Bread dough rises due to the formation of gas.
Explanation:
If we observe the outer part of the bread, which contains tiny hole like structure, due to the gas formation inside the bread dough.In the bread dough rising process, the industries are using the yeast, which makes the gas that separates the protein particles in the bread move apart and makes the dough of the bread rise.In this reaction, the yeast utilizing the carbohydrate to make a gas namely carbon-di-oxide gas which makes the dough to riseThe collapse point is an indicator of sufficient Bread dough rise.
Discussion;
When yeast is added to water and flour to create dough, it eats up the sugars in the flour and in turn produces carbon dioxide gas and ethanol in a process termed fermentation.
The elastic protein, gluten in the dough traps the carbon dioxide gas, preventing it from escaping.
In essence, the bread volume increases as the dough rises.
The collapse point is an indicator of sufficient dough rise.
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A small toy car draws a 0.50-mA current from a 3.0-V NiCd (nickel-cadmium) battery. In 10 min of operation, (a) how much charge flows through the toy car, and (b) how much energy is lost by the battery? 4. (Resistance and Ohm’s law, Prob. 17.16, 1.0 point) How
Answer:
0.3 Coulomb charge flows through the toy car.
0.9 Joules of energy is lost by the battery
Explanation:
[tex]Current(I)=\frac{charge(Q)}{Time (T)}[/tex]
a)
Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)
milli Ampere = 0.001 Ampere
Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s
1 min = 60 seconds
Charge flows through the toy car be Q
[tex]0.50\times 0.001 A=\frac{Q}{600 s}[/tex]
[tex]Q=0.50 \times 0.001 A\times 600 s=0.3 C[/tex]
0.3 Coulomb charge flows through the toy car.
b)
[tex]Heat(H)=Voltage(V)\times Current(I)\times Time(T)[/tex]
Voltage of the battery = V = 3.0 V
Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)
Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s
[tex]H=V\times I\time T=3.0 V\times 0.50 \times 0.001 A\times 600 s[/tex]
H = 0.9 Joules
0.9 Joules of energy is lost by the battery
if 12 molecules of methane reacted with plenty of oxygen, we should expect to produce____ molecules of car on dioxide and _____ molecules of water
Answer:
12 molecules of carbon dioxide
24 molecules of water
Explanation:
Given parameters:
Number of molecules of methane = 12 molecules
Unknown:
Number of molecules of carbon dioxide = ?
Number of molecules of water = ?
Solution:
The given amount of methane is the limiting reagent in this reaction. By this, we can ascertain the amount of products that will be formed and the extent of the reaction.
We first write the balanced chemical equation for this reaction;
CH₄ + 2O₂ → CO₂ + 2H₂O
From balanced equation;
1 mole of methane produced 1 mole of carbon dioxide
12 molecules of methane will produce 12 molecules of carbon dioxide
Also;
1 mole of methane produced 2 moles of water;
12 molecules of methane will produce (12 x 2)molecules = 24 molecules of water
When 12 molecules of methane react with excess oxygen, 12 molecules of carbon dioxide and 24 molecules of water are produced, following a 1:2:1:2 ratio of methane to oxygen to carbon dioxide to water.
Explanation:If 12 molecules of methane (CH4) react with plenty of oxygen (O2), the balanced chemical equation for the combustion of methane shows a 1:2:1:2 ratio. This means for every one methane molecule, two oxygen molecules are consumed, and one carbon dioxide (CO2) molecule and two water (H2O) molecules are produced. Therefore, 12 molecules of methane reacting with excess oxygen will produce 12 molecules of carbon dioxide and 24 molecules of water.
This equation states that for every 1 molecule of methane and 2 molecules of oxygen that react, we will produce 1 molecule of carbon dioxide and 2 molecules of water.
When solid (NH4)(NH2CO2)(NH4)(NH2CO2) is introduced into an evacuated flask at 25 ∘C∘C, the total pressure of gas at equilibrium is 0.116 atmatm. What is the value of KpKp at 25 ∘C∘C?
Answer:
the value of Kp at 25°C is 2.37 × 10⁻⁴atm³
Explanation:
Given that Kp = 0.116atm
NH₄(NH₂CO₂)(s) ⇄ 2NH₃(g) + CO₂(g)
2x atm xatm
Pt = PNH₃ PCO₂
0.116 = 2x + x
0.116 = 3x
x = 0.116/3
x = 0.039 atm
PNH₃ = 2x =2(0.039) = 0.078 atm
PCO₂ = x = 0.039 atm
Now,
KP = PNH₃² × PCO₂
= 0.078² × 0.039 atm³
= 2.37 × 10⁻⁴ atm³
the value of Kp at 25°C is 2.37 × 10⁻⁴atm³
A solution of toluene in 401 g of cyclohexane has a boiling point of 90.3 °C. How many moles of toluene are in the solution? (For cyclohexane Kb = 2.92 °C/m, Tb = 80.9 °C)
Answer:
There are 1.29 moles of toluene in the solution.
