Answer:
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
Explanation:
Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.
The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:
Q = c * m * ΔT
Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).
In this case:
[tex]c=4.18 \frac{J}{g*K}[/tex]m= 1.30 kg= 1,300 g (1 kg=1,000 g)ΔT= 34.2 °C - 22.4 °C= 11.8 °C= 11.8 °K Being a temperature difference, it is independent if they are degrees Celsius or degrees Kelvin. That is, the temperature difference is the same in degrees Celsius or degrees Kelvin.Replacing:
[tex]Q=4.18 \frac{J}{g*K}*1,300 g*11.8 K[/tex]
Q= 64,121.2 J
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
Predict the order of acid strengths in the following series of cationic
species: CH3CH2NH3
+, CH3CH=NH2
+ and CH3C-NH+
Answer:
CH3C-NH+> CH3CH=NH2 >CH3CH2NH3+
Explanation:
The acid strength has to do with the ease of the loss of hydrogen ion from the cationic specie. Hydrogen ion will easily be lost from any specie which contains an atom, group of atoms or bond which withdraws electrons along the chain of the N-H bond.
The pi bond system is known to be highly electronegative and withdraws electrons along the chain hence a withdrawal of electron density along the chain which makes the hydrogen ion to be easily lost from a system which contains a pi bond along the chain. A triple bond is more electronegative than a double bond, hence the answer above.
You have to prepare a pH 3.65 buffer, and you have the following 0.10M solutions available: HCOOH, CH3COOH, H3PO4, HCOONa, CH3COONa, and NaH2PO4. How many mL of HCOOH and HCOONa would you use to make approximately a liter of the buffer?
Answer:
550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M
Explanation:
It is possible to find the pH of a buffer by using H-H equation:
pH = pKa + log [A⁻]/[HA]
For the formic buffer (HCOOH/HCOONa):
pH = 3.74 + log [HCOONa]/[HCOOH]
As you need a buffer of pH 3.65:
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)Where [HCOONa]/[HCOOH] can be taken as the moles of each specie.
As molarity of both solutions is 0.10M (0.10mol / L) and you need 1L of solution, total moles of the buffer are:
0.10 moles = [HCOONa] + [HCOOH] (2)Replacing (2) in (1):
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 molesAnd moles of HCOONa are:
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 molesAs concentration of the solutions is 0.1M, the volume you need to add of both solutions is:
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
The number should be considered like 550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M.
Calculation of mL:Here we used the H-H equation:
pH = pKa + log [A⁻]/[HA]
Now
For the formic buffer (HCOOH/HCOONa):
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
Now
need a buffer of pH 3.65:
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)
here [HCOONa]/[HCOOH] can be considered as the moles of each specie.
Now the total moles should be
0.10 moles = [HCOONa] + [HCOOH] (2)
Now
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 moles
And moles of HCOONa should be
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 moles
Now
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
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A 1 liter solution contains 0.436 M hypochlorous acid and 0.581 M potassium hypochlorite. Addition of 0.479 moles of barium hydroxide will: (Assume that the volume does not change upon the addition of barium hydroxide.)
