Answer:
The molar solubility of distilled water is [tex]$1.25 \times 10-3$[/tex] (Option B)
Explanation:
Given:
[tex]Pbl_2 -7.9 \times 10^-^9.[/tex]
Expression for solubility constant is,
[tex]$K_{s p}=\left[P b^{2+}\right]\left[I^{-}\right]^{2}$[/tex]
The given equation is,
[tex]$K_{s p}=7.9 \times 10^{-9}$[/tex]
Equation is to calculated as,
[tex]$7.9 \times 10^{-9}=(s) \times(2 s)^{2}$[/tex]
[tex]$7.9 \times 10^{-9}=4 s^{3}$[/tex]
[tex]$s=1.25 \times 10^{-3} \mathrm{~mol} / L$[/tex]
The solubility product in distilled water is [tex]$s=1.25 \times 10^{-3} \mathrm{~mol} / L$[/tex].
Learn more about bonds refer:
https://brainly.com/question/819068The correct answer is C.)[tex]5.0x10^-^4.[/tex]
To determine the molar solubility of PbI2 in distilled water, we need to use the solubility product constant (Ksp) for PbI2, which is given as 7.9x10^-9.
The chemical equation for the dissociation of PbI2 in water is:
[tex]\[ PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq) \][/tex]
The solubility product expression for this equilibrium is:
[tex]\[ Ksp = [Pb^{2+}][I^-]^2 \][/tex]
Let's denote the molar solubility of PbI2 as s. This means that the concentration of Pb^2+ ions will be s, and the concentration of I^- ions will be 2s (since 2 moles of I^- are produced for every mole of PbI2 that dissolves). We can now express the Ksp equation in terms of s:
[tex]\[ Ksp = s \cdot (2s)^2 \][/tex]
[tex]\[ Ksp = s \cdot 4s^2 \][/tex]
[tex]\[ Ksp = 4s^3 \][/tex]
Now we can solve for s using the given Ksp value:
[tex]\[ 7.9 \times 10^{-9} = 4s^3 \][/tex]
[tex]\[ s^3 = \frac{7.9 \times 10^{-9}}{4} \][/tex]
[tex]\[ s^3 = 1.975 \times 10^{-9} \][/tex]
[tex]\[ s = \sqrt[3]{1.975 \times 10^{-9}} \][/tex]
[tex]\[ s \approx 5.82 \times 10^{-4} \][/tex]
Since we are looking for the molar solubility, we can round this value to the appropriate number of significant figures, which is two significant figures based on the Ksp value given [tex](7.9x10^-9[/tex] has two significant figures). Therefore, the molar solubility of PbI2 in distilled water is approximately:
[tex]\[ s \approx 5.8 \times 10^{-4} \][/tex]
The closest answer choice to this calculated value is C.) 5.0x10^-4.
The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes,
(1) NO(g) + NO
2
(g)
→
N
2
O
3
(g) ;
Δ
H
o
r
x
n
= -39.kJ
(2) NO(g) + NO
2
(g) + O
2
(g)
→
N
2
O
5
(g) ;
Δ
H
o
r
x
n
= -112.5 kJ
(3) 2NO
2
(g)
→
N
2
O
4
(g) ;
Δ
H
o
r
x
n
= -57.2 kJ
(4) 2NO(g) + O
2
(g)
→
2NO
2
(g) ;
Δ
H
o
r
x
n
= -114.2 kJ
(5) N
2
O
5
(s)
→
N
2
O
5
(g) ;
Δ
H
o
s
u
b
l
= 54.1 kJ
Calculate the heat of reaction for N
2
O
3
(g) + N
2
O
5
(s)
→
2N
2
O
4
(g)
Δ
H = _ _ _ _ _ kJ
Answer:
The heat of the given reaction is -23.0 kJ.
Explanation:
We are given with ;
[tex]NO(g) + NO_2(g)\rightarrow N_2O_3(g) \Delta H_{1,rxn}=-39.kJ[/tex]..[1]
[tex]NO(g) + NO_2(g) + O_2(g)\rightarrow N_2O_5(g), \Delta H_{2,rxn} = -112.5 kJ [/tex]..[2]
[tex]2NO_2(g) \rightarrow N_2O_4(g) ,\Delta H_{3,rxn} = -57.2 kJ [/tex]..[3]
[tex]2NO(g) + O_2(g)\rightarrow 2NO_2(g), \Delta H_{4,rxn} = -114.2 kJ[/tex]..[4]
[tex]N_2O_5(s)\rightarrow N_2O_5(g) ,\Delta H_{5,sub}= 54.1 kJ[/tex]..[5]
To find heat of reaction:
[tex]N_2O_3(g) + N_2O_5(s)\rightarrow 2N_2O_4(g),\Delta H_{6,rxn} = ?[/tex]..[6]
Using Hess's law:
[5] +2 × [3] + [4] - [2] - [1] = [6]
[tex]\Delta H_{6,rxn}=\Delta H_{5,sub}+2\times \Delta H_{3,rxn}+\Delta H_{4,rxn}-\Delta H_{2,rxn}-\Delta H_{1,rxn}[/tex]
[tex]\Delta H_{6,rxn}=54.1 kJ+(2\times (-57.2 kJ))+(-114.2 kJ)-(-112.5 kJ)-(-39.0 kJ)[/tex]
[tex]\Delta H_{6,rxn}=-23.0 kJ[/tex]
The heat of the given reaction is -23.0 kJ.
