Explanation:
Alpha decay: In this process, alpha particles is emitted when a heavier nuclei decays into lighter nuclei. The alpha particle released has a charge of +2 units.
The general alpha decay reaction is given as:
[tex]_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha[/tex]
[tex]_{92}^{233}\textrm{U}\rightarrow _{90}^{229}\textrm{Th}+_2^4\alpha[/tex]
[tex]_{90}^{229}\textrm{Th}\rightarrow _{88}^{225}\textrm{Ra}+_2^4\alpha[/tex]
When Uranium-233 undergoes alpha decay it splits into an alpha particle and thorium-229.Thorium-229 further undergoes alpha decay to produce radium-225 along with an alpha particle.
Determine the number of 3s electrons in na.
What acid and what base would you choose to prepare the salt potassium perchlorate (kclo4)?
Answer : The perchloric acid and potassium hydroxide base is used to prepare the salt of potassium perchlorate, [tex](KClO_4)[/tex]
Explanation :
when the perchloric acid react with the potassium hydroxide as a base to form a salt of potassium perchlorate, [tex](KClO_4)[/tex]
The balanced chemical reaction will be,
[tex]HClO_4+KOH\rightarrow KClO_4+H_2O[/tex]
By the stoichiometry, we can say that 1 mole of perchloric acid react with the 1 mole of potassium hydroxide base to give 1 mole of potassium perchlorate and 1 mole of water as a product.
Hence, the perchloric acid and potassium hydroxide base is used to prepare the salt of potassium perchlorate, [tex](KClO_4)[/tex]
True or false if you refuse to submit a breath you’re in a blood test is an assumable for evidence in a DUI criminal proceedings in Florida
Answer:
True
Explanation:
The molar mass of carbon dioxide is 44.0 g/mol. a mass of 150.0 grams of carbon dioxide is equivalent to how many moles?
A) 3.00 mol
B) 3.41 mol
C) 29.3 mol
D) 106 mol
The molar mass of carbon dioxide is 44.0 g/mol. a mass of 150.0 grams of carbon dioxide is equivalent to 3.41 mol of carbon dioxide.
What is mole?In the International System of Units, the mole is the unit of substance quantity (SI). How so many elementary units of a certain substance are present in an item or sample is determined by the quantity of that material. There are precisely 6.022×10²³ elementary entities in a mole.
For instance, although having differing volumes and weights, 10 moles containing water along with ten moles of mercury both contain the same quantity of material, and the mercury includes exactly 1 atom for every molecule of water.
M(CO₂)=44.0 g/mol
m(CO₂)=150.0 g
n(CO₂)=m(CO₂)/M(CO₂)
n(CO₂)=150.0/44.0=3.41 mol
Therefore, 3.41 mol of carbon dioxide is there.
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If the ΔG value of a reaction is negative, what is true of the ΔG value of the reverse reaction?
A. ΔG > 0
B. ΔG < 0
C. ΔG = 0
D. ΔG fluctuates
What are three physical properties often analyzed to identify covalent compounds?
Covalent compounds can be identified by their low melting and boiling points, non-conductivity of electricity, and solubility in nonpolar solvents.
Explanation:Covalent compounds, also referred to as molecular compounds, are usually identified through several physical properties such as their low melting and boiling points, non-conductivity of electricity, and their solubility in nonpolar solvents.
Low melting and boiling points: Covalent compounds don't have strong intermolecular forces, resulting in low melting and boiling points.
Non-conductivity of electricity: Since most covalent compounds don't dissociate into ions in solutions, they typically do not conduct electricity.
Solubility in nonpolar solvents: Due to the 'like soluble like' rule, covalent compounds tend to be soluble in nonpolar solvents rather than polar solvents such as water.
