Answer:
68%
Explanation:
Since we need a percentage we can use any number we want for our initial value.
5(1/2)^900/1620 = 3.40
(3.40 / 5)*100 = 68%
To make sure lets use a different initial amount
1(1/2)^900/1620 = 0.68
(0.68/1) * 100 = 68%
To solve this question, we'll assume the initial amount of radium-226 to be 1.
Now, we shall proceed to obtaining the percentage of radium-226 that will after 900 years. This can be obtained as illustrated below:
Step 1Determination of the number of half-lives that has elapsed.
Half-life (t½) = 1620 years
Time (t) = 900 years
Number of half-lives (n) =?[tex]n = \frac{t}{t_{1/2}}\\\\n = \frac{900}{1620}\\\\n = \frac{5}{9}[/tex]
Step 2:Determination of the amount remaining
Initial amount (N₀) = 1
Number of half-lives (n) = 5/9
Amount remaining (N) =?[tex]N = \frac{N_{0} }{2^{n}}\\\\N = \frac{1}{2^{5/9}}[/tex]
N = 0.68Step 3Determination of the percentage remaining.
Initial amount (N₀) = 1
Amount remaining (N) = 0.68
Percentage remaining =?Percentage remaining = N/N₀ × 100
Percentage remaining = 0.68/1 × 100
Percentage remaining = 68%Therefore, the percentage amount of radium-226 that remains after 900 years is 68%
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One hundred fifty joules of heat are removed from a heat reservoir at a temperature of 150 K. What is the entropy change of the reservoir (in J/K)?
Answer:
ΔS surrounding (entropy change of the reservoir) = -1 J/K
Explanation:
Given:
Change in heat (ΔH) = 150 joules
Temperature (T) = 150 K
Find:
ΔS surrounding (entropy change of the reservoir)
Computation:
ΔS surrounding (entropy change of the reservoir) = - ΔH / T
ΔS surrounding (entropy change of the reservoir) = - 150 / 150
ΔS surrounding (entropy change of the reservoir) = -1 J/K
Which molecule is butane?
Answer:
option c see image... and yw
The molecule shown in option C is butane.
What is butane?The organic compound with four carbon atoms and ten hydrogen atom lie in linear chain without any functional group is called butane.It is belongs to alkane family.What is alkane?It is acyclic saturated hydrocarbons organic compounds.It is also known as paraffin.An alkanes contain carbon atoms and hydrogen atoms bonded with single bond.It general formula is [tex]C_{n} H_{n+2}[/tex] .To learn more about butane,
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How to do q solution, qrxn, moles of Mg , and delta Hrxn?
Answer:
14, 508J/K
ΔHrxn =q/n
where q = heat absorbed and n = moles
Explanation:
m = mass of substance (g) = 0.1184g
1 mole of Mg - 24g
n moles - 0.1184g
n = 0.0049 moles.
Also, q = m × c × ΔT
Heat Capacity, C of MgCl2 = 71.09 J/(mol K)
∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)
= 14, 508 J/K/kg
ΔT= (final - initial) temp = 38.3 - 27.2
= 11.1 °C.
mass of MgCl2 = 95.211 × 0.1184 = 11.27
⇒ q = 11.27g × 11.1 °C × 14, 508 j/K/kg
= 1,7117.7472 J °C-1 g-1
∴ ΔHrxn = q/n
=1,7117.7472 ÷ 0.1184
= 14, 508J/K
For which one of the following reactions will the enthalpy change be approximately equal to the internal energy change?
A. H2 + I2 → 2HI
B. PCl5(g) → PCl3(g) + Cl2
C. 2H2O2 → 2H2O2 + O2
D. C(s) + O2(g) → CO2(g)
Answer: A. [tex]H_{2}_{(g)}+I_{2}_{(g)}=>2HI_{(g)}[/tex] and D.[tex]C_{(s)}+O_{2}_{(g)}=>CO_{2}_{(g)}[/tex]
Explanation: The relationship between internal energy change and enthalpy change during a chemical reaction occurs according to the following formula:
[tex]\Delta H=\Delta E+\Delta(PV)[/tex]
So, for changes in enthalpy and internal energy to be equal volume or pressure has to be constant, i.e., zero.
Change in the number of moles of gas during the reaction can make the difference between [tex]\Delta H[/tex] and [tex]\Delta E[/tex] be larger, so for them to be equal and pressure constant, number of moles must be the same in reagents and products.
Analysing each reaction above:
Reaction A has the same number of moles in reagents and products, so enthalpy change and internal energy change will be equal;
Reactions B and C don't have the same number of moles at both sides, so enthalpy and energy will be different.
