26.62 grams of HBr would be present in 355 mL of a 7.5% m/v HBr solution.
Concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution.
There are various methods of expressing the concentration of a solution.
Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.
Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.
Given,
HBr = 7.5% m/ v
This means 7.5g of HBr in 100 ml of the solution.
1 ml of the solution has 0.075g
355 ml of the solution will have = 0.075 × 355 = 26.62g of HBr
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what is the PH scale of 0.02m of hydrochloric acid
Answer:
Explanation:
The pH of 0.02 M hydrochloric acid is approximately 1.7.
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How many calories are in 4,180 joules?
Answer:
To convert joules to calories, you can use the conversion factor:
1 calorie = 4.184 joules
To find out how many calories are in 4,180 joules, divide the given value by the conversion factor:
4,180 joules / 4.184 joules per calorie = 0.9 calories (approximately)
Therefore, there are approximately 0.9 calories in 4,180 joules.
Is this reaction endothermic or exothermic?
Exothermic
Endothermic
Not enough information
Both
Answer:
Una reacción exotérmica es aquella cuyo valor de entalpía es negativo, es decir, el sistema desprende o libera calor al entorno (ΔH < 0). Una reacción endotérmica es aquella cuyo valor de entalpía es positivo, es decir, el sistema absorbe calor del entorno (ΔH > 0).
Explanation:
The tomato is dropped. What is the velocity, v
, of the tomato when it hits the ground? Assume 86.0 %
of the work done in Part A is transferred to kinetic energy, E
, by the time the tomato hits the ground.
Express your answer with the appropriate units.
To determine the tomato's velocity when it hits the ground, we need more information. Specifically, we need the height from which the tomato was dropped and the tomato mass.
Without these details, it is impossible to calculate velocity accurately. The velocity of an object when it hits the ground depends on factors such as the height of the fall, the mass of the object, and any forces acting on it during the fall (such as air resistance).
If you can provide the necessary information, I can help you calculate the velocity of the tomato when it hits the ground.
Calculate the volume of O2, at STP, required for the complete combustion of 125g octane (C8H18) to CO2 and H20
306.178 liters is the volume of O2 at STP, required for the complete combustion of 125g octane ([tex]C_{8}H_{18}[/tex]) to [tex]CO_{2}[/tex] and H20
To calculate the volume of [tex]O_{2}[/tex] required for the complete combustion of octane ([tex]C_{8}H_{18}[/tex]) to [tex]CO_{2}[/tex] and [tex]H_{2}O[/tex] at STP (Standard Temperature and Pressure), we need to consider the stoichiometry of the balanced chemical equation.
The balanced equation for the combustion of octane is:
[tex]C_{8}H_{18}[/tex] + 12.5[tex]O_{2}[/tex] -> 8[tex]CO_{2}[/tex] + 9[tex]H_{2}O[/tex]
From the equation, we can see that 1 mole of octane requires 12.5 moles of [tex]O_{2}[/tex] to completely combust. The molar mass of octane ([tex]C_{3}H_{18}[/tex]) is approximately 114.22 g/mol.
To calculate the moles of octane, we divide the given mass by the molar mass:
Moles of octane = 125 g / 114.22 g/mol ≈ 1.093 mol
Since the molar ratio between octane and [tex]O_{2}[/tex] is 1:12.5, the moles of [tex]O_{2}[/tex]required can be calculated as:
Moles of [tex]O_{2}[/tex] = 1.093 mol * 12.5 ≈ 13.663 mol
Now, we can use the ideal gas law, PV = nRT, to calculate the volume of [tex]O_{2}[/tex] at STP. At STP, the temperature is 273 K, and the pressure is 1 atm.