"In the absence of an adequate supply of oxygen, yeasts obtains metabolic energy by fermentation of glucose to produce ethanol. C6H12O6(s) LaTeX: \longrightarrow⟶ 2 C2H5OH(l) + 2 CO2(g) Use the standard enthalpies of formation to calculate ΔH for this reaction" Substance ΔHo glucose(s) -304.5 kcal/mol CO2(g) -93.9 kcal/mol C2H5OH(l) -66.4 kcal/mol
Answer: [tex]\Delta H=-16.5 kcal[/tex]
Explanation:
The balanced chemical reaction is,
[tex]C_6H_{12}O_6(s)\longrightarrow 2C_2H_5OH(l)+2CO_2(g)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\Delta H=[(n_{C_2H_5OH}\times \Delta H_{C_2H_5OH})+(n_{CO_2}\times \Delta H_{CO_2})]-[(n_{C_6H_{12}O_6}\times \Delta H_{C_6H_{12}O_6})][/tex]
where,
n = number of moles
Now put all the given values in this expression, we get
[tex]\Delta H=[(2\times -66.4)+(2\times -93.9)]-[(1\times -304.5)][/tex]
[tex]\Delta H=-16.5kcal[/tex]
Therefore, the enthalpy change for this reaction is -16.5 kcal
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e– → Sn(s). What is the concentration of Sn2+ if Zn2+ is 2.5 × 10−3 M and the cell emf is 0.660 V? The standard reduction potentials are given below Zn+2(aq) + 2 e− Sn2+(aq) + 2 e– →→ Zn(s) Sn(s) E∘red E∘red == −0.76 V −0.136 V Consider a voltaic cell where the anode half-reaction is and the cathode half-reaction is . What is the concentration of if is and the cell emf is 0.660 ? The standard reduction potentials are given below 9.0 × 10−3 M 3.3 × 10−2 M 6.9 × 10−4 M 7.6 × 10−3 M 1.9 × 10−4 M
the concentration of [tex]\(Sn^{2+}\)[/tex] is approximately [tex]\(3.3 \times 10^{-2} \, \text{M}\)[/tex].
To solve this problem, we can use the Nernst equation, which relates the cell potential [tex](\(E_{\text{cell}}\))[/tex] to the standard cell potential [tex](\(E^{\circ}_{\text{cell}}\))[/tex] and the concentrations of the species involved in the redox reaction:
[tex]\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0592}{n} \log \left( \frac{[\text{cathode product}]^m}{[\text{anode product}]^n} \right) \][/tex]
Given the standard reduction potentials, we can determine that the number of electrons involved in the half-reactions is [tex]\(n = 2\)[/tex]. The given cell potential is [tex]\(E_{\text{cell}} = 0.660 \, \text{V}\)[/tex]. We need to find the concentration of [tex]\(Sn^{2+}\)[/tex].
First, let's write the Nernst equation for the given cell:
[tex]\[ 0.660 \, \text{V} = (E^{\circ}_{\text{cell}}) - \frac{0.0592}{2} \log \left( \frac{[\text{Sn}^{2+}]}{[\text{Zn}^{2+}]} \right) \][/tex]
We're given the standard reduction potentials, which are [tex]\(E^{\circ}_{\text{cell}} = -0.76 \, \text{V}\)[/tex] for the zinc half-reaction and [tex]\(E^{\circ}_{\text{cell}} = -0.136 \, \text{V}\)[/tex] for the tin half-reaction.
Plugging in the values and solving for [tex]\([\text{Sn}^{2+}]\)[/tex]:
[tex]\[ 0.660 \, \text{V} = (-0.76 \, \text{V}) - \frac{0.0592}{2} \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) \][/tex]
[tex]\[ 0.660 \, \text{V} = (-0.76 \, \text{V}) - 0.0296 \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) \][/tex]
[tex]\[ 0.660 + 0.76 = 0.0296 \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) \][/tex]
[tex]\[ 0.0296 \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) = 0.76 + 0.660 \][/tex]
[tex]\[ \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) = \frac{1.42}{0.0296} \][/tex]
[tex]\[ \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) = 47.973 \][/tex]
[tex]\[ \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} = 10^{47.973} \][/tex]
[tex]\[ [\text{Sn}^{2+}] = (2.5 \times 10^{-3}) \times 10^{47.973} \][/tex]
[tex]\[ [\text{Sn}^{2+}] \approx 3.3 \times 10^{45} \, \text{M} \][/tex]
Therefore, the concentration of [tex]\(Sn^{2+}\)[/tex] is approximately [tex]\(3.3 \times 10^{-2} \, \text{M}\)[/tex].
Enthalpy of formation (kJ/mol) C6H12O6(s)-1260 O2 (g)0 CO2 (g)-393.5 H2O (l)-285.8 Calculate the enthalpy of combustion per mole of C6H12O6. C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l)
Answer : The enthalpy of combustion per mole of [tex]C_6H_{12}O_6[/tex] is -2815.8 kJ/mol
Explanation :
Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
The equilibrium reaction follows:
[tex]C_6H_{12}O_6(s)+6O_2(g)\rightleftharpoons 6CO_2(g)+6H_2O(l)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})][/tex]
We are given:
[tex]\Delta H^o_f_{(C_6H_{12}O_6(s))}=-1260kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol[/tex]
Therefore, the enthalpy of combustion per mole of [tex]C_6H_{12}O_6[/tex] is -2815.8 kJ/mol
Answer:
muahhh
Explanation:
A sample of NI3 is contained in a piston and cylinder. The samples rapidly decomposes to form nitrogen gas and iodine gas, and releases 3.30 kJ of heat and does
950 J of work. What is ΔE?
Answer:
ΔE is -4250 J
Explanation:
Step1: Data given
3.30 kJ of heat is released
There is 950 J of work done
Step 2: Calculate ΔE
ΔE = q + w
⇒ with q = heat of energy released = 3.3 kJ = 3300 J ( negative because the heat is released)
⇒ work done = 950 J ( also negative )
ΔE = -3300J-950J= -4250 J
ΔE is -4250 J
Part A Add single electrons and/or electron pairs as needed to complete the electron-dot symbol for astatine, At. To return the atom to its original state, use the reset button. Click to select an electron-pair or single electron, then click on the element symbol to add electrons. Click the reset button to clear all electrons and start over. View Available Hint(s) AANNN
Answer:
7 valence electrons
Explanation:
Astatine has the atomic number 85. Thus, its electron configuration is:
[Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁵
As we can see, in the last level (6) it has 2 + 5 = 7 electrons, that is, astatine has 7 electrons in its valence shell. In the Lewis dot structure (attached) we write 3 pairs of electrons and 1 unpaired electron around the symbol of At.
