Answer:
rate = [NO]²[H₂]
Explanation:
2NO + H2 ⟶N2 + H2O2 (slow)
H2O2 + H2 ⟶2H2O (fast)
From the question, we are given two equations.
In chemical kinetics; that is the study of rate reactions and changes in concentration. The rate law is obtained from the slowest reaction.
This means that our focus would be on the slow reaction. Generally the rate law is obtained from the concentrations of reactants in a reaction.
This means our rate law is;
rate = [NO]²[H₂]
What would happen to the rate of a reaction with rate law rate = k [NO]2[Hz] if
the concentration of NO were doubled?
Answer:
The rate would have doubled
Explanation:
At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with X Benzene
CHECK COMPLETE QUESTION BELOW
At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with Xbenzene = 0.580.
Answer:
The total vapor pressure is [tex]81.3 mmHg[/tex]
Explanation:
We will be making use of Dalton and Raoults equation in order to calculate the total pressure,
Which is [tex]PT= (PA × XA) +(PB ×XB)[/tex]
PT= total vapor pressure
From the question
Benene's Mole fraction = 0.580
then to get Mole fraction of toluene we will substract the one of benzene from 1. because total mole fraction is always 1.
= (1 - 0.580) = 0.420
Vapor pressure of benzene given = 183 mmHg
Vapor pressure of toluene given= 59.2 mmHg
If we substitute those value into above equation, we have
PT=(183×0.580)+(59.2×0.420)
=81.3mmHg
Therefore,, the total vapor pressure of the solution is 81.3 mmHg
Based on relative bond strengths, classify these reactions as endothermic (energy absorbed) or exothermic (energy released).
Strongest Bond
A-B
A-A
B-B
C-C
B-C
A-C
1. A2 + C2 rightarrow 2AC
2. B2 + C2 rightarrow 2BC
3. A + BC rightarrow AB + C
4. A2 + B2 rightaarrow 2AB
5. AB + C rightarrow AC + B
a. endothermic
b. exothermic
Answer:
1. Exothermic 2. Exothermic 3. Endothermic 4. Endothermic 5. Exothermic.
Explanation:
1. An A-A and a C-C bond results in 2 A-C bonds which are lower than the A-A and C-C bonds so this reaction is exothermic.
2. A B-B bond and a C-C bond results in 2 B-C bonds which are lower than the first 2 bonds so this reaction is also exothermic.
3. There is no bond for single A, a single B-C bond results in a A-B bond and a C molecule. A-B bond is stronger than the B-C bond so the reaction absorbed energy along the way. This shows that it is endothermic.
4. An A-A bond and a B-B bond results in 2 A-B bonds which are stronger than the first two bonds so this reaction is also endothermic.
5. An A-B bond and a C molecule result in an A-C bond and a B molecule. A-C bond is weaker than the A-B bond so there is energy released. This reaction is exothermic.
I hope this answer helps.
Which substance has the highest boiling point?
Select the correct answer below:
A) ethyl ether
B) ethyl alcohol
C) water
D) these all have the same boiling point
Answer:
D) these all have the same boiling point
What is the pOH of a solution at 25.0∘C with [H3O+]=4.8×10−6 M?
Answer:
8.68
Explanation:
pOH = 8.68
all you need is contained in the sheet
Answer:
Approximately [tex]8.68[/tex].
Explanation:
The [tex]\rm pOH[/tex] of a solution can be found from the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] with the following equation:
[tex]\displaystyle \rm pOH = -\log_{10} \rm \left[OH^{-}\right][/tex].
On the other hand, the ion-product constant of water, [tex]K_{\text{w}}[/tex], relates the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] of a solution to its hydronium ion concentration [tex]\rm \left[{H_3O}^{+}\right][/tex]:
[tex]K_\text{w} = \rm \left[{H_3O}^{+}\right] \cdot \rm \left[OH^{-}\right][/tex].
