Answer:
The required specific enthalpy for oxygen = 8277.34 J/mol
Explanation:
Given that :
Pressure = 2.4 atm
Temperature = 310 K
specific internal energy U = 5700 J/mol
To find the specific enthalpy using the formula:
H = U + PV
where;
H is known as the specific enthalpy
Recall that from ideal gas equation
PV = nRT
For specific enthalpy, it is constant that n = 1
Thus;
PV = RT
replacing that into the equation (H = U + PV), we have:
here;
R = 8.314 J/mol K (constant)
H = U + RT
H = 5700 J/mol +( 8.314 J/mol K × 310 K)
H = 5700 J/mol + ( 2577.34 J/mol)
H = 8277.34 J/mol
which of the following illustrates a reversible change a cooking corn be rusting c frying egg and the boiling water
Helppp
What do you need to know in order to find the mass of 3.00 moles of carbon?
Answer:
36g
Explanation:
you need to know the equation mass=moles*mr (in this case mr of carbon which is 12)
so 3*12=36g
hope this helps :)
Phosphine, PH3, a reactive and poisonous compound, reacts with oxygen as follows: 4PH3(g) 8O2(g) - P4O10(s) 6H2O(g) If you need to make 6.5 moles of P4O10, how many moles of PH3 is required for the reaction
Answer: 26 moles of [tex]PH_3[/tex] are required for the reaction.
Explanation:
We are given:
Moles of [tex]P_4O_{10}[/tex] = 6.5 moles
The given chemical reaction follows:
[tex]4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)[/tex]
By the stoichiometry of the reaction:
If 1 mole of [tex]P_4O_{10}[/tex] is produced by 4 moles of [tex]PH_3[/tex]
So, 6.5 moles of [tex]P_4O_{10}[/tex] will be produced by = [tex]\frac{4}{1}\times 6.5=26mol[/tex] of [tex]PH_3[/tex]
Hence, 26 moles of [tex]PH_3[/tex] are required for the reaction.
a polluted lake is 0.300 μg (micrograms) per liter of water, what is the total mass of mercury in the lake, in kilograms, if the lake has a surface area of 15.0 square miles and an average depth of 27.0 feet?
Answer:
95.9 kg
Explanation:
First we convert 15.0 mi² to m²:
15.0 mi² * ([tex]\frac{1609.34 m}{1mi}[/tex])² = 3.88x10⁷ m²Then we convert 27.0 ft to m:
27.0 ft * [tex]\frac{0.3048m}{1ft}[/tex] = 8.23 mNow we calculate the total volume of the lake:
3.88x10⁷ m² * 8.23 m = 3.20x10⁸ m³Converting 3.20x10⁸ m³ to L:
3.20x10⁸ m³ * [tex]\frac{1000L}{1m^3}[/tex] = 3.20x10¹¹ LNow we calculate the total mass of mercury in the lake, using the given concentration:
0.300 μg / L * 3.20x10¹¹ L = 9.59x10¹⁰ μgFinally we convert μg to kg:
9.59x10¹⁰ μg * [tex]\frac{1kg}{1x10^9ug}[/tex] = 95.9 kgHypochlorous acid decays in the presence of ultraviolet radiation. Assume that degradation occurs accord- ing to first-order kinetics and the rate of degradation was measured to be 0.12 day−1 (at a particular sun- light intensity and temperature). Given this, how long does it take for the concentration of hypochlorous acid to reach nondetectable levels (0.05 mg · L−1) if the initial concentration were 3.65 mg · L−1?
Answer:
35.75 days
Explanation:
From the given information:
For first-order kinetics, the rate law can be expressed as:
[tex]\mathsf{In \dfrac{C}{C_o} = -kt}[/tex]
Given that:
the rate degradation constant = 0.12 / day
current concentration C = 0.05 mg/L
initial concentration C₀ = 3.65 mg/L
[tex]\mathsf{In( \dfrac{0.05}{3.65})= -(0.12) t}[/tex]
㏑(0.01369863014) = -(0.12) t
-4.29 = -(0.12)
t = -4.29/-0.12
t = 35.75 days
g When aqueous solutions of and are mixed, a solid forms. Determine the mass of solid formed when 140.7 mL of 0.1000 M is mixed with an excess of an aqueous solution of .