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Compare bond lengths in butane and t butylcyclohexane
Answer:
The lengths of the C-C bonds increases with the decrease in the resistance of said bond, for example, a triple bond has a shorter length than in the case of a single bond. Butane has the single bonds and the CC bond is hybridized with sp3 hybridization, however in the butylcyclohexane structure the CC bond is also sp3 and the angle is 120°, however the angle shown is equal to 109.5°, so there is a certain angular tension and it is very unstable with respect to butane
Explanation:
During hydrolysis, ___________ must be added before the bonds can be broken. (look at flash animation on hydrolysis)
The correct answer is water.
During hydrolysis, _____water______ must be added before the bonds can be broken.
For example, sucrose undergoes hydrolysis to break into glucose and fructose. Here sucrose is a disaccharide and glucose and fructose are monosaccharides.
Sucrose + H₂O → glucose + fructose
Esters are formed by the combination of carboxylic acid and alcohol.
But the hydrolysis of ester causes the release of carboxylic acid and alcohol.
RCOOR'(ester) + H₂O → RCOOH(carboxylic acid) + R'OH(alcohol)
________ is the science of obtaining reliable measurements from photographs.
When a colorless aqueous solution of lead nitrate is combined with a colorless aqueous solution of sodium iodide a bright yellow precipitate is formed. what is the chemical formula for the precipitate?
The chemical formula of the bright yellow precipitate is PbI₂ (lead iodide).
What is the balanced chemical equation?A chemical equation is the representation of a chemical reaction which consists of reactants participating, formed products, and an arrow indicating the direction of the chemical reaction.
The equation that has the number of atoms of substances equal on either side of the chemical equation is known as a balanced chemical equation.
The law of conservation of mass has to be followed by a balanced chemical equation, according to which, the total mass of the elements on the reactant side must be equal to the total mass of elements on the product side.
The chemical equation of the reaction of lead nitrate and an aqueous solution of sodium iodide:
[tex]Pb(NO_3)_2(aq) + 2NaI \longrightarrow 2NaNO_3(aq) + PbI_2 (s)[/tex]
The bright yellow precipitate formed in the above chemical reaction has the chemical formula PbI₂.
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How many grams of h3po4 are in 521 ml of a 9.30 m solution of h3po4?
Final answer:
To find the mass of H3PO4 in 521 ml of a 9.30 M solution, convert the volume to liters, multiply by the molarity to get moles, and then convert moles to grams using the molar mass of H3PO4. Approximately 474.6434 grams of H3PO4 are in 521 ml of a 9.30 M solution.
Explanation:
The student has asked how many grams of H3PO4 are in 521 ml of a 9.30 M solution. To find this, we use the formula:
Calculate the number of moles of H3PO4 by multiplying the molarity (M) of the solution by the volume of the solution in liters.
Convert the moles of H3PO4 to grams using the molar mass of H3PO4 (which is approximately 98 g/mol).
First, we convert 521 ml to liters (521 ml * 0.001 L/ml = 0.521 L). Next, we calculate the moles of H3PO4 (9.30 M * 0.521 L = 4.8433 mol). Finally, we multiply the number of moles by the molar mass of H3PO4 to get the mass in grams (4.8433 mol * 98 g/mol = 474.6434 g).
Therefore, there are approximately 474.6434 grams of H3PO4 in 521 ml of a 9.30 M solution.
A sample of hydrated sodium thiosulfate has a mass of 6.584 g. After it is heated, it has a mass of 4.194 g. What is the percentage by mass of water in the hydrate?
36.30%
[only 20 chars]
of the following, ____ has the lowest boiling point
cl2
o2
n2
h2
Solve an equilibrium problem (using an ice table) to calculate the ph of each solution: part a a solution that is 0.195 m in hc2h3o2 and 0.125 m in kc2h3o2
To calculate the pH of the given solution, use an ICE table to determine the concentrations of HC2H3O2 and C2H3O2-. Substitute these concentrations into the Ka expression to calculate the Ka value. Finally, use the definition of pH to determine the pH of the solution.
Explanation:To calculate the pH of a solution that is 0.195 M in HC2H3O2 and 0.125 M in KC2H3O2, we can use an ICE table. First, we set up the ICE table to determine the concentrations of HC2H3O2 and C2H3O2- at equilibrium. Once we have these concentrations, we substitute them into the Ka expression to calculate the Ka value. Finally, we use the definition of pH to determine the pH of the solution.