Answer:
Exceed the buffer capacity and Raise the pH by several units
Explanation:
Options are:
Raise the pH slightly
Lower the pH slightly
Raise the pH by several units
Lower the pH by several units
Not change the pH
Exceed the buffer capacity
The hypochlorous acid, HClO, is in equilibrium with Hypochlorite ion (From potassium hypochlorite, ClO⁻) producing a buffer. Using H-H equation, pH of initial buffer is:
pH = pKa + log [ClO⁻] / [HClO]
pKa for hypochlorous acid is 7.53
pH = 7.53 + log [0.581M] / [0.436M]
pH = 7.65
Barium hydroxide reacts with HClO producing more ClO⁻, thus:
Ba(OH)₂ + 2HClO → 2ClO⁻ + 2H₂O
As 0.479 moles of Barium hdroxide are added. For a complete reaction you require 0.479mol * 2 = 0.958 moles of HClO
As you have just 0.436 moles (Volume = 1L),
The addition will:
Exceed the buffer capacityThe Ba(OH)₂ that reacts is:
0.436 moles HClO * (1mole (Ba(OH)₂ / 2 mol HClO) = 0.218 moles Ba(OH)₂ and will remain:
0.479 mol - 0.218 mol = 0.261 moles Ba(OH)₂
As 1 mole of Ba(OH)₂ contains 2 moles of OH⁻, moles of OH⁻ and molarity is:
0.261 moles* 2 = 0.522 moles OH⁻ = [OH⁻]
pOH = -log [OH⁻]
pOH = 0.28
And pH = 14 - pOH:
pH = 13.72
Thus, after the addition the pH change from 7.65 to 13.62:
Raise the pH by several units
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
The ceramic glaze on a red-orange Fiestaware plate is U2O3 and contains 50.1 grams of 238U, but very little 235U. (a) What is the activity of the plate (in Ci)
Answer:
The correct answer is 1.68 × 10⁻⁵ Ci
Explanation:
The activity of the uranium is determined by using the formula,
R = 0.693 N/t1/2 -------------- (i)
The number of atoms is, N = nNA
Here, NA is the Avogadro number and n is the number of moles. The value of n is m/M, that is, mass/molecular mass. Now the value of N becomes,
N = (m/M) NA
The m or mass of uranium given is 50.1 grams, and the molecular mass is 238 g/mol, now putting the values we get,
N = (50 g/238 g) (6.023 × 10²³) = 1.26 × 10²³
The half-life of 238U from year to second is,
t1/2 = (4.468 × 10⁸ year) (3.16 × 10⁷ s/ 1 year) = 1.412 × 10¹⁶ s
Substituting the values of t1/2 as 1.412 × 10¹⁶, and 1.26 × 10²³ for N in equation (i) we get,
R = 0.639 (1.26 × 10²³) / 1.412 × 10¹⁶ s
= 6.18 × 10⁶ Bq (2.7027 × 10⁻¹¹ Ci/1 Bq)
= 1.68 × 10⁻⁵ Ci
Hence, the activity of the plate is 1.68 × 10⁻⁵ Ci
11mg of cyanide per kilogram of body weight is lethal for 50% of domestic chickens. How many grams per kilogram of body weight is a lethal dose for 50% of domestic chickens?
Answer:
[tex]0.033g[/tex]
Explanation:
Hello,
In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:
[tex]11mg\longrightarrow 1kg\\\\x\ \ \ \ \ \ \longrightarrow 3kg[/tex]
Thus, we obtain:
[tex]x=\frac{3kg*11mg}{1kg}\\ \\x=33mg[/tex]
That in grams is:
[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]
Regards.
How many cups are in five gallons?
Answer:
In 5 US liquid gallons, there are 80 cups.
Explanation:
To get from gallons to cups, just multiply the amount of gallons you have by 16.
What are the conditions that are favorable for extensive solid solubility of one element in another (Hume-rothery rules)
Answer:
Atomic radius less than 15%, similar structure and same valency.
Explanation:
The conditions that are favorable for extensive solid solubility of one element in another are the following.
The atomic radius of the solute and solvent atoms must be less than 15%. The structure of both solute and solvent are similar. Solubility completes when both have same valency. Valency means number of electrons in the outermost shell. If both solute and solvent has same number of electrons so it will be completely soluble in each other.
The conditions that are favorable for extensive solid solubility of one element in another is the same size, electrongativity and valency.
What is Hume - Rothery rules?Hume - Rothery rules are the sets of some important rules which gives idea about the desired condition for the formation of solid solution.
Following main points are described in this rule:
Difference between the size of the solute and the solvent should be less than 15%.Electronegativity difference between the solute and solvents should be small.And they both should have same valency, means same no. of electrons in the outermost shell.Hence size, electronegativity and valency are the conditions.
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What happens when an atom of sulfur combines with two atoms of chlorine to produce SCl2?
Answer:
Each chlorine atom shares a pair of electrons with the sulfur atom. ... Each chlorine atom shares all its valence electrons with the sulfur atom
When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are: potassium hydrogen sulfate (aq) potassium hydroxide (aq) potassium sulfate (aq) water (l)\
Answer:
Explanation:
Answer:
1, 1, 1, 1
Explanation:
potassium hydrogen sulfate + potassium hydroxide ⟶ potassium sulfate + water(l)
KHSO₄ + KOH ⟶ K₂SO₄ + H₂O
1. Put a 1 in front of the most complicated-looking formula (K₂SO₄?):
KHSO₄ + KOH ⟶ 1K₂SO₄ + H₂O
2. Balance S:
We have fixed 1 S on the right. We need 1 S on the left. Put a 1 in front of KHSO₄ to fix it.