Explanation:
m = ∆Tb/Kb
m is molality of the solution
∆Tb is the change in boiling point of cyclohexane = 90.3 ° C - 80.9 °C = 9.4 °C
Kb is the boiling point elevation constant of cyclohexane = 2.92 °C/m
m = 9.4/2.92 = 3.22 mol/kg
Number of moles of toluene = molality × mass of cyclohexane in kilogram = 3.22 × 401/1000 = 1.29 moles
Phosphine, PH₃, is a colorless, toxic gas that is used in the production of semiconductors as well as in the farming industry. When heated, phosphine decomposes into phosphorus and hydrogen gases.
4 PH₃(g) ⟶ P₄(g) + 6 H₂(g)
This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial pressure of hydrogen gas that is present after 70.5 s if a 2.20 L vessel containing 2.29 atm of phosphine gas is heated to 953 K.
1.72 atm of hydrogen gas is created when 2.29 atm of phosphine breaks down at 953 K for 70.5 seconds.
Explanation:
The first step in this calculation is to use the first-order decay equation to find the remaining moles of phosphine. For a first-order reaction with a half-life of 35.0 s, after 70.5 s (or two half-lives), 25% of the phosphine would remain. To find the remaining moles of phosphine, we multiply 2.29 atm by 25%, yielding 0.57 atm.
From the balanced equation for the decomposition of phosphine (4PH3 -> P4 + 6H2), we see that 1 mole of phosphine yields 3/2 moles of H2. Thus, the number of moles of H2 produced is 75% of the original moles of PH3 (1.72 atm).
The sum of the partial pressures of phosphine and hydrogen is 0.57 + 1.72 = 2.29 atm, which is the final pressure in the 2.20 L vessel after 70.5 s.
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Sodium chloride as a compound does not truly exist in the ocean. True or False
Describes the difference between the formulas for nitrogen monoxide and nitrogen dioxide?
Answer:
nitrogen monoxide: NO
nitrogen dioxide: NO₂
Explanation:
Nitrogen monoxide is composed by 1 atom of O (prefix "mono-") and 1 atom of N. Nitrogen dioxide is composed by 2 atoms of O (prefix "di-") and 1 atom of N. As the oxigen atom in oxides has the valency -2 (it shares 2 electrons), the nitrogen has valency +2 in NO and +4 in NO₂.
When powdered zinc is heated with sulfur, a violent reaction occurs, and zinc sulfide forms: Some of the reactants also combine with oxygen in air to form zinc oxide and sulfur dioxide. When 83.2 g of Zn reacts with 52.4 g of S8, 104.4 g of ZnS forms. (a) What is the percent yield of ZnS? (b) If all the remaining reactants combine with oxygen, how many grams of each of the two oxides form?
Answer:
Explanation:
The detailed steps and step by step calculations is as shown in the attached file.
To calculate the percent yield of ZnS, use the given masses of Zn and S8 and compare to the theoretical yield. For the oxides formed, determine the moles of remaining Zn and S8 using the balanced chemical equation.
Explanation:To calculate the percent yield of ZnS, we need to compare the actual yield (104.4 g) to the theoretical yield. The theoretical yield can be calculated using the stoichiometry of the reaction. The balanced chemical equation is: Zn + S8 → ZnS. The molar ratio between Zn and ZnS is 1:1, which means that the moles of ZnS formed will be equal to the moles of Zn used. Convert the given masses of Zn and S8 to moles using their molar masses, then compare the moles to calculate the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction. Once you have determined the limiting reactant and calculated the theoretical yield, you can calculate the percent yield using the formula: (actual yield / theoretical yield) x 100.
For part (b), if all the remaining reactants combine with oxygen, we need to determine the remaining moles of Zn and S8 after the reaction with ZnS is complete. Convert these moles to masses using their molar masses, and then use the balanced chemical equation to determine the moles of each oxide formed. The balanced chemical equation for the reaction of zinc with oxygen is: 2Zn + O2 → 2ZnO. The molar ratio between Zn and ZnO is 2:2, so the moles of ZnO formed will be equal to the moles of Zn remaining. The balanced chemical equation for the reaction of sulfur with oxygen is: S8 + 8O2 → 8SO2. The molar ratio between S8 and SO2 is 1:8, so the moles of SO2 formed will be 8 times the moles of S8 remaining. Convert the moles of each oxide formed to masses using their molar masses.
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Hydrogenotrophy is Choose one: A. an award given to fuel companies that have the most fuel-efficient vehicles. B. the oxidation of water during photosynthesis to liberate electrons, protons, and oxygen gas. C. the use of hydrogen gas as an electron donor. D. the generation of hydrogen gas by methanogens in syntrophy.
