Answer:
768g
Explanation:
We can use to formula [tex]N(A) = N_0(\frac{1}{2})^\frac{t}{t_{1/2}}[/tex] . Here, N(A) is the final amount. N0 is the initial amount. t is the time elapsed, and [tex]t_{1/2}[/tex] is the half life. Plugging in, we get the answer above.
The half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, 800 gram were in the original sample.
What is half life?The half-life (symbol t12) is the amount of time it takes for a volume (of material) to be reduced to half of its original value. In nuclear physics, the phrase is typically used to indicate how rapidly unstable atoms experience radioactive decay or even how long stable nuclei survive.
The phrase is sometimes used more broadly to describe any form of exponential (or, in rare cases, non-exponential) decay. The biological ½ of medications and other compounds in the human body, for example, is referred to in the medical sciences. In exponential growth, the inverse of half-life is doubling time.
ln P = kt + C
P = amount of I-137 at time t
C = constant
k = 1/time
t = time
1st condition:
P = Po, t = 0 days
2nd condition: (half-life)
P = 0.5Po, t = 8.07 days
3rd condition:
P = 25 grams, t = 40.35 days
Po = 800 grams
mass of I-137 = 800 gram
Therefore, 800 gram were in the original sample.
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3 points
18) A student determines the density of gold to be 20.9g/L. The true
density of gold is 19.30g/L. What is the student's percent error?round
answer to 2 significant figures *
Answer:
The answer is 8.29 %Explanation:
The percentage error of a certain measurement can be found by using the formula
[tex]P(\%) = \frac{error}{actual \: \: number} \times 100\% \\ [/tex]
From the question
actual density = 19.30g/L
error = 20.9 - 19.3 = 1.6
We have
[tex]p(\%) = \frac{1.6}{19.3} \times 100 \\ = 8.290155440...[/tex]
We have the final answer as
8.29 %Hope this helps you
How many liters of chlorine gas at 25°C and 0.950 atm can be produced by the reaction of 12.0 g of MnO2 with excess HCl(aq) according to the following chemical equation?
MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)
Answer:
3.55 L.
Explanation:
We'll begin by calculating the number of mole in 12 g of MnO2. This can be obtained as follow:
Molar mass of MnO2 = 55 + (16×2)
= 55 + 32
= 87 g/mol
Mass of MnO2 = 12 g
Mole of MnO2 =...?
Mole = mass /Molar mass
Mole of MnO2 = 12 / 87
Mole of MnO2 = 0.138 mole
Next, we shall determine the number of mole Cl2 produced from the reaction. This is illustrated below:
The balanced equation for the reaction is given below:
MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)
From the balanced equation above,
1 mole of MnO2 reacted to produce 1 mole of Cl2.
Therefore, 0.138 mole of MnO2 will also produce 0.138 mole of Cl2.
Finally, we shall determine the volume of Cl2 gas obtained from the reaction. This can be obtained as shown below:
Temperature (T) = 25 °C = 25 °C + 273 = 298 K
Pressure (P) = 0.950 atm
Number of mole (n) = 0.138 mole
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =.?
PV = nRT
0.950 × V = 0.138 × 0.0821 × 298
Divide both side by 0.950
V = (0.138 × 0.0821 × 298) / 0.950
V = 3.55 L
Therefore, 3.55 L of chlorine gas were obtained from reaction.
If it takes 26.0 mL of 0.0250 M potassium dichromate to titrate 25.0 mL of a solution containing Fe2 , what is the molar concentration of Fe2
Answer:
Explanation:
moles of potassium dichromate = .0250 x .026 = 65 x 10⁻⁵ moles
1 mole of potassium dichromate reacts with 6 moles of Fe⁺²
65 x 10⁻⁵ moles of potassium dichromate will react with
6 x 65 x 10⁻⁵ moles of Fe⁺²
= 390 x 10⁻⁵ moles
390 x 10⁻⁵ moles are contained in 25 mL of solution
molarity of solution = 390 x 10⁻⁵ / 25 x 10⁻³
= 15.6 x 10⁻² M .
The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).
Complete Question
The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).