At [tex]25 \; ^\circ \rm C[/tex], [tex]K_{\text{w}} \approx 1.0 \times 10^{-14}[/tex]. For this particular [tex]25 \; ^\circ \rm C[/tex] solution, [tex]\rm \left[{H_3O}^{+}\right] = 4.8 \times 10^{-6}\; \rm mol \cdot L^{-1}[/tex].Hence the [tex]\rm \left[OH^{-}\right][/tex] of this solution:
[tex]\begin{aligned}\left[\mathrm{OH}^{-}\right] &= \frac{K_\text{w}}{\rm \left[{H_3O}^{+}\right]} \\ &= \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-6}}\; \rm mol\cdot L^{-1} \approx 2.08333 \times 10^{-9}\; \rm mol\cdot L^{-1}\end{aligned}[/tex].
Therefore, the [tex]\rm pOH[/tex] of this solution would be:
[tex]\begin{aligned}\rm pOH &= -\log_{10} \rm \left[OH^{-}\right] \\ &\approx -\log_{10} \left(4.8 \times 10^{-6}\right) \approx 8.68\end{aligned}[/tex].
Note that by convention, the number of decimal places in [tex]\rm pOH[/tex] should be the same as the number of significant figures in [tex]\rm \left[OH^{-}\right][/tex].
For example, because the [tex]\rm \left[{H_3O}^{+}\right][/tex] from the question has two significant figures, the [tex]\rm \left[OH^{-}\right][/tex] here also has two significant figures. As a result, the [tex]\rm pOH[/tex] in the result should have two decimal places.
Ortho and para hydrogen are....... a). molecular form. b). Nuclear form. c) allotropic form. d). All
Ortho and para hydrogen are nuclei forms
how many moles of H2 and N2 can be formed by the decomposition of 0.145 mol of ammonia, NH3 ?
Answer:
Explanation:
The moles of H2 and N2 are as follows respectively, 0.3915mol of H2 and 0.1305 mol of N2.
0.2175,0.0725 moles of [tex]H_2[/tex] and [tex]N_2[/tex] can be formed by the decomposition of 0.145 mol of ammonia, [tex]NH_3[/tex]
The reaction for the decomposition of ammonia is as follows:-
[tex]N_2+3H_2 \rightarrow 2NH_3[/tex]
Calculate the mole of [tex]H_2[/tex] and [tex]N_2[/tex] as follows:-
[tex]Mole\ of \ H_2=0.145\ mol\ NH_3\times\frac{3\ mol\ H_2}{2\ mol\ NH_3} \\\\=0.2175\ mol\ H_2[/tex]
[tex]Mole\ of \ N_2=0.145\ mol\ NH_3\times\frac{1\ mol\ N_2}{2\ mol\ NH_3} \\\\=0.0725\ mol\ H_2[/tex]
Hence, the number of moles of [tex]H_2[/tex] and [tex]N_2[/tex] are 0.2175 mol, and 0.0725 mol.
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The number of moles of H2O which contains 4g of oxygen?
Answer:
16G = 1 mole ; then 4G = how many moles? 4/16 = 0.25 mole; That means 4 grams of oxygen is 0.25 moles.
Explanation:
A mole of water molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.
If you had a cup full of methanol and a pool full of methanol, would the mass change?
Answer:
the mass does not change
Explanation:
2.Which of the alcohols listed below would you expect to react most rapidly with PBr3?A)CH3CH2CH2CH2CH2CH2OHB)(CH3CH2)2CH(OH)CH2CH3C)(CH3CH2)2CHOHCH3D)(CH3CH2)3COHE)(CH3CH2)2C(CH3)OH
Answer:
A) CH3CH2CH2CH2CH2CH2OH
Explanation:
For this question, we have the following answer options:
A) CH3CH2CH2CH2CH2CH2OH
B) (CH3CH2)2CH(OH)CH2CH3
C) (CH3CH2)2CHOHCH3
D) (CH3CH2)3COH
E) (CH3CH2)2C(CH3)OH
We have to remember the reaction mechanism of the substitution reaction with [tex]PBr_3[/tex]. The idea is to generate a better leaving group in order to add a "Br" atom.
