A student placed 11.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 40.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution
Answers
Answer:
There is 0.92 g of glucose in 100 mL of the final solution.
Explanation:
Initially, 11.5 g of glucose is added to the volumetric flask
Water is then added to 100 mL Mark,
The flask was then shaken until the solution was uniform.
The shaking of the mixture makes the concentration of glucose to become uniform all through the solution.
At this point, the concentration of this solution in g/mL is (11.5/100) = 0.115 g/mL
A 40.0 mL sample of this glucose solution was diluted to 0.500 L.
40.0 mL of the already mixed solution is then diluted to 0.500 L.
The mass of glucose in 40.0 mL of the mixed solution with concentration 0.115 g/mL is then given as
Mass = (conc in g/mL) × (volume) = 0.115 × 40 = 4.6 g
So, this mass is then diluted to 0.500 L mark.
New concentration = (mass)/(conc In mL) = (4.6/500) = 0.0092 g/mL
How many grams of glucose are in 100. mL of the final solution
Mass = (conc in g/mL) × (Volume in mL) = 0.0092 × 100 = 0.92 g
Hope this Helps!!!
Answer:
0.459 gram
Explanation:
Find the attachment
Related Questions
The pH at 25 °C of an aqueous solution of the sodium salt of p-monochlorophenol (NaC6H4ClO) is 11.05. Calculate the concentration of C6H4ClO- in this solution, in moles per liter. Ka for HC6H4ClO is equal to 6.6×10-10.
Answers
Answer:
Approximately [tex]8.3 \times 10^{-2}\; \rm mol \cdot L^{-1}[/tex].
Explanation:
The [tex]K_a[/tex] in this question refers the dissociation equilibrium of [tex]\rm HC_6H_4ClO[/tex] as an acid:
[tex]\rm HC_6H_4ClO\, (aq) \rightleftharpoons H^{+} \, (aq) + C_6H_4ClO^{-}\, (aq)[/tex].
[tex]\displaystyle K_a\left(\mathrm{HC_6H_4ClO}\right) = \frac{\left[\mathrm{H^{+}}\right] \cdot \left[\mathrm{C_6H_4ClO^{-}}\right]}{\left[\mathrm{HC_6H_4ClO}\right]}[/tex].
However, the question also states that the solution here has a [tex]\rm pH[/tex] of [tex]11.05[/tex], which means that this solution is basic. In basic solutions at [tex]\rm 25\;^\circ C[/tex], the concentration of [tex]\rm H^{+}[/tex] ions is considerably small (typically less than [tex]10^{-7}\;\rm mol \cdot L^{-1}[/tex].) Therefore, it is likely not very appropriate to use an equilibrium involving the concentration of [tex]\rm H^{+}[/tex] ions.
Here's the workaround: note that [tex]\rm C_6H_4ClO^{-}\, (aq)[/tex] is the conjugate base of the weak acid [tex]\rm HC_6H_4ClO\, (aq)[/tex]. Therefore, when [tex]\rm C_6H_4ClO^{-}\, (aq)[/tex] dissociates in water as a base, its [tex]K_b[/tex] would be equal to [tex]\displaystyle \frac{K_w}{K_a} \approx \frac{10^{-14}}{K_a}[/tex]. ([tex]K_w[/tex] is the self-ionization constant of water. [tex]K_w \approx 10^{-14}[/tex] at [tex]\rm 25\;^\circ C[/tex].)
In other words,
[tex]\begin{aligned} & K_b\left(\mathrm{C_6H_4ClO^{-}}\right) \\ &= \frac{K_w}{K_a\left(\mathrm{HC_6H_4ClO}\right)} \\ &\approx \frac{10^{-14}}{6.6 \times 10^{-10}} \\ & \approx 1.51515 \times 10^{-5}\end{aligned}[/tex].
And that [tex]K_b[/tex] value corresponds to the equilibrium:
[tex]\rm C_6H_4ClO^{-}\, (aq) + H_2O\, (l) \rightleftharpoons HC_6H_4ClO\, (aq) + OH^{-}\, (aq)[/tex].
[tex]\displaystyle K_b\left(\mathrm{C_6H_4ClO^{-}}\right) = \frac{\left[\mathrm{HC_6H_4ClO}\right]\cdot \left[\mathrm{OH^{-}}\right]}{\left[\mathrm{C_6H_4ClO^{-}}\right]}[/tex].
The value of [tex]K_b[/tex] has already been found.
The [tex]\rm OH^{-}[/tex] concentration of this solution can be found from its [tex]\rm pH[/tex] value:
[tex]\begin{aligned}& \left[\mathrm{OH^{-}}\right] \\ &= \frac{K_w}{\left[\mathrm{H}^{+}\right]} \\ & = \frac{K_w}{10^{-\mathrm{pH}}} \\ &\approx \frac{10^{-14}}{10^{-11.05}} \\ &\approx 1.1220 \times 10^{-3}\; \rm mol\cdot L^{-1} \end{aligned}[/tex].
To determine the concentration of [tex]\left[\mathrm{HC_6H_4ClO}\right][/tex], consider the following table:
[tex]\begin{array}{cccccc}\textbf{R} &\rm C_6H_4ClO^{-}\, (aq) & \rm + H_2O\, (l) \rightleftharpoons & \rm HC_6H_4ClO\, (aq) & + & \rm OH^{-}\, (aq) \\ \textbf{I} & (?) & \\ \textbf{C} & -x & & + x& & +x \\ \textbf{E} & (?) - x & & x & & x\end{array}[/tex]
Before hydrolysis, the concentration of both [tex]\mathrm{HC_6H_4ClO}[/tex] and [tex]\rm OH^{-}[/tex] are approximately zero. Refer to the chemical equation. The coefficient of [tex]\mathrm{HC_6H_4ClO}[/tex] and [tex]\mathrm{HC_6H_4ClO}[/tex] are the same. As a result, this equilibrium will produce [tex]\rm OH^{-}[/tex] and [tex]\mathrm{HC_6H_4ClO}[/tex] at the exact same rate. Therefore, at equilibrium, [tex]\left[\mathrm{HC_6H_4ClO}\right] \approx \left[\mathrm{OH^{-}}\right] \approx 1.1220 \times 10^{-3}\; \rm mol\cdot L^{-1}[/tex].