The question is incomplete, the complete question is:
When aqueous solutions of NaCl and [tex]Pb(NO_3)_2[/tex] are mixed, a solid forms. Determine the mass of solid formed when 140.7 mL of 0.1000 M NaCl is mixed with an excess of an aqueous solution of
Answer: The mass of lead chloride produced is 1.96 g
Explanation:
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{ \text{Volume of solution (mL)}}[/tex] .....(1)
Given values:
Molarity of NaCl = 0.1000 M
Volume of the solution = 140.7 mL
Putting values in equation 1, we get:
[tex]0.1000=\frac{\text{Moles of NaCl}\times 1000}{140.7}\\\\\text{Moles of NaCl}=\frac{0.1000\times 140.7}{1000}=0.01407mol[/tex]
The chemical equation for the reaction of NaCl and lead nitrate follows:
[tex]Pb(NO_3)_2(aq)+2NaCl(aq)\rightarrow PbCl_2(s)+2NaNO_3(aq)[/tex]
By the stoichiometry of the reaction:
If 2 moles of NaCl produces 1 mole of lead chloride
So, 0.01407 moles of NaCl will produce = [tex]\frac{1}{2}\times 0.01407=0.007035mol[/tex] of lead chloride
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(2)
Molar mass of lead chloride = 278.1 g/mol
Plugging values in equation 2:
[tex]\text{Mass of lead chloride}=(0.007035mol\times 278.1g/mol)=1.96g[/tex]
Hence, the mass of lead chloride produced is 1.96 g
HCIO4 is identified as what acid
Classify each phrase according to whether it applies to photophosphorylation, oxidative phosphorylation, or both
Photophosphorylation Oxidative phosphorylation Both
1. occurs in plants produces ATP
2. occurs in chloroplasts
3. occurs in mitochondria
4. involves a larger electrical component
5. involves a smaller electrical component
6. involves a proton gradient
Answer:
1. Both
2. Phosphorylation
3. Both
4. Phosphorylation
5. Oxidative.
6. Both
Explanation:
Phosphorylation only occurs in chloroplast and it involves larger electrical component. Both Phosphorylation and oxidative occurs in mitochondria and it involves proton gradient. They occur in plants to produce ATP. Oxidative involves in smaller electrical component.
Photophosphorylation is a process that captures the solar energy from the sun to transform it into chemical energy. It occurs in the chloroplast of a plant cell.
What are photophosphorylation and oxidative phosphorylation?Photophosphorylation is a process of converting solar energy from the sun to ATP needed by plants and other organisms for cellular function and activity. This process takes place in the chloroplast of the plant cell and requires electrical components.
Oxidative Phosphorylation is the process of producing ATP with the help of oxygen and enzymes hence, occurs in aerobic cells. It does not need a larger electrical component.
Both phosphorylation and oxidative phosphorylation occurs in the mitochondria of plants cells and involves a proton gradient for the formation of ATP.
Therefore, oxidative phosphorylation option 5. involves a smaller electrical component, phosphorylation option 2. occurs in the chloroplast, and option 4. needs a larger electrical component.
Learn more about phosphorylation here:
https://brainly.com/question/1870229
The compound sodium hydrogen sulfate is a strong electrolyte. Write the reaction when solid sodium hydrogen sulfate is put into water:
Answer:
NaHSO₄(s) --H₂O--> Na⁺(aq) + HSO₄⁻(aq)
Explanation:
Sodium hydrogen sulfate is a strong electrolyte, that is, when dissolved in water it completely dissociates into the cation sodium and the anion hydrogen sulfate. The corresponding chemical equation is:
NaHSO₄(s) --H₂O--> Na⁺(aq) + HSO₄⁻(aq)
Which of the following could not be a resonance structure of CH3NO2?
a)
H
H-C-NO
H
b)
H .0:
H-C-N
H
c)
H:03
H-C-NC2
H:06
d)
H
H-C=N
H :9-H
e) Both c and d
Answer:
the answer is b.CH3NO2 I guess I'm correct
QUESTIONS :
1.
Many of the flavours and smells of fruits are esters. A learner prepared an ester with a sme
Ilke banana in the school laboratory using pentanol and ethanoic acid. She set up the
apparatus as shown in the diagram below.
PAPER TOWEL DIPPED
-WATER BATH
IN COLD WATER
PEITANOL ETHANOIC
ACID+ 4 DROPS OF
SULPHURIC ACID
1.1 Which property of sulphuric acid makes it suitable to use as a catalyst for the
preparation of esters?
1.2 Why do we heat the test tube in a water bath and not directly over a flame?
1.3 With reference to the characteristic smells of esters, name TWO examples where
esters are used in different industries.
1.4 State ONE function of the wet paper towel in the opening of the test tube.
1.5 Write down the IUPAC name of an ester fomed.
Answer:
See explanation
Explanation:
Esterification is a reaction that involves the combination of an alkanoic acid and an alkanol. The product is always a sweet smelling substance.