Answer:
C. the use of hydrogen gas as an electron donor.
Explanation:
Hydrogenotrophy is the convertion of hydrogen gas to other compounds as part of its metabolism.
When solid (NH4)(NH2CO2) is introduced into an evacuated flask at 25 ∘C, the total pressure of gas at equilibrium is 0.116 atm. What is the value of Kp at 25 ∘C? Kp =
Answer : The value of [tex]K_p[/tex] is, [tex]2.32\times 10^{-4}[/tex]
Explanation :
For the given chemical reaction:
[tex](NH_4)(NH_2CO_2)(s)\rightleftharpoons 2NH_3(g)+CO_2(g)[/tex]
The expression of [tex]K_p[/tex] for above reaction follows:
[tex]K_p=(P_{NH_3})^2\times P_{CO_2}[/tex] ........(1)
[tex](NH_4)(NH_2CO_2)(s)\rightleftharpoons 2NH_3(g)+CO_2(g)[/tex]
Initial: 0 0
At eqm: 2x x
As we are given that:
Total pressure of gas at equilibrium = 0.116 atm
2x + x = 0.116 atm
3x = 0.116 atm
x = 0.0387 atm
Putting values in expression 1, we get:
[tex]K_p=(2x)^2\times (x)[/tex]
[tex]K_p=(2\times 0.0387)^2\times (0.0387)[/tex]
[tex]K_p=2.32\times 10^{-4}[/tex]
Thus, the value of [tex]K_p[/tex] is, [tex]2.32\times 10^{-4}[/tex]
Polar molecules have a partial positive charge on one side and a partial negative charge on the other side, a separation of charge called a dipole. Which of the following molecules has this kind of a dipole?a. IonicB. Covalent
Answer:Covalent
Explanation:
Covalent bonds are formed by sharing of electrons between two atoms. The shared electrons are normally situated between the nuclei of the both atoms. However, atoms of some elements posses an usual ability to attract the shared electrons of a bond towards itself. We say that such atoms have a high electronegativity. High electronegativity leads to the existence of polar covalent bonds since the shared electron pair is now closer to one of the bonding atoms than the other. The atom to which the electron pair is closer becomes partially negative while the other becomes partially positive. This is only possible in a covalent bond where electrons are shared between bonding atoms. The charge separation is known as a dipole.
Note that ionic bonds involve a complete transfer of electrons leading to the formation of ions and is not applicable here.
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The relationship between electricity and magnetism is called
a. current.
b. electromagnetism.
c. a solenoid.
d. voltage.
Answer:
B
Explanation:
The interaction between magnetism and electricity is called electromagnetism. The movement of a magnet can generate electricity. The flow of electricity can generate a magnetic field.
Answer: B: electromagnetism
Explanation:
A changing magnetic field produces an electric current in a wire or conductor.
How many grams of NaCl would need to be added to 1001 g of water to increase the boiling temperature of the solution by 1.500 °C? (Kb for water is 0.5100 °C/m)
To find the amount of NaCl needed to increase the boiling temperature of water by 1.500°C, one must understand molal boiling point elevation and van't Hoff factor. Using the formula and given constants, we calculate the grams of NaCl required through the steps of determining molality, calculating the moles of NaCl, and then converting to grams.
Explanation:To calculate the amount of NaCl needed to raise the boiling temperature of water by 1.500°C, we first understand the concept of molal boiling point elevation and van't Hoff factor. Given that the Kb for water is 0.5100°C/m, we use the formula ΔTb = i*Kb*m, where ΔTb is the boiling point elevation, i is the van't Hoff factor for NaCl (which is 2 because NaCl disassociates into Na+ and Cl- ions), Kb is the ebullioscopic constant for the solvent (water), and m is the molality of the solution.
To find the molality (m), we rearrange the equation as m = ΔTb / (i*Kb). Substituting the known values, we get m = 1.500 / (2*0.5100) = 1.4706 mol/kg. Knowing the molality and the mass of the solvent (water), we can then calculate the moles of NaCl required, which is molality * mass of solvent in kg. Finally, converting moles of NaCl to grams using its molar mass (58.44 g/mol), gives us the total grams of NaCl needed. This step-by-step process provides a clear link between the theoretical concepts and their practical application.
Approximately 85.78 grams of [tex]NaCl[/tex] would need to be added to 1001 grams of water to increase the boiling temperature of the solution by [tex]1.500 \°C[/tex].