Reaction D, although reagent side have 2 compounds, carbon is solid, so reaction have the same number of moles in both sides. Enthalpy and Energy will be equal.
A laboratory technician combines 35.9 mL of 0.258 M chromium(II) chloride with 35.8 mL 0.338 M potassium hydroxide. How many grams of chromium(II) hydroxide can precipitate
Answer:
0.52 g of chromium(II) hydroxide, Cr(OH)2.
Explanation:
We'll begin by calculating the number of mole of chromium (ii) chloride, CrCl2 in 35.9 mL of 0.258 M chromium(II) chloride solution.
This can be obtained as follow:
Molarity of CrCl2 = 0.258 M
Volume = 35.9 mL = 35.9/1000 = 0.0359 L
Mole of CrCl2 =?
Molarity = mole /Volume
0.258 = mole of CrCl2 /0.0359
Cross multiply
Mole of CrCl2 = 0.258 x 0.0359
Mole of CrCl2 = 0.0093 mole
Next, we shall determine the number of mole of potassium hydroxide, KOH in 35.8 mL 0.338 M potassium hydroxide solution.
This can be obtained as follow:
Molarity of KOH = 0.338 M
Volume = 35.8 mL = 35.8/1000 = 0.0358 L
Mole of KOH =.?
Molarity = mole /Volume
0.338 = mole of KOH /0.0358
Cross multiply
Mole of KOH = 0.338 x 0.0358
Mole of KOH = 0.0121 mole.
Next, we shall write the balanced equation for the reaction. This is given below:
2KOH + CrCl2 → Cr(OH)2 + 2KCl
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2 to produce 1 mole of Cr(OH)2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2.
Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of CrCl2.
From the calculations made above, we can see that only 0.00605 mole out of 0.0093 mole of CrCl2 is needed to react completely with 0.0121 mole of KOH.
Therefore, KOH is the limiting reactant.
Next, we shall determine the number of mole of Cr(OH)2 produced from the reaction.
In this case, we shall be using the limiting reactant because it will give the maximum yield of Cr(OH)2.
The limiting reactant is KOH and the number of mole of Cr(OH)2 produced can be obtained as illustrated below:
From the balanced equation above,
2 mole of KOH reacted to produce 1 mole of Cr(OH)2.
Therefore, Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of Cr(OH)2.
Finally, we shall convert 0.00605 mole of Cr(OH)2 to grams.
This is illustrated below:
Mole of Cr(OH)2 = 0.00605 mole
Molar mass of Cr(OH)2 = 52 + 2(16 + 1) = 52 + 2(17) = 86 g/mol
Mass of Cr(OH)2 =..?
Mole = mass /Molar mass
0.00605 = mass of Cr(OH)2/86
Cross multiply
Mass of Cr(OH)2 = 0.00605 x 86
Mass of Cr(OH)2 = 0.52 g
Therefore, 0.52 g of chromium(II) hydroxide, Cr(OH)2 was produced.
whts the ph of po4 9.78
Answer:
4.22
Explanation:
We know from the question, that the pOH of the solution is 9.78. Now the pOH is defined as -log [OH^-].
If the pOH of a solution is given, one may obtain the pH of such solution from the formula;
pH + pOH =14
Hence we can write;
pH = 14-pOH
pH = 14 - 9.78 = 4.22
Hence the pH of the solution is 4.22.
The new hybrid car can get 51.5 km/gal. It has a top speed of 40000.00 cm/min and is 4m long. How fast can the car go in m/hr?
Answer:
The anawer of this question is 0.024 m/h
Explanation:
Other explanations of the question are additional.
what mass of aluminum nitrate do you need to prepare 3.58L of a 1.77M Solution?
The condensation of chromosomes is inducted in?
Answer:
Chromosome condensation, the landmark event at the onset of prophase, often begins in isolated patches of chromatin at the nuclear periphery. Later, chromosome condense into two threads termed sister chromatids that are closely paired along their entire lengths.
Explanation:
hope dis help & may I have brainly plz
What is the value of the equilibrium constant, K, for a reaction for which ∆G° is equal to –5.20 kJ at 50°C?
Answer:
6.93
Explanation:
Step 1: Given data
Standard Gibbs free energy (∆G°): -5.20 kJTemperature (T): 50°CEquilibrium constant (K): ?Step 2: Convert the temperature to the Kelvin scale
We will use the following expression.
K = °C + 273.15
K = 50°C + 273.15
K = 323 K
Step 3: Calculate K
We will use the following expression.