Calculate the equilibrium concentration of [tex]\left[\mathrm{C_6H_4ClO^{-}}\right][/tex] from [tex]K_b\left(\mathrm{C_6H_4ClO^{-}}\right)[/tex]:
[tex]\begin{aligned} & \left[\mathrm{C_6H_4ClO^{-}}\right] \\ &= \frac{\left[\mathrm{HC_6H_4ClO}\right]\cdot \left[\mathrm{OH^{-}}\right]}{K_b}\\&\approx \frac{\left(1.1220 \times 10^{-3}\right) \times \left(1.1220 \times 10^{-3}\right)}{1.51515\times 10^{-5}}\; \rm mol \cdot L^{-1} \\ &\approx 8.3 \times 10^{-2}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
Draw the structure(s) of the organic product(s) predicted when this compound reacts with NaBH4, ethanol then H3O . Use the wedge and dash tools to indicate stereochemistry as appropriate.
Answers
Answer:
what subject is this and how can u draw the structure :)
The reaction involves reduction of the given compound using NaBH4, followed by solvolysis with ethanol and protonation with H3O. Organic molecules structures are often simplified using the skeletal structure method. The stereochemistry of compounds can be represented using wedge and dash notation.
Explanation:The given reaction involves the reduction of an organic compound using NaBH4, followed by the treatment with ethanol and then H3O. When a carbonyl compound reacts with NaBH4, the borohydride group of NaBH4 (it is a weak reducing agent with H-) adds to the carbonyl carbon. This process, known as reduction, transforms the carbonyl group to an alcohol. Following reduction, the compound is undergoing solvolysis in ethanol and then protonated with H3O. Due to the restriction of the platform, displaying the structural representation of the organic product is limited.
The wedge and dash notation can be used to represent the stereochemistry of the compounds. Solid wedges represent bonds coming up out of the plane whereas dashed lines represent bonds going into the plane. These techniques assist chemists to define molecular structures in three dimensions. Skeletal structure, a common way of simplifying complex organic structures in chemistry, represents carbons as each end of a line or bend in a line.
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The equilibrium constant for the following reaction is 6.30 at 723K. 2NH3 <=> N2(g) + 3H2(g) If an equilibrium mixture of the three gases in a 10.2 L container at 723 K contains 0.221 mol of NH3(g) and 0.315 mol of N2(g), the equilibrium concentration of H2 is _________ M. (Enter numbers as numbers, no units. For example, 300 minutes would be 300. For letters, enter A, B, or C. Enter numbers in scientific notation using e# format. For example 1.43×10-4 would be 1.43e-4.)
Answers
Answer:
0.458 M
Explanation:
Let's consider the following reaction at equilibrium.
2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)
The concentrations at equilbrium are:
[NH₃] = 0.221 mol / 10.2 L = 0.0217 M
[N₂] = 0.315 mol / 10.2 L = 0.0309 M
[H₂] = ?
We can find the concentration of H₂ at equilibrium using the equilibrium constant (K).
K = [N₂] × [H₂]³ / [NH₃]²
[H₂] = ∛(K × [NH₃]²/ [N₂])
[H₂] = ∛(6.30 × 0.0217²/ 0.0309) = 0.458 M
Answer:
The equilibrium concentration of H2 is 0.4579 M
Explanation:
Step 1: Data given
The equilibrium constant for the following reaction is 6.30 at 723K
Volume = 10.2 L
Number of moles NH3 = 0.221 moles
Number of moles N2 = 0.315 moles
Step 2: The balanced equation
2NH3 <=> N2(g) + 3H2(g)
Step 3: Calculate concentration
Concentration = moles / volume
Concentration NH3= 0.221 moles / 10.2 L
Concentration NH3 = 0.0217 M
Concentration N2 = 0.315 moles / 10.2 L
Concentration N2 = 0.0309 M
Step 4: Define Kc
Kc = [H2]³[N2] / [NH3]²
6.30 = [H2]³ * (0.0309) / (0.0217)²
[H2]³ = 0.09601 M
[H2] = 0.4579 M
The equilibrium concentration of H2 is 0.4579 M
After treating cyclohexanone with LDA, he then added tert-butyl bromide to the same reaction vessel, which one would normally do for a one-pot two-step reaction sequence. However, none of the desired compound formed. Instead, he isolated three organic compounds in a 1:1:1 ratio. What is the structure of each of these organic compounds?
Answers
Answer:
2-Tert-butyl-cyclohexanone, 2,6-Di-Tert-Butylcyclohexanone and Di-isopropyl-amine.
Explanation:
The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)(1)2NO(g) + O2(g) → 2NO2(g)(2)3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)(3)Use the values of ΔH°f to calculate the value of ΔH° for each of the preceding reactions.ΔH°f, kJ/molNH3(g)-46NO(g)90.H2O(g)-242NO2(g)34H2O(l)-286HNO3(aq)-207ΔH°1 = kJΔH°2 = kJΔH°3 = kJWrite the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations. (Water is also a product.)(Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
Answers
The enthalpy change (ΔH°) for each reaction can be calculated using the given enthalpy of formation (ΔH°f) values. ΔH°1 = (4ΔH°f(NO) + 6ΔH°f(H2O)) - (4ΔH°f(NH3) + 5ΔH°f(O2)). ΔH°2 = (2ΔH°f(NO2)) - (2ΔH°f(NO) + ΔH°f(O2)). ΔH°3 = (2ΔH°f(HNO3) + ΔH°f(NO)) - (3ΔH°f(NO2) + ΔH°f(H2O)).