Using the molar volume of an ideal gas at STP (22.4 L/mol), the volume of [tex]O_{2}[/tex] required is:
Volume of [tex]O_{2}[/tex] = Moles of [tex]O_{2}[/tex] * Molar volume = 13.663 mol * 22.4 L/mol ≈ 306.178 L
Therefore, approximately 306.178 liters of [tex]O_{2}[/tex] at STP would be required for the complete combustion of 125 grams of octane ([tex]C_{8}H_{18}[/tex]) to [tex]CO_{2}[/tex] and [tex]H_{2}O[/tex]
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PLEASE SOLVE ALL FOUR BOXES IF POSSIBLE AND COLOR CODE THEM!! I REALLY NEED THE HELP!! GIVING LOTS OF POINTS!!
NEED HELP HURRY PLEASE
1) The higher the temperature and the concentration, the more the Collison between the reactants and the more they react. A catalyst only speeds up the rate of reaction
2) The Collison frequency, orientation and the activation energy
3) At equilibrium, the rate of the forward and the reverse reactions are the same.
4) Increasing the concentration of the reactants would shift the equilibrium to the right
5) Decreasing the products would shift the equilibrium position to the right.
What is the collision theory?Chemistry has a hypothesis known as the collision theory that describes how chemical reactions take place. According to this, reacting particles (atoms, molecules, or ions) must collide in order for a reaction to occur. The theory offers a framework for comprehending the variables that affect the likelihood and pace of chemical reactions.
The influence of variables including temperature, concentration, surface area, and the presence of catalysts on the pace of chemical reactions is explained in part by the collision hypothesis.
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10 examples of elimination reaction
Answer:
1. Dehydration of alcohols
2. Dehydrohalogenation of alkyl halides
3. Decarboxylation of carboxylic acids
4. Pyrolysis of esters
5. Deamination of amino acids
6. Dealkylation of ethers
7. Dehalogenation of aryl halides
8. Dehydration of amides
9. Dehydrogenation of alkanes
10. Dehydrogenation of alkenes.
Explanation:
A balanced chemical equation has equal numbers of atoms of each type on both sides of the equation. This illustrates the principle of
Answer:
conservation of mass
A gas has a volume of 550 mL at a temperature of -55 °C. The volume of the gas at 30 °C is
Blank 1:
mL.
The combined gas law equation is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
The volume of the gas at 30 °C is approximately 760.67 mL.
To determine the volume of the gas at 30 °C, we can use the combined gas law equation, which relates the initial and final conditions of temperature and volume for a gas.
The combined gas law equation is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 and P2 are the initial and final pressures, respectively
V1 and V2 are the initial and final volumes, respectively
T1 and T2 are the initial and final temperatures in Kelvin, respectively
We need to convert the temperatures from Celsius to Kelvin by adding 273.15 to each value.
Given:
V1 = 550 mL
T1 = -55 °C = 218.15 K
T2 = 30 °C = 303.15 K
Assuming the pressure remains constant, we can rearrange the equation to solve for V2:
V2 = (P1 * V1 * T2) / (P2 * T1)
Since the pressure is not specified in the problem, we can assume it remains constant, allowing us to cancel out the pressure terms. Thus, the final equation becomes:
V2 = (V1 * T2) / T1
Plugging in the given values:
V2 = (550 mL * 303.15 K) / 218.15 K
Simplifying the calculation, we find:
V2 ≈ 760.67 mL
Therefore, the volume of the gas at 30 °C is approximately 760.67 mL.
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Which formula represents an isomer of CH3− CH2− COOH?
CH3−C−OH−CH3
CH3−CO−O−CH3
CH3−CO−CO−CH3
CH3−CH2−CO2−CH3
The formula that represents an isomer of CH3−CH2−COOH is D. CH3−CH2−CO2−CH3.
An isomer is a compound that has the same molecular formula but differs in the arrangement or connectivity of its atoms. In this case, the molecular formula is CH3−CH2−COOH, which represents the carboxylic acid called propanoic acid.
Option D, CH3−CH2−CO2−CH3, is an isomer of propanoic acid. It represents methyl propanoate, an ester formed by the reaction of propanoic acid with methanol. In this isomer, the -COOH group of propanoic acid is replaced with -COOCH3 group, indicating the presence of an ester functional group.