Sulphuric acid acts as a catalyst in this reaction because it is a dehydrating agent thereby pushing the equilibrium position towards the right by the removal of water molecules.
The test tubs is heated in a water bath and not directly moved the flame because the alcohol is flammable. Also heating in a water bath helps to separate the reaction mixture from the newly formed ester.
Esters are used in industries that produces soaps and perfumes. There is a great need for the use of fragrances which are ester compounds in these industries.
The wet paper towel in the opening of the test tube cools the top of the test tube. It usually serves as a kind of condenser preventing an excess loss of vapour from the reaction mixture.
The reaction of pentanol and ethanoic acid yields pentyl ethanoate according to IUPAC nomenclature.
A pressure cooker contains 5.68 L of air at a temperature of 390 4K if the absolute pressure of the air in the pressure cooker is 205 Pa how many moles of air are in the cooker
Answer:
3.59x10⁻⁴ mol
Explanation:
Assuming ideal behaviour we can solve this problem by using the PV=nRT formula, where:
P = 205 PaV = 5.68 Ln = ?R = 8314.46 Pa·L·mol⁻¹·K⁻¹T = 390.4 KWe input the data given by the problem:
205 Pa * 5.68 L = n * 8314.46 Pa·L·mol⁻¹·K⁻¹ * 390.4 KAnd solve for n:
n = 3.59x10⁻⁴ molWhen 1.00 g of coal is burned in a bomb calorimeter, the temperature increases by 1.48°C. If the heat capacity of the calorimeter is 21.6 kJ/°C, determine the heat (in GJ) produced by combustion of a ton of coal.
Answer:
32.0 kJ
General Formulas and Concepts:
Thermochemistry
Specific Heat Formula: q = mcΔT
q is heat (in J) m is mass (in g) c is specific heat (in J/g °C) ΔT is change in temperature (in °C)Explanation:
Step 1: Define
Identify variables
[Given] m = 1.00 g
[Given] ΔT = 1.48 °C
[Given] c = 21.6 kJ/g °C
[Solve] q
Step 2: Find Heat
Substitute in variables [Specific Heat Formula]: q = (1.00 g)(21.6 kJ/g °C)(1.48 °C)Multiply [Cancel out units]: q = (21.6 kJ/°C)(1.48 °C)Multiply [Cancel out units]: q = 31.968 kJStep 3: Check
Follow sig fig rules and round. We are given 3 sig figs.
31.968 kJ ≈ 32.0 kJ
Discuss the four impure forms of carbon
Various structures, or allotropes, of carbon, are precious stone, graphite, and fullerenes. In jewel, every carbon iota is attached to four other carbon iotas, shaping an unbending construction that makes precious stones hard.
54.56 g of water at 80.4 oC is added to a calorimeter that contains 47.24 g of water at 40 oC. If the final temperature of the system is 59.4 oC, what is the calorimeter constant (C calorimeter)
Answer:
49.5J/°C
Explanation:
The hot water lost some energy that is gained for cold water and the calorimeter.
The equation is:
Q(Hot water) = Q(Cold water) + Q(Calorimeter)
Where:
Q(Hot water) = S*m*ΔT = 4.184J/g°C*54.56g*(80.4°C-59.4°C) = 4794J
Q(Cold water) = S*m*ΔT = 4.184J/g°C*47.24g*(59.4°C-40°C) = 3834J
That means the heat gained by the calorimeter is
Q(Calorimeter) = 4794J - 3834J = 960J
The calorimeter constant is the heat gained per °C. The change in temperature of the calorimeter is:
59.4°C-40°C = 19.4°C
And calorimeter constant is:
960J/19.4°C =
49.5J/°C
propose a synthetic route for the synthesis of a named alkanal starting with ethyl formate and grignard reagent.
Answer:
See explanation and image attached
Explanation:
A Grignard reagent is an alkyl magnesium halide. If it reacts with ethyl formate, an intermediate is formed as shown.
This intermediate can undergo water hydrolysis to form a diol, ethanol and MgBrOH.
Oxidation of the diol obtained now yields the corresponding alkanal which in this case is ethanal.
The scheme of the reaction is shown in the image attached to this answer.
4) The initial rate of the reaction between substances P and Q was measured in a series of
experiments and the following rate equation was deduced.
rate = k[P]?[Q]
Complete the table of data below for the reaction between P and Q
*Help asap please*
Answer:
The initial rate of the reaction between substances P and Q was measured in a series of
experiments and the following rate equation was deduced.