Explanation:In order to determine the enthalpy change (ΔH°) for each reaction, we need to use the given enthalpy of formation (ΔH°f) values. The enthalpy change for each reaction can be calculated by subtracting the sum of the enthalpy of formation of the reactants from the sum of the enthalpy of formation of the products. Using this method, we can calculate the enthalpy change for each of the preceding reactions as follows:
ΔH°1 = (4ΔH°f(NO) + 6ΔH°f(H2O)) - (4ΔH°f(NH3) + 5ΔH°f(O2))
ΔH°2 = (2ΔH°f(NO2)) - (2ΔH°f(NO) + ΔH°f(O2))
ΔH°3 = (2ΔH°f(HNO3) + ΔH°f(NO)) - (3ΔH°f(NO2) + ΔH°f(H2O))
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The ΔH° for each reaction in the Ostwald process is calculated with a specific formula, and the overall ΔH° is found by summing the individual values. The final reaction is 4NH3(g) + 5O2(g) → 4HNO3(aq) with ΔH° = -1192 kJ.
Explanation:In The Ostwald process, the ΔH° for each reaction is calculated with the formula: ΔH° = ΣΔH°f(products) - ΣΔH°f(reactants). So, for the first reaction, 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g), we calculate ΔH°1 = [4(90) + 6(-242)] - [4(-46) + 5(0)] = -906 kJ. The second reaction, 2NO(g) + O2(g) → 2NO2(g), gives ΔH°2 = [2(34)] - [2(90) + 0] = -112 kJ. Lastly, for 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g), ΔH°3 = [2(-207) + 90] - [3(34) + -286] = -174 kJ.
Now, adding all three reactions, we get the overall reaction: 4NH3(g) + 5O2(g) → 4HNO3(aq), with the overall ΔH° = -1192 kJ.
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For most solids at room temperature, the specific heat is determined by oscillations of the atom cores in the lattice (each oscillating lattice site contributes 3kT of energy, by equipartition), as well as a contribution from the mobile electrons (if it's a metal). At room temperature the latter contribution is typically much smaller than the former, so we will ignore it here. In other words, you can reasonably estimate the specific heat simply by counting the number of atoms! Use this fact to estimate the specific heat of copper (atomic mass = 63.6), given that the specific heat of aluminum (atomic mass = 27.0) is 900 J/kg-K.
Answers
Answer:
The specific heat of copper is [tex]C= 392 J/kg\cdot ^o K[/tex]
Explanation:
From the question we are told that
The amount of energy contributed by each oscillating lattice site is [tex]E =3 kT[/tex]
The atomic mass of copper is [tex]M = 63.6 g/mol[/tex]
The atomic mass of aluminum is [tex]m_a = 27.0g/mol[/tex]
The specific heat of aluminum is [tex]c_a = 900 J/kg-K[/tex]
The objective of this solution is to obtain the specific heat of copper
Now specific heat can be defined as the heat required to raise the temperature of 1 kg of a substance by [tex]1 ^o K[/tex]
The general equation for specific heat is
[tex]C = \frac{dU}{dT}[/tex]
Where [tex]dT[/tex] is the change in temperature
[tex]dU[/tex] is the change in internal energy
The internal energy is mathematically evaluated as
[tex]U = 3nk_BT[/tex]
Where [tex]k_B[/tex] is the Boltzmann constant with a value of [tex]1.38*10^{-23} kg \cdot m^2 /s^2 \cdot ^o K[/tex]
T is the room temperature
n is the number of atoms in a substance
Generally number of atoms in mass of an element can be obtained using the mathematical operation
[tex]n = \frac{m}{M} * N_A[/tex]
Where [tex]N_A[/tex] is the Avogadro's number with a constant value of [tex]6.022*10^{23} / mol[/tex]
M is the atomic mass of the element
m actual mass of the element
So the number of atoms in 1 kg of copper is evaluated as
[tex]m = 1 kg = 1 kg * \frac{10000 g}{1kg } = 1000g[/tex]
The number of atom is
[tex]n = \frac{1000}{63.6} * (6.0*0^{23})[/tex]
[tex]= 9.46*10^{24} \ atoms[/tex]
Now substituting the equation for internal energy into the equation for specific heat
[tex]C = \frac{d}{dT} (3 n k_B T)[/tex]
[tex]=3nk_B[/tex]
Substituting values
[tex]C = 3 (9.46*10^{24} )(1.38 *10^{-23})[/tex]
[tex]C= 392 J/kg\cdot ^o K[/tex]
A particular reaction has an activation energy, Ea, of 117 kJ/mol. If the rate constant for the reaction is 0.00289 s −1 at 590 °C, at what temperature(in°C) would the rate constant be 0.492 s −1?
Answers
Answer:
986.9 °C
Explanation:
To find the temperature we can use the Arrhenius equation:
[tex] k = Ae^{(-E_{a}/RT)} [/tex] (1)
Where:
k: is the reaction rate coefficient
A: is the pre-exponential factor
Ea: is the activation energy
R: is the gas constant
T: is the absolute temperature
With the values given:
T = 590 °C = 863 K; Ea = 117 kJ/mol; k = 0.00289 s⁻¹,
we can find the pre-exponential factor, A:
[tex]A = \frac{k}{e^{(-E_{a}/RT)}} = \frac{0.00289 s^{-1}}{e^{(-117 \cdot 10^{3} J*mol^{-1}/(8.314 J*K^{-1}*mol^{-1}*863 K))}} = 3.49 \cdot 10^{4} s^{-1}[/tex]
Now, we can find the temperature by solving equation (1) for T:
[tex] ln(\frac{k}{A}) = -\frac{E_{a}}{RT} [/tex]
[tex] T = -\frac{E_{a}}{R*ln(\frac{k}{A})} = -\frac{117 \cdot 10^{3} J*mol^{-1}}{8.314 J*K^{-1}*mol^{-1}*ln(\frac{0.492 s^{-1}}{3.49 \cdot 10^{4} s^{-1}})} = 1259.9 K = 986.9 ^{\circ} C [/tex]
Therefore, at 986.9 °C, the rate constant will be 0.492 s⁻¹.
I hope it helps you!