(a) before addition of any HCl (b) after addition of 25.0 mL of HCl
Answer:
a The value is [tex]pH =12.81[/tex]
b [tex]pH = 11.9[/tex]
Explanation:
From the question we are told that
The first pKb value for B is [tex]pK_b_1 = 2.10[/tex]
The second pKb value for B is [tex]pK_b_2 = 7.54[/tex]
The volume is [tex]V = 50.0 mL =[/tex]
The concentration of B is [tex][B] = 0.60 M[/tex]
The concentration of [tex]C_A = 0.60 M[/tex]
Generally the reaction equation showing the first dissociation of B is
[tex]\ce{B_{(aq) } + H_2O _{(l)} <=> BH^+ _{(aq)} + OH^- _{(aq)} }[/tex]
Here the ionic constant for B is mathematically represented as
[tex]K_i = \frac{[BH^+] [OH^-]}{[B]}[/tex]
Let denot the concentration of [BH^+] as z and since [tex][BH^+] = [OH^-][/tex] then [tex][OH^-][/tex] is also z
So [B] = 0.60 - z
Here [tex]K_i[/tex] is ionic constant for the first reaction of a dibasic base B and the value is
[tex]K_i = 7.94 *10^{-3}[/tex]
So
[tex] 7.94 *10^{-3}= \frac{z^2}{ 0.60 - z}[/tex]
=> [tex]z^ 2 + 0.00794 z - 0.00476[/tex]
using quadratic formula to solve this equation
[tex]z = 0.0651[/tex]
Hence the concentration of [tex]OH^{-}[/tex] is [tex][OH^-] =0.0651[/tex]
Generally [tex]pOH = -log [OH^-][/tex]
=> [tex]pOH = -log (0.065)[/tex]
=> [tex]pOH = 1.187 [/tex]
Generally the pH is mathematically represented as
[tex]pH = 14 - 1.187[/tex]
[tex]pH =12.81[/tex]
Generally the volume of [tex]HCl[/tex] at the second dissociation of the base B is [tex] 50 mL [/tex]
The volume of the [tex]HCl[/tex] half way to the first dissociation of the base is 25mL
Now the pOH at half way to the first dissociation of the base is
[tex]pOH = -log(K_i)[/tex]
=> [tex]pOH = -log(0.00794)[/tex]
=> [tex]pOH = 2.100[/tex]
Generally the pH after addition of 25.0 mL of HCl is
[tex]pH = 14 - 2.100[/tex]\
=> [tex]pH = 11.9[/tex]
The first dissociation's equation is as follows:
[tex]B(aq) + H_2O(l) \leftrightharpoons BH^{+} (aq) + OH^{-}(aq) \\\\[/tex]
Constant of base ionization
[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 7.94\times 10^{-3} = \frac{x\times x}{(0.95- x)} \\\\\to 7.94\times 10^{-3} = \frac{x^2}{(0.95- x)} \\\\\to x^2=7.94\times 10^{-3} (0.95-x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x = 0.0830\ M\\\\[/tex]
So,
[tex]\to [OH^{-}] = 0.0830\ M\\\\[/tex]
The second dissociation of the base equation is
[tex]BH^{+}\ (aq) + H_20\ (l) \leftrightharpoons BH_2^{2+}\ (aq) + OH^{-}\ (aq) \\\\[/tex]
Constant of base ionization
[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 3.2 \times 10^{-8} =\frac{y \times (0.0830+y)}{(0.0830- y)}\\\\[/tex]
[tex]\to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y = 3.2\times 10^{-8}[/tex]
So,
[tex]\to [OH^{-}] = 0.0830\ M \\\\\to pOH = 1.08 \\\\\to pH = 14.00 - pOH = 12.92\\\\[/tex]
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A sample of an ideal gas has a volume of 2.23 L at 289 K and 1.05 atm. Calculate the pressure when the volume is 1.08 L and the temperature is 304 K. P= atm
Answer:
2.28 atm
Explanation:
V₁ = 2.33L, V₂ = 1.08L
T₁ = 289K, T₂ = 304K
P₁ = 1.05 atm, P₂ = ?
Where V₁ and V₂ are initial and final volume respectively
T₁ and T₂ are initial and final temperature respectively
P₁ and P₂ are initial and final pressure respectively
The formula to be used here is the general gas equation:
P₁V₁/T₁=P₂V₂/T₂
1.05 × 2.23/289 = P₂ × 1.08/304
P₂ × 1.08 × 289 = 1.05 × 2.23 × 304
P₂ = (1.05 × 2.23 ×304) ÷ (1.08 × 289)
P₂ = 711.82 ÷ 312.12
P₂ = 2.28 atm
Stephen learned that there are two forces that keep the moon in orbit around Earth. How do these forces keep the moon from flying off into space?
A. Gravity keeps the moon in motion, and inertia attracts the moon toward Earth.
B. Gravity attracts the moon toward Earth, and inertia keeps the moon in motion.
C. Gravity attracts the moon toward Earth, and the distance keeps it from going further away.
D. Mass weighs the moon down so it stays close to Earth, and inertia keeps the moon in motion.
Answer:
b
Explanation:
Answer:
Gravity attracts the moon Earth, and Inertia keeps the moon in motion.