To solve this problem, we will use the concept of boiling point elevation, which states that the boiling point of a solvent is increased by an amount proportional to the molal concentration of the solute. The proportionality constant is known as the ebullioscopic constant (Kb). The formula to calculate the boiling point elevation is:
[tex]\[ \Delta T_b = i \cdot K_b \cdot m \][/tex]
where:
- [tex]\( \Delta T_b \)[/tex] is the increase in boiling point temperature,
- [tex]\( i \)[/tex] is the van 't Hoff factor (the number of moles of particles in solution per mole of solute; for [tex]NaCl[/tex], [tex]\( i = 2 \)[/tex] because it dissociates into two ions, [tex]Na^+ and Cl^-)[/tex],
- [tex]\( K_b \)[/tex] is the ebullioscopic constant for the solvent (given as [tex]0.5100 \°C[/tex]/m for water),
- [tex]\( m \)[/tex] is the molality of the solution (moles of solute per kilogram of solvent).
Given:
- [tex]\( \Delta T_b = 1.5000 °C \)[/tex]
- [tex]\( K_b = 0.5100 °C/m \)[/tex]
- [tex]\( i = 2 \)[/tex] for [tex]NaCl[/tex]
- Mass of water (solvent) = 1001 g (which is approximately 1 kg, since 1000 g = 1 kg)
First, we will rearrange the formula to solve for the molality (m):
[tex]\[ m = \frac{\Delta T_b}{i \cdot K_b} \][/tex]
Substitute the given values:
[tex]\[ m = \frac{1.5000 °C}{2 \cdot 0.5100 °C/m} \][/tex]
[tex]\[ m = \frac{1.5000 °C}{1.0200 °C/m} \][/tex]
[tex]\[ m \approx 1.4695 m \][/tex]
Now that we have the molality, we can calculate the number of moles of [tex]NaCl[/tex] needed to achieve this molality in 1 kg of water:
[tex]\[ \text{moles of NaCl} = m \cdot \text{mass of solvent in kg} \][/tex]
[tex]\[ \text{moles of NaCl} = 1.4695 \text{ m} \cdot 1 \text{ kg} \][/tex]
[tex]\[ \text{moles of NaCl} \approx 1.4695 \][/tex]
The molar mass of [tex]NaCl[/tex] is approximately 58.44 g/mol. To find the mass of [tex]NaCl[/tex] needed, we multiply the number of moles by the molar mass:
[tex]\[ \text{mass of NaCl} = \text{moles of NaCl} \cdot \text{molar mass of NaCl} \][/tex]
[tex]\[ \text{mass of NaCl} \approx 1.4695 \text{ mol} \cdot 58.44 \text{ g/mol} \][/tex]
[tex]\[ \text{mass of NaCl} \approx 85.7797 \text{ g} \][/tex]
Of the reactions below, only ________ is not spontaneous. Mg (s) 2HCl (aq) MgCl2 (aq) H2(g) 2Ag (s) 2HNO3 (aq) 2AgNO3 (aq) H2 (g) 2Ni (s) H2SO4 (aq) Ni2SO4 (aq) H2 (g) 2Al (s) 6HBr (aq) 2AlBr3 (aq) 3H2 (g) Zn (s) 2HI (aq) ZnI2(aq) H2 (g)
Answer:
2Ag (s) + 2HNO₃ (aq) 2AgNO₃ (aq) + H₂ (g)
Explanation:
2Ag (s) + 2HNO₃ (aq) 2AgNO₃ (aq) + H₂ (g)
Ag is below H₂ in reactivity series. Therefore Ag does not spontaneously replace H₂ from any compound.
Answer:
The answer to the question is
2Ag (s) +2HNO₃ (aq) →2AgNO₃ (aq)+ H₂ (g)
Explanation:
The position of a cell on the reduction potential table determines if a reduction reaction will be spontaneous
The higher the positive reduction potential the higher the spontaneity of the reduction reaction. that is if the E⁰ cell is positive the cell is a spontaneous voltaic cell, however if the E⁰ cell is negative the cell is a electrolytic non-spontaneous cell
On the standard potential table silver has a low positive potential of
Ag⁺ + e⁻ → Ag = 0.799
It is an oxidizing agent tending to scarcely release electrons, therefore for the reaction to take place, there has to be some external supply of electromotive force.
Potassium chlorate decomposes to potassium chloride and oxygen. If 20.8 g of potassium chlorate decomposes, how many liters of oxygen will form at STP?
The number of liters of oxygen formed at STP when 20.8 g of potassium chlorate decomposes is 5.70 L.
Explanation:Potassium chlorate decomposes according to the following balanced equation:
2KClO3 → 2KCl + 3O2
The molar mass of KClO3 is 122.55 g/mol.
To calculate the number of moles of KClO3, we divide the given mass (20.8 g) by the molar mass:
20.8 g / 122.55 g/mol = 0.1697 mol
According to the balanced equation, for every 2 moles of KClO3, 3 moles of O2 are produced.
Therefore, the number of moles of O2 produced is:
0.1697 mol × (3/2) = 0.2546 mol
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies approximately 22.4 liters.