A heat flux meter attached to the inner surface of a 3-cm-thick refrigerator door indicates a heat flux of 25 W/m² through the door. Also, the temperatures of the inner and the outer surfaces of the door are measured to be 7°C and 15°C, respectively. Determine the average thermal conductivity of the refrigerator door. Answer: 0.0938 W/m. C
Answer:
0.0938 W/m.°C
Explanation:
The heat is flowing by the refrigerator door by conduction, and can be calculated by the equation:
q = k*ΔT/L
Where q is the heat flux (25 W/m²), k is the thermal conductivity of the refrigerator, ΔT is the variation of the temperature (inner - outer), and L is the thickness of the door (0.03 m).
25 = k*(15-7)/0.03
8k = 0.75
k = 0.0938 W/m.°C
Almost all amino acids are chiral. The body only uses L amino acids, and proteins are chains of amino acids that in the most simple sense fold up into a single structure by themselves when they are formed. Do you think a cell could live with all D (the opposite of L) amino acids, explain your logic. When I ask 'live', I mean could the organism/cell be composed completely of D amino acids and live (This is just a thought question, don't bother trying to find the answer anywhere…)
Answer:
An organism that is completely composed D amino acids cannot survive.
Explanation:
Most of amino acids in all organism are present in L conformation.As result in all organism the enzymes are specific for L amino acids but not for D amino acids.
Bacterial cell wall contain some D amino acids such as D glutamate,D alanine but not entirely composed of D amino acids.
In Haber’s process, 30 moles of hydrogen and 30 moles of nitrogen react to make ammonia. If the yield of the product is 50%, what is the mass of nitrogen remaining after the reaction?
Answer:
Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams
Explanation:
Step 1: Data given
Number of moles hydrogen = 30 moles
Number of moles nitrogen = 30 moles
Yield = 50 %
Molar mass of N2 = 28 g/mol
Molar mass of H2 = 2.02 g/mol
Molar mass of NH3 = 17.03 g/mol
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: Calculate limiting reactant
For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3
Hydrogen is the limiting reactant.
The 30 moles will be completely be consumed.
N2 is in excess. There will react 30/3 =10 moles
There will remain 30 -10 = 20 moles (this in the case of a 100% yield)
In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.
Step 4: Calculate moles of NH3
There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)
For a 50% yield there will be produced, 10 moles of NH3
Step 5: Calculate the mass of NH3
Mass of NH3 = mol NH3 * Molar mass NH3
Mass of NH3 = 20 moles * 17.03
Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)
Step 6: Calculate actual mass
50% yield = actual mass / theoretical mass
actual mass = 0.5 * 340.6
actual mass = 170.3 grams
Step 7: The mass of nitrogen remaining
There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain
Mass of nitrogen = 25 moles * 28 g/mol
Mass of nitrogen = 700 grams
Answer:
25 moles
Explanation:
1 mole of nitrogen reacts with 3 moles of hydrogen to give 2 moles of ammonia. 30 moles of hydrogen will react with 10 moles of nitrogen to give 20 moles of ammonia. As the actual yield is 50%, ammonia formed is 10 moles, the amount of nitrogen reacted is 5 moles, and the amount of hydrogen reacted is 15 moles. The mass of the remaining hydrogen is 15 moles and of the remaining nitrogen is 25 moles.
(a)-Use Lewis symbol store present there action that occurs between Ca and F atoms. (b)-What is the chemical formula of the most likely product? (c)-How many electrons are transferred? (d)-Which atom loses electrons in the reaction?
The Lewis symbol for Ca is Ca²⁺, while F is F⁻. The Ca atom loses two electrons, and the F atom gains one electron. The most likely product is CaF₂.
Explanation:(a) The Lewis symbol for Ca is Ca2+, while the Lewis symbol for F is F-. The action that occurs between Ca and F atoms is a redox reaction, where one atom loses electrons (oxidation) and the other gains electrons (reduction).
(b) The chemical formula of the most likely product is CaF2 since one Ca atom can bond with two F atoms.
(c) In the reaction, two electrons are transferred, with Ca losing two electrons and each F atom gaining one electron.
(d) Ca atom loses electrons in the reaction.
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A saturated solution of lead(II) chloride, PbCl2, was prepared by dissolving solid PbCl2 in water. The concentration of Pb2+ ion in the solution was found to be 1.62×10−2 M . Calculate Ksp for PbCl2.
Answer:
Ksp for PbCl2 is 1.7 *10^-5
Explanation:
Step 1: Data given
The concentration of Pb2+ ion in the solution was found to be 1.62 *10^−2 M
Step 2: The balanced equation
PbCl2 ⇔Pb^2+ + 2Cl -
Step 2: ICE chart
The initial concentration of Pb2+ is 0 M
There will react X M and 2X of Cl-
At the equillibrium there is X M of Pb^2+ and 2X M of Cl-
The concentration of Pb2+ ion in the solution was found to be 1.62 *10^−2 M
Step 3: Calculate Ksp
Since PbCl2 is solid, it doesn't aply for Ksp
Ksp = [Pb^2+][Cl-]²
Ksp = X*(2X)²
Ksp = 4X³
⇒ X = 1.62 *10^-2 M
Ksp = 4*( 1.62 *10^-2)³
Ksp =1.7 *10^-5
Ksp for PbCl2 is 1.7 *10^-5
For fatty acids with the same number of carbon atoms, how does the melting point change as the number of double bonds in the fatty acid changes?
a. The melting point of the fatty acid decreases as the number of double bonds increases.
b. There is no relationship between the melting point of a fatty acid and the number of double bonds.
c. The melting point of the fatty acid is unchanged as the number of double bonds changes.
d. The melting point of the fatty acid decreases as the number of double bonds decreases.
The answer is a. "The melting point of the fatty acid decreases as the number of double bonds increases" because when there are double bonded carbons it cancels the polarity of the carboxyl in the opposite direction, increasing that effect as double bonded carbons add to the molecule, giving the net dipole moment to the molecule. As the net dipole moment of the molecule decrease, the melting point of the fatty acid also decreases.
In addition, when the fatty acid is major than 8 carbons and have a double bond that prevents the fatty acid from crystal lattice formation, so the melting point will be lower also for this reason in this case.
Final answer:
The melting point of a fatty acid decreases as the number of double bonds increases because double bonds create kinks in the fatty acid chains, preventing tight packing and weakening intermolecular forces. The correct answer is option a.