Answer:
The temperature is 1259.4 K (986.4°C)
Explanation:
Step 1: Data given
The activation energy Ea, = 117 kJ/mol
the rate constant for the reaction is 0.00289 s −1 at 590 °C
The new rate constant = 0.492 s −1
Step 2: Calculate the temperature
ln(kX / k 590°C) = Ea/R * (1/T1 - 1/T2)
⇒with kX is the rate constant at the new temperature = 0.492 /s
⇒with k 590 °C = the rate constant at 590 °C = 0.00289 /s
⇒with Ea = the activation energy = 117000 J/mol
⇒with R = the gas constant = 8.314 J/mol*K
⇒with T1 = 590 °C = 863 K
⇒ with T2 = the new temperature
ln (0.492 / 0.00289) = 117000/8.314 *(1/863 - 1/T2)
5.137 =14072.6 * (1/863 - 1/T2)
3.65*10^-4 = (1/863 - 1/T2)
3.65*10^-4 = 0.001159 - 1/T2
0,000794 = 1/T2
T2 = 1259.4 K
The temperature is 1259.4 K or 986.4 °C
At a certain temperature the rate of this reaction is first order in HI with a rate constant of :7.21/s2HI(g) = H2(g) + I2(g)Suppose a vessel contains HI at a concentration of 0.440M. Calculate the concentration of HI in the vessel 0.210 seconds later. You may assume no other reaction is important.
Answers
Answer:
C HI = 0.0968 M
Explanation:
2HI(g) → H2(g) + I2(g)- ra = K(Ca)∧α = - δCa/δt∴ a: HI(g)
∴ order (α) = 1
∴ rate constant (K) = 7.21/s
∴ Initial concenttration HI(g) (Cao) = 0.440 M
∴ t = 0.210 s ⇒ Ca = ?
⇒ - δCa/δt = KCa
⇒ - ∫δCa/Ca = K*∫δt
⇒ Ln(Cao/Ca) = K*t
⇒ Cao/Ca = e∧(K*t)
⇒ Cao/e∧(K*t) = Ca
⇒ Ca = (0.440)/e∧((7.21*0,21))
⇒ Ca = 0.440/4.54533
⇒ Ca = 0.0968 M
A sealed copper container with a mass of 0.3 kg is filled with 1.5 mole of helium gas. Initially, the helium gas is at a temperature of 124 oC and the copper container is at 20 oC. The helium-copper system is thermally isolated. Note that the specific heat of copper is 386 J/(kgK) and the molar specific heat of helium is 12.5 J/(molK). Find the equilibrium temperature of the system.
Answers
Answer:
[tex]T = 34.493\,^{\textdegree}C[/tex]
Explanation:
The equilibrium temperature of the gas-container system is:
[tex]Q_{Cu} = -Q_{g}[/tex]
[tex](0.3\,kg)\cdot\left(386\,\frac{J}{kg\cdot ^{\textdegree}C} \right)\cdot (T-20\,^{\textdegree}C) = (1.5\,mol)\cdot \left(12.5\,\frac{J}{mol\cdot ^{\textdegree}C} \right)\cdot (124\,^{\textdegree}C-T)[/tex]
[tex]\left(115.8\,\frac{J}{^{\textdegree}C} \right)\cdot (T-20\,^{\textdegree}C) = \left(18.75\,\frac{J}{^{\textdegree}C} \right)\cdot (124\,^{\textdegree}C-T)[/tex]
[tex]134.55\cdot T = 4641[/tex]
[tex]T = 34.493\,^{\textdegree}C[/tex]
Provide a preliminary design o fa n air stripping column t o remove toluene from ground water. Levels of toluene range from 0.1 to 2.1 mglL and this must be reduced to 50 f.LglL. A hydrogeologic study of the area indicates that a flow rate of 1 1 0 gal/min is required to ensure that contamination not spread. Laboratory investigations have determined the overall transfer constant, KLa = 0.020 s-' . Use a column diameter of 2.0 feet and an air-to-water ratio of 15. Specifically determine: liquid loading rate, stripping factor, and height of the tower. Provide a sketch of the unit indicating all required appurtenances.
Answers
Answer:
Explanation:
Find attach the solution
What is the molar concentration of sodium ions in a 0.650 M Na3PO4 solution?
Answers
The molar concentration of sodium ions in a 0.650M Na₃PO₄ solution is equal to 1.95M.
What is the molar concentration?The concentration of the solution can be determined when we have the molecular formula and its molecular weight. We can easily determine the molar concentration of a substance in a solution.
The concentration of the solution calculated as the number of moles of a solute in a liter of a solution is called molarity or molar concentration.
The Molar concentration of the solution is determined in the following way.
Molarity (M) = Moles (n) of solute/Volume of the Solution ( in L)
Given, the molar concentration of the Na₃PO₄ solution = 0.650 M
The dissociation of Na₃PO₄ into ions can be represented as:
Na₃PO₄ → 3 Na⁺ + PO₄³⁻
From the above equation, we can say that one mole of the Na₃PO₄ gives 3 moles of sodium ions.
The molar concentration of sodium ions in Na₃PO₄ solution = 3× 0.650
The molar concentration of Na⁺ions = 1.95M
Therefore, the molar concentration of Na⁺ions in Na₃PO₄ solution is 1.95M.
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Mole ratios for a solving a stoichiometry problem are obtained from the
1. Molar Mass
2. Periodic Table
3. Balanced Equation
4. Total Mass of the Products
Answers
g Identify which of the following statements about human glycogen debranching enzyme are true based on the HPLC results. The (α‑1→6) glucosidase catalytic center is in the C‑terminal half. The transferase catalytic center can hydrolyze α‑1,6 glycosidic bonds. The oligo‑(α1→4)‑(α1→4) glucanotransferase catalytic center is in the C‑terminal half. Oligo‑(α1→4)‑(α1→4) glucanotransferase activity creates a substrate for (α‑1→6) glucosidase. Based on the peaks in the HPLC charts, what do you think is the most likely substrate for the oligo‑(α1→4)‑(α1→4) glucanotransferase catalytic center of glycogen debranching enzyme? maltotetraosyl and glucosyl residues 6‑O‑α‑glucosyl cyclomaltoheptaose maltosyl and maltotriosyl residues maltoheptaosyl and maltooctaosyl residues cyclomaltoheptaose (β‑cyclodextrin)
Answers
The most likely substrates for the oligo-(α1→4)-(α1→4) glucanotransferase catalytic center of glycogen debranching enzyme are maltosyl and maltotriosyl residues.