Therefore, the number of liters of O2 produced at STP is:
0.2546 mol × 22.4 L/mol = 5.70 L
a 125g bar of aluminum at 22 degrees celsius. determine the final temperature of the aluminum, if the amount of energy applied is equal to 3600 calories. the specific heat of aluminum is .90 j/gc
Answer : The final temperature of the aluminum is, [tex]155.9^oC[/tex]
Explanation :
Formula used :
[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]
where,
q = heat = 3600 cal = 15062.4 J (1 cal = 4.184 J)
m = mass of aluminum = 125 g
c = specific heat of aluminum = [tex]0.90J/g^oC[/tex]
[tex]T_{final}[/tex] = final temperature = ?
[tex]T_{initial}[/tex] = initial temperature = [tex]22^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]15062.4J=125g\times 0.90J/g^oC\times (T_{final}-22)^oC[/tex]
[tex]T_{final}=155.9^oC[/tex]
Thus, the final temperature of the aluminum is, [tex]155.9^oC[/tex]
4. The volume of a sample of a gas at STP is 200.0 ml. If the pressure is increased to 4.00 atmospheres (temperature constant), what is the new volume?
Answer : The value of new volume is, 50.0 mL
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex]
or,
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1[/tex] = initial pressure at STP = 1 atm
[tex]P_2[/tex] = final pressure = 4.00 atm
[tex]V_1[/tex] = initial volume at STP = 200.0 mL
[tex]V_2[/tex] = final volume = ?
Now put all the given values in the above equation, we get:
[tex]1atm\times 200.0mL=4.00atm\times V_2[/tex]
[tex]V_2=50.0mL[/tex]
Therefore, the value of new volume is, 50.0 mL
. The molar heat of vaporization of acetone, C3H6O, is 30.3 kJ/mol at its boiling point. How many kilojoules of heat would be liberated by the condensation of 5.00 g of acetone?
2.61 kilojoules of heat would be liberated by the condensation of 5.00 g of acetone
Explanation:
To convert grams into moles
[tex]moles = grams \times \frac{1 mole}{grams}[/tex]
We have 5.00 g acetone
[tex]moles = 5 \times \frac{1}{58.1}[/tex]
[tex]moles = 0.0861[/tex]
Heat liberated = moles [tex]\times[/tex] heat of vapourization
=0.0861 mol x 30.3 kJ/mol
= 2.61 kJ
Therefore, 2.61 kilojoules of heat would be liberated by the condensation of 5.00 g of acetone
A recent study has revealed that chlorinated hydrocarbons, gasoline and other volatile organic compounds (VOC's) have become significant pollutants in ________ from ________.
Answer:
ground water / leaking storage tanks
20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator would be most suitable for this titration
20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator would be most suitable for this titration? The pKa of CH3CH2NH3+ is 10.75.
Answer:
Bromocresol green, color change from pH = 4.0 to 5.6
Explanation:
The equation for the reaction is as follows:
[tex]C_2H_5NH_{2(aq)[/tex] + [tex]H^+_{(aq)[/tex] ⇄ [tex]C_2H_5NH_{3(aq)}^+[/tex]
Given that concentration of [tex]C_2H_5NH_{2(aq)[/tex] = 10%
i.e 10 g of [tex]C_2H_5NH_{2(aq)[/tex] in 100 ml solution
molar mass = 45.08 g/mol
number of moles = [tex]\frac{10}{45.08}[/tex]
= 0.222 mol
Molarity of [tex]C_2H_5NH_{2(aq)[/tex] = 0.222 × [tex]\frac{1000}{100}mL[/tex]
= 2.22 M
However, number of moles of [tex]C_2H_5NH_{2(aq)[/tex] in 20 mL can be determined as:
number of moles of [tex]C_2H_5NH_{2(aq)[/tex] = 20 mL × 2.22 M
= [tex]44*10^{-3} mole[/tex]
Concentration of [tex]C_2H_5NH_{2(aq)[/tex] = [tex]\frac{44*10^{-3}*1000}{20}[/tex]
= 2.22 M
Similarly, The pKa Value of [tex]C_2H_5NH_{2(aq)[/tex] is given as 10.75
pKb value will be: 14 - pKa
= 14 - 10.75
= 3.25
Finally, the pH value at equivalence point is:
pH= [tex]\frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C][/tex]
pH = [tex]\frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22][/tex]
pH = 5.21
∴
The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6.
For titration of the weak base ethylamine with a strong acid like HCl, the pH at the equivalence point will be less than 7, so an indicator that changes color in an acidic pH range such as methyl orange would be most suitable.
Explanation:The appropriate indicator in this case would depend on the equivalence point of the titration. Ethylamine, CH3CH2NH2, is a weak base, and it's being titrated with HCl, a strong acid. The reaction between them generates water, and ethylammonium chloride which functions as a weak acid. This setup indicates that the pH at the equivalence point will be less than 7.