Options A, B, and C do not represent isomers of CH3−CH2−COOH. Option A, CH3−C−OH−CH3, represents dimethyl ether, an entirely different compound. Option B, CH3−CO−O−CH3, represents dimethyl carbonate, which also has a different structure. Option C, CH3−CO−CO−CH3, represents a compound known as methyl propanoate, which is not an isomer but the same compound as option D. Therefore, option D is correct.
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i. ii. iii. questions on the basis of it. Name 'B' and 'E'? ody. B What happens in 'A'? What type of blood (oxygenated or deoxygenated) flows in 'E' and 'F'? iv. What is the name of the blood circulation that occurs in between the heart and 'A'? The table shows recipients and donors of following questions on the basis of it.
It involves the transport of oxygenated blood from the heart to the rest of the body, and the return of deoxygenated blood from the body to the heart.
The given table shows the blood types of recipients and donors. A person's blood type is determined by the presence or absence of specific antigens on the surface of their red blood cells. In humans, there are four main blood groups: A, B, AB, and O.
Blood group B: The individuals having blood group B have B antigens present on the surface of their red blood cells.
Blood vessel E: The blood vessel E is pulmonary artery.
The chamber 'A' is the right atrium of the heart. The right atrium receives deoxygenated blood from the body and pumps it into the right ventricle.
Blood vessel E is pulmonary artery that carries deoxygenated blood from the right ventricle of the heart to the lungs.
Blood vessel F is pulmonary vein that carries oxygenated blood from the lungs to the left atrium of the heart.
The blood circulation that occurs in between the heart and the chamber 'A' (right atrium) is systemic circulation.
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Which of the following equations correctly model a nuclear fusion reaction? Justify your reasoning
Answer:
Explanation:
c) fusion combines into one product plus energy
compare and contrast galileo's and newton's views on inertia using the graphic organizer below
Galileo studied the earth's principles of motion and established the laws of gravity. Newton established the laws of motion for good at the beginning and connected them to Kepler's principles of planetary motion. Before Newton, no one had successfully proven how the motions of celestial bodies affected the laws of physics on Earth.
Galileo claimed that a rolling ball will continue to move with a steady speed if left "alone." The idea of force is the only distinction between Galileo's claim and Newton's first law of motion. Galileo was not yet familiar with the idea of force, and it was Newton who eventually clarified its nature.
Isaac Newton's renowned three principles of motion were the culmination of Galileo's theories.
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61. Given the following information:
Ag2 CrO4(s)=2Agt (aq) + CrO4²- (aq)
Ag+ (aq) + e- Ag(s)
find the standard reduction potential at 25°C for the half-reaction
Ksp = 1 × 10-12
E = +0.799 V
Ag2 CrO4(s) + 2e¯ 2Ag(s) + CrO4²- (aq)
Q = Ksp = 1 × 10^(-12).
Substituting the values into the Nernst equation, we have:
0.799 V = E° - (RT/2F) * ln(1 × 10^(-12))
Now, solving for E°:
E° = 0.799 V + (RT/2F) * ln(1 × 10^(-12))
The value of R is the ideal gas constant, T is the temperature in Kelvin, and F is the Faraday constant.
To find the standard reduction potential at 25°C for the half-reaction Ag2CrO4(s) + 2e¯ → 2Ag(s) + CrO4²-(aq), we can use the Nernst equation, which relates the standard reduction potential (E°) to the equilibrium constant (K) and the reaction quotient (Q).
The Nernst equation is given as follows:
E = E° - (RT/nF) * ln(Q)
Given information:
Ksp = 1 × 10^(-12)
E = +0.799 V (standard reduction potential of Ag+ to Ag)
Since the reaction involves the dissolution of Ag2CrO4(s), the reaction quotient Q can be expressed as [Ag+]²/[CrO4²-].
Since the stoichiometry of the reaction is 2:1 for Ag2CrO4 to Ag+, we can say that [Ag+]² = Ksp.