Explanation:The glycogen debranching enzyme has two activities; it acts as a glucosidase and a glucanotransferase. The (α‑6) glucosidase activity hydrolyzes α-1,6 glycosidic bonds, whereas the oligo-(α1→4)-(α1→4) glucanotransferase activity shifts α-1,4-linked glucose chains from one branch to another, often creating a substrate that the glucosidase can act upon. Based on the mechanism of glycogenolysis, the enzyme glycogen phosphorylase releases glucose units from the linear chain until a few are left near the branching point, and it is here that the glucan transferase action is relevant. The glucan transferase shifts the remaining α-1,4 linked glucose units, leaving a single α-1,6 linked glucose that the glucosidase can then release. Hence, the most likely substrates for the oligo-(α1→4)-(α1→4) glucanotransferase catalytic center are maltosyl and maltotriosyl residues, as these are the short α-1,4 linked glucose chains that are left after phosphorylase action.
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How many moles of gas are present in 1.13 L of gas at 2.09 atm and 291 K?
Answers
n = 0.0989 moles
Explanation:n = PV / RT
P = 2.09atm
V = 1.13L
R = 0.08206
T = 291K
Plug the numbers in the equation.
n = (2.09atm)(1.13L) / (0.08206)(291K)
n = 0.0989 moles
Final answer:
Using the ideal gas law PV = nRT, and rearranging for n (moles), we plug in the given values of pressure, volume, and temperature alongside the ideal gas constant to calculate the number of moles of gas present. There are approximately 0.0988 moles of gas present in 1.13 L of gas at [tex]\(2.09 \, \text{atm}\) and \(291 \, \text{K}\).[/tex]
Explanation:
To calculate the number of moles of gas present, we can use the ideal gas law equation:
[tex]\[ PV = nRT \][/tex]
First, we need to convert the given pressure to atm and the volume to liters if they are not already in those units.
Given:
- [tex]\( P = 2.09 \, \text{atm} \)[/tex]
-[tex]\( V = 1.13 \, \text{L} \[/tex]
- [tex]\( T = 291 \, \text{K} \)[/tex]
Now, we can rearrange the ideal gas law equation to solve for \( n \):
[tex]\[ n = \frac{PV}{RT} \][/tex]
[tex]\[ n = \frac{(2.09 \, \text{atm})(1.13 \, \text{L})}{(0.0821 \, \text{atm} \cdot \text{L} / \text{mol} \cdot \text{K})(291 \, \text{K})} \][/tex]
[tex]\[ n = \frac{2.3617}{23.9211} \][/tex]
[tex]\[ n \approx 0.0988 \, \text{mol} \][/tex]
Therefore, There are approximately 0.0988 moles of gas present in 1.13 L of gas at [tex]\(2.09 \, \text{atm}\) and \(291 \, \text{K}\).[/tex]
The phosphorylation of glucose to glucose 6-phosphate Group of answer choices is so strongly exergonic that it does not require a catalyst. is an exergonic reaction not coupled to any other reaction. is an endergonic reaction that takes place because it is coupled to the exergonic hydrolysis of ATP. is an exergonic reaction that is coupled to the endergonic hydrolysis of ATP.
Answers
Answer:
The phosphorylation of glucose to glucose-6-phosphate is endergonic reaction that is coupled to the exergonic hydrolysis of ATP.
Explanation:
In glycosis, the first reaction that takes place is the phosphorylation of glucose to glucose-6-phosphate by the enzyme hexokinase. This is an exergenic reaction. This is a coupled reaction in which phosphorylation of glucose is coupled to ATP hydrolysis. The free energy of ATP hydrolysis fuels glucose phosphorylation.
What is the full ionic equation for CuCl2 and (NH4)2SO4?
Answers
Answer:
Explanation:
No reaction if there is no solid then the ions just stay there brah. All the ions remain in solution. We are all gonna make it!
Why does a higher concentration make a reaction faster?
There are more collisions per second and the collisions are of greater energy.
Collisions occur with greater energy.
There are more collisions per second.
Answers
Higher concentration make a reaction faster due to faster collisions. The correct answer is "There are more collisions per second."
When the concentration of reactants is increased, it leads to a higher number of particles or molecules in the reaction mixture. As a result, there are more frequent collisions between the reactant particles per unit time.
In a chemical reaction, the reactant particles must collide with sufficient energy and proper orientation for a successful reaction to occur. By increasing the concentration, the chances of successful collisions increase because there are more particles available to collide with each other.
However, it's important to note that while a higher concentration increases the frequency of collisions, it does not necessarily guarantee that all collisions will result in a reaction. The energy of the collisions and proper orientation are also essential factors for a successful reaction.
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Interpret the following Arterial Blood Gases 1. pH 7.33 PaCO2 60 HCO3 34 A. Normal ABG values B. Respiratory acidosis without compensation C. Respiratory acidosis with partial compensation D. Respiratory acidosis with full compensation 2. pH 7.48 PaCO2 42 HCO3 30 A. Metabolic acidosis without compensation B. Respiratory alkalosis with partial compensation C. Respiratory alkalosis with full compensation D. Metabolic alkalosis without compensation 3. pH 7.38 PaCO2 38 HCO3 24 A. Respiratory alkalosis B. Normal C. Metabolic Alkalosis D. None of the above 4. pH 7.21 PaCO2 60 HCO3 24 A. Normal B. Respiratory acidosis without compensation C. Metabolic acidosis with partial compensation D. Respiratory acidosis with complete compensation 5. pH 7.48 PaCO2 28 HCO3 20 A. Respiratory alkalosis with partial compensation B. Respiratory alkalosis with complete compensation C. Metabolic alkalosis without compensation D. Metabolic alkalosis with complete compensation
Answers
Answer: 1) C; 2)D; 3)B; 4)B; 5) A
Explanation:Interpreting the following Arterial Blood gases, we have
1. pH 7.33 PaCO2 60 HCO3 34----Respiratory acidosis with partial compensation----C
2. pH 7.48 PaCO2 42 HCO3 30------. Metabolic alkalosis without compensation----D
3. pH 7.38 PaCO2 38 HCO3 24 ----- Normal---B
4. pH 7.21 PaCO2 60 HCO3 24------ Respiratory acidosis without compensation-----B
5. pH 7.48 PaCO2 28 HCO3 20 ----Respiratory alkalosis with partial compensation
The Arterial blood gas interpretation from analysis shows the pH and the partial pressures of oxygen and carbon dioxide in the arterial blood of an individual which can detect how well the lungs are functioning thereby making a physician make a diagnosis, estimate the severity of a condition and profer treatment.
