Answer:
22.1
Explanation:
Step 1: Calculate the number of formula units of lithium nitrate
A 5.08 L solution contains 2.68 × 10²¹ formula units per liter.
5.08 L × 2.68 × 10²¹ formula units/1 L = 1.36 × 10²² formula units
Step 2: Calculate the number of nitrate ions
Lithium nitrate dissociates completely in water according to the following equation.
LiNO₃(aq) ⇒ Li⁺(aq) + NO₃⁻(aq)
The molar ratio of LiNO₃ to NO₃⁻ is 1:1. The number of nitrate ions is 1/1 × 1.36 × 10²² = 1.36 × 10²².
Then,
log (1.36 × 10²²) = 22.1
The number of nitrate ions are present in the following aqueous solution is 13.6x10²¹, and log(base 10) value of this is 22.2.
What are strong electrolytes?
Those salts which are completely dissociate into their ions in the solution form will known as strong electrolyte.
Lithium nitrate is a strong electrolyte and it shows complete dissociation as:
LiNO₃ → Li⁺ + NO₃⁻
Given that,
1 liter of solution containing = 2.68 x 10²¹ formula unit
5.08 L of solution containing = 5.08 x 2.68 x 10²¹ = 13.6x10²¹ formula unit
So, number of nitrate ion in given solution = 13.6x10²¹ formula unit
Log(13.6x10²¹) = 22.1
Hence, required values are 13.6x10²¹ and 22.1.
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150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reaction is completed; calculate the concentration of nitrate ions in solution. Assume that the total volume of the solution is 3.0 x 10^2 mL
Answer:
[tex]0.175\; \rm mol \cdot L^{-1}[/tex].
Explanation:
Magnesium chloride and silver nitrate reacts at a [tex]2:1[/tex] ratio:
[tex]\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)[/tex].
In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:
[tex]\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}[/tex].
The precipitate silver chloride [tex]\rm AgCl[/tex] is insoluble in water and barely ionizes. Hence, [tex]\rm AgCl\![/tex] isn't rewritten as ions.
Net ionic equation:
[tex]\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}[/tex].
Calculate the initial quantity of nitrate ions in the mixture.
[tex]\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}[/tex].
Since nitrate ions [tex]\rm {NO_3}^{-}[/tex] do not take part in any reaction in this mixture, the quantity of this ion would stay the same.
[tex]n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol[/tex].
However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be [tex](1/2)[/tex] of the concentration in the original solution.
[tex]\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.
Answer:
[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.
Explanation:
Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.
Analysis:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷
C(i) 0.115M 0 0
ΔC -x +x +x
C(eq) 0.115M - x x x
≅ 0.115M
Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M
= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.
In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from
[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.
________________________________________________________
NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.
The concentration of hydroxide ion of given solution is 4.3 x 10⁻¹¹M.
How we calculate the [OH⁻]?We can calculate the concentration of hydroxide ions as follow:
[OH⁻][H⁺] = 10⁻¹⁴
Given chemical reaction with ICE table shown as below:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq)
Initial: 0.115 0 0
Change: -x +x +x
Equilibrium: 0.115-x +x +x
Given that, Ka = 4.3 x 10⁻⁷
Equilibrium constant for this reaction is written as:
Ka = [H⁺][HCO₃⁻]/[H₂CO₃]
4.3 x 10⁻⁷ = x² / 0.115
x = 2.32 x 10⁻⁴M = [H⁺]
Now we calculate the concentration of hydroxide ion as:
[OH⁻][H⁺] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴ / 2.32 x 10⁻⁴ = 4.3 x 10⁻¹¹M
Hence, value of [OH⁻] is 4.3 x 10⁻¹¹M.
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3. Oxalic acid, C H20, is a toxic substance found in rhubarb leaves. When mixed with sufficient
quantities of a strong base, this weak diprotic acid loses two protons to form a polyatomic ion
called oxalate, C2022. Write a balanced equation that describes the reaction between oxalic acid
and sodium hydroxide.