Explanation:
A student determines that the mass of a sample of aluminum is 8.3 g. How many moles are in the sample?
Answer:
0.307
Explanation:
use the formula of n=mass /molar mass
Use the data in the Successive Ionization Energies and Electron Affinity tables to determine the following. (Assume the values in the Successive Ionization Energies table are given to the ones place.) (a) the electron affinity of Ar2+ kJ/mol (b) the electron affinity of Si + kJ/mol (c) the ionization energy of Cl − kJ/mol (d) the ionization energy of Cl kJ/mol (e) the electron affinity of Cl + kJ/mol
Answer:
Use the data in the Successive Ionization Energies and Electron Affinity tables to determine the following. (Assume the values in the Successive Ionization Energies table are given to the ones place.)
(a)
the electron affinity of Ar2+
kJ/mol
(b)
the electron affinity of S+
kJ/mol
(c)
the ionization energy of Cl−
kJ/mol
(d)
the ionization energy of Cl
kJ/mol
(e)
the electron affinity of Cl+
kJ/mol
Successive Ionization Energies in Kilojoules per Mole for the Elements in Period 3 General increase 13 14 Element 11 12 I. 16
Explanation:
When converting an “ordinary” number that is greater than 1 to scientific notation, how many non-zero digits are to the LEFT of the decimal point when you are finished?
Answer: 0 I think
Explanation:
pretty sure its zero because i learned it last year but im in middle school so you might want to look it up.
If equal volumes of a strong base and a weaker acid are mixed together, what would you expect the pH of the resulting salt to be
Answer:
Above 7
Explanation:
The equivalence point of any titration can be read off from the appropriate titration curve.
A titration curve is a plot of the pH of analyte against the volume of titrant added.
For a strong base and weak acid, the equivalence point lies above 7.
The pH of the resulting salt to be pH> 7 .
What does Equivalence point tell?The equivalence point of any titration can be read off from the appropriate titration curve. A titration curve is a plot of the pH of analyte against the volume of titrant added. It is a point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution. At the equivalence point in an acid-base titration.
For a strong base and weak acid, the equivalence point lies above 7.
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An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 35.0 atm and releases 74.6 kJ of heat. Before the reaction, the volume of the system was 8.20 L . After the reaction, the volume of the system was 2.80 L . Calculate the total internal energy change, ΔE, in kilojoules.
Answer:
ΔU = −55.45 kJ
Explanation:
From first law of thermodynamics in chemistry, we have;
ΔU = Q + W
where;
ΔU is change in internal energy
Q is the net heat transfer
W is the net work done
We are given;
Q = 74.6 kJ
But Q will be negative since heat is released
Thus;
ΔU = -74.6 kJ + W
We are given;
Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²
Volume before reaction; Vi = 8.2 L = 0.0082 m³
Volume after reaction; V_f = 2.8 L = 0.0028 m³
Now,
W = -P(V_f - V_i)
W = - 3546375(0.0028 - 0.0082)
W = 19.15 KJ
Thus;
ΔU = Q + W
ΔU = -74.6 kJ + 19.15 KJ =
ΔU = −55.45 kJ
Consider the balanced equation below. Upper P Upper C l Subscript 3 Baseline + Upper C l Subscript 2 Baseline right arrow Upper P Upper C l Subscript 5. What is the mole ratio of PCl3 to PCl5? 1:1 2:1 3:5 5:3
Answer : The mole ratio of [tex]PCl_3[/tex] to [tex]PCl_5[/tex] is 1 : 1.
Explanation :
Balanced chemical reaction : It is a chemical reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.
The balanced chemical reaction is:
[tex]PCl_3+Cl_2\rightarrow PCl_5[/tex]
By the stoichiometry of the reaction we can say that 1 mole of [tex]PCl_3[/tex] reacts with 1 mole of [tex]Cl_2[/tex] to give 1 mole of [tex]PCl_5[/tex].
From this we conclude that the mole ratio of [tex]PCl_3[/tex] to [tex]PCl_5[/tex] is 1 : 1.
Hence, the mole ratio of [tex]PCl_3[/tex] to [tex]PCl_5[/tex] is 1 : 1.
Answer:
A
Explanation:
The solubility of silver(I)phosphate at a given temperature is 2.43 g/L. Calculate the Ksp at this temperature. After you calculate the Kspvalue, take the negative log and enter the (pKsp) value with 2 decimal places.