Therefore, Q = Ksp = 1 × 10^(-12).
Substituting the values into the Nernst equation, we have:
0.799 V = E° - (RT/2F) * ln(1 × 10^(-12))
Now, solving for E°:
E° = 0.799 V + (RT/2F) * ln(1 × 10^(-12))
The value of R is the ideal gas constant, T is the temperature in Kelvin, and F is the Faraday constant.
Please note that without specific values for temperature (T) and the ideal gas constant (R), the exact standard reduction potential at 25°C cannot be determined.
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Explain how temperature, concentration and a
catalyst will affect the rate of a reaction.
What are the three points of collision theory
that are required for a reaction to happen?
Answer:
Temperature, concentration, and a catalyst can all affect the rate of a chemical reaction.
Temperature: Increasing the temperature generally increases the rate of a reaction. This is because higher temperatures provide more kinetic energy to the reactant particles, causing them to move faster and collide more frequently. With increased collision frequency, the chances of successful collisions with sufficient energy to overcome the activation energy barrier and proceed with the reaction are also increased. As a result, the reaction rate typically increases with temperature.
Concentration: Increasing the concentration of reactants generally increases the rate of a reaction. When the concentration of reactant particles is higher, they become more crowded, increasing the likelihood of collisions between reactant particles. With more collisions occurring, there is a higher probability of successful collisions leading to a reaction. Therefore, higher reactant concentrations generally result in a higher reaction rate.
Catalyst: A catalyst is a substance that increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy. Catalysts themselves are not consumed during the reaction and do not undergo any permanent changes. They work by providing an alternative route that requires less energy for the reactants to reach the transition state. This lowers the activation energy barrier, making it easier for the reaction to occur. By providing an alternative pathway, catalysts increase the rate of reaction without being consumed in the process.
Regarding collision theory, the three key points required for a reaction to happen are:
Collision: Reactant particles must collide with each other for a reaction to occur. Collisions bring the reactant particles in close proximity, allowing them to interact and potentially form new chemical bonds.
Energy: Colliding particles must possess enough energy, equal to or greater than the activation energy, for the reaction to take place. Activation energy is the minimum energy required for the reactant particles to break existing bonds and initiate the formation of new bonds. Only collisions with sufficient energy can overcome the activation energy barrier and lead to a reaction.
Orientation: In addition to sufficient energy, the collision between reactant particles must occur with the correct orientation. This means that the particles must collide in a way that allows the necessary atoms or groups to come into contact and form new bonds. If the collision occurs with an incorrect orientation, the particles may simply bounce off each other without any reaction taking place.
In summary, according to collision theory, for a reaction to happen, reactant particles must collide with sufficient energy and the correct orientation. Temperature and concentration affect the rate of reaction by influencing collision frequency, while a catalyst provides an alternative reaction pathway with lower activation energy.
3. You need to prepare 100.0 mL of a 0.050 M solution of calcium chloride (CaCl₂).
a. How many moles of CaCl₂ are needed?
b. How many grams of CaCl₂ are needed?
Answer:
How many moles of CaCl2 are needed would be 0.005 moles.
How many grams of CaCl2 are needed would be 0.555 grams.
The pressure of a sample of helium in a 1L container is .988 atm. What is the new pressure if the sample is placed in a 2L container?
Answer:
The new pressure will be 0.494 atm.
Explanation:
We can use the ideal gas law to solve this problem:
PV = nRT
where:
P is the pressureV is the volumen is the number of moles of gasR is the ideal gas constantT is the temperatureWe know that the initial pressure is 0.988 atm, the initial volume is 1 L, and the temperature is constant.
We also know that the final volume is 2 L.
We can solve for the final pressure as follows:
[tex]P_2 = \frac{P_1V_1}{V_2}[/tex]
substituting value
[tex]P_2 = \frac{0.988\: atm*1 L}{2 L}[/tex]
[tex]P_2 = 0.494 atm[/tex]
Therefore, the new pressure will be 0.494 atm.