What type of particles will a hot object have if it has more kinetic energy?
Answers
Answer:
celulares por que tienen una fuente de energiq
Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The KspKsp of Zn(OH)2Zn(OH)2 is 3×10−153×10−15 and the KfKf of Zn(OH)2−4Zn(OH)42− is 2×10152×1015.
Answers
Answer:
pH = 13.09
Explanation:
Zn(OH)2 --> Zn+2 + 2OH- Ksp = 3X10^-15
Zn+2 + 4OH- --> Zn(OH)4-2 Kf = 2X10^15
K = Ksp X Kf
= 3*2*10^-15 * 10^15
= 6
Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M
Zn(OH)₂ + 2OH⁻(aq) --> Zn(OH)₄²⁻(aq)
Initial: 0 0.3 0
Change: -2x +x
Equilibrium: 0.3 - 2x x
K = Zn(OH)₄²⁻/[OH⁻]²
6 = x/(0.3 - 2x)²
6 = x/(0.3 -2x)(0.3 -2x)
6(0.09 -1.2x + 4x²) = x
0.54 - 7.2x + 24x² = x
24x² - 8.2x + 0.54 = 0
Upon solving as quadratic equation, we obtain;
x = 0.089
Therefore,
Concentration of (OH⁻) = 0.3 - 2x
= 0.3 -(2*0.089)
= 0.122
pOH = -log[OH⁻]
= -log 0.122
= 0.91
pH = 14-0.91
= 13.09
Nitrate salts (NO3), when heated, can produce nitrites (NO2) plus oxygen (O2). A sample of potassium nitrate is heated, and the 02 gas produced is collected in a 700 mL flask. The pressure of the gas in the flask is 2.7 atm, and the temperature is recorded to be 329 K. The value of R= 0.0821 atm L/(mol K) How many moles of O2 gas were produced? moles After a few hours, the 700 mL flask cools to a temperature of 293K. What is the new pressure due to the O2 gas?
Answers
Answer: a) 0.070 moles of oxygen were produced.
b) New pressure due to the oxygen gas is 2.4 atm
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 2.7 atm
V = Volume of gas = 700 ml = 0.7 L
n = number of moles = ?
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature = 329 K
[tex]n=\frac{PV}{RT}[/tex]
[tex]n=\frac{2.7atm\times 0.7L}{0.0821 L atm/K mol\times 329K}=0.070moles[/tex]
Thus 0.070 moles of oxygen were produced.
When the 700 mL flask cools to a temperature of 293K.
[tex]PV=nRT[/tex]
[tex]P=\frac{nRT}{V}[/tex]
[tex]P=\frac{0.070\times 0.0821\times 293}{0.7}[/tex]
[tex]P=2.4atm[/tex]
The new pressure due to the oxygen gas is 2.4 atm
A silver nitrate (AgNO3) solution is 0.150 M. 100.0 mL of a diluted silver nitrate solution is prepared using 10.0 mL of the more concentrated solution (which is then diluted with distilled water to 100.0 mL). Use the dilution equation to find the molarity of silver nitrate in the diluted solution.
Answers
Answer: The molarity of silver nitrate in the diluted solution is 0.015 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
According to the dilution law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of concentrated solution = 0.150 M
[tex]V_1[/tex] = volume of concentrated solution = 10.0 ml
[tex]M_2[/tex] = molarity of diluted solution = ?
[tex]V_2[/tex] = volume of diluted solution = 100.0 ml
[tex]0.150\times 10.0=M_2\times 100.0[/tex]
[tex]M_2=0.015M[/tex]
Thus the molarity of silver nitrate in the diluted solution is 0.015 M
Final answer:
The molarity of the diluted silver nitrate solution, prepared by diluting 10.0 mL of a 0.150 M solution to a final volume of 100.0 mL, is 0.015 M when calculated using the dilution equation.
Explanation:
The question is asking how to calculate the molarity of a diluted silver nitrate solution using the dilution equation. When 10.0 mL of a 0.150 M silver nitrate solution is diluted to 100.0 mL, the molarity of the diluted solution can be found using the equation M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
Using the dilution equation:
To find M2 (the molarity of the diluted solution), rearrange the equation to M2 = (M1V1) / V2.
The molarity of the diluted silver nitrate solution would be 0.015 M (calculated as (0.150 M * 10.0 mL) / 100.0 mL).
Dinitrogen monoxide gas is collected at 19.0 °C in an evacuated flask with a measured volume of 30.0 L. When all the gas has been collected, the pressure in the flask is measured to be 0.500 atm . Calculate the mass and number of moles of dinitrogen monoxide gas that were collected. Be sure your answer has the correct number of significant digits.
Answers
Answer:
0.625 mol
27.5 g
Explanation:
Given data
Gas: N₂OMolar mass (M): 44.01 g/molTemperature (T): 19.0°C + 273.15 = 292.2 KVolume (V): 30.0 LPressure (P): 0.500 atmWe can find the moles of dinitrogen monoxide gas using the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 0.500 atm × 30.0 L/0.0821 atm.L/mol.K × 292.2 K
n = 0.625 mol
Then, we can find the mass corresponding to 0.625 moles.