Explanation:
For fatty acids with the same number of carbon atoms, the melting point changes as the number of double bonds in the fatty acid changes. The correct option is (a) - the melting point of the fatty acid decreases as the number of double bonds increases. This decrease in melting point occurs because unsaturated fatty acids, which have one or more double bonds, tend to have kinks or bends at the location of these bonds.
These kinks inhibit the fatty acids from packing closely together, resulting in weaker intermolecular Van der Waals forces (specifically London dispersion forces) and thus a lower melting point. In contrast, saturated fatty acids without double bonds have straight chains and can pack tightly together, leading to stronger intermolecular forces and higher melting points.
A good illustration of this concept is the substantial difference in melting points between palmitoleic acid, which contains a cis-double bond and melts over 60℃ lower than its saturated counterpart, palmitic acid. When considering the physical properties of fatty acids, it's important to note that more saturated fatty acids are typically more solid at room temperature due to their higher melting points, while unsaturated fatty acids are usually liquid.
How much heat is absorbed/released when 25.00 g of NH3(g) reacts in the presence of excess O2(g) to produce NO(g) and H2O(l) according to the following chemical equation? 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) ΔH° = 1168 kJ
428.7 kJ of heat are absorbed when 25.00 g of NH₃(g) reacts in the presence of excess O₂(g) to produce NO(g) and H₂O(l).
Let's consider the following thermochemical equation.
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(l) ΔH° = 1168 kJ
The standard enthalpy of the reaction is positive, which means that the reaction is endothermic, that is, heat is absorbed.
We will convert 25.00 g of NH₃ to moles using its molar mass (17.03 g/mol).
[tex]25.00 g \times \frac{1mol}{17.03g} = 1.468 mol[/tex]
1168 kJ of heat are absorbed when 4 moles of NH₃ react. The heat absorbed when 1.468 moles of NH₃ react is:
[tex]1.468 mol \times \frac{1168kJ}{4mol} = 428.7 kJ[/tex]
428.7 kJ of heat are absorbed when 25.00 g of NH₃(g) reacts in the presence of excess O₂(g) to produce NO(g) and H₂O(l).
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Chemical reactions are defined as the reaction in which two or more reactants combine to form single or more products. The heat can be absorbed or evolved in the reaction. The heat absorbed is 428.7kJ when ammonia reacts with excess oxygen.
Given that,
Mass of ammonia = 25 gMolar mass of ammonia = 17.03 g/molThe chemical equation between ammonia and oxygen is:
[tex]\text {4 NH}_3 \;+\; \text {5 O}_2\;\rightarrow\; \text {4 NO + 6 H}_2\text {O}\;\;\;\;\;\Delta\text H&=1168 \text {kJ}[/tex]Now, the enthalpy of the reaction is positive, such that the reaction is endothermic. In endothermic reaction heat is absorbed.
Now, converting the mass of ammonia into moles, we get:
[tex]25.00\text {g} \;\times\;\dfrac{1 \text{mol}}{17.03\text{g}} &= 1.468\; \text {mol}[/tex]1168 kJ of heat is absorbed when 4 moles of ammonia react with oxygen. The heat absorbed in 1.468 moles, will be:
[tex]1.468\;\text{mol}\;\times\;\dfrac{1168\;\text{kJ}}{4\;\text{mol}}&=428.7\;\text{kJ}[/tex]Therefore, when 25 grams of ammonia reacts in the presence of oxygen 428.7 kJ will be absorbed. Hence, it is an endothermic reaction.
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A reaction will be spontaneous only at low temperatures if both ΔH and ΔS are negative. For a reaction in which ΔH = −320.1 kJ/mol and ΔS = −86.00 J/K · mol.
Determine the temperature (in °C) below which the reaction is spontaneous.
Answer:
This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C
Explanation:
Step 1: Data given
ΔH = −320.1 kJ/mol
ΔS = −86.00 J/K · mol.
Step 2: Calculate the temperature
ΔG<0 = spontaneous
ΔG= ΔH - TΔS
ΔH - TΔS <0
-320100 - T*(-86) <0
-320100 +86T < 0
-320100 < -86T
320100/86 > T
3722.1 > T
The temperature should be lower than 3722.1 Kelvin (= 3448.9 °C)
We can prove this with Temperature T = 3730 K
-320100 -3730*(-86) <0
-320100 + 320780 = 680 this is greater than 0 so it's non spontaneous
T = 3700 K
-320100 -3700*(-86) <0
-320100 + 318200 = -1900 this is lower than 0 so it's spontaneous
The temperature is quite high because of the big difference between ΔH and ΔS.
This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C
Match each weather phenomena listed below to its appropriate scale of motion. longwave ridges and troughs microscale a high pressure system persists over the central plains for 4 days synoptic scale a supercell thunderstorm that lasts for over an hour global scale A turbulent eddy mesoscale
Answer:
Longwave ridges = Global scale
High pressure system persists over the central plains for 4 days = Synoptic scale
Supercell thunderstorm that lasts for over an hour = mesoscale
A turbulent eddy = microscale
Explanation:
Global scale has a range of the entire earth
synoptic scale has a range of about 100-1000km and can last fro days to weeks
mesoscale has a range of 4-100km and lasts for a day maximum
microscale is the least of all atmospheric motions.
A solution of the primary standard potassium hydrogen phthalate, (204.22 g/mol), was prepared by dissolving 0.4877 g of in about 50 mL of water. The solution was titrated with an solution and mL were needed to reach the phenolphthalein end point. What is the molarity of the solution?
Answer: The molarity of KOH solution is 0.0663 M.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of KHP = 0.4877 g
Molar mass of KHP = 204.22 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of KHP}=\frac{0.4877g}{204.22g/mol}=0.0024mol[/tex]
The chemical reaction for the formation of chromium oxide follows the equation:
[tex]KHC_8H_4O_4(aq.)+KOH\rightarrow K_2C_8H_4O_4(aq.)+H_2O(l)[/tex]
By Stoichiometry of the reaction:
1 mole of KHP reacts with 1 mole of KOH.
So, 0.0024 moles of KHP will react with = [tex]\frac{1}{1}\times 0.0024=0.0024mol[/tex] of KOH.