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Which of the following does not apply to obtaining a career in chemistry?
•There is only one path to get a career in chemistry.
•There are many paths to getting a career in chemistry.
•Start by talking to teachers and counselors about different career options.
•Talk to students already enrolled in the programs that you are interested in.
ANSWER: There is only one path to get a career in chemistry.
EXPLANATION: There are many paths of career in Chemistry and not only one. It would not be justified to say that there is only one path to get a career in chemistry. Careers in chemistry would include Biochemistry, Forensic Scientist, Research Scientist, Chemical Engineer, Chemical Plant Operator and even a Science teacher or Chemistry professor in a school or university.
a) Diamond and graphite are two different forms of pure elemental carbon with densities of 3.51 g/cc and 2.25 g/cc respectively. What volume would be occupied by a 0.50 g diamond? What volume would be occupied by a 0.50 g piece of graphite?
What is the ph of a 2.7 m solution of hclo4?
The concentration of H+ ions in the solution of 2.7 molar HClO₄ is 2.7 molar. Thus the pH of the acid is -0.43.
What is PH?pH of a solution is the measure of its H+ ions concentration. It determines how much acidic or basic the solution is. Mathematically it is the negative logarithm of H+ ion concentration.
A pH of 7 indicates that the solution is neutral and a pH below 7 is acidic and pH of above 7 is for basic solution. Thus strong acids such as chloric acid HClO₄ , have a pH very less. But it depends on the concentration of solute in the solution.
Given that the molarity of the acid is 2.7 molar. It furnish equal number of moles H+ ions and anions. Thus the concentration of H+ ion is 2.7 M.Now the pH is calculated as follows:
pH = -log [H+]
= -log (2.7)
= -0.43
Hence, the pH of the solution is -0.43.
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When does a scientists create the conclusion
How much energy would be released if 1.0 g of material were completely converted into energy?
Final answer:
The energy released from converting 1.0 g of mass into energy is 9 × 10¹³ joules (J), using the equation E = mc², where c is the speed of light.
Explanation:
According to Einstein's famous equation E = mc², where E represents energy, m is mass, and c is the speed of light in a vacuum, the energy released from completely converting 1.0 g of mass into energy is tremendously large. Since the speed of light, c, is approximately 3 × 10⁸ meters per second, and the mass m is 1.0 g (which is 1/1000 of a kilogram), the calculation is E = (1.0 g / 1000) × (3 × 10⁸ m/s)². This results in an energy release of 9 × 10¹³ joules (J), which is equivalent to about twice the energy released by the atomic bomb dropped on Hiroshima.
Find the missing part of this equation
a 0.175 m weak acid acid solution has ph of 3.25 find ka for the acid
To find the Ka value for the weak acid, we can use the given pH and concentration of the acid solution. The Ka value is calculated using the equation Ka = ([H3O+][A-])/[HA].
Explanation:To find the value of Ka for the weak acid, we can use the given pH and concentration of the acid solution. We know that the pH is a measure of the concentration of H3O+ ions in a solution, so we can use the pH to calculate the [H3O+] concentration. From the given pH of 3.25, we can determine that the [H3O+] concentration is 10^(-pH). So, [H3O+] = 10^(-3.25) M.
Now, the equilibrium equation for the dissociation of the weak acid is HA(aq) + H2O(l) -> H3O+(aq) + A-(aq). Since we know the [H3O+] concentration, we can assume that the [HA] concentration is equal to the [H3O+] concentration. So, [HA] = [H3O+] = 10^(-3.25) M.
The Ka value is calculated using the equation Ka = ([H3O+][A-])/[HA]. Substituting the given values, Ka = (10^(-3.25)^2)/10^(-3.25). Simplifying this expression gives us the value of Ka for the acid.