1KHSO₄ + KOH ⟶ 1K₂SO₄ + H₂O
3. Balance K:
We have fixed 2 K on the right and 1 K on the left. We need 1 more K on the left. Put a 1 in front of KOH.
1KHSO₄ + 1KOH ⟶ 1K₂SO₄ + H₂O
4. Balance O
We have fixed 4 O on the right and 5 O on the left. We need 1 more O on the right. Put a 1 in front of H₂O.
1KHSO₄ + 1KOH ⟶ 1K₂SO₄ + 1H₂O
Every formula has a coefficient. The equation should be balanced.
5. Check that atoms balance:
[tex]\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{K} & 2 &2\\\text{H} & 2 & 2\\\text{S} & 1 & 1\\\text{O}&5&5\\\end{array}[/tex]
It checks.
The coefficients are 1, 1, 1, 1.
Write the Arrhenius Base reaction for the following:
Sr(OH)2
Answer:
Explanation:
Sr(OH)2 (aq) ⇔ Sr+2 (aq) + 2OH- (aq)
A chamber contains equal molar amounts of He, Ne, Ar, and Kr. If the total chamber pressure is 1 atm, then the partial pressure (in atm) of Kr is:
Answer:
The correct answer is 0.25 atm.
Explanation:
As mentioned in the given question that the chamber comprises equal molar concentrations of He, Ne, Ar and Kr gas. So, let us assume that the moles of all the gases will be x then the total number of moles will be 4x.
The formula for calculating mole fraction is,
Mole fraction = mole of the substance/total moles
The mole fraction of Kr = x/4x = 0.25
The total pressure given in the chamber is 1 atm. Therefore, the partial pressure will be,
Partial pressure = mole fraction * Total pressure
Partial pressure = 0.25 * 1 = 0.25 atm.
The partial pressure of Kr is 0.25 atm.
We are told that there is an equal molar amounts of He, Ne, Ar, and Kr. If we decide to say, let the molar amount of each gas be x, the total number of moles of all the gases will now be; x + x+ x + x = 4x
Let us recall that partial pressure of a gas can be obtained form the formula;
Partial pressure = mole fraction × total pressure
Mole fraction of Kr = x/4x = 1/4
Total pressure = 1 atm
Partial pressure of Kr = 1/4 × 1 = 0.25 atm
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In the experiment, you heated a sample of metal in a water bath and quickly removed and dried the sample prior to inserting it into the cold water "system". Why did you need to dry the sample prior to placing it into the cold water "system"
a. The extra water might react with the metal which would ruin the sample.
b. Any residual hot water would have added to the heat transferred from the metal to the cold water "system".
c. The wet metal would not transfer any heat thus causing inaccuracy in you measurements.
d. The metal would oxidize in the presence of water thus ruining the sample
Answer:
b
Explanation:
Provided the experiment is about the transfer of heat from one medium to another, any residual hot water on the metal would have added to the heat transferred from the metal to the cold water system.
The residual hot water on the metal posses its own heat and when such is transferred along with the metal into the cold water, the heat from the residual hot water will interfere with the measurement of the actual heat transferred to the cold water by the metal. Hence, the accuracy of the result would be impacted.
The correct option is b.
People take antacids, such as milk of magnesia, to reduce the discomfort of acid stomach or heartburn. The recommended dose of milk of magnesia is 1 teaspoon, which contains 500 mg of Mg(OH)2. What volume of HCl solution with a pH of 1.25 can be neutralized by 1 dose of milk of magnesia
Answer:
[tex]V_{HCl}=0.208L=208mL[/tex]
Explanation:
Hello,
In this case, since the chemical reaction is:
[tex]2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O[/tex]
We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:
[tex]n_{HCl}=2*n_{Mg(OH)_2}[/tex]
In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:
[tex]n_{Mg(OH)_2}=500mg*\frac{1g}{1000mg}*\frac{1mol}{58.3g} =0.00858mol[/tex]
Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:
[tex][H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M[/tex]
Then, since the concentration and the volume define the moles, we can write:
[tex][HCl]*V_{HCl}=2*n_{Mg(OH)_2}[/tex]
Therefore, the neutralized volume turns out:
[tex]V_{HCl}=\frac{2*0.00858mol}{0.0562\frac{mol}{L} }\\ \\V_{HCl}=0.208L=208mL[/tex]
Best regards.