Answer:
COOHCOOH + 2OH⁻ ⇄ C₂O₄²⁻ + 2H₂O
Explanation:
The reaction of oxalic acid with a strong base like sodium hydroxide is the following:
COOHCOOH + OH⁻ ⇄ COOHCOO⁻ + H₂O (1)
In this first reaction, the oxalic acid loses one proton. In a second reaction with NaOH, the ion COOHCOO⁻ loses its second proton to form ion oxalate as follows:
COOHCOO⁻ + OH⁻ ⇄ C₂O₄²⁻ + H₂O (2)
The general reaction between oxalic acid and NaOH is (eq 1 + eq 2):
COOHCOOH + 2OH⁻ ⇄ C₂O₄²⁻ + 2H₂O
I hope it helps you!
Nitrous oxide, NO, decomposes exothermically into nitrogen gas and oxygen gas. When graphing [NO] versus time, a straight line can be drawn through the experimental points. From this information, determine the reaction order.
Answer:
Zero-Order
Explanation:
The exothermic decaying of nitrous oxide at 575° C will lead to [tex]N_{2} and O_{2}[/tex] as follows:
[tex]2N_{2}O[/tex] → [tex]2N_{2}(g) + O_{2} (g)[/tex]
Hot platinum wire in the above reaction would function as a catalyst in the zero-order. However, if the reaction is considered in the gaseous phase, it will be more inclined towards second-order.
In the given scenario([tex]2N_{2}O[/tex] → [tex]2N_{2}(g) + O_{2} (g)[/tex]), the reactant molecules of Nitrous oxide are restricted to the ones which have linked themselves to the catalyst's surface. Once this limited surface is filled, the extra molecules of gas would remain vacant until the previously attached molecules with the surface are decayed entirely.
The Nernst equation at 20oC is:
Eion= 58 millvolts/z. [log10 (ion)out/(ion)in]
Calculate the equilibrium potential for Cl- if the concentration of Cl- outside of the cell is 100 and the concentration inside of the cell is 10 mmol/liter.
a. 58 millivolts
b. +58 millivolts
c. -116 millivolts
d. 0
Answer:
a. -58 millivolts
Explanation:
The given Nernst equation is:
[tex]E_{ion} = 58 millivolts /z \Big[ log_{10} \Big( \dfrac{[ion]_{out}}{[ion]_{in}}\Big) \Big]}[/tex]
The equilibrium potential given by the Nernst equation can be determined by using the formula:
[tex]E_{Cl^-} = \dfrac{2.303*R*T}{ZF} \times log \dfrac{[Cl^-]_{out}} {[Cl^-]_{in}}[/tex]
where:
gas constant(R) = 8.314 J/K/mol
Temperature (T) = (20+273)K
= 298K
Faraday constant F = 96485 C/mol
Number of electron on Cl = -1
[tex]E_{Cl^-} = \dfrac{2.303*8.314*298} {(-1)*(96845)} \times log \dfrac{100} {10}[/tex]
[tex]E_{Cl^-} = - 0.05814 \ volts[/tex]
[tex]\mathsf{E_{Cl^-} = - 0.05814 \times 1000 \ milli volts}[/tex]
[tex]\mathsf{E_{Cl^-} \simeq - 58\ milli volts}[/tex]
does anyone know how to solve this and what the answer would be?
Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.
At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.
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For the following reaction, 5.29 grams of water are mixed with excess diphosphorus pentoxide. The reaction yields 13.3 grams of phosphoric acid . diphosphorus pentoxide(s) + water(l) phosphoric acid(aq). What is the theoretical yield of phosphoric acid?
Answer:
19.2 g
Explanation:
Step 1: Write the balanced equation
P₂O₅(s) + 3 H₂O(l) ⇒ 2 H₃PO₄
Step 2: Calculate the moles corresponding to 5.29 g of H₂O
The molar mass of H₂O is 18.02 g/mol.
5.29 g × 1 mol/18.02 g = 0.294 mol
Step 3: Calculate the theoretical yield of phosphoric acid, in moles
The molar ratio of H₂O to H₃PO₄ is 3:2. The theoretical yield of H₃PO₄ is 2/3 × 0.294 mol = 0.196 mol
Step 4: Calculate the mass corresponding to 0.196 moles of H₃PO₄
The molar mass of H₃PO₄ is 97.99 g/mol.