Calculate the equilibrium constant k for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 k. express your answer numerically using two significant figures. hints
The equilibrium constant for isomerization reaction is [tex]\boxed{9.615}[/tex]
Further Explanation:
The standard Gibbs free energy change in a reaction [tex]\left( {{{\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }}} \right)[/tex] is the difference of sum of the standard free energies of formation of product molecules and sum of standard free energies of formation of reactant molecules at the standard conditions. The formula used to calculate the value of standard Gibbs free energy change for a reaction [tex]\left( {{{\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }}} \right)[/tex] is as follows:
[tex]\Delta\text{G}_{\text{rxn}}^{\circ}=\sum\text{n}\Delta\text{G}_{\text{f}(\text{products})}^{\circ}-\sum\text{m}\Delta\text{G}_{\text{f}(\text{reactants})}^{\circ}[/tex]
Here, n is the stoichiometric coefficients of products, and m are the stoichiometric coefficients of reactants in a balanced chemical equation.
The formula to determine the relationship between change in standard Gibbs free energy [tex]\left( \Delta{\text{G}^{\circ}} \right)[/tex] and equilibrium constant [tex]\left({\text{K}}\right)[/tex] is given as follows:
[tex]{\Delta }}{{\text{G}}^{{^\circ }}} = - {\text{RTlnK}}[/tex] ......(1)
Here,
[tex]\Delta{\text{G}^{\circ}[/tex] is the standard Gibbs free energy change.
[tex]{\text{R}[/tex] is the gas constant.
[tex]{\text{T}}[/tex] is the temperature in Kelvin.
[tex]{\text{K}}[/tex] is the equilibrium constant.
The isomerization of glucose-1-phosphate to fructose-6-phosphate occurs in 2 steps:
The reaction of step 1 is as follows:
[tex]{\text{glucose - 1 - phosphate}} \to {\text{glucose - 6 - phosphate}}[/tex]
......(2)
[tex]\Delta{\text{G}^{\circ}_{1}[/tex] for equation (2) is [tex]- 7.28\;{\text{kJ/mol}}[/tex]
The reaction of step 2 is as follows:
[tex]{\text{fructose - 6 - phosphate}} \to {\text{glucose - 6 - phosphate}}[/tex]
......(3)
[tex]\Delta{\text{G}^{\circ}_{2}[/tex] for equation (3) is [tex]- 1.67\;{\text{kJ/mol}}[/tex]
Reverse the reaction of step 2.
[tex]{\text{glucose - 6 - phosphate}} \to {\text{frutcose - 6 - phosphate}}[/tex]
......(4)
[tex]\Delta{\text{G}^{\circ}_{3}[/tex] for equation (4) is [tex]+ 1.67\;{\text{kJ/mol}}[/tex]
Add equation (1) and (3) to get the final equation.
[tex]{\text{glucose - 1 - phosphate}} \to {\text{frutcose - 6 - phosphate}}[/tex]
To calculate [tex]\Delta {\text{G}}_{{\text{rxn}}}^{^\circ }}[/tex], add [tex]\Delta{\text{G}^{\circ}_{1}[/tex] and [tex]\Delta{\text{G}^{\circ}_{3}[/tex] as follows:
[tex]\Delta{\text{G}^{\circ}_{\text{rxn}}=\Delta{\text{G}^{\circ}_{1}+\Delta{\text{G}^{\circ}_{3}[/tex] ......(5)
Substitute [tex]- 7.28\;{\text{kJ/mol}}[/tex] for [tex]\Delta{\text{G}^{\circ}_{1}[/tex] and [tex]+ 1.67\;{\text{kJ/mol}}[/tex] for [tex]\Delta{\text{G}^{\circ}_{3}[/tex] in equation (5).