The [tex]PBr_3[/tex] attacks the "OH" generation new a bond to P (O-P bonds are very strong), due to this new bond we will have a better leaving group that can remove the oxygen an allow the attack of the Br atom to generating a new C-Br bond. This is made by an Sn2 reaction. Therefore we will have a faster reaction with primary substrates. In this case, the only primary substrate is molecule A. So, "CH3CH2CH2CH2CH2CH2OH" will react faster.
See figure 1
I hope it helps!
What does the third quantum number (m) describe?
A. Which energy level the electron is in
B. What type of orbital the electron is in
C. What direction the electron is spinning
D. The specific orbital within a sublevel
The third quantum number (m) describes the specific orbital within a sublevel. The correct answer is option D.
Quantum numbers are a set of four numbers that describe the properties of an electron in an atom. They specify the energy, position, and orientation of an electron in an atom.
The third quantum number (m) is also called the magnetic quantum number. It describes the orientation of the orbital in space. Each orbital within a sublevel has a different orientation in space, and the magnetic quantum number tells us which orbital we are talking about.
The value of m can range from -l to +l, where l is the second quantum number (the angular momentum quantum number). The value of l determines the number of orbitals in a sublevel, and the value of m determines the specific orbital within that sublevel.
In conclusion, the third quantum number (m) describes the specific orbital within a sublevel. It tells us how the orbital is oriented in space, and it can have values ranging from[tex]\rm -l \ to +l[/tex], where l is the second quantum number.
Option D is the correct answer.
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An equilibrium mixture of N2, H2, and NH3 at 700 K contains 0.036 M N2 and 0.15 M H2. At this temperature, Kc for the reaction N2(g) + 3 H2(g)<=> 2NH3(g) is 0.29.
What is the concentration of NH3?
Answer:
5.94×10¯³
Explanation:
The following data were obtained from the question:
Concentration of N2, [N2] = 0.036 M
Concentration of H2, [H2] = 0.15 M
Equilibrium constant (Kc) = 0.29 M
Concentration of NH3, [NH3] =.…?
The equation for the reaction is given below:
N2(g) + 3H2(g) <=> 2NH3(g)
Thus, we can determine the concentration of NH3 by using the equilibrium expression for the reaction.
This is illustrated below:
The equilibrium constant for a reaction is simply defined as the ratio of the concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.
The equilibrium constant for the above equation is given below:
Kc = [NH3]² / [N2] [H2]³
Inputting the value of Kc, [N2], and [H2] the value of [NH3]can be obtained as follow:
Kc = [NH3]² / [N2] [H2]³
0.29 = [NH3]²/ 0.036 × 0.15³
Cross multiply
[NH3]² = 0.29 × 0.036 × 0.15³
[NH3]² = 3.5235×10¯⁵
Take the square root of both side
[NH3] = √(3.5235×10¯⁵)
[NH3] = 5.94×10¯³
Therefore, the concentration of NH3, [NH3] is 5.94×10¯³ M.
1.) A sample of neon gas at a pressure of 0.646 atm and a temperature of 242 °C, occupies a volume of 515 mL. If the gas is cooled at constant pressure until its volume is 407 mL, the temperature of the gas sample will be ________°C.
2.) A sample of argon gas at a pressure of 0.633 atm and a temperature of 261 °C, occupies a volume of 694 mL. If the gas is heated at constant pressure until its volume is 796 mL, the temperature of the gas sample will be___________°C.
3.) 0.962 mol sample of carbon dioxide gas at a temperature of 20.0 °C is found to occupy a volume of 21.5 liters. The pressure of this gas sample ismm ____________ Hg.
Answer:1 )T2=134°C 2) T2=339.48°C. 3)
P=817.59 mmHg.
Explanation:
1.Given ;
pressure, P1 of neon gas = 0.646 atm
temperature, T1 =242oC + 273=515oC
Volume, V1 =515ml
Volume V2= 407ml
temperature , T 2= ?