acid-catalyzed hydration of 1-methylcyclohexene gives two alcohols. The major product does not undergo oxidation, while the minor product will undergo oxidation. Explain
Answer:
Major product does not undergo oxidation since it is a tertiary alcohol whereas minor product undergoes oxidation to ketone as it is secondary alcohol.
Explanation:
Hello,
In this case, given the attached picture, the hydration of the 1 methylcyclohexene yields to alcohols; 1-methylcyclohexan-1-ol and 1-methylcyclohexan-2-ol. Thus, since the OH in the 1-methylcyclohexan-1-ol (major product) is bonded to a tertiary carbon (bonded with other three carbon atoms) it is not able to increase the number of oxygen bonds (oxidation) as it already attained the octet whereas the 1-methylcyclohexan-2-ol (minor product) is able to undergo oxidation to ketone as the carbon bonded to it is secondary (bonded with other two carbon atoms), so one extra bond the oxygen is allowed to be formed to carbonyl.
Best regards.
Complete and balance the molecular equation for the reaction of aqueous sodium sulfate, Na2SO4, and aqueous barium nitrate, Ba(NO3)2. Include physical states.molecular equation:Na_{2}SO_{4}(aq) + Ba(NO_{3})_{2}(aq) ->Na2SO4(aq)+Ba(NO3)2(aq)⟶Enter the balanced net ionic equation for this reaction. Include physical states.net ionic equation:
Answer:
1. The balanced molecular equation is given below:
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)
2. The net ionic equation is given below:
SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)
Explanation:
1. The balanced molecular equation
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)
The above equation can be balance as follow:
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)
There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NaNO3 as shown below:
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)
Now, the equation is balanced.
2. The bal net ionic equation.
This can be obtained as follow:
Na2SO4(aq) + Ba(NO3)2(aq) —>
In solution, Na2SO4 and Ba(NO3)2 will dissociate as follow:
Na2SO4(aq) —> 2Na^+(aq) + SO4^2-(aq)
Ba(NO3)2(aq) —> Ba^2+(aq) + 2NO3^-(aq)
Na2SO4(aq) + Ba(NO3)2(aq) —>
2Na^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2NO3^-(aq) —> BaSO4(s) + 2Na^+(aq) + 2NO3^-(aq)
Cancel the spectator ions i.e Na^+ and NO3^- to obtain the net ionic equation.
SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)
Gaseous indium dihydride is formed from the elements at elevated temperature:
In(g)+H2(g)⇌InH2(g),Kp=1.48 at 973 K
The partial pressures measured in a reaction vessel are
PIn =0.0540atm
PH2= 0.0250atm
PInH2 =0.0780atm
Calculate Qp and give equal partial pressure for In, H2, and InH2.
Answer:
The reaction given is:
In (g) + H₂ (g) ⇔ InH₂ (g), the Kp is 1.48 at 973 K.
The partial pressures measured in the reaction vessel is Partial pressure of In is 0.0540 atm, partial pressure of H₂ is 0.0250 atm, and the partial pressure of InH₂ is 0.0780 atm. By using the table given in the attachment below, the value of PInH₂ is (0.078-x), PIn is (0.054 + x), and the value of PH2 is (0.025 + x).
Kp = PInH₂/PIn × PH₂ = (0.078 - x) / (0.054 +x) (0.025 + x)
1.48 = (0.078 - x) / (0.054 +x) (0.025 + x)
x = 0.06689
Now the partial pressures of In, H₂ and InH₂ will be,
PH₂ = 0.025 + x = 0.025 + 0.0668 = 0.0918 atm
PIn = 0.054 + 0.0668 = 0.1208 atm
PInH₂ = 0.078 - 0.0668 = 0.0112 atm
Now the Qp or the reaction quotient will be,
Qp = (0.078) / (0.054) (0.025) = 57.78.