[tex]\begin{aligned}\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }} &= - 7.28\;{\text{kJ/mol + }} + 1.67\;{\text{kJ/mol}}\\{\text{}}&= - 5.61\;{\text{kJ/mol}}\\\end{aligned}[/tex]
For equilibrium constant (K), rearrange equation (1)
[tex]{\text{K}}={\text{e}}\frac{-\Delta{\text{G}}^{\circ}}{\text{RT}}[/tex] ......(6)
Substitute [tex]- 5.61\;{\text{kJ/mol}}[/tex] for [tex]\Delta{\text{G}^{\circ}[/tex],[tex]8.314\;{\text{J/mol}} \cdot {\text{K}}[/tex] for R and [tex]298\;{\text{K}}[/tex] for T in equation (6)
[tex]\begin{aligned} {\text{K}}&= {{\text{e}}^{\frac{{ - \left( { - 5.61\;{\text{kJ/mol}}} \right)}}{{\left( {8.314\;{\text{J/mol}} \cdot {\text{K}}} \right)\left( {\frac{{{\text{1J}}}}{{1000{\text{kJ}}}}} \right)\left( {298\;{\text{K}}} \right)}}}}\\&= {{\text{e}}^{2.2634}}\\&= 9.615\\\end{aligned}[/tex]
The equilibrium constant for the reaction is 9.615.
Learn more:
1. The change in standard gibbs free is for a reaction: https://brainly.com/question/10838453
2. Determination of the equilibrium constant for pure water: https://brainly.com/question/3467841
Answer details:
Grade: Senior Secondary School
Subject: Chemistry
Chapter: Chemical Equilibrium
Keywords: Standard Gibbs free energy, equilibrium, constant, glucose-1-phosphate and fructose-6-phosphate.
When hydrochloric acid and zinc were combined, the flask was sealed and thus no gas could escape from the flask. why did the pressure decrease as the temperature decreased?
Explanation:
According to the the ideal gas law, PV= nRT.
This means that pressure is directly proportional to temperature.
So, when HCl and zinc are combined in a flask then it will lead to the formation of zinc chloride and hydrogen gas.
The reaction equation will be as follows.
[tex]HCl + Zn \rightarrow ZnCl_{2} + H_{2}[/tex]
Since, the flask is sealed hence, hydrogen gas will not be able to move out of the flask.
So, when it will behave ideally then due to directly proportional relation between pressure and temperature there will occur a decrease in temperature with decrease in pressure.
The feed to a batch process contains equimolar quantities of nitrogen and methane. write an expression for the kilograms of nitrogen in terms of the total moles n(mol) of this mixture.
How Does The Modern Periodic Table Arrange Elements? By Atomic Mass By Atomic Number
By Number Isotopes
By Electron Mass
Benzene and its derivatives undergo a type of substitution reaction in which a hydrogen atom is replaced by a substituent, but the stable aromatic benzene ring is regenerated at the end of the mechanism.
Which of the following is an indication that a substance has undergone a chemical change?
No new product has been formed.
The color of the substance has not changed.
The original constitute has not changed.
The molecular structure has changed.
Answer:
D.
The molecular structure has changed.
Explanation:
I got it right on Plato.
The main difference between heat and temperature is the temperature is Solely dependent on the
Answer:the answer is the Velocity of molecules in the body
Explanation:hope this helped have a great day
Pressure and volume are inversely related. When the pressure on a gas is doubled, what happens to the volume
Answer: Volume decreases to half of original volume
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)
[tex]{P_1V_1}={P_2V_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = p
[tex]P_2[/tex] = final pressure of gas = 2p
[tex]V_1[/tex] = initial volume of gas = v
[tex]V_2[/tex] = final volume of gas = ?
Now put all the given values in the above equation, we get the final pressure of gas.
[tex]{p\times v}=2p\times V_2[/tex]
[tex]V_2=\frac{v}{2}[/tex]
Therefore, the final volume of the gas will become half of initial volume.
HELP PLEASE. STRUGGLING. Which of the following is true in an acidic solution? A)The molarity of hydroxide is double the molarity of hydronium. B)The molarity of hydroxide is equal to the molarity of hydronium. C)The molarity of hydroxide is lower than the molarity of hydronium. D)The molarity of hydroxide is higher than the molarity of hydronium.