distinguished from
each other by the number of protons (atomic number) in their
nucleus.
Answers
Answer:
Elements
Explanation:
An element is defined by the number of protons in its nucleus.
In a lab, a scientist mixes a purple powder-like substance and a clear liquid together. The scientist observes that the two substances begin combining together, and he notices a gas begin to bubble out of the mixture. Did the scientist witness a chemical reaction? Why or why not?
Answers
The scientist witnessed a chemical reaction because a gas was evolved when the two substances were mixed.
A chemical reaction is the same as a chemical change. When a chemical change occurs, atoms recombine to form new substances. Certain occurrences may accompany a chemical reaction such as;
Change in colorEvolution of gasAppearance of a solid in an aqueous phase reactionSudden change in the temperature of the system.When the scientist mixed purple powder-like substance and a clear liquid together, he noticed that a gas began to bubble out of the mixture. This implies that new substance has been formed so a chemical reaction has taken place.
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what have you realized about your self after drawing the kite
Answers
draw the chemical reaction that occurs when the benzoic acid reacts with the naoh
Answers
The chemical reaction between benzoic acid (C₆H₅COOH) and NaOH can be represented as C₆H₅COOH + NaOH → C₆H₅COONa + H₂O.
This reaction involves the displacement of the hydrogen atom from the carboxylic acid group in benzoic acid by the sodium atom from sodium hydroxide. As a result, sodium benzoate (C₆H₅COONa) and water (H₂O) are formed.
The reaction between benzoic acid and NaOH is classified as an acid-base reaction, where the acidic benzoic acid reacts with the basic NaOH to produce a salt (sodium benzoate) and water.
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1.Why is oil nonpolar?
2.Why is water polar?
Answers
The molecules of oil are nonpolar because their charges are balanced and therefore these molecules are not attracted by other in solution, while water molecules are polar because their atoms are arranged to form positive and negative poles in solution.
Why is oil nonpolar and water polar?Oil is nonpolar and water polar due to the active forces between molecules, in the case of non polar substances the charges are balanced while polar molecules contain negative and positive poles.
Therefore, with this data, we can see that oil is nonpolar and water is polar due to the charges of their atoms.
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what is the volume of 72g of copper has a density of 9/cm 3
Answers
v=72/9=8
so the density is 8g/ml
Answer:
8cm³
Explanation:
density = mass/volume
or volume = mass/density
= 72/9
= 8 cm³