0.625 mol × 44.01 g/mol = 27.5 g
Answer:
0.626 moles of dinitrogen monoxide gas was collected, this is 27.6 grams
Explanation:
Step 1: Data given
Temperature of Dinitrogen monoxide gas = 19.0 °C = 292 K
Volume of the flask = 30.0 L
Pressure in the flask = 0.500 atm
Step 2: Calculate moles of Dinitrogen monoxide gas
p*V = n*R*T
⇒with p = the pressure of the gas = 0.500 atm
⇒with V = the volume of the flask = 30.0 L
⇒with n = the number of moles of N2O gas = TO BE DETERMINED
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 292 K
n = (p*V) / (R*T)
n = (0.500 * 30.0) / (0.08206 * 292)
n = 0.626 moles
Step 3: Calculate mass N2O gas
Mass N2O gas = moles N2O * molar mass N2O
Mass N2O = 0.626 moles * 44.013 g/mol
Mass N2O = 27.55 grams ≈ 27.6 grams
0.626 moles of dinitrogen monoxide gas was collected, this is 27.6 grams
Consider the following initial rate data (at 273 K) for the decomposition of a substrate (substrate 1) which decomposes to product 1 and product 2: [Substrate 1] (M) Initial Rate (M/s) 0.4 0.183 0.8 0.183 2 0.183 Determine the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M.
Answers
Answer:
15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.
Explanation:
A → B + C
The rate law of the reaction will be :
[tex]R=k[A]^x[/tex]
Initial rate of the reaction when concentration of the substrate was 0.4 M:
[tex]0.183 M/s=k[0.4 M]^x[/tex]..[1]
Initial rate of the reaction when concentration of the substrate was 0.8 M:
[tex]0.183 M/s=k[0.8 M]^x[/tex]...[2]
[1] ÷ [2] :
[tex]\frac{0.183 M/s}{0.183 M/s}=\frac{k[0.4 M]^x}{k[0.8 M]^x}[/tex]
x = 0
The order of the reaction is zero.
For the value of rate constant ,k:
[tex]0.183 M/s=k[0.4 M]^x[/tex]..[1]
x = 0
[tex]0.183 M/s=k[0.4 M]^0[/tex]
k= 0.183 M/s
The half life of the zero order kinetics is given by :
[tex]t_{1/2}=\frac{[A_o]}{2k}[/tex]
Where:
[tex][A_o][/tex] = Initial concentration of A
k = Rate constant of the reaction
So, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M:
[tex]t_{1/2}=\frac{2.77 M}{0.183 M/s}=15.1 s[/tex]
15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.
The half-life for the decomposition of substrate 1 when the initial concentration is 2.77 M is approximately 18 minutes.
Explanation:The half-life of a reaction is the time required for one-half of a given amount of reactant to be consumed. In this case, we have initial rate data for the decomposition of substrate 1 with varying concentrations. To determine the half-life of substrate 1, we need to find the time it takes for the concentration to decrease to half its initial value.
Based on the given data, we can see that the initial rate of the reaction is constant at 0.183 M/s for different initial concentrations of substrate 1 (0.4 M, 0.8 M, and 2 M). Since the initial concentration of substrate 1 is given as 2.77 M, it will take the same amount of time as the other initial concentrations to reach half its initial concentration.
Therefore, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M is the same as the half-life observed when the initial concentration is 0.4 M, 0.8 M, or 2 M, which is approximately 18 minutes.
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Define the end point of a titration. Select one: a. It is a synonym for equivalence point. b. It is when a change that indicates equivalence is observed in the analyte solution. c. It is the point at which the pH no longer changes. Next
Answers
Answer:
b. It is when a change that indicates equivalence is observed in the analyte solution.
Explanation:
The end point of a titration is the point at which the indicator undergoes the change noticeable by our senses. Ideally, the equivalence point and end point coincide; but this does not usually happen in practice, because the indicator does not always change perceptibly at the same moment in which the equivalence point is reached and also for the change of the indicator, some of the reagent used in the evaluation is usually necessary .
The difference between the end point and the equivalence point of a valuation is called the valuation error or end point error.
At this point the pH changes the color of the indicator. nvhdurn
Final answer:
The end point of a titration is the moment observed (option b) when a change, due to an indicator or sensor, suggests that stoichiometric equivalence has been reached. It is an experimental determination and serves as the best estimate of the theoretical equivalence point.
Explanation:
The end point of a titration is not a synonym for the equivalence point. Rather, it represents the moment during a titration when a change indicating equivalence is observed in the analyte solution (option b). This point is typically detected through a change in color due to an indicator or through a change detected by a sensor. It is important to note that the end point is an experimental value, our best estimate of the equivalence point, and that any difference between the end point and the theoretical equivalence point is a source of determinate error.
In an acid-base titration, the end point is often observed when a pH indicator changes color, signifying that a stoichiometric amount of titrant has been added to the analyte. In redox and complexation titrations, it can be detected by changes in the solution conditions that are measurable by indicators and sensors.
A 1.000 kg sample of nitroglycerine, C3H5N3O9, explodes and releases gases with a temperature of 1985°C at 1.100 atm. What is the volume of gas produced? 4 C3H5N3O9(s) → 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g) A 1.000 kg sample of nitroglycerine, C3H5N3O9, explodes and releases gases with a temperature of 1985°C at 1.100 atm. What is the volume of gas produced? 4 C3H5N3O9(s) → 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g) 4730 L 5378 L 3525 L 742.2 L
Answers
Answer:
742.2 L
Explanation:
First we must find the number of moles of nitroglycerine reacted.
Molar mass of nitroglycerine= 227.0865 g/mol
Mass of nitroglycerine involved = 1×10^3 g
Number of moles of nitroglycerine= 1×10^3g/227.0865 g/mol
n= 4.40361 moles
T= 1985°C + 273= 2258K
P= 1.100atm
R= 0.082atmLmol-1K-1
Using the ideal gas equation:
PV= nRT
V= nRT/P
V= 4.40361× 0.082× 2258/1.1
V= 742 L
Considering the reaction stoichiometry and ideal gas law, the volume of gas produced is 5375.626 L.