[tex]rate = k[P]^{2} [Q][/tex]
Complete the table of data below for the reaction between P and Q
Explanation:
Given rate of the reaction is:
[tex]rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }[/tex]
Substitute the given values in this formulae to get the [P], [Q] and rate values.
From the first row,
the value of k can be calulated:
[tex]k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4[/tex]
Second row:
2. Rate value:
[tex]rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1[/tex]
3.Third row:
[tex][Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}[/tex]
4. Fourth row:
[tex][P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}[/tex]
2 AICI3 + 3 Ca - 3 CaCl2 + 2 Al
You react aluminum chloride with calcium metal. You want to produce 40.00 grams of aluminum. How many grams of calcium do
you need?
Answer:
50 gram calcium do you need
Explanation:
please make me brainlist answer
Gaseous ethane (CH,CH,) will react with gaseous oxygen (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H,0). Suppose 4.21 g of
ethane is mixed with 31. 9 of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has
the correct number of significant digits.
Answer: The mass of [tex]CO_2[/tex] produced is 12.32 g
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
For ethane:Given mass of ethane = 4.21 g
Molar mass of ethane = 30 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of ethane}=\frac{4.21g}{30g/mol}=0.140mol[/tex]
For oxygen gas:Given mass of oxygen gas = 31.9 g
Molar mass of oxygen gas= 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of oxygen gas}=\frac{31.9g}{32g/mol}=0.997mol[/tex]
The chemical equation for the combustion of ethane follows:
[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]
By stoichiometry of the reaction:
If 2 moles of ethane reacts with 7 moles of oxygen gas
So, 0.140 moles of ethane will react with = [tex]\frac{7}{2}\times 0.140=0.49mol[/tex] of oxygen gas
As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, ethane is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 2 moles of ethane produces 4 moles of [tex]CO_2[/tex]
So, 0.140 moles of ethane will produce = [tex]\frac{4}{2}\times 0.140=0.28mol[/tex] of [tex]CO_2[/tex]
We know, molar mass of [tex]CO_2[/tex] = 44 g/mol
Putting values in above equation, we get:
[tex]\text{Mass of }CO_2=(0.28mol\times 44g/mol)=12.32g[/tex]
Hence, the mass of [tex]CO_2[/tex] produced is 12.32 g
20ml of water is mixed with 40gm of fine powder. Calculate the concentration of the solution obtained.
Answer:
[tex]\%m=66.7\%[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the concentration of the solution obtained, by knowing 20 mL of water are the same to 20 g and therefore the mass of the solution is 40g+20g=60g.
Next, we apply the following equation to obtain the required concentration:
[tex]\%m=\frac{40g}{60g} *100\%\\\\\%m=66.7\%[/tex]
Regards!
Of the below gases, which would deviate most from ideal gas behavior? CO O2 NH3 SF4
Answer:
For gases such as hydrogen, oxygen, nitrogen, helium, or neon, deviations from the ideal gas law are less than 0.1 percent at room temperature and atmospheric pressure. Other gases, such as carbon dioxide or ammonia, have stronger intermolecular forces and consequently greater deviation from ideality.
Explanation:
Someone please help me with this
Answer:
I think A should be the answer because oxygen is the chemical change of carbon.
Consider the reaction between an alcohol and tosyl chloride, followed by a nucleophile. Write the condensed formula of the expected main organic product.
CH3CH2CH2OH---------- 2.CI 1.TsCl,pyridine__________
Answer:
Consider the reaction between an alcohol and tosyl chloride, followed by a nucleophile. Write the condensed formula of the expected main organic product.
CH3CH2CH2OH---------- 2.CI 1.TsCl,pyridine__________
Explanation:
Given alcohol is propanol.
When it reacts with TsCl, the hydrogen in -OH group is replaced with tosyl group.
Pyridine is a weak base and it neutralizes the HCl (acid) formed during the reaction.
The reaction is shown below:
Which of the choices below has more heat being transferred as thermal energy from one place to another?
A. A bowl of ice water
B. A pot of boiling water
Answer:
B
Explanation:
So, a pot of boliling is hot right? of course, since it is hot thermal energy will be transferred from one place to another. I don't know if this is correct but I just wanted to give it a try.
which type of chemical bond would be formed between two elements having electron configuration of 1s2 2s2 2p6 3s2 and 1s2 2s2 2p4
NCl3 + 3H20 - NH3 + 3HCIO
How many grams of ammonia can be produced from 1.33 grams of nitrogen trichloride?