Answer:
Kps = 3.07 x 10⁻⁸
pKsp= 7.51
Explanation:
First, we calculate the molar solubility of silver(I)phosphate (Ag₃PO₄) from the solubility in g/L by using its molar mass (418.6 g/mol):
2.43 g/L x 1 mol/418.6 g = 5.8 x 10⁻³ mol/L= s
Now, we have to write the ICE chart for the aqueous equilibrium of Ag₃PO₄ as follows:
Ag₃PO₄(g) ⇄ 3 Ag⁺(aq) + PO₄³⁻
I 0 0
C +3s +s
E 3s s
Ksp = [Ag⁺]³[PO₄³⁻]= (3s)³s= 27s⁴
Since s=5.8 x 10⁻³ mol/L, we calculate Ksp:
Ksp= 27(5.8 x 10⁻³ mol/L)⁴= 3.07 x 10⁻⁸
The pKsp value is:
pKsp= - log Ksp = -log (3.07 x 10⁻⁸) = 7.51
What does this diagram represent?
Answer:
Linear molecule with two domains
Explanation:
What are the two ways
that heat is measured?
Answer:
heat is mesured in calories and also joules
Explanation:
Sodium carbonate, also known as soda ash, is used in glassmaking. It is obtained from a reaction between sodium chloride and calcium carbonate; calcium chloride is the other product. Calculate the percent yield of sodium carbonate if 92.6 g is collected when 112. g of sodium chloride reacts with excess calcium carbonate.
Answer:
The percentage yield of sodium carbonate is 91.47%
Explanation:
we start by writing the reaction equation:
2NaCl + CaCO3 ——-> Na2CO3 + CaCl2
From the reaction we can see that 2 moles of sodium chloride produced 1 mole of sodium carbonate
Let us calculate the actual number of moles of sodium chloride produced from 112 g of it
Mathematically,
number of moles = mass/molar mass
Molar mass of sodium chloride is 23 + 35.5 = 58.5 g/mole
So the number of moles of sodium chloride produced will be 112/58.5 = 1.91 moles
The number of moles of sodium carbonate produced is half of this = 1.91/2 = 0.955
The mass of sodium carbonate produced from 0.955 moles of it will be;
number of moles * molar mass
The molar mass of sodium carbonate is 106 g/mol
So the number of moles is = 0.955 * 106 = 101.23 g
Mathematically;
percentage yield = actual yield/theoretical yield * 100%
Percentage yield = 92.6/101.23 * 100% = 91.47%
One way to represent a substance is with a chemical formula. In the formula CO2, what do the symbols Cand o refer to?
Answer:
C is for carbon and O is for oxygen
twelve grams of sodium chloride wwere dissolved in 52 ml (52g) of distilled water, calculate the % sodium chloride in the solution
Answer:
The mass of sodium chloride in the mixture is 18.75%
Explanation:
Here, we want to calculate the percentage of sodium chloride in the mixture.
The total mass of the mixture is 52 g + 12 g = 64 g
So the percentage mass of sodium chloride will be;
mass of sodium chloride/ Total mass * 100%
That will be: 12/64 * 100 = 18.75%
what is so unique about water that hydrogen bonding becomes possible
Answer:
Water molecules are polar, so they form hydrogen bonds. This gives water unique properties, such as a relatively high boiling point, high specific heat, cohesion, adhesion and density.
EXPLANATION :
So it is unusual for water to be a liquid at room temperature! Water is liquid at room temperature so it's able to move around quicker than it is as solid, enabling the molecules to form fewer hydrogen bonds resulting in the molecules being packed more closely together.
Atomic radius is....
O The tendency for an atom to attract electrons
The energy required to remove an electron
O The energy required to add an electron
O The distance from the nucleus to the last orbital
What volume of 1.27 M HCl is required to prepare 197.4 mL of 0.456 M HCl
Answer:
70.88 mL volume of 1.27 M of HCl is required.
Explanation:
Given data:
Initial volume = ?
Initial molarity = 1.27 M
Final volume = 197.4 mL
Final molarity = 0.456 M
Solution:
Formula:
M₁V₁ = M₂V₂
Now we will put the values in formula.
1.27 M × V₁ = 0.456 M × 197.4 mL
V₁ = 0.456 M × 197.4 mL/1.27 M
V₁ = 90.014M.mL/1.27 M
V₁ = 70.88 mL
70.88 mL volume of 1.27 M of HCl is required.
1
An atom of element Q contains 19 electrons, 19 protons and 20 neutrons.
What is Q?