To calculate the molarity of KOH, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
We are given:
Moles of KOH = 0.0024 moles
Volume of solution = 36.21 mL = 0.03621 L (Assuming) (Conversion factor: 1L = 1000 mL)
Putting values in above equation, we get:
[tex]\text{Molarity of KOH }=\frac{0.0024mol}{0.03621L}=0.0663M[/tex]
Hence, the molarity of KOH solution is 0.0663 M.
In which instance is a gas most likely to behave as an ideal gas?A.) At low temperatures, because the molecules are always far apartB.) When the molecules are highly polar, because IMF are more likelyC.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apartD.) At high pressures, because the distance between molecules is likely to be small in relation to the size of the molecules
Answer:
C.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apart.
Explanation:
For gas to behave as an ideal gas there are 2 basic assumptions:
The intermolecular forces (IMF) are neglectable.The volume of the gas is neglectable in comparison with the volume of the container.In which instance is a gas most likely to behave as an ideal gas?
A.) At low temperatures, because the molecules are always far apart. FALSE. At low temperatures, molecules are closer and IMF are more appreciable.
B.) When the molecules are highly polar, because IMF are more likely. FALSE. When IMF are stronger the gas does not behave as an ideal gas.
C.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apart. TRUE.
D.) At high pressures, because the distance between molecules is likely to be small in relation to the size of the molecules. FALSE. At high pressures, the distance between molecules is small and IMF are strong.
An air/gasoline vapor mix in an automobile cylinder has an initial temperature of 180 ∘C and a volume of 13 cm3 . If the mixture is heated to 587 ∘C with the pressure and amount held constant, what will be the final volume of the gas in cubic centimeters? Express your answer in cubic centimeters to three significant figures.
Answer:
The final volume will be 24.7 cm³
Explanation:
Step 1: Data given:
Initial temperature = 180 °C
initial volume = 13 cm³ = 13 mL
The mixture is heated to a fina,l temperature of 587 °C
Pressure and amount = constant
Step 2: Calculate final volume
V1/T1 = V2/T2
with V1 = the initial volume V1 = 13 mL = 13*10^-3
with T1 = the initial temperature = 180 °C = 453 Kelvin
with V2 = the final volume = TO BE DETERMINED
with T2 = the final temperature = 587 °C = 860 Kelvin
V2 = (V1*T2)/T1
V2 = (13 mL *860 Kelvin) /453 Kelvin
V2 = 24.68 mL = 24.7 cm³
The final volume will be 24.7 cm³
What is the concentration in molarity of a solution made using 10.0 grams of KCl in 300.0 mL of water?
Please help immediately!!! :(
Answer:
From Molarity=concentration/molar mass
Concentration=10/0.3dm³
33.33g/dm³
Molar mass=H2O=18g/mol
Molarity=1.852mol/dm³
Answer:
0. 446 mol L−1
Explanation:
For Molarity, we must know the following things
• the number of moles of solute present in solution
• the total volume of the solution
we know the mass of one mole of potassium chloride = 74.55 g
Number of moles = Given Mass of substance / Mass of one mole
No of moles = 10 / 74.55
= 0.134 moles
Now we know that molarity is expressed per liter of solution. Since you dissolve 0.134 moles of potassium chloride in 300. mL of solution, you can say that 1.0 L will contain
For 300 ml of solution, no of moles are = 0.134 moles
For 1 ml of solution, no of moles are = 0.134/300 moles
For 1(1000) ml of solution, no of moles are= 0.134/300 x 1000
= 0.446 moles/ L
Answer is =0. 446 mol L−1
Find the equilibrium partial pressures of A and B for each of the following different values of Kp.?Consider the following reaction:A(g) = 2B(g)Find the equilibrium partial pressures of A and B for each of the following different values of Kp. Assume that the initial partial pressure of B in each case is 1.0 atm and that the initial partial pressure of A is 0.0 atm. Make any appropriate simplifying assumptions.Kp = 1.4?Kp = 2.0 * 10^-4?Kp = 2.0 * 10^5?
Answer:
For Kp = 1,4; [tex]P_{[A] = 0,22[/tex], [tex]P_{[B] = 0,56atm[/tex]
For Kp = 2,0x10⁻⁴; [tex]P_{[A] = 0,495[/tex], [tex]P_{[B] = 0,01atm[/tex]
For Kp = 2,0x10⁵; [tex]P_{[A] = 5x10^{-6}[/tex], [tex]P_{[B] = 0,99999atm[/tex]
Explanation:
For the reaction:
A(g) ⇄ 2B(g)
kp is: [tex]kp = \frac{P_{[B]}^2}{P_{[A]}}[/tex]
If initial pressure of B is 1,0atm and initial pressure of A is 0,0atm the equilibrium pressures are:
[tex]P_{[A] = 0,0atm + X[/tex]
[tex]P_{[B] = 1,0atm - 2X[/tex]
Replacing for Kp= 1,4:
[tex]1,4 = \frac{(1-2X)^2}{X}[/tex]
1,4X = 4X² - 4X + 1
0 = 4X² - 5,4X + 1
Solving for X:
X = 0,22 -Right answer-
X = 1,13
Replacing:
[tex]P_{[A] = 0,22[/tex]
[tex]P_{[B] = 1,0atm - 0,44atm = 0,56atm[/tex]
For Kp= 2,0x10⁻⁴:
[tex]2,0x10^{-4} = \frac{(1-2X)^2}{X}[/tex]
2,0x10^{-4}X = 4X² - 4X + 1
0 = 4X² - 4,0002X + 1
Solving for X:
X = 0,495atm
Replacing:
[tex]P_{[A] = 0,495atm[/tex]
[tex]P_{[B] = 1,0atm - 0,99atm = 0,01atm[/tex]
For Kp= 2,0x10⁵:
[tex]2,0x10^5 = \frac{(1-2X)^2}{X}[/tex]
2,0x10^5X = 4X² - 4X + 1
0 = 4X² - 2,00004x10^5X + 1
Solving for X:
X = 5x10⁻⁶ -Right answer-
Replacing:
[tex]P_{[A] = 5x10^{-6}[/tex]
[tex]P_{[B] = 1,0atm - 0,00001atm = 0,99999atm[/tex]
I hope it helps!
To determine the equilibrium partial pressures of A and B for the given reaction, one must solve a quadratic equation derived from the equilibrium expression involving the given Kp and initial conditions, considering the stoichiometry and the relationship between the changes in partial pressures of A and B as the reaction reaches equilibrium.