Solution;
And at constant pressure, the volume cools at V2=407 mL at T2=?
From ideal gas equation, PV=nRT
V/T=constant
therefore
V1/V2=T1/T2 = T2=(V2 xT1)/V1
T2=(407 mL x 515 K)/515 mL= 407K.
T2= 407K -273= 134°C. recall 0°C=273 K)
2..Given ;
pressure, P1 of neon gas = 0.633 atm
temperature, T1 =261oC + 273=534oC
Volume, V1 =694ml
Volume V2= 796ml
temperature , T 2= ?
Solution;
And at constant pressure, the volume expands at V2=796mL at T2=?
From ideal gas equation, PV=nRT
V/T=constant
therefore
V1/V2=T1/T2 = T2=(V2 xT1)/V1
T2=(796 mL x 534 K)/694mL= 612.48K.
T2= 612.48K -273= 339.48°C. recall 0°C=273 K
3
Given;
moles of CO2= n=0.962 mol,
temperature T=20°C=20+273 K =293 K,
volume V=21.5 L,
gas constant R at L·mmHg/mol·K= 62.3637 L mmHg mol^-1 K^-1
Using ideal gas equation PV=nRT
P=nRT/V
P=(0.962 mol)x(62.3637mmHg mol^-1 K^-1)x(293 K)/(21.5L)
P=817.59 mmHg.
What class of organic product results when 1-heptyne is treated with a mixture of mercuric acetate in aqueous sulfuric acid, and then HOCH2CH2OH with catalytic sulfuric acid
Answer:
2-methyl-2-pentyl-1,3-dioxolane
Explanation:
In this case, we have two reactions:
First reaction:
1-heptyne + mercuric acetate -------> Compound A
Second reaction:
Compound A + HOCH2CH2OH -------> Compound C
First reaction
In the first reaction, we have as a main functional group a triple bond. We have to remember that mercuric acetate in sulfuric acid will produce a ketone. The carbonyl group (C=O) would be placed in the most substituted carbon of the triplet bond (in this case, carbon 2). With this in mind, we will have as a product: heptan-2-one. (See figure 1).
Second reaction
In this reaction, we have as reagents:
-) Heptan-2-one
-) Ethylene-glycol [tex]HOCH_2CH_2OH[/tex]
-) Sulfuric acid [tex]H_2SO_4[/tex]
When we put ethylene-glycol with a ketone or an aldehyde we will form a cyclic acetal. In this case, this structure would be formed on carbon 2 forming 2-methyl-2-pentyl-1,3-dioxolane. (See figure 2).
I hope it helps!
Consider the following reaction at 298K.
I2 (s) + Pb (s) = 2 I- (aq) + Pb2+ (aq)
Which of the following statements are correct?
Choose all that apply.
ΔGo > 0
The reaction is product-favored.
K < 1
Eocell > 0
n = 2 mol electrons
B-
Answer:
Eªcell > 0; n = 2
Explanation:
The reaction:
I2 (s) + Pb (s) → 2 I- (aq) + Pb2+ (aq)
Is product favored.
A reaction that is product favored has ΔG < 0 (Spontaneous)
K > 1 (Because concentration of products is >>>> concentration reactants).
Eªcell > 0 Because reaction is spontaneous.
And n = 2 electrons because Pb(s) is oxidizing to Pb2+ and I₂ is reducing to I⁻ (2 electrons). Statements that are true are:
Eªcell > 0; n = 2Classify the following oxides as acidic, basic, amphoteric, or neutral:
(a) Select the acidic oxides:
A. CO₂
B. N₂O₅
C. Al₂O₃
D. NO
E. SO₃
(b) Select the basic oxides:
A. CaO
B. CO
C. SO₃
D. K₂O
E. BaO
(c ) Select the amphoteric oxides:
A. K₂O
B. Al₂O₃
C. CaO
D. CO₂
E. SnO₂
d) Select the neutral oxides:
A. CO
B. NO
C. SNO₂
D. N₂O₅
E. BaO
Answer:
(a).