0.196 mol × 97.99 g/mol = 19.2 g
PLEASE HELP I NEED THIS ASAP
Select all that apply.
The spectrum of Star S is compared to a reference hydrogen spectrum. What can be concluded about Star S?
Star S has radial motion.
Star S has transverse motion.
Star S is moving toward Earth.
Star S is moving away from Earth.
Answer:
I say Star S has radial motion
Explanation:
I'm not sure if it right but let me know if you have any other questions
write down the different uses of water that you know about
Answer:
The various uses of water :
1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.
2. Water is used as a universal solvent.
3. water maintains the temperature of our body.
4. Water helps in digestion in our body.
5 .water is used in factories and industries.
6. Water is used to grow plants , vegetables and crops.
Answer:
The various use of water are;
I) Cooking.
ii) Drinking
III) Bathing
iv) Generating hydro- electricity
v) Construction work etc
A student was asked to determine the percent mass of sodium nitrate in a mixture of sodium nitrate (NaNO3) and calcium carbonate (CaCO3). The mass of the mixture used was 3.2 g. The student extracted NaNO3 from the mixture with water and separated the insoluble CaCO3 from the solution by filtration. After evaporating the filtrate, the student recovered and dried the NaNO3, and found that it weighted 0.45 g. The student dried the insoluble residue of CaCO3 and found that it weighted 2.23 g. Calculate the percent mass of NaNO3 in the mixture and round your final answer to the correct number of significant figures.
Answer:
自分の仕事をする translate to english
Explanation:
Consider the following titration for these three questions:
1.00 L of 2.00 M HCl is titrated with 2.00 M NaOH.
a. How many moles of acid are equal to one equivalent in this titration?
b. How many moles of HCl are found in solution at the halfway point of the titration?
c. How many liters of base will be needed to reach the equivalence point of the titration?
Answer:
a. 1 mole of acid is equal to one equivalent.
b. 1.00 moles of HCl are found.
c. 1L of 2.00M NaOH is needed to reach the equivalence point
Explanation:
HCl reacts with NaOH as follows:
HCl + NaOH → NaCl + H2O
Where 1 mole of HCl reacts with 1 mole of NaOH. The reaction is 1:1
a. As the reaction is 1:1, 1 mole of acid is equal to one equivalent
b. The initial moles of HCl are:
1.00L * (2.00moles HCl / 1L) = 2.00 moles of HCl
At the halfway point, the moles of HCl are the half, that is:
1.00 moles of HCl are found
c. At equivalence point, we need to add the moles of NaOH needed for a complete reaction of the moles of HCl. As the moles of HCl are 2.00 and the reaction is 1:1, we need to add 2.00 moles of NaOH, that is:
2.00moles NaOH * (1L / 2.00mol) =
1L of 2.00M NaOH is needed to reach the equivalence point
list two uses of H2SO4
For each of the sites specified in the molecules, select whether the site is nucleophilic, electrophilic, or neither. Compound A: The indicated site is a carbon in cyclohexane which is bonded to a bromine and a hydrogen. The indicated carbon in compound A is nucleophilic. neither electrophilic nor nucleophilic. electrophilic. Compound B: The indicated site is the double bond in cyclohexene, a 6 carbon ring with an internal alkene. The indicated bond in compound B is nucleophilic. electrophilic. neither electrophilic nor nucleophilic. Compound C: The indicated site is a carbon double bonded to oxygen, and bonded to O C H 3 and ethyl. The indicated carbon in compound C is neither electrophilic nor nucleophilic. nucleophilic. electrophilic. Compound D: THe indicated site is a carbon bonded to a methyl, two hydrogens and a carbon. There is a nitrogen atom two bonds away. The indicated carbon in compound D is neither electrophilic nor nucleophilic. electrophilic. nucleophilic. Compound E: The indicated site is an oxygen bonded to a carbon and a hydrogen. The indicated oxygen in compound E is neither electrophilic nor nucleophilic. electrophilic. nucleophilic.