Explanation:When considering the equilibrium partial pressures of A and B for the reaction A(g) = 2B(g), we need to calculate how the initial conditions and the equilibrium constant (Kp) affect the final partial pressures at equilibrium. Starting with an initial partial pressure of B as 1.0 atm and A as 0.0 atm, and using the equilibrium expression Kp = (PB)^2/PA, we can plug in different values of Kp to solve for the unknowns. Assuming the reaction has proceeded to equilibrium, we can express any changes in partial pressure of A as -x and of B as +2x because for each mole of A reacting, 2 moles of B are formed.
For example, with Kp = 1.4, let's assume 'x' is the change in partial pressure of B; we then construct the equation Kp = (1 + 2x)^2/(0.0 + x). Solving the quadratic equation derived would give us the value of 'x', which in turn provides the equilibrium partial pressures PA and PB. We would need to solve a similar quadratic equation for different values of Kp such as 2.0 * 10^-4 and 2.0 * 10^5 following the same approach. It's important to note that some values might yield non-physical solutions, like negative pressures; those are dismissed since partial pressures must be positive.
Use the following information on Cr to determine the amounts of heat for the three heating steps required to convert 126.3 g of solid Cr at 1760°C into liquid Cr at 2060°C. mp = 1860°C bp = 2672°C Enter in kJ. Useful data: \Delta HΔ Hfus = 20.5 kJ/mol; \Delta HΔ Hvap = 339 kJ/mol; c(solid) 44.8 J/g°C; c(liquid) = 0.94 J/g°C
Answer : The amount of heat required is, 639.3 KJ
Solution :
The conversions involved in this process are :
[tex](1):Cr(s)(1760^oC)\rightarrow Cr(s)(1860^oC)\\\\(2):Cr(s)(1860^oC)\rightarrow Cr(l)(1860^oC)\\\\(3):Cr(l)(1860^oC)\rightarrow Cr(l)(2060^oC)[/tex]
Now we have to calculate the enthalpy change.
[tex]\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = enthalpy change or heat required = ?
m = mass of Cr = 126.3 g
[tex]c_{p,s}[/tex] = specific heat of solid Cr = [tex]44.8J/g^oC[/tex]
[tex]c_{p,l}[/tex] = specific heat of liquid Cr = [tex]0.94J/g^oC[/tex]
n = number of moles of Cr = [tex]\frac{\text{Mass of Cr}}{\text{Molar mass of Cr}}=\frac{126.3g}{52g/mole}=2.428mole[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 20.5 KJ/mole = 20500 J/mole
Now put all the given values in the above expression, we get
[tex]\Delta H=[126.3g\times 44.8J/g^oC\times (1860-(1760))^oC]+2.428mole\times 20500J/mole+[126.3g\times 0.94J/g^oC\times (2060-1860)^oC][/tex]
[tex]\Delta H=639342.4J=639.3KJ[/tex] (1 KJ = 1000 J)
Therefore, the amount of heat required is, 639.3 KJ
To determine the amounts of heat for each heating step, we need to calculate the heat required to raise the temperature from solid Cr at 1760°C to its melting point at 1860°C, the heat of fusion to convert the solid Cr at its melting point to liquid Cr, and the heat required to raise the temperature from liquid Cr at 1860°C to the final temperature of 2060°C.
Explanation:To determine the amounts of heat for each heating step, we need to calculate the heat required to raise the temperature from solid Cr at 1760°C to its melting point at 1860°C, the heat of fusion to convert the solid Cr at its melting point to liquid Cr, and the heat required to raise the temperature from liquid Cr at 1860°C to the final temperature of 2060°C.
Step 1: Calculate the heat required to raise the temperature from 1760°C to 1860°C using the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.Step 2: Calculate the heat of fusion using the formula Q = nΔHfus, where Q is the heat, n is the number of moles, and ΔHfus is the heat of fusion.Step 3: Calculate the heat required to raise the temperature from 1860°C to 2060°C using the formula Q = mcΔT.By summing up the heats calculated in each step, you can determine the total amount of heat required to convert the given mass of solid Cr at 1760°C to liquid Cr at 2060°C.
Learn more about Heat calculation here:https://brainly.com/question/32376262
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Organic Chem Rxn Question
Using the reagents below, list in order (by letter, no period) those necessary to prepare trans-1,2-dibromocyclopentane from cyclopentane.
Note: Not all spaces provided may be needed. Type "na" in any space where you have no reagent. (There are 3 spaces)
a. Br2, heat, light
b. HBr
c. Br2
d. H2O
e. HBr, HOOH
f. EtOH
g. NaOEt, EtOH, heat
Answer:
a, g, c
Explanation:
The conversion of the stable cyclopentane into Trans-1, 2dibromocyclopentane will require three step reactions.
The first is to convert the compound into a cyclopentene, through the addition of Bromine water under heat and photons (light). So option A is the first in the order. This will generate 1 bromocyclopentane through halogenation of the alkane. Secondly, a hot and strong base should be added like the NaOEt, EtOH to remove the added bromine and one atom of hydrogen from the resulting 1 bromocyclopentane in the previous reaction. This will yield cyclopentene, thus making the compound more electrophilic. So option g is required. Thirdly, bromine molecules will be added (C) to take up their places at the two electrophilic regions of the compound to produce Trans-1, 2dibromocyclopentane.
Final answer:
To synthesize trans-1,2-dibromocyclopentane from cyclopentane, the correct reagent is Br2 alone (option c). Other reagents like heat and light could lead to radical mechanisms, which are not appropriate for this specific transformation.
Explanation:
To prepare trans-1,2-dibromocyclopentane from cyclopentane, you need to add bromine (Br2) across the double bond in a trans configuration. The best way to achieve this without any other rearrangements or products is to use Br2 alone (option c). When Br2 is added to a double bond, it will typically add in an anti-fashion, leading to the trans product. Using Br2 with heat/light (option a) or with solvents like EtOH (option f) would likely lead to radical mechanisms or other side products.
Therefore, the correct order of reagents to prepare trans-1,2-dibromocyclopentane from cyclopentane would be just: c. The other two spaces would be filled with 'na' as no additional reagents are needed for this transformation.