∆G° = -R × T × ln K
-5.20 × 10³ J = -(8.314 J/mol.K) × 323 K × ln K
K = 6.93
HELP ASAS 15 POINTS
When using the process of evaporation to separate a mixture, what is left behind in the evaporating dish?
A. None of these.
B. The liquid evaporates and the solid is left in the dish.
C. The mixture does not separate, and the entire mixture evaporates.
D. The mixture does not separate, and the entire mixture remains in the dish.
Answer:
liquid will be evaporated while solid remains
The pH of an acid solution is 5.82. Calculate the Ka for the monoprotic acid. The initial acid concentration is 0.010 M.
Answer:
The answer is
[tex]Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3} [/tex]Explanation:
The Ka of an acid when given the pH and concentration can be found by
[tex]pH = - \frac{1}{2} log(Ka) - \frac{1}{2} log(c) [/tex]where
c is the concentration of the acid
From the question
pH = 5.82
c = 0.010 M
Substitute the values into the above formula and solve for Ka
We have
[tex]5.82 = - \frac{1}{2} log(Ka) - \frac{1}{2} log(0.010) [/tex][tex] - \frac{1}{2} log(Ka) = 5.82 + 1[/tex][tex] - \frac{1}{2} log(Ka) = 6.82[/tex]Multiply through by - 2
[tex] log(Ka) = - 13.64[/tex]Find antilog of both sides
We have the final answer as
[tex]Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3} [/tex]Hope this helps you
1. What's the concentration of hydronium ions if a water-base solution has a temperature of 25°C (Kw = 1.0x10-14), with a concentration of
hydroxide ions of 2.21x10-6 M? A. 3.1x10-6 M
B. 4.52 X10-9 M
C. 2.8x10-8 M
D. 1.6x10-9 M
Answer:
B. 4.52 X10-9 M
Explanation:
Our goal for this question is to calculate the concentration of hydronium ions [tex]H^+[/tex] produced by water in a vessel with a concentration of hydroxide ions of [tex]2.21X10^-^6~M[/tex]. So, our first approach can be the ionization reaction of water:
[tex]H_2O_(_l_)~->~H^+~_(_a_q_)~+~OH^-~_(_a_q_)[/tex]
If we write the Keq expression for this reaction we will have:
[tex]Keq=[H^+][OH^-][/tex]
Now, water is the universal solvent, so, Keq has a special name. In the equilibrium problems for water we have to use "Kw" instead of "Keq":
[tex]Kw=[H^+][OH^-][/tex]
From this equation, we know the Kw value () and the concentration of the hydroxide ions ([2.21X10^-^6~M]). If we replace these values into the equation we can solve for [tex][H^+][/tex]:
[tex]1.0X10^-^1^4=[H^+][2.21X10^-^6~M][/tex]
[tex][H^+]=\frac{1.0X10^-^1^4}{2.21X10^-^6}=4.52X^-^9[/tex]
I hope it helps!
Answer:
B. 4.52 * 10^-9M
Explanation:
did the test
Nitric oxide reacts with oxygen to give nitrogen dioxide, an important reaction in the Ostwald process for the industrial synthesis of nitric acid: 2NO(g)+O2(g)⇌2NO2(g)
Part A If Kc=6.9×105 at 227 ∘C,
what is the value of Kp at this temperature? Express your answer using two significant figures. Kp =
Part BIf Kp=1.3×10−2 at 1000 K, what is the value of Kc at 1000 K? Express your answer using two significant figures. Kc =
Answer:
Kp=1.68×10⁴∆1.7×10⁴
Kc=1.06∆1.1
Explanation:
Value of Kp at 227°C is 2.86×10² and value of Kc at 1000 K is 1.56.
How are Kp and Kc related?Kp and Kc are related by the formula Kp=Kc(RT).For part A , Kp is calculated as,
Kp=6.9×10⁵×8.314×500=28.683×10² and for part B Kc is calculated as,
Kc=1.3×10[tex]^-2[/tex]/(8.314×1000)=1.56
Kc and Kp are equilibrium constants of a mixture of ideal gases. Kp is equilibrium constant when concentrations at equilibrium are in atmospheric pressure and Kc is equilibrium constant when concentrations are in molarity. The relation is only valid for gaseous mixtures. The relation between these two parameters is obtained through ideal gas equation.
Kc and Kp of reaction change with temperature of reaction but remain unaffected by change in concentration , pressure and presence of catalyst.
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What is the relationship between concentration and rate of reaction?
not related
directly proportional
Inversely proportional
partially related
Explanation:
The answer is directly proportional, because when there is more concentration their will more reactants to react fast diring the chemical reaction which increases the rate of chemical reaction.