Draw a structure for an alcohol that exhibits a molecular ion at M+ = 88 and that produces fragments at m/z = 73, m/z = 70 and m/z = 59.
Answer:
3-pentanol
Explanation:
In this case, we have alcohol as the main functional group (OH) with a molecular ion at 88. If the molecular ion is 88 the molar mass is also 88 g/mol therefore the formula for the unknown molecule is [tex]C_5H_1_2O[/tex].
Additionally, if the mass spectrum shows the molecular ion peak we can not have tertiary alcohols (tertiary alcohols often do not show M+ at all). So, the structures only can be primary and secondary structures.
With this in mind, our options are:
-) 1-pentanol
-) 2-pentanol
-) 3-pentanol
Now we can analyze each structure:
-) 1-pentanol
The structure must explain all the fragments produced (73, 70, and 59). In this primary alcohol, we will have an alpha cleavage (the red bond would be broken). If this has to happen, we will have fragments at 31 and 57. These fragments dont fit with the reported ones, therefore this is not a possible structure (See figure 1).
-) 2-pentanol
On this structure, we will have also an alpha cleavage (red bond). In this rupture we will have fragments at 45 and 43, these m/z values dont fit with the reported ones, therefore this is not a possible structure (See figure 1).
-) 3-pentanol
In this structure, we have the "OH" bonded to carbon three. So, we can analyze each fragment:
-) m/z 59
This fragment, can be explained as an alpha cleavage. But, in this case we have two ruptures that can produce the same ion. The carbons on both sides of the C-OH bond.
-) m/z 71
This fragment, can be explained as a loss of water (M-18) in which we have the production of a carbocation in the carbon where we previously have the C-OH bond.
-) m/z 73
This fragment, can be explained as a beta cleavage. But, in this case, also we have two ruptures that can produce the same ion. The methyl groups on each end molecule.
See figure 2
I hope it helps!
Assuming that you start with 21.4 g of ammonia gas and 18.0 g of sodium metal and assuming that the reaction goes to completion, determine the mass (in grams) of each product.
Answer:
[tex]m_{NaNH_2}=30.42gNaNH_2[/tex]
[tex]m_{H_2}=0.783gH_2[/tex]
Explanation:
Hello,
In this case, the reaction between sodium and ammonia is:
[tex]2Na+2NH_3\rightarrow 2NaNH_2+H_2[/tex]
Thus, as we know the initial masses of both sodium and ammonia, we should first identify the limiting reactant, for which we firstly compute the available moles of sodium:
[tex]n_{Na}=18.0gNa*\frac{1molNa}{23.0gNa}=0.783molNa[/tex]
And the moles of sodium consumed by 21.4 g of ammonia (2:2 mole ratio):
[tex]n_{Na}^{\ consumed}=21.4gNH_3*\frac{1molNH_3}{17gNH_3} *\frac{2molNa}{2molNH_3} =1.26molNa[/tex]
In such a way, since less moles of sodium are available than consumed by ammonia, we can say, sodium is the limiting reactant. Furthermore, the mass of both sodium amide (39 g/mol) and hydrogen gas (2 g/mol) that are produced turn out:
[tex]m_{NaNH_2}=0.783molNa*\frac{2molNaNH_2}{2molNa}*\frac{39gNaNH_2}{1molNaNH_2}=30.42gNaNH_2[/tex]
[tex]m_{H_2}=0.783molNa*\frac{1molH_2}{2molNa}*\frac{2gH_2}{1molH_2}=0.783gH_2[/tex]
Best regards.
The half-life of radium-226 is 1620 years. What percentage of a given amount of the radium will remain after 900 years
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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A quantity of liquid methanol, CH 3OH, is introduced into a rigid 3.00-L vessel, the vessel is sealed, and the temperature is raised to 500K. At this temperature, the methanol vaporizes and decomposes according to the reaction CH 3OH(g) CO(g) + 2 H 2(g), K c= 6.90×10 –2. If the concentration of H 2 in the equilibrium mixture is 0.426M, what mass of methanol was initially introduced into the vessel?