A calcium
B potassium
С
strontium
D
yttrium
Answer:
b)Potassium is the right answerAnswer:
B. Potassium
Explanation:
The element with 19 electrons, 19 protons, and 20 neutrons is potassium
Which pair of elements would most likely have a similar arrangement of outer
electrons and have similar chemical behaviors?
boron and aluminum
helium and fluorine
carbon and nitrogen
chlorine and oxygen
Answer:
Boron and Aluminum
Explanation:
If you write the electron configuration for boron and aluminum, you get:
[tex]1s^22s^22p^1[/tex] for boron and [tex]1s^22s^22p^63s^23p^1[/tex] for aluminum. Both have 3 valance electrons and has 2 electrons in a s-orbital and 1 in a p-orbital. These valance electron similarities are based on the column/group the elements are. Therefore, Boron and Aluminum have similar chemical behaviours and similar arrangement of outer/valance electrons.
Which of the following is a good definition of matter?
O A. Anything that is made up of light and gravity
O B. Anything that has mass and takes up space
O C. Anything that produces heat and mass
O D. Anything that has energy and creates heat
Answer:
B
Explanation:
I did the question before and got it right.
H₂C=CH-CH₂-CH=CH₂
how many single and double bonds
Answer:
2 double bonds 2 single bonds
Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be synthesized using a two-step thermal process. In the first step, phosphorus and oxygen react to form diphosphorus pentoxide: P4(l)+5O2(g-2 P20s(g) In the second step, diphosphorus pentoxide and water react to form phosphoric acld P20(9)+3 H200 2H,PO40) Write the net chemical equation for the production of phosphoric acid from phosphorus, oxygen and water.
Answer:
P₄(l) + 5 O₂(g) + 6 H₂O(l) ⇒ 4 H₃PO₄(aq)
Explanation:
Phosphoric acid is synthesized using a two-step thermal process.
In the first step, phosphorus and oxygen react to form diphosphorus pentoxide. The corresponding chemical equation is:
P₄(l) + 5 O₂(g) ⇒ 2 P₂O₅(g)
In the second step, diphosphorus pentoxide and water react to form phosphoric acid. The corresponding chemical equation is:
P₂O₅(g) + 3 H₂O(l) ⇒ 2 H₃PO₄(aq)
We can get the net chemical equation by adding the first step, the second step multiplied by 2, and canceling what is repeated on both sides.
P₄(l) + 5 O₂(g) + 2 P₂O₅(g) + 6 H₂O(l) ⇒ 2 P₂O₅(g) + 4 H₃PO₄(aq)
P₄(l) + 5 O₂(g) + 6 H₂O(l) ⇒ 4 H₃PO₄(aq)
Determine the volume of 15.5 g of a substance with a density of 6.89 g/ml
Answer:
The answer is 2.25 mLExplanation:
The volume of a substance when given the density and mass can be found by using the formula
[tex]volume = \frac{mass}{density} \\ [/tex]
From the question
mass = 15.5 g
density = 6.89 g/mL
We have
[tex]volume = \frac{15.5}{6.89} \\ = 2.249637155...[/tex]
We have the final answer as
2.25 mLHope this helps you
Hydrogen has 3 isotopes. Hydrogen 1, Hydrogen 2 and Hydrogen 3. What is the difference between these 3 is isotopes
Answer:
Number of neutrons
Explanation:
All have one single proton. Hydrogen has no neutrons. Hydrogen 2 or deuterium has 1 neutron. Hydrogen 3 or tritium has 2 neutrons.
A volume of 80.0 mL of a 0.690 M HNO3 solution is titrated with 0.790 M KOH. Calculate the volume of KOH required to reach the equivalence point. Express your answer to three significant figures and include the appropriate units.
Given :
A volume of 80.0 mL of a 0.690 M [tex]HNO_3[/tex] solution is titrated with 0.790 M KOH.
To Find :
The volume of KOH required to reach the equivalence point.
Solution :
We know, at equivalent point :
moles of [tex]HNO_3[/tex] = moles of KOH
[tex]M_{HNO_3}V_{HNO_3}=M_{KOH}V_{KOH}\\\\0.690\times 80 = 0.790\times V_{KOH}\\\\V_{KOH}=\dfrac{0.690\times 80 }{ 0.790}\ ml\\\\V_{KOH}=69.87\ ml[/tex]
Therefore, volume of KOH required is 69.87 ml.
Hence, this is the required solution.
Which accurately represents these building blocks of matter from the smallest to the largest?
atom -- molecule or compound
O molecule -- atom - element
compound - molecule -- element
molecule atom or element
Answer:
A - Atom ---> molecule or compound.