The balanced reaction is:
4 C₃H₅N₃O₉(s) → 12 CO₂(g) + 10 H₂O(g) + 6 N₂(g) + O₂(g)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
C₃H₅N₃O₉(s): 4 moles O₂(g): 1 moles CO₂(g): 12 moles H₂O(g): 10 molesN₂(g): 6 molesThen 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas [12 moles of CO₂(g) + 10 moles of H₂O(g) + 6 moles of N₂(g) + 1 mole O₂(g)]
Being the molar mass of nitroglycerine C₃H₅N₃O₉(s) 227 g/mole, then the amount of moles that 1 kg (1000 g) of the compound contains can be calculated as:
[tex]1000 gramsx\frac{1 mole}{227 grams}= 4.405 moles[/tex]
Then you can apply the following rule of three: if by stoichiometry 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas, 4.405 moles of C₃H₅N₃O₉(s) will produce how many moles of gas?
[tex]amount of moles of gas=\frac{4.405 moles of nitroglycerinex29 moles of gas}{4 moles of nitroglycerine}[/tex]
amount of moles of gas= 31.93625 moles
On the other hand, an ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, you know:
P= 1.1 atmV= ?n= 31.93625 molesR= 0.082[tex]\frac{atmL}{molK}[/tex]T= 1985 C= 2258 K (being 0 C=273 K)Replacing:
1.1 atm× V= 31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K
Solving:
V= (31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K) ÷ 1.1 atm
V= 5375.625 L
Finally, the volume of gas produced is 5375.626 L.
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Answers
Answer:
[tex]\large \boxed{\text{103 kPa}}[/tex]
Explanation:
We can use the Ideal Gas Law — pV = nRT
Data:
V = 66.8 L
m = 77.8 g
T = 25 °C
Calculations:
(a) Moles of N₂
[tex]\text{Moles of N}_{2} = \text{77.8 g N}_{2} \times \dfrac{\text{1 mol N}_{2}}{\text{28.01 g N}_{2}} = \text{2.778 mol N}_{2}[/tex]
(b) Convert the temperature to kelvins
T = (25 + 273.15) K = 298.15 K
(c) Calculate the pressure
[tex]\begin{array}{rcl}pV & =& nRT\\p \times \text{66.8 L} & = & \text{2.778 mol} \times \text{8.314 kPa$\cdot$ L$\cdot$K$^{-1}$mol$^{-1}\times$ 298.15 K}\\66.8p & = & \text{6886 kPa}\\p & = & \textbf{103 kPa}\end{array}\\\text{The pressure in the bag is $\large \boxed{\textbf{103 kPa}}$}[/tex]
True or False: Ionic compounds are called Electrolytes because they
dissociate(break apart) in water to form a solution that conducts electric
current. *
Answers
The statement is true as all Ionic compounds are called Electrolytes because they dissociate(break apart) in water to form a solution that conducts electric current.
Explanation:
An ionic compound also called an electrolyte. Ionic compounds have capability to get dissociate into water to give ions(cation and anion) which conduct electric current. This process is called conductivity.
Some insoluble compounds that are not dissolved in water still they form ions which conduct electricity.
The cations and anions released are responsible for carrying current.
Strong acids and strong bases are the strong electrolytes as they dissociate to give more ions as NaOH, HCl.
The cations move to the cathode and anions towards the anode. Cation and anion movement in the solution is electric current.
Ionic compounds are called electrolytes because they dissociate in water to form a solution that conducts electric current.
Explanation:True. Ionic compounds are called electrolytes because they dissociate (break apart) in water to form a solution that conducts electric current.
When an ionic compound like NaCl is dissolved in water, the positive sodium ions and negative chloride ions separate from each other and are surrounded by water molecules. These separated ions can move freely in the solution and carry electric charge, allowing the solution to conduct electricity.
Other examples of electrolytes include compounds like potassium chloride (KCl) and magnesium sulfate (MgSO4). These compounds also dissociate in water to produce solutions that can conduct electric current.
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Problem PageQuestion Suppose a rhodium atom in the oxidation state formed a complex with one iodide anion and five ammonia molecules. Write the chemical formula of this complex.
Answers
Answer:
It forms an octahedral complex
Rh[NH3]5I
Explanation:
Or see structure attached below
Answer:
[ Rhl(NH₃)₅]²⁺
Explanation:
Symbol of Rhodium atom = Rh
charge of Rhodium in complex = +3
charge on Iodine in complex = -1
Charge on NH₃= 0 (neutral)
Therefore, the total charge on the complex is given as follows;
Charge on complex= +3-1 + 0
= +2
Therefore, the required formula of the complex = [ Rhl(NH₃)₅]²⁺
[Note: Inside the co-ordination complex,metal is written first and followed by negative and neutral ligands respectively and charge on complex is written on the right top of the complex. therefore, the chemical formular of the given complex is given as follows; [ Rhl(NH₃)₅]²⁺
For many purposes we can treat ammonia (NH) as an ideal gas at temperatures above its boiling point of -33.C. Suppose the temperature of a sample of ammonia gas is raised from -21.0°C to 23.0°C, and at the same time the pressure is changed. If the initial pressure was 4.6 atm and the volume decreased by 50.0%, what is the final pressure? Round your answer to 2 significant digits. atm OP x 5 ?
Answers
The final pressure of the ammonia gas, given that the the volume decreased by 50.0%, is 11 atm
How to calculate the final pressure f the ammonia gas?
To solve this question in the most simplified way, we shall begin by listing out the given data from the question. This is shown below:
Initial temperature of ammonia gas (T₁) = -21 °C = -21 + 273 = 252 KFinal temperature of ammonia gas (T₂) = 23 °C = 23 + 273 = 296 KInitial volume of ammonia gas (V₁) = VInitial pressure of ammonia gas (P₁) = 4.6 atmFinal volume of ammonia gas (V₂) = 50% of V₁ = 0.5VFinal pressure of ammonia gas (P₂) = ?The final pressure of the ammonia gas can be calculated as shown below:
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\\\frac{4.6\ \times\ V}{252} = \frac{P_2\ \times\ 0.5V}{296}\\\\252\ \times\ P_2\ \times\ 0.5V = 4.6\ \times\ V\ \times\ 296\\\\P_2 = \frac{4.6\ \times\ V\ \times\ 296}{252\ \times\ 0.5V} \\\\P_2 = 11\ atm[/tex]
Thus, the final pressure is 11 atm