Answer:
74.3g of methanol were introduced into the vessel
Explanation:
In the equilibrium:
CH₃OH(g) ⇄ CO(g) + 2H₂(g)
Kc is defined as the ratio between concentrations in equilibrium of :
Kc = 6.90x10⁻² = [CO] [H₂]² / [CH₃OH]
Some methanol added to the vessel will react producing H₂ and CO. And equilibrium concentrations must be:
[CH₃OH] = ? - X
[CO] = X
[H₂] = 2X
Where ? is the initial concentration of methanol
As [H₂] = 2X = 0.426M; X = 0.213M
[CH₃OH] = ? - 0.213M
[CO] = 0.213M
[H₂] = 0.426M
Replacing in Kc to solve equilibrium concentration of methanol:
6.90x10⁻² = [0.213] [0.426]² / [CH₃OH]
[CH₃OH] = 0.560
As:
[CH₃OH] = ? - 0.213M = 0.560M
? = 0.773M
0.7733M was the initial concentration of methanol. As volume of vessel is 3.00L, moles of methanol are:
3.00L * (0.773 mol / L) = 2.319 moles methanol.
Using molar mass of methanol (32.04g/mol), initial mass of methanol added was:
2.319 moles * (32.04g / mol) =
74.3g of methanol were introduced into the vesselWhat would be the voltage (Ecell) of a voltaic cell comprised of Cd(s)/Cd2+(aq) and Zr(s)/Zr4+(aq) if the concentrations of the ions in solution were [Cd2+] = 0.5 M and [Zr4+] = 0.5 M at 298K?
Answer:
1.05 V
Explanation:
Since;
E°cell= E°cathode- E°anode
E°cathode= -0.40 V
E°anode= -1.45 V
E°cell= -0.40-(-1.45) = 1.05 V
Equation of the process;
2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)
n= 8 electrons transferred
From Nernst's equation;
Ecell = E°cell - 0.0592/n log Q
Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]
Since log 1=0
Ecell= E°cell= 1.05 V
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
What is the electron configuration for the transition metal ion in each of the following compounds?
[Ni(H2O)6]Br2
[Cr(H2O)4(NO2)2]I
Answer:
1)Ni=1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d10 called full-filled
2)Cr=1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5 called half-filled
A buffer is prepared such that [H2PO4-] = 0.095M and [HPO42-] = 0.125M? What is the pH of this buffer solution? (pKa = 7.21 for H2PO4-)
Answer:
pH of the buffer is 7.33
Explanation:
The mixture of the ions H₂PO₄⁻ and HPO₄²⁻ produce a buffer (The mixture of a weak acid, H₂PO₄⁻, with its conjugate base, HPO₄²⁻).
To find pH of a buffer we use H-H equation:
pH = pka + log [A⁻] / [HA]
Where A⁻ is conjugate base and HA weak acid.
For the H₂PO₄⁻ and HPO₄²⁻ buffer:
pH = pka + log [HPO₄²⁻] / [H₂PO₄⁻]
Computing values of the problem:
pH =7.21 + log [0.125M] / [0.095M]
pH = 7.33
pH of the buffer is 7.33
Draw the structure for the organic radical species produced by reaction of the compound with a chlorine atom. Assume reaction occurs at the weakest C-H bond.
Answer:
See explanation
Explanation:
The reaction of chlorine with the pictured compound will occur via free radical mechanism. The stability of the free radical formed will depend on its structure.
The order of stability of free radicals is methyl < primary < secondary < tertiary. Hence a tertiary carbon free radical is the most stable.
Looking at the compound, the radical will form at the position shown in the image attached since it will lead to a secondary free radical which is more stable.
The structure that should be drawn is shown below.
The reaction of chlorine:It should be within the pictured compound that will arise via a free radical mechanism. The stability should be based on the structure. The stability of the order of free radicals should be methyl < primary < secondary < tertiary. Thus, a tertiary carbon free radical should be most stable.
Here look at the compound, the radical should form at the position that should be shown in the image that resulted in the secondary free radical i.e. more stable.