The nature of attack on sites in a molecule depends on the nature of such sites. The following are the nature of the sites mentioned in the question:
1) The indicated carbon in compound A is electrophilic.
2) The indicated bond in compound B is nucleophilic.
3) The indicated carbon in compound C is electrophilic.
4) The indicated carbon in compound D is neither electrophilic nor nucleophilic.
5) The indicated oxygen in compound E is nucleophilic.
The terms "electrophilic" and "nucleophilic" are very common in chemistry.
An electrophilic center is usually positively charged, has a positive dipole or is electron deficient hence it attacks negative centers. The term itself means "electron loving". That actually means that it has an affinity for negative charges.
The -I inductive effect of the bromine atom in the carbon in compound A makes that carbon atom to be electrophilic. Also, the carbonyl bond and the O C H 3 attached to the carbon in compound C also makes it electrophilic.
The term "nucleophilic" literately means "nucleus loving". That means a specie that has a high affinity for positive charges. This specie must be electron rich.
The carbon atom in compound B has a double bond which is electron rich and can attack any positive center hence it is nucleophilic. Also, the oxygen atom in E bears two lone pairs of electrons which can attack any positive center in a molecule hence the oxygen atom is also nucleophilic.
In compound D, the carbon atom is bonded to a methyl, two hydrogens and a carbon. There is a nitrogen atom two bonds away. There is no +I or -I inductive effect on this carbon atom because the nitrogen atom is far away. Therefore, the indicated carbon in compound D is neither electrophilic nor nucleophilic.
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What is the equilibrium expression for the reaction below?
Caco (s)
Cao(s) + CO (g)
A.
B.
[Cacoz]
[Cao]
[Caco.]
[Cao]+[CO,
[Cao][COL]
[Caco:]
C.
D. [co]
Answer:
D. [CO₂]
Explanation:
Let's consider the following equation at equilibrium.
CaCO₃(s) ⇄ CaO(s) + CO₂(g)
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients. It only includes gases and aqueous species.
Kc = [CO₂]
# of protons
# of neutrons
# of electrons
Atomic Number
Mass Number
18
17
35
17
37
6
8
6
6
15
Answer:
35
Explanation:
is the answer for your question
The correct geometry around oxygen in CH3OCH3 is
(a). linear. (b). bent. C). tetrahedral/(a). trigonal planar
Explanation:
the force of the lone pairs from the bottom would cancel out the force of the lone pairs from the top. Thus, the molecule will be linear.
How many grams of CO2 are formed if 44.7 g C5H12 is mixed with 108 g O2?
Explanation:
here's the answer to your question
3. Suppose you wanted to design an experiment to test the composition of a mixture that includes sodium phenoxide (NaC6H5O). You know that this solid mixture contains both the NaC6H5O and some inert NaCl, but do not know how much of each is present. You decide to test the composition by titrating with 0.100-M HCl. a. If a 1.000-g sample is 25% NaC6H5O by mass, how many mL of 0.100-M HCl would be required to reach the equivalence point of the titration
Answer:
21.5mL of a 0.100M HCl are required
Explanation:
The sodium phenoxide reacts with HCl to produce phenol and NaCl in a 1:1 reaction.
To solve this question we need to find the moles of sodium phenoxide. These moles = Moles of HCl required to reach equivalence point and, with the concentration, we can find the needed volume as follows:
Mass NaC6H5O:
1.000g * 25% = 0.250g NaC6H5O
Moles NaC6H5O -116.09g/mol-
0.250g NaC6H5O * (1mol/116.09g) = 2.154x10⁻³ moles = Moles of HCl required
Volume 0.100M HCl:
2.154x10⁻³ moles HCl * (1L/0.100mol) = 0.0215L =
21.5mL of a 0.100M HCl are requiredThe half life for the decay of carbon-14 is 5.73 times 10^3 years.
Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of wood from an archeological dig is measured to be 77.The activity in a similar-sized sample of fresh wood is measured to be 85.Calculate the age of the artifact. Round your answer to 2 significant digits.