So, we can state that the relationship between them are directly proportional.
Hope it helps...
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many
The given question is incomplete.
The complete question is:
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction?
Answer: 4 grams of methane were needed for the reaction
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
{tex]CH_4+2O_2\rightarrow CO_2+H_2O[/tex]
Given: mass of oxygen = 16 g
Mass of carbon dioxide = 11 g
Mass of water = 9 g
Mass of products = Mass of carbon dioxide + mass of water = 11 g +9 g = 20 g
Mass or reactant = mass of methane + mass of oxygen = mass of methane + 16 g
As mass of reactants = mass of products
mass of methane + 16 g= 20 g
mass of methane = 4 g
Thus 4 grams of methane were needed for the reaction
If my primary DNA strand is ATACCGCAA
a write the complimentary DNA strand
Answer:
TATGGCGTT
Explanation:
Complimentary base pairs:
A-T
C-G
Use the other letter for complimentary strands
Question 6
Which compound is more soluble in water?
O 3-propyl-2-octanol
1-methyl-3-pentanol
O2-methyl-1-ethanol
02-ethyl-3-heptanol
mouthing
Answer:
ethyl - 3- heptanol mouthing
The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 3.11 g of water boils at atmospheric pressure?
Answer:
The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.
Explanation:
A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.
To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:
H: 1 g/moleO: 16 g/molethe molar mass of water is:
H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole
So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?
[tex]moles of water=\frac{3.11 grams*1 mole}{18 gramos}[/tex]
moles of water= 0.1728
Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?
[tex]heat=\frac{0.1728 moles*40.66 kJ}{1 mole}[/tex]
heat= 7.026 kJ
The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.
Consider the following reaction: 2Fe2+(aq) + Cu2+ --> 2Fe3+(aq) + Cu. When the ion concentrations change to the point where the reaction comes to equilibrium, what would be the cell voltage?
Answer:
At equilibrium, the cell voltage is zero volts.
Explanation:
During an electrochemical reaction, electrical energy is produced. The reaction continues to produce electrical energy until a point is reached in which the reaction attains equilibrium.
Before the reaction attains equilibrium, the cell voltage continues to decrease progressively as the reaction progresses. At equilibrium, the cell voltage becomes zero and the read out voltmeter records 0 V.
Hence, at equilibrium, the cell voltage is zero volts.
Which of the following reasons is why spent fuel rods are stored in a pool of water?
Answers
A.
Water recovers useful energy from the continuing reaction.
B.
Water increases the speed of the chain reaction in the fuel rods.
C.
Water acts as a radiation shield to reduce the radiation levels.
D.
Water dissolves the nuclear waste in the rods so it is easier to handle.
Answer:
C. water acts as a radiation shield to reduce the radiation level
What volume (in mL) needs to be added to 69.6 mL of 0.0887 M MgF2 solution to make a 0.0224 M MgF2 solution
Answer:
The correct answer is 206 ml.
Explanation:
Based on the given information, the molarity or M₁ of MgF₂ solution is 0.0887 M, the molarity or M₂ of the final solution given is 0.0224 M. The initial volume of V₁ of the solution is 69.6 ml, for finding the final volume of V₂ of the solution, the formula to be used is,
M₁V₁ = M₂V₂
Now putting the values in the formula we get,
0.0887 × 69.6 = 0.0224 M × V₂
V₂ = 0.0887 × 69.6 / 0.0224
V₂ = 275.6 ml
Therefore, the volume in ml added to the initial volume of 69.6 ml to make the molarity of the solution 0.0224 will be,
= 275.6 ml - 69.6 ml = 206 ml
If you have a polyatomic anion of Ammonium (NH41+), how many valence electrons must your Lewis Structure have?
Answer:
One can draw the 3-dimensional structure of an atom once they have the Lewis Structure of an atom. The 3-dimensional geometrical structure of ammonium, NH4+ is referred to as Tetrahedral. ... But the + sign decrees that NH4+ has 8 valence shell electrons, due to the positive ion.
Explanation:
Aqueous potassium nitrate (KNO3) and solid silver bromide are formed by the reaction of aqueous potassium bromide and aqueous silver nitrate (AgNO3). Write a balanced chemical equation for this reaction
Answer:
For the mentioned reaction, the balanced chemical equation is:
KBr (aq) + AgNO3 (s) ⇒ KNO3 (aq) + AgBr (s)
The number written in front of the ion, atoms, and molecules in a chemical reaction so that each of the elements on both the sides of reactants and products of the equation gets balanced is known as the stoichiometric coefficient.