How many grams of PtBr4 will dissolve in 250.0 mL of water that has 1.00 grams of KBr dissolved in it
Answer:
[tex]m_{PtBr_4}=0.306gPtBr_4[/tex]
Explanation:
Hello,
In this case, since the solubility product of platinum (IV) bromide is 8.21x10⁻⁹, and the dissociation is:
[tex]PtBr_4(s)\rightleftharpoons Pt^{4+}(aq)+4Br^-(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Pt^{4+}][Br^-]^4[/tex]
Thus, since the salt is added to a solution initially containing 1.00 grams of potassium bromide, there is an initial concentration of bromide ions:
[tex][Br^-]_0=\frac{1.00gKBr*\frac{1molKBr}{119gKBr}*\frac{1molBr^-}{1molKBr} }{0.250L}=0.0336M[/tex]
Hence, in terms of the molar solubility [tex]x[/tex], we can write:
[tex]8.21x10^{-9}=(x)(0.0336+4x)^4[/tex]
In such a way, solving for [tex]x[/tex], we obtain:
[tex]x=0.00238M[/tex]
Which is the molar solubility of platinum (IV) bromide. Then, since its molar mass is 514.7 g/mol, we can compute the grams that get dissolved in the 250.0-mL solution:
[tex]m_{PtBr_4}=0.00238\frac{molPtBr_4}{1L}*0.250L *\frac{514.7gPtBr_4}{1molPtBr_4} \\\\m_{PtBr_4}=0.306gPtBr_4[/tex]
Best regards.
What are the correct half reactions for the following reaction: Cu2+ + Mg -> Cu + Mg2+
Answer:
Cu2 + 2Mg-> 2Cu+ Mg2
Explanation:
Balance the equation and make sure both the reactant and the products are the same
Hope it will be helpful
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
What is a balanced equation?A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total oxidation numbers is the same for both the reactants and the products.
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.
Learn more about balanced equations here:
brainly.com/question/15052184
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A 1.0 L buffer solution is 0.250 M HC2H3O2 and 0.050 M LiC2H3O2. Which of the following actions will destroy the buffer?
A. adding 0.050 moles of NaOH
B. adding 0.050 moles of LiC2H3O2
C. adding 0.050 moles of HC2H3O2
D. adding 0.050 moles of HCl
E. None of the above will destroy the buffer.
Answer:
D
Explanation:
Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.
The action that destroys the buffer is option c. adding 0.050 moles of HCl.
What is acid buffer?It is a solution of a weak acid and salt.
Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.
The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this, there will be only acid in the solution.
Since
moles of HC2H3O2 = 1*0.250 = 0.250
moles of NaC2H3O2 = 1*0.050 = 0.050.
moles of HCl is added = 0.050
Now
The reaction between HCl and NaC2H3O2
[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]
Now
BCA table is
NaC2H3O2 HCl HC2H3O2
Before 0.050 0.050 0.250
Change -0.050 -0.050 +0.050
After 0 0 0.300
Now, the solution contains the acid (HC2H3O2 ) only.
Therefore addition of 0.050 moles of HCl will destroy the buffer.
Learn more about moles here: https://brainly.com/question/24817060
Determine whether each chemical substance would remain the same color or turn pink in the presence of phenolphthalein.
Answer:
See the answer below
Explanation:
The complete question can be seen in the attached image.
Phenolphthalein is an indicator that is often utilized in an acid-base reaction to indicate the endpoints of such reactions due to its ability to change color from pink/colorless to colorless/pink depending on if the final solution is acidic or basic.
Phenolphthalein is usually colorless in acidic solutions and appears pink in basic solutions. The more basic or alkaline a solution is, the stronger the pink color of phenolphthalein. Hence;
1. Ammonia with a pH of 11 is basic, phenolphthalein will turn pink.
2. Battery acid with a pH of 1 is acidic, it will remain colorless.
3. Lime juice with a pH of 2 is acidic, it will remain colorless.
4. Mashed avocado with a pH of 6.5 is acidic, it will remain colorless.
5. Seawater with a pH of 8.5 is basic, it will turn pink.
6. Tap water with a pH of 7 is neutral, it will remain colorless
Phenolphthalein is a chemical compound with the formula[tex]C_{20}H_{14}O_4[/tex]. Phenolphthalein is often used as an indicator in acid-base titrations. For this application, it turns colorless in acidic solutions and pink in basic solutions
Phenolphthalein works as in:-
Colorless in acidPink in baseAccording to the question, There are 5 solutions having different ph and the indication only turns basic solution to pink
The indicator only turn the basic solution pink and these solutions are as follows,
AmmoniaSea waterTap water.Hence, these are the answer.
For more information, refer to the link:-