Because the pH at the equivalence point will be in the acidic range (below 7), we should select an indicator which changes color around this pH. A good choice would be methyl orange, which changes color from red to yellow across a pH range of approximately 3.1 to 4.4.
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Cytotoxic t cells can attack target cells with which chemical weapons?
Answer:
secrete cytotoxic substance which triggers apoptosis of target cell.
Explanation:
Cytotoxic T cells have cell surface receptor which recognizes the antigen present on the receptor of target cell. This interaction initiates the process of killing of target cell.
After interaction cytotoxic t cell release cytotoxic substance called granzyme and perforin. Granzyme triggers apoptosis through the activation of caspases or by making the release of cytochrome c and activation of the apoptosome.
Perforin make pores in the cell and its action is similar to complement membrane attack complex. Therefore cytotoxic substances are released by Tc cells which trigger apoptosis of target cell.
The size of a population increases if the number of individuals added to the population is equal to the number of individuals leaving the population, is this true
Answer:
The answer to the question
The size of a population increases if the number of individuals added to the population is equal to the number of individuals leaving the population, is this true?
is No
Explanation:
The size of a population increases if the number of individuals added to the population is more than the number of individuals leaving the population
In biology, a population is the total number of organism of a particular species or group living together at a given geographical location. It is the number of inhabitants of a place.
Population growth is the increase in the number count of the specie or group in a population.
Given that ammonia is a gas at room temperature, what can you infer about the relative strengths of the intermolecular forces between ammonia molecules and between water molecules
Answer:
The answer to the question is;
The inter-molecular forces of water are stronger than those of hydrogen.
Explanation:
Ammonia is a compressible gas at room temperature with molecules free to move about and so fill up the volume of the container in which it is placed due to the weaker inter-molecular Van der Waals forces such as Keesom, Debye and London dispersion forces holding the particles of ammonia together in a given volume of the compound.
The inter-molecular forces between water molecules is hydrogen binding and dipole moments due to the strongly electronegative oxygen and hydrogen which tends to move the electrons towards the oxygen creating a charge imbalance that causes the hydrogen surrounding the water molecule to aggregate to neutralize the the charge imbalance forming the bases for the strong hydrogen bonds.
Therefore water is a liquid at room temperature while ammonia is a gas due to the difference in strength of their inter-molecular forces.
The intermolecular forces in water are stronger than the intermolecular forces in ammonia.
The intermolecular forces hold substances together in a particular state of matter. There are three states of matter;
SolidLiquid GasThe strongest degree of intermolecular interaction occurs between matter in the solid state. weaker intermolecular interactions occur in the liquid state and the weakest intermolecular interaction occurs in the gaseous state.
Since ammonia is a gas at room temperature, it has weaker intermolecular interaction between its molecules compared to molecules of water at room temperature:
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A weather balloon is inflated to a volume of 28.6 L at a pressure of 737 mmHg and a temperature of 26.8 ∘C. The balloon rises in the atmosphere to an altitude where the pressure is 385 mmHg and the temperature is -16.3 ∘C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.
Answer: Volume of the balloon at this altitude is 46.9 L
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 737 mm Hg
[tex]P_2[/tex] = final pressure of gas = 385 mm Hg
[tex]V_1[/tex] = initial volume of gas = 28.6 L
[tex]V_2[/tex] = final volume of gas = ?
[tex]T_1[/tex] = initial temperature of gas = [tex]26.8^oC=273+26.8=299.8K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]-16.3^oC=273-16.3=256.7K[/tex]
Now put all the given values in the above equation, we get:
[tex]\frac{737\times 28.6}{299.8}=\frac{385\times V_2}{256.7}[/tex]
[tex]V_2=46.9L[/tex]
Thus the volume of the balloon at this altitude is 46.9 L
How many grams of potassium fluoride can form if 4.00 grams of potassium are reacted with 3.00 grams of fluorine gas according to the reaction: 2K (s) + F2 (g) → 2KF (s)
Answer:
We can for 5.93 grams potassium fluoride
Explanation:
Step 1: Data given
Mass of potassium = 4.00 grams
Mass of fluorine = 3.00 grams
Molar mass potassium = 39.10 g/mol
Molar mass fluorine gas =38.00 g/mol
Step 2: The balanced equation
2K (s) + F2 (g) → 2KF (s)
Step 3: Calculate moles potassium
Moles potassium = 4.00 grams / 39.10 g/mol
Moles potassium = 0.102 moles
Step 4: Calculate moles F2
Moles F2 = 3.00 grams / 38.00 g/mol
Moles F2 = 0.0789 moles
Step 5: Calculate limiting reactant
Potassium is the limiting reactant. There will react 0.102 moles
Fluorine gas is in excess. There will react 0.102/ 2 = 0.051 moles
There will remain 0.0789 - 0.051 = 0.0279 moles
Step 6: Calculate moles potassium fluoride
For 2 moles potassium we need 1 mol fluorine to produce 2 moles potassium fluoride
For 0.102 moles K we need 0.102 moles KF
Step 7: Calculate mass KF
Mass KF = moles KF * molar mass KF
Mass KF = 0.102 moles * 58.10 g/mol
Mass KF = 5.93 grams
Final answer:
By calculating the moles of potassium and fluorine used in the reaction from their given masses, and considering the stoichiometry of the balanced equation, it's determined that 4.59 grams of potassium fluoride can be produced.
