Answer:
a) 2.9 mol
b) 19 atm
Explanation:
Step 1: Given data
Volume of the container (V): 3.7 LMass of CO gas (m): 82 gTemperature (T): 298 KStep 2: Calculate the number of moles (n) corresponding to 82 g of CO
The molar mass of CO is 28.01 g/mol.
82 g × 1 mol/28.01 g = 2.9 mol
Step 3: Calculate the pressure (P) inside the container
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T / V
P = 2.9 mol × (0.0821 atm.L/mol.K) × 298 K / 3.7 L = 19 atm
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 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
define molecular formula?
A molecular formula is an expression that states the number and type of atoms present in a molecule of a substance.
Example : H2O (water) There are 2 atoms of Hydrogen and 1 atom of Oxygen in this substance
You have 4 litres of a 3.0 mol/L solution of NaCl in a chemical store room.
How many moles of NaCl are present?
Answer:
12
Explanation:
nNaCl= 4x3=12
HCIO4 is identified as what acid
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].
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
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
Rank the solutions below in order of increasing acidity. (Drag and drop into the appropriate area)
0.01 M CH3COOH
0.1 M NaOH
0.01 M H2SO4
3 M NH3
0.1 M HCl
Answer:
0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl
Explanation:
Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.
Then, the more concentrated acid or base will be more acidic or basic.
CH3COOH. Weak acid
NaOH. Strong base
H2SO4. Strong acid
NH3. Weak base.
HCl. Strong acid
The less acid (More basic):
0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HClStrong base, weak base, weak acid, diluted strong acid, undiluted strong acid
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:
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.
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
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
A straight chain hydrocarbon with the formula C5H8_____
Answer:
I has 2 double carbon carbon bonds
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
Can someone help me with this one
Answer:
Easy my dude let me help you out
A.In
B.27
C.73
D.49
E.56
F.56
G.114
H.180
Also with protons and electrons they equal the same atomic number
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 type of chemical bond would be formed between two elements having electron configuration of 1s2 2s2 2p6 3s2 and 1s2 2s2 2p4
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!
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
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 molg The activation energy for diffusion depends on the strength of atomic bonds. Higher strength bonds result in higher activation energies. How do you predict the activation energy for self diffusion will be related to melting temperature
Answer:
Yes, the activation energy for self diffusion is related to melting temperature.
Explanation:
The activation energy for self diffusion is related to melting temperature because bonds result due to activation energies and we know that bonds are break down when the temperature is increases and as a result the solid substance melts and change into liquid state. When the bonds between atoms are broken down with the use of heat energy the state of matter changes so we can say that activation energy is related to temperature.
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
Balance the equation by ion electron method
Answer:i believe you are to decompose the formula (i think)
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
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.
Someone please help me with this
Answer:
I think A should be the answer because oxygen is the chemical change of carbon.
Atoms are found to move from one lattice position to another at the rate of 300,000 jumps/s at 500 0C when the activation energy for their movement is 10,000 cal/mol. Calculate the jump rate at 400 0C.
Answer:
1
Explanation:
1