Thus, the list in order would be: c, na, na.
Be sure to answer all parts. Enter your answers in scientific notation.
The following values are the only allowable energy levels of a hypothetical one-electron atom:
E6 = −2.0×10−19 J
E5 = −7.0×10−19 J
E4 = −11.0×10−19 J
E3 = −15.0×10−19 J
E2 = −17.0×10−19 J
E1 = −20.0×10−19 J
(a) If the electron were in the n = 5 level, what would be the highest frequency (and minimum wavelength) of radiation that could be emitted?
Frequency-
Wavelength-
(b) If the electron were in the n = 1 level, what would be the shortest wavelength (in nm) of radiation that could be absorbed without causing ionization?
Answer:
a) f = 3.02x10¹⁵ s⁻¹, and λ = 99.4 nm.
b) 99.4 nm
Explanation:
a) The energy of radiation is given by:
E = h*f
Where h is the Planck constant (6.626x10⁻³⁴ J.s), and f is the frequency. To have the highest frequency, the energy must be the highest too, because they're directly proportional. So we must use E = -E1 = 20x10⁻¹⁹ J
20x10⁻¹⁹ = 6.626x10⁻³⁴xf
f = 3.02x10¹⁵ s⁻¹
The wavelenght is the velocity of light (3.00x10⁸ m/s) divided by the frequency:
λ = 3.00x10⁸/3.02x10¹⁵
λ = 9.94x10⁻⁸ m = 99.4 nm
b) To have the shortest wavelength, it must be the highest energy and frequency, so it would be the same as the letter a) 99.4 nm.
Arizona was the site of a 400,000-acre wildfire in June 2002. How much carbon dioxide (CO2) was produced into the atmosphere by that fire? [Hints: Assume that the density of carbon on the acreage was 10 kg/m2 and that 50% of the biomass burned. In addition, 10,000 m2 = 2.47 acre].
Answer:
2.97 × 10¹³ g
Explanation:
First, we have to calculate the biomass the is burned. We can establish the following relations:
2.47 acre = 10,000 m² 10 kg of C occupy an area of 1 m²50% of the biomass is burnedThe biomass burned in the site of 400,000 acre is:
[tex]400,000acre\times\frac{10,000m^{2} }{2.47acre} \times \frac{10kgC}{m^{2} } \times 50\% = 8.10 \times 10^{9} kgC[/tex]
Let's consider the combustion of carbon.
C(s) + O₂(g) ⇒ CO₂(g)
We can establish the following relations:
The molar mass of C is 12.01 g/mol1 mole of C produces 1 mole of CO₂The molar mass of CO₂ is 44.01 g/molThe mass of produced is CO₂:
[tex]8.10 \times 10^{12}gC \times \frac{1molC}{12.01gC} \times \frac{1molCO_{2}}{1molC} \times \frac{44.01gCO_{2}}{1molCO_{2}} =2.97 \times 10^{13} gCO_{2}[/tex]
Be sure to answer all parts. Write a balanced equation and Kb expression for the following Brønsted-Lowry base in water: benzoate ion, C6H5COO−. Include the states of all reactants and products in your equation. You do not need to include states in the equilibrium expression. Balanced equation: ⇌ Kb expression:
Answer:
The balanced reaction is:-
[tex]C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}[/tex]
[tex]K_b[/tex] expression is:-
[tex]K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}[/tex]
Explanation:
Benzoate ion is the conjugate base of the benzoic acid. It is a Bronsted-Lowry base and the dissociation of benzoate ion can be shown as:-
[tex]C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}[/tex]
The expression for dissociation constant of benzoate ion is:
[tex]K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}[/tex]
Final answer:
The balanced chemical equation for the reaction of benzoate ion, C6H5COO−, in water is C6H5COO−(aq) + H2O(l) ⇌ C6H5COOH(aq) + OH−(aq), and the Kb expression is Kb = [C6H5COOH][OH−] / [C6H5COO−].
Explanation:
The question involves writing a balanced equation for the reaction of the benzoate ion, C6H5COO−, as a Brønsted-Lowry base in water, and then writing the Kb expression for the reaction. In water, the benzoate ion accepts a proton (H+) from water (‘H2O’), forming benzoic acid (C6H5COOH) and hydroxide ions (OH−). The balanced chemical reaction is as follows:
C6H5COO−(aq) + H2O(l) ⇌ C6H5COOH(aq) + OH−(aq)
The Kb expression, which represents the base ionization constant, is written without including the states of the substances. It is derived from the concentrations of the products over the concentration of the reactant, not including water due to its constant concentration in dilute solutions. The Kb expression is:
Kb = [C6H5COOH][OH−] / [C6H5COO−]
When 1.98g of a hydrocarbon is burned in a bomb calorimeter, the temperature increases by 2.06∘C. If the heat capacity of the calorimeter is 69.6 J∘C and it is submerged in 944mL of water, how much heat (in kJ) was produced by the hydrocarbon combustion?
Answer:
8.3 kJ
Explanation:
In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:
q for water:
q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g
c: specific heat of water = 4.186 J/gºC
ΔT : change in temperature = 2.06 ºC
so solving for q :
q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J
For calorimeter
q calorimeter = C x ΔT where C: heat capacity of calorimeter = 69.6 ºC
ΔT : change in temperature = 2.06 ºC
q calorimeter = 69.60J x 2.06 ºC = 143.4 J
Total heat released = 8,140 J + 143.4 J = 8,2836 J
Converting into kilojoules by dividing by 1000 we will have answered the question:
8,2836 J x 1 kJ/J = 8.3 kJ
The heat produced by the combustion of 1.98g of the hydrocarbon is 0.143 kJ.
The volume of water was not needed for this particular calculation.
To determine the heat produced by burning 1.98g of a hydrocarbon in a bomb calorimeter, we can use the formula :q = Ccal × ΔTWhere:
Ccal is the heat capacity of the calorimeterΔT is the change in temperatureGiven:
Mass of hydrocarbon: 1.98gTemperature increase: 2.06°CHeat capacity of calorimeter: 69.6 J/°CVolume of water: 944 mL (not needed for this calculation)Let's plug in the values:
q = 69.6 J/°C × 2.06°Cq = 143.376 JSince we need the heat in kJ :q = 143.376 J / 1000 = 0.143376 kJThe heat produced by the hydrocarbon combustion is 0.143 kJ.Identify two structural features of purines and Pyrimidines
Purines
A. contain only three ring nitrogen atoms.
B. contain one heterocyclic ring.
C. contain four ring nitrogen atoms.
D. contain only two ring nitrogen atoms.
E.contain two heterocyclic rings.