Explanation:
The question asks how many grams of potassium fluoride can form when 4.00 grams of potassium react with 3.00 grams of fluorine gas according to the balanced chemical equation 2K (s) + F2 (g) → 2KF (s). To solve this, we first find the molar mass of potassium (K) and fluorine (F2), which are approximately 39.10 g/mol and 38.00 g/mol, respectively. Then, we calculate the moles of K and F2 available by dividing the given masses by their respective molar masses. With potassium: 4.00 g / 39.10 g/mol = 0.102 moles of K; with fluorine: 3.00 g / 38.00 g/mol = 0.079 moles of F2. As the equation shows a 2:1 ratio, fluorine limits the reaction. Therefore, we can form 0.079 moles of KF. The molar mass of KF is approximately 58.10 g/mol (39.10 g/mol for K + 19.00 g/mol for F). Thus, the mass of KF that can form is 0.079 moles × 58.10 g/mol = 4.59 grams.
at atmoshperic pressure, a balloon contains 2.00L of nitrogen of gas. How would the volume change if the Kelvin temperature were only 75 percent of its original value
Answer: The percent change in volume will be 25 %
Explanation:
To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
where,
[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.
[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.
We are given:
[tex]V_1=2L\\T_1=T_1\\V_2=?\\T_2=75\% \text{ of }T_1=0.75\times T_1[/tex]
Putting values in above equation, we get:
[tex]\frac{2L}{T_1}=\frac{V_2}{0.75\times T_1}\\\\V_2=\frac{2\times 0.75\times T_1}{T_1}=1.5L[/tex]
Percent change of volume = [tex]\frac{\text{Change in volume}}{\text{Initial volume}}\times 100[/tex]
Percent change of volume = [tex]\frac{(2-1.5)}{2}\times 100=25\%[/tex]
Hence, the percent change in volume will be 25 %
For the following reaction at 600. K, the equilibrium constant, Kp, is 11.5. PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g) Suppose that 2.010 g of PCl5 is placed in an evacuated 555 mL bulb, which is then heated to 600. K. (a) What would be the pressure of PCl5 if it did not dissociate? WebAssign will check your answer for the correct number of significant figures. atm (b) What is the partial pressure of PCl5 at equilibrium? WebAssign will check your answer for the correct number of significant figures. atm (c) What is the total pressure in the bulb at equilibrium? WebAssign will check your answer for the correct number of significant figures. atm (d) What is the degree of dissociation of PCl5 at equilibrium?
Answer:
a) pPCl5 = 0.856 atm
b)pPCl5 = 0.0557 atm
pCl2 = pCl3 = 0.800 atm
c) Ptotal = 1.66 atm
d) 93.5
Explanation:
Step 1: Data given
Temperature = 600 K
Kp = 11.5
Mass of PCl5 = 2.010 grams
Volume of the bulb = 555 mL = 0.555 L
The bulb is heated to 600 K
Step 2: The balanced equation
PCl5(g) ⇄ PCl3(g) + Cl2(g)
Step 3:
a) pv = nrt
⇒with p = the pressure = TO BE DETERMINED
⇒with V = the volume = 0.555 L
⇒ with n =the number of moles PCl5 = 2.010 grams / 208.24 g/mol = 0.00965 moles
⇒ with R = the gas constant = 0.08206 L*atm/mol*K
⇒ with T = the temperature = 600K
p = (0.00965 *0.08206*600)/0.555
pPCl5 = 0.856 atm
b)
The initial pressures
pPCl5 = 0.856 atm
pCl2 = pCl3 = 0 atm
For 1 mol PCl5 we'll have PCl3 and 1 mol Cl2
The pressure at the equilibrium
pPCl5 = (0.856 -x) atm
pCl2 = pCl3 = x atm
Kp = pPCl3 * pCl2/pPCl5
11.5 = x*x / (0.856 - x)
11.5 = x²/(0.856- x)
x = 0.8003
pPCl5 = (0.856 -x) atm = 0.0557 atm
pCl2 = pCl3 = x atm = 0.800 atm
c) Since x = 0.8003 and PCl3 and PCl2 are x
Ptotal = 0.8003 + 0.8003 +0.0557 = 1.66 atm
d)
The degree of dissociation = (x / initial pressure PCl5)
(0.8003/0.856) * 100 = 93.5
The limiting reactant can be described as: Entry field with incorrect answer the amount actually obtained from a reaction the substance that is depleted first and stops a reaction the maximum amount that can be produced in a reaction the substance left over at the end of the reaction the ratio of the amount produced to the maximum possible
Answer:
The limiting reactant can be described as the substance that is depleted first and stops
Explanation:
Imagine you have hydrogen and oxygen to produce water.