Answer:
790 years
Explanation:
Given that;
0.693/t1/2 = 2.303/t log [A]o/[A]
So;
t1/2 =half life of carbon-14
t= age of the sample
[A]o= activity of the living sampoke
[A] = activity at time t
0.693/5.73 ×10^3 = 2.303/t log 85/77
1.21 × 10^-4 = 2.303/t log 1.1
1.21 × 10^-4 = 0.0953/t
t= 0.0953/1.21 × 10^-4
t= 790 years (to 2sf)
The forensic technician at a crime scene has just prepared a luminol stock solution by adding 17.0 \({\rm g}\) of luminol into a total volume of 75.0 \(\rm mL\) of \(\rm H_2O\). What is the molarity of the stock solution of luminol
Answer:
1.28 M
Explanation:
Step 1: Given data
Mass of luminol (solute): 17.0 g
Volume of water: 75.0 mL (this is also the volume of solution)
Step 2: Calculate the moles corresponding to 17.0 g of luminol
The molar mass of luminol is 177.16 g/mol.
17.0 g × 1 mol/177.16 g = 0.0960 mol
Step 3: Calculate the molarity of the solution
We will use the definition of molarity
M = moles of solute / liters of solution
M = 0.0960 mol / 0.0750 L = 1.28 M
Write a chemical equation for LiOH(aq) showing how it is an acid or a base according to the Arrhenius definition.
Answer:
LiOH(aq) → Li⁺(aq) + OH⁻(aq).
You are titrating 24.3 mL of 2.00 M HCl with 1.87 M NaOH. How much NaOH do you expect to have added when you reach the equivalence point?
26.0 mL
15.4 mL
13.4 mL
Answer:
26mL
Explanation:
NaOH+HCl= NaCl+H2O
nHCl=0.0243*2=0.0486
nNaOh=nHCl
VNaOH=0.0486/1.87=0.026l=26ml
Answer:
26.0 mL
Explanation:
Benzoyl chloride undergoes hydrolysis when heated with water to make benzoic acid. Reaction scheme of benzoyl chloride with water and heat over the arrow, and benzoic acid and hydrochloric acid as products. Calculate the molar mass of the reactant and product. Report molar masses to 1 decimal place.
Answer:
The molar mass of benzoic acid is 122.1 g/mol
The molar mass of hydrochloric acid = 36.5 g/mol
Explanation:
Benzoyl chloride is an organic compound with the molecular formula C₆H₅COCl. It is an acyl chloride since is it an organic derivative of a carboxylic acid. Acyl chlorides have the general molecular formula, R-COCl, where R is a side chain.
The R group of benzoyl chloride is the benzyl group C₆H₅. It reacts with water (hydrolysis) to produce hydrochloric acid and benzoic acid. The equation of the reaction is given below:
C₆H₅COCl + H₂O → C₆H₅CO₂H + HCl
The molar mass of benzoic acid as well as of hydrochloric acid is calculated from the sum of the masses of the atoms of the elements present in the compound thus:
Molar mass of carbon = 12.0107 g
Molar mass of hydrogen = 1.00784 g
Molar mass of oxygen = 15.999 g
Molar mass of chlorine = 35.453 g
Molar mass of benzoic acid, C₆H₅CO₂H containing 7 moles of atoms of carbon, 6 moles of atoms of hydrogen and 2 moles of atoms of oxygen = 7 × 12.0107 + 6 × 1.00784 + 2 × 15.999 = 122.1 g
Therefore, the molar mass of benzoic acid is 122.1 g/mol
Molar mass of hydrochloric acid, HCl, containing 1 mole of atoms of hydrogen and 1 mole of atoms of chlorine = 1 × 1.00784 + 1 × 35.453 = 36.5 g
Therefore, the molar mass of hydrochloric acid = 36.5 g/mol
Kc is 1.67 × 1020 at 25°C for the formation of iron(III) oxalate complex ion: Fe3+(aq) + 3 C2O42-(aq) ⇌ [Fe(C2O4)3]3-(aq). If 0.0800 M Fe3+ is initially mixed with 1.00 M oxalate ion, what is the concentration of Fe3+ ion at equilibrium?