» E. SO₃, sulphur trioxide.
(b).
» A. CaO, Calcium oxide.
» D. K₂O, potassium oxide.
» E. BaO, barium oxide.
(c).
» B. Al₂O₃, Aluminium oxide.
» E. SnO₂, tin (IV) oxide.
(d).
» A. CO, carbon monoxide.
» B. NO, nitrogen monoxide.
Explanation:
[tex]{ \underline{ \sf{ \blue{christ \:† \: alone }}}}[/tex]
The following reactions all have K < 1. 1) a. C6H5COO- (aq) + C6H5OH (aq) → C6H5COOH (aq) + C6H5O- (aq) b. F- (aq) + C6H5OH (aq) → C6H5O- (aq) + HF (aq) c. C6H5COOH (aq) + F- (aq) → HF (aq) + C6H5COO- (aq) Arrange the substances based on their relative acid strength.
Answer:
the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]
Explanation:
Given that :
a . [tex]\mathsf{C_6H_5COO^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5COOH _{(aq)} + C_6H_5O^- _{(aq)}}[/tex]
b. [tex]\mathsf{ F^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5O^- _{(aq)} + HF _{(aq)} }[/tex]
c. [tex]\mathsf{C_6H_5COOH _{(aq)} + F^- _{(aq)} \to HF _{(aq)} + C_6H_5COO^- _{(aq)} }[/tex]
Acid strength is the ability of an acid to dissociate into a proton and an anion. Take for instance.
HA ↔ H⁺ + A⁻
The acid strength of the following compounds above are:
[tex]\mathsf{C_6H_5OH _{(aq)} }[/tex] = 1.00 × 10⁻¹⁰
[tex]\mathsf{HF _{(aq)} }[/tex] = 6.6 × 10⁻⁴
[tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] = 6.3 × 10⁻⁵
As the acid dissociation constant increases the relative acid strength also increases.
From above, the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]
[tex]\mathsf{C_6H_5COO^- }[/tex], [tex]\mathsf{C_6H_5O^- _{(aq)}}[/tex] and F⁻ are Bronsted- Lowry acid
Bronsted- Lowry acid are molecule or ion that have the ability to donate a proton.
Sighting along the C2-C3 bond of 2-methylbutane, the least stable conformation (Newman projection) has a total energy strain of ______kJ/mol
Answer:
21 KJ/mol
Explanation:
For this question, we have to start with the linear structure of 2-methylbutane. With the linear structure, we can start to propose all the Newman projections keep it in mind that the point of view is between carbons 2 and 3 (see figure 1).
Additionally, we have several energy values for each interaction present in the Newman structures:
-) Methyl-methyl gauche: 3.8 KJ/mol
-) Methyl-H eclipse: 6.0 KJ/mol
-) Methyl-methyl eclipse: 11.0 KJ/mol
-) H-H eclipse: 4.0 KJ/mol
Now, we can calculate the energy for each molecule.
Molecule A
In this molecule, we have 2 Methyl-methyl gauche interactions only, so:
(3.8x2) = 7.6 KJ/mol
Molecule B
In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:
(11)+(6)+(4) = 21 KJ/mol
Molecule C
In this molecule, we have 1 Methyl-methyl gauche interaction only, so:
3.8 KJ/mol
Molecule D
In this molecule, we have three Methyl-H eclipse interaction, so:
(6*3) = 18 KJ/mol
Molecule E
In this molecule, we have 1 Methyl-methyl gauche interaction only, so:
3.8 KJ/mol
Molecule F
In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:
(11)+(6)+(4) = 21 KJ/mol
The structures with higher energies would be less stable. In this case, structures B and F with an energy value of 21 KJ/mol (see figure 2).
I hope it helps!