From the mentioned balanced equation, the stoichiometric coefficient before KBr is 1, AgNO3 is 1, KNO3 is 1, as well as before AgBr is also 1. Thus, it is clear that 1 mole of potassium bromide reacts with 1 mole of silver nitrate to produce 1 mole of potassium nitrate and 1 mole of silver bromide.
Please help due tomorrow
Answer:
OPTION C
Explanation:
AS IN HUMANS THE DNA COMES FROM BOTH MALE AND FEMALE-SPERM AND OVA . SO 23 CHROMOSOMES FROM FATHER AND 23 CHROMOSOME FROM MOTHER
g Use the References to access important values if needed for this question. A researcher took 2.592 g of a certain compound containing only carbon and hydrogen and burned it completely in pure oxygen. All the carbon was changed to 7.851 g of CO2, and all the hydrogen was changed to 4.018 g of H2O . What is the empirical formula of the original compound
Answer:
Empirical formula is: C₂H₅
Explanation:
The chemical equation of burning of a compound that conatins only Carbon and Hydrogen is:
CₓHₙ + O₂ → XCO₂ + n/2H₂O
That means the moles of CO₂ produced are the moles of Carbon in the compound and moles of hydrogen are twice moles of water. Empirical formula is the simplest ratio between moles of each element in the compound. Thus, finding molse of C and moles of H we can find empirical formula:
Moles C and H:
Moles C = Moles CO₂:
7.851g CO₂ ₓ (1mol / 44g) = 0.1784 moles CO₂ = Moles C
Moles H = 2 Moles H₂O
4.018g H₂O ₓ (1mol / 18.01g) = 0.2231 * 2 = 0.4417 moles H
Ratio C:H
The ratio between moles of hydrogen and moles of Carbon are:
0.4417 moles H / 0.1784 moles C = 2.5
That means there are 2.5 moles of H per mole of Carbon. As empirical formula must be given only in whole numbers,
Empirical formula is: C₂H₅A student mixes 1.0 mL of aqueous silver nitrate, AgNO3 (aq), with 1.0 mL of aqueous sodium chloride, NaCl (aq), in a clean test tube. What will the student observe
Answer:
AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.
Explanation:
If a student mixes 1.0 mL of aqueous silver nitrate AgNO3 (aq) with 1.0 mL of aqueous sodium chloride, NaCl (aq), in a clean test tube.
The sodium chloride is being acidified with dilute trioxonitrate (V) acid. Then a few drops of silver trioxonitrate(V) is added afterwards. A white precipitate of silver chloride, which dissolves readily in aqueous ammonia indicates the presence of sodium chloride.
The reaction proceeds as follows:
[tex]\mathtt{AgNO_{3(aq)} + NaCl _{(aq)} \to AgCl _{(s)} + NaNO_3_{(aq)}}[/tex]
From the reaction between AgNO3 (aq) and NaCl (aq), AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.
Which processes occur during the second stage of technological design? Check all that apply.
designing a solution
studying relevant information
rebuilding and retesting
reporting a solution
defining criteria of success
identifying a problem
building a prototype
In which pair do both compounds exhibit predominantly ionic bonding? A) KCl and CO2 B) SO2 and BaF2 C) F2 and N2O D) N2O3 and Rb2O E) NaF and SrO
Answer:
E) NaF and SrO
Explanation:
The ionic bonding occurs between atoms with a great difference in electronegativity. This usually happens between a metal and a non-metal.
In which pair do both compounds exhibit predominantly ionic bonding?
A) KCl and CO₂. NO. C and O are non-metals and present covalent bonding.
B) SO₂ and BaF₂. NO. S and O are non-metals and present covalent bonding.
C) F₂ and N₂O. NO. Both compounds contain non-metals and present covalent bonding.
D) N₂O₃ and Rb₂O. NO. N and O are non-metals and present covalent bonding.
E) NaF and SrO. YES. Na and Sr are metals while F and O are non-metals.
During which phase do the centromeres split, allowing the two linked chromatids to separate?
Answer:
Anaphase
Explanation:
The centromere splits during the anaphase of the cell division. Thus, allowing the two linked chromatids to separate.
A typical chromosome is made up of two sister chromatids joined together by a structure known as the centromere. During cell division - at the metaphase stage - the chromosomes align at the equator of the cell, forming the metaphase plate. The spindle from the opposing ends of the cell engages each chromosome at the kinetochore of the centromere.
At the anaphase stage, the centromere splits, leading to the separation of the sister chromatids of each chromosome. The sister chromatids of the same then start migrating in the opposite direction as a result of the shortening of the spindle fiber.