Answer:
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Select the correct relationship among the concentrations of species present in a 1.0 M aqueous solution of the weak acid represented by HA. A. [H2O] > [HA] > [A-] > [H3O ] > [OH-] B. [H2O] > [A-] ~ [H3O ] > [HA] > [OH-] C. [HA] > [H2O] > [A-] > [H3O ] > [OH-] D. [H2O] > [HA] > [A-] ~ [H3O ] > [OH-] E. [HA] > [H2O] > [A-] ~ [H3O ] > [OH-]
Answer:
D
Explanation:
We have to bear in mind that the acid is a weak acid. A weak acid does not dissociate completely in solution. We will have more concentration of undissociated acid than A^- and H3O^+ and OH^- in the system at equilibrium.
Being a weak acid, there is maximum concentration of water molecules followed by that of undissiociated acid.
Hence, for this solution, the concentration of ions in solution follows the order;
[H2O] > [HA] > [A-] ~ [H3O ] > [OH-]
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 6.90kg of water at 34.7 degrees C . During the reaction 57.1kJ of heat flows out of the bath and into the flask.
Calculate the new temperature of the water bath. You can assume the specific heat capacity of water under these conditions is 4.18J.g^(-1).K^(-1) . Round your answer to significant digits.
Answer:
[tex]T_2= 36.7 \textdegree C[/tex]
Explanation:
Mass of Water [tex]m_w=6.90kg[/tex]
Temperature [tex]T=34.7 degrees[/tex]
Heat Flow [tex]H=57.1kJ[/tex]
Specific heat capacity of water [tex]\mu= 4.18J.g^(-1).K^(-1)[/tex]
Generally the equation for Final Temperature is mathematically given by
[tex]M*\mu *T_1 + Q = M*\mu *T_2[/tex]
[tex]T_2=\frac{M*\mu *T_1 + Q }{M*\mu}[/tex]
Therefore
[tex]T_2=\frac{6.90*4.18*34.7 + 57.1}{6.90*4.18}[/tex]
[tex]T_2= 36.7 \textdegree C[/tex]
work 10
HOMEWORK ASSIGNMENTS Content
Detai
Question 1
6.25 Points
р А.
71
Calculate AE of a gas for a process in which the gas evolves 24 J of heat and has 9 of work done on it.
A
✓
A -33)
B +33)
Gradin
-220)
D) +15)
E -15)
Question 2
6.25 Points
Answer
A
Explanation:
due to high specific heat capacity it loses heat and has low temperature
A mixture of argon and neon gases at a total pressure of 874 mm Hg contains argon at a partial pressure of 662 mm Hg. If the gas
mixture contains 12.0 grams of argon, how many grams of neon are present?
Answer:
6.684g
Explanation:
Here, we can use the mole ratio of the gases to calculate.
We know that the mole ratio of the gases equate to their number of moles.
Firstly, we calculate the number of moles of the oxygen gas. The number of moles of the oxygen gas is equal to the mass of the oxygen gas divided by the molar mass of the oxygen gas. The molar mass of the oxygen gas is 32g/mol
Thus, the number of moles produced is 5.98/32 = 0.186875
Where do we move from here?
We know that if we place the partial pressure of oxygen over the total pressure, this would be equal to the number of moles of oxygen divided by the total number of moles. Now let’s do this.
449/851 = 0.186875/n
n =(0.186875 * 851)/449
n = 0.3542
Now we do the same for argon to get the number of moles of argon.
Firstly, we use dalton’s partial pressure law to get the partial pressure of argon. In the simplest form, the partial pressure of argon is the total pressure minus the partial pressure of oxygen.
P = 851 - 449 = 402 mmHg
We now use the mole ratio relation.
402/851 = n/0.3542
n = (402 * 0.3542) / 851
n = 0.1673
Since we now know the number of moles of argon, we can use this multiplied by the atomic mass of argon to get the mass.
The atomic mass of argon is 39.948 amu
The mass is thus 39.948 * 0.1673 = 6.684g
Which type of organic compound is shown below?
A. Carboxylic acid
B. Ester
C. Amine
D. Alcohol
Answer:
I think its A maybe am not sure
all question are compulsory
Answer:
is this question or you just asking I can't understand.