A mixture of gases includes helium, neon, argon and krypton. The partial pressures are as follows:
Partial pressure, mm Hg
He 152
Ne 164
Ar 126
Kr 169
What is the total pressure of the mixture after removing the argon component? Report your answer rounded
to the nearest mm Hg but without units
Answer:
the answer to your question is ar 126
15.Vicinal coupling is:A)coupling between 1H nuclei attached to adjacent C atoms.B)coupling between 1H nuclei in an alkene.C)coupling between 1H nuclei attached to the same C atom.D)coupling between 1H nuclei in an alkane.
Answer:
A)coupling between 1H nuclei attached to adjacent C atoms.
Explanation:
The word ‘vicinal’ in chemistry means three bonds from the functional groups. The two functional groups are in a relationship with the atoms in adjacent position to them.
The 1H nuclei consists of two Hydrogen nucleus which acts as the functional groups. They are however attached and in a relationship with the adjacent C atoms. This makes option A the right choice.
Please tell the answer
Answer:
see the photo
Explanation:
it was the answer
What is the pH of 10.0 mL solution of 0.75 M acetate after adding 5.0 mL of 0.10 M HCl (assume a Ka of acetic acid of 1.78x10-5)
Answer:
5.90
Explanation:
Initial moles of CH3COO- = 10.0/1000 x 0.75 = 0.0075 mol
Moles of HCl added = 5.0/1000 x 0.10 = 0.0005 mol
CH3COO- + HCl => CH3COOH + Cl-
Moles of CH3COO- left = 0.0075 - 0.0005 = 0.007 mol
Moles of CH3COOH formed = moles of HCl added = 0.0005 mol
pH = pKa + log([CH3COO-]/[CH3COOH])
= -log Ka + log(moles of CH3COO-/moles of CH3COOH)
= -log(1.78 x 10^(-5)) + log(0.007/0.0005)
= 5.90
Answer:
The correct answer is 5.895.
Explanation:
The reaction will be,
CHCOO⁻ + H+ ⇔ CH₃COOH
Both the HCl and the acetate are having one n factor.
The millimoles of CH₃COO⁻ is,
= Volume in ml × molarity = 10 × 0.75 = 7.5
The millimoles of HCl = Volume in ml × molarity = 5 × 0.1 = 0.5
Therefore, 0.5 will be the millimoles of CH₃COOH formed, now the millimoles of the CH₃COO⁻ left will be, 7.5-0.5 = 7.0
The volume of the solution is, 10+5 = 15 ml
The molarity of CH₃COO⁻ is, millimoles / volume in ml = 7/15
The molarity of CH₃COOH is 0.5/15
pH = pKa + log[CH₃COO⁻]/[CH₃COOH]
= 4.74957 + 1.146
= 5.895
When HCl is added to water, the [H3O+] = 0.6 M. What is the [OH-]?
What is the pH of the solution?
Answer:
pH=0.22.
[tex][OH^-]=1.66x10^{-14}M[/tex]
Explanation:
Hello,
In this case, since the pH is directly computed from the given concentration of hydronium ions:
[tex]pH=-log([H_3O^+])=-log(0.6M)=0.22[/tex]
It is widely known that the pH and POH are directly related via:
[tex]pH+pOH=14[/tex]
Therefore, the pOH is:
[tex]pOH=14-pH\\\\pOH=14-0.22=13.78[/tex]
Thus, the concentration of hydroxyl ions are computed from the pOH:
[tex]pOH=-log([OH^-]}\\\\[/tex]
[tex][OH^-]=10^{-pOH]=10^{-13.78}[/tex]
[tex][OH^-]=1.66x10^{-14}M[/tex]
Regards.
How many moles of barium sulfate are produced from 0.100 mole of barium chloride?
Answer:
0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.
Explanation:
Barium chloride and sodium sulfate react according to the following balanced reaction:
BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagent and products participate in the reaction:
BaCl₂: 1 moleNa₂SO₄: 1 moleBaSO₄: 1 moleNaCl : 2 molesThen you can apply the following rule of three: if 1 mole of BaCl₂ produces 1 mole of BaSO₄, 0.100 mole of BaCl₂ how many moles of BaSO₄ does it produce?