Answer:
0.189 g
Explanation:
Step 1: Write the balanced equation
NCl₃ + 3 H₂O ⇒ NH₃ + 3 HCIO
Step 2: Calculate the moles corresponding to 1.33 g of NCl₃
The molar mass of NCl₃ is 120.36 g/mol.
1.33 g × 1 mol/120.36 g = 0.0111 mol
Step 3: Calculate the moles of NH₃ produced from 0.0111 moles of NCl₃
The molar ratio of NCl₃ to NH₃ is 1:1. The moles of NH₃ produced are 1/1 × 0.0111 mol = 0.0111 mol.
Step 4: Calculate the mass corresponding to 0.0111 moles of NH₃
The molar mass of NH₃ is 17.03 g/mol.
0.0111 mol × 17.03 g/mol = 0.189 g
Please help me, it’s my last try
Answer:
Group 1A: alkali metals, or lithium family.
Group 2A: alkaline earth metals, or beryllium family.
Group 7A: the manganese family.
Group 8A: the iron family.
Explanation:
Answer:
1A: Alkali Metals
2A: Alkaline Earth Metals
7A: Halogens
8A: Noble Gases
A sample of 0.2140 g of an unkown substance monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.950 M NaOH. The acid required 27.4 mL of base to reach the equivalence point. After 15.0 mL of base had been added in the titration, the pH was found to be 6.50. What is the Ka for the unknown acid?
Solution :
The equation is :
[tex]$HA (aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)$[/tex]
The number of the moles of HA os 0.00285, and the volume is 25 mL.
15 mL of the 0.0950 M NaOH is added.
The total volume of a solution is V = 25 mL + 15 mL = 40 mL
The pH of the solution is 6.50
Calculating the [tex]K_a[/tex] of HA
[tex]$HA(aq) \rightleftharpoons A^-(aq)+H^+$[/tex]
[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]
Let s calculate the concentration of HA and NaOH
[tex]$[HA] = \frac{^nH_A}{V}$[/tex]
[tex]$=\frac{0.00285 \ mol}{0.04 \ L}$[/tex]
= 0.07125 M
[tex]$[NaOH]= \frac{0.015L \times 0.0950 M}{V}$[/tex]
[tex]$=\frac{0.001425 mol}{0.04L}$[/tex]
= 0.0356 M
[tex]$HA(aq) \ \ + \ \ NaOH(aq) \ \ \rightleftharpoons NaA(aq) \\ + \ \ H_2O(aq)$[/tex]
Initial conc. (M) 0.07125 M 0.0356 M 0 M
Change in conc. (M) -0.0356 M -0.0356 M + 0.0356 M
Equilibrium conc. (M) 0.03565 M 0 M 0.0356 M
Therefore, the concentration of HA and the NaA at the equilibrium are [HA] = 0.03565 M and [NaA]= 0.0356 M
0.0356 M of NaA dissociates completely into 0.0356 M [tex]Na^+[/tex] and 0.0356 M [tex]A^-[/tex]
Now for [tex][H^+][/tex]
[tex]$[H^+] = 10^{-pH}$[/tex]
[tex]$=10^{-6.5}$[/tex]
[tex]$=3.16 \times 10^{-7}$[/tex]
Calculating the value of [tex]K_a[/tex],
[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]
[tex]$=\frac{0.0356 \times 3.16 \times 10^{-7}}{0.03565}$[/tex]
[tex]$=3.16\times 10^{-7}$[/tex]
Therefore the the value of [tex]K_a[/tex] for the unknown acid is [tex]$3.16\times 10^{-7}$[/tex].
A sample of oxygen gas occupies a volume of 2.,0cm3 at pressure of 700K pa. what will be pressure of the same sample occupies a volume of 150cm, assume temperature remains constant
Answer:
The pressure will be 933.33 Kpa
Explanation:
Given that:
Volume V₁ = 200 cm³ (note, there is a mistake in the volume. It is supposed to be 200 cm³)
Pressure P₁ = 700 Kpa
Pressure P₂ = ??? (unknown)
Volume V₂ = 150 cm³
Temperature = constant
Using Boyle's law:
PV = constant
i.e.
P₁V₁ = P₂V₂
700 Kpa × 200 cm³ = P₂ × 150 cm³
P₂ = (700 Kpa × 200 cm³)/150 cm³
P₂ = 933.33 Kpa
Which of the following ionization energies indicates an atom is most likely to gain electrons and form an anion or not form an ion at all?
Group of answer choices
578 kJ/mol
9460 kJ/mol
496 kJ/mol
786 kJ/mol
Answer:
Explanation:
578kj/mol