Pyrimidines
A.contain two heterocyclic rings.
B. contain only three ring nitrogen atoms.
C. contain only two ring nitrogen atoms.
D. contain four ring nitrogen atoms
E. contain one heterocyclic ring.
Answer:
The answers are:
Purines:
C. contain four ring nitrogen atoms.
E. contain two heterocyclic rings.
Pyrimidines:
C. contain only two ring nitrogen atoms.
E. contain one heterocyclic ring.
Explanation:
Purines and Pyrimidines are nitrogenous bases which are the building blocks of nucleic acids (DNA and RNA).
Purines are composed by two fused heterocyclic rings, one of them is a 6-ring and the other is a 5-ring. Each ring contains two nitrogen atoms which form part of the ring. Thus, the nitrogen positions in purines are: 1', 3', 7' and 9'. Depending on the functional groups bonded to the two-ring structure, a purine base can be Guanidine (G) or Adenine (A).
The structure of Pyrimidines is a single heterocycle ring wich contains two nitrogen atoms in positions 1' and 3'. Depending of the functional groups, they can be: Cytosine (C), Thymidine (T) and Uracil (U, which is found in RNA).
Glucose is present in blood plasma at a concentration of approximately 80 mg/dL. A typical 70 kg human has approximately 5 L blood, which is 55% plasma. Using the 44.96 mg/mL glucose concentration of Sprite that you calculated in the previous question, how many mL of Sprite does a human need to drink to reach a plasma glucose concentration of 80 mg/dL, assuming none is mobilized? (Note: all extracellular glucose is about 10 times this level, which is not important for this calculation.(Please pay attention to the units mg/dL)
Answer:
A human need to drink 48.93 mL of sprite to reach a plasma glucose concentration of 80 mg/dL.
Explanation:
Volume of blood in 7 Kg human = 5 L
Percentage of plasma in blood = 55%
Volume of plasma in 5 L blood = [tex]\frac{55}{100}\times 5 L=2.75 L=27.5 dL[/tex] (1 L = 10 dL)
Concentration of glucose in plasma = 80 mg/dL
Amount of glucose present in 27.5 dL : 80 mg /dL × 27.5 dL= 2200 mg
Let the volume of sprite with 2200 mg glucose be x
Concentration of glucose in sprite = 44.96 mg/mL
[tex]x\times 44.96 mg/mL=2200 mg[/tex]
[tex]x=\frac{2200 mg}{44.96 mg/mL}=48.93 mL[/tex]
A human need to drink 48.93 mL of sprite to reach a plasma glucose concentration of 80 mg/dL.
A stock solution contains a mixture of ~100 ppm chloride, fluoride, nitrite, bromide, nitrate and phosphate anions. In order to prepare 1 L of 100 ppm nitrite stock solution, you weigh out 150.0 mg of NaNO2. The actual concentration of nitrite would be:______.
Answer:
[tex]150~ppm[/tex]
Explanation:
The first step is to draw the ionization reaction of [tex]NaNO_2[/tex], so:
[tex]NaNO_2~->~Na^+~+~NO_2^-[/tex]
The molar ratio between [tex]NaNO_2[/tex] and [tex]NO_2^-[/tex] is 1:1
The next step is the calculation of the concentration. We have to remember that the formula of ppm is:
[tex]ppm=\frac{mg}{L}[/tex]
In this case we will have a mass of 150 mg and a volume of 1 L so:
[tex]ppm=\frac{150~mg}{1~L}=~150ppm[/tex]
What evidence is there to suggest that the Earth is composed of tectonic plates?
Question 1 options:
The presence of a Ring of “Fire, hot spots, and the locations of earthquakes are some kinds of evidence that indicate plates exist and move. Data from seafloor spreading is another kind of data. The presence of fossils indicates that two land masses that once were close are now far apart. Mountain ranges that once were part of the same range are now on different continents. All of these indicates plates exist and move.
The tectonic plates are cracked up due to the frequency of earthquakes and scientists can see and measure these cracks. The cracks are the proof that the earth is made up of tectonic plates.
Answer:
All of these indicate plates exist and move.
Explanation:
The Ring of Fire, hot spots, and the locations of earthquakes draw a pattern on the surface of geologic activity underground. Seafloor spreading is proof of tectonic plates because it increases the area of some oceans and decreases the area of others. Magma from the mantle pushes the plates apart. The case of matching fossils in parts of South America and Africa is consistent with the theory that at one point the continents were joined together. According to seafloor spreading, the Atlantic Ocean is young and is still growing. The same principle applies to mountain ranges now found on two continents that used to be on the same one. It is thought that the Appalachian mountains were connected to Europe and Africa.Evidence supporting the existence of tectonic plates includes geological features such as the Ring of Fire, earthquake locations, and seafloor spreading. Fossil distribution across continents and the formation of mountain ranges also support the theory. Hot spots show how islands form and move, further confirming plate tectonics.
There is substantial evidence to suggest that the Earth is composed of tectonic plates. Some of the most compelling pieces of evidence include the presence of the Ring of Fire, the occurrence of hot spots, and the locations of earthquakes. These features highlight areas where plates interact. Additionally, data from seafloor spreading, such as the formation of mid-ocean ridges and the discovery of symmetrical patterns of magnetic stripes on either side of these ridges, supports this theory.
The presence of fossils provides further proof. For example, identical fossils of the fern Glossopteris have been found on continents that are now widely separated, indicating that these landmasses were once connected. Similarly, mountain ranges that were once part of the same range are now found on different continents. These phenomena can be explained by the movement of tectonic plates.
Moreover, the study of hot spots, such as those under Hawaii, shows how islands are formed and then move away from the hot spot due to plate movement. These various lines of evidence converge to form a comprehensive picture of how tectonic plates shape the Earth's surface.