[tex]amount of moles of BaSO_{4} =\frac{0.100 mole of BaCl_{2}* 1 mole of BaSO_{4} }{1 mole of BaCl_{2}}[/tex]
amount of moles of BaSO₄= 0.100
0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.
Explain your reasoning. Match each explanation to the appropriate blanks in the sentences on the right.
1. the atomic radius decreases
2. the number of gas molecules decreases
3. molar mass and structure complexity decreases
4. structure complexity decreases
5. molar mass decreases
6. each phase (gas, liquid, solid) becomes more ordered
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as______.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as_______.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as_______.
Answer:
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.
Explanation:
Hello.
In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.
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Beginning with Na, record the number of energy levels, number of protons, and atomic radius for each element in period 3.
Answer:
Sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon are the elements of third period.
Explanation:
There are three energy levels in sodium atom. It has 11 electrons revolving around the nucleus. the atomic radius of sodium atom is 227 ppm. Magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon has also three energy levels like sodium because all these elements belongs to third period. There are 12 electrons present in magnesium, 13 in aluminium, 14 in silicon, 15 in phosphorus, 16 in sulfur, 17 in chlorine, and 18 electrons in argon. The atomic radius of magnesium atom is 173 ppm. The atomic radius of aluminium atom is 143 ppm. The atomic radius of silicon atom is 111 ppm. The atomic radius of phosphorus atom is 98 ppm. The atomic radius of sulfur atom is 87 ppm. The atomic radius of chlorine atom is 79 ppm and the atomic radius of argon atom is 71 ppm.
Explain the working principle of a clinical thermometer
Answer:
A clinical thermometer is a thermometer used to measure human body temperature. Most made in the 20th century are mercury-in-glass thermometers. They are accurate and sensitive, having a narrow place where the mercury level rises very fast. A kink in the tube stops the mercury level from falling on its own.
A person has a standing order for 6 doses of acetaminophen in a 24 hour period for fevers
acetaminophen requires what type of reporting?
ut of
Select one:
a. Routine
estion
b. Emergent
C. Certain Time
d. Urgent
4
A change in the type of seizure the person usually experiences that causes them to stop brea
Acetaminophen is used to treat patients with aches and fever. It is safe to give a dose of 4 grams in 24 hours period.
It is an effective medicine which does not require much time to provide relief to the patient.
Usually tylenol is the best medicine which consists of Acetaminophen in appropriate quantity to provide relief from pain and fever.
The reporting for the doses introduced to the patient must be routine since it is not an emergency case.
The correct answer is a. Routine
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A man weighs 185 lb. What is his mass in grams?
Please show work.
Thank you
Answer:
83914.52 grams
Explanation:
Given that,
Weight of a man is 185 lb
We need to find his weight in grams
For this, we must know the relation between lb and grams.
We know that,
1 lb = 453.592 grams
To find the mass of man in grams, the step is :
185 lb = (453.592 × 185) grams
= 83914.52 grams
So, the mass of a man is 83914.52 grams.
A solution contains 2.2 × 10-3 M in Cu2+ and 0.33 M in LiCN. If the Kf for Cu(CN)42- is 1.0 × 1025, how much copper ion remains at equilibrium?
Answer:
[Cu²⁺] = 2.01x10⁻²⁶
Explanation:
The equilibrium of Cu(CN)₄²⁻ is:
Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻
And Kf is defined as:
Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴
As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:
[Cu²⁺] = 0
[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M
[Cu(CN)₄²⁻] = 2.2x10⁻³
Some [Cu²⁺] will be formed and equilibrium concentrations will be:
[Cu²⁺] = X
[CN⁻] = 0.3212M + 4X
[Cu(CN)₄²⁻] = 2.2x10⁻³ - X
Where X is reaction coordinate
Replacing in Kf equation:
1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴
1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0
Solving for X:
X = 2.01x10⁻²⁶
As
[Cu²⁺] = X
[Cu²⁺] = 2.01x10⁻²⁶