The reaction is: 2H₂(g) + O₂(g) → 2H₂O(g)
You have 1 mol of each reactant. As you see ratio is 2:1, so the limiting reactant is the hydrogen.
You know by stoichiometry, that 2 moles of H₂ need 1 mol of O₂ to react
If I have 1 mol of H₂, I will need the half of moles of O₂, so 0.5 moles. It is ok because I have 1 mole, as I need the half, then half a mole will remain unreacted. This is what is called excess reagent,
If I make to react 1 mol of oxgen I need 2 moles of H₂. As I have 1 mol, of course I will need 2 moles but the thing is I have 1 mol.
This is the limiting reactant. I do not have enough of reactant so the reaction will happen until I complete to use it, that's why we can say that is depleted first and stops.
In a chemical reaction, when you have data of both reactants you can determine the limiting. Otherwise the excersise must tell that one ractant is in excess, to work with the limiting. Limiting reactant is the first step to work with the reaction, all the operations must be done by it. You do not use the reagent in excess
Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required?
Answer:
42 liters
Explanation:
Let the volume of 2% sulfuric acid required be X while that of 12% sulfuric acid required be Y.
According to the concentration: 0.02X + 0.12Y = 0.05(X + Y).......equation 1
According to the volume: X + Y = 60...............equation 2
Substitute equation 2 into equation 1.
0.02X + 0.12(Y) = 0.05(60)
2X + 12Y = 300
X + 6Y = 150..............................equation 3
From equation 2, Y = 60 -X...................equation 4
Substitute equation 4 into equation 3.
X + 6(60-X) = 150
X + 360 - 6X = 150
-5X = -210
X = 42
Hence, the volume of 2% sulfuric acid required is 42 liters.
an aqeous solution of oxalic acid h2c2o4 was prepared by dissolving a 0.5842g of solute in enough water to make a 100 ml solution a 10 ml aliquot of this solution was then transferred to a volumetric flask and diluted to a final volume of 250 ml?propanoic
The question is incomplete, here is the complete question:
An aqeous solution of oxalic acid was prepared by dissolving a 0.5842 g of solute in enough water to make a 100 ml solution a 10 ml aliquot of this solution was then transferred to a volumetric flask and diluted to a final volume of 250 ml. How many grams of oxalic acid are in 100. mL of the final solution?
Answer: The mass of oxalic acid in final solution is 0.0234 grams
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex] ......(1)
Given mass of oxalic acid = 0.5842 g
Molar mass of oxalic acid = 90 g/mol
Volume of solution = 100 mL
Putting values in equation 1, we get:
[tex]\text{Molarity of oxalic acid solution}=\frac{0.5842\times 1000}{90\times 100}\\\\\text{Molarity of oxalic acid solution}=0.0649M[/tex]
To calculate the molarity of the diluted solution, we use the equation:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated oxalic acid solution
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted oxalic acid solution
We are given:
[tex]M_1=0.0649M\\V_1=10mL\\M_2=?M\\V_2=250.0mL[/tex]
Putting values in above equation, we get:
[tex]0.0649\times 10=M_2\times 250.0\\\\M_2=\frac{0.0649\times 10}{250}=0.0026M[/tex]
Now, calculating the mass of glucose by using equation 1, we get:
Molarity of oxalic acid solution = 0.0026 M
Molar mass of oxalic acid = 90 g/mol
Volume of solution = 100 mL
Putting values in equation 1, we get:
[tex]0.0026=\frac{\text{Mass of oxalic acid solution}\times 1000}{90\times 100}\\\\\text{Mass of oxalic acid solution}=\frac{0.0026\times 90\times 100}{1000}=0.0234g[/tex]
Hence, the mass of oxalic acid in final solution is 0.0234 grams
The question involves dilution of an oxalic acid solution and subsequent calculation of the molarity for titration, which is a typical procedure in a college-level chemistry course.
Explanation:The student's question pertains to the dilution of an aqueous solution of oxalic acid and the calculation of molarity after dilution and titration. Starting with a mass of 0.5842g of oxalic acid dissolved to make a 100 ml solution, a 10 ml aliquot is further diluted to 250 ml. The goal is to understand the change in concentration as a result of the dilution process and to apply this understanding to various problems presented, such as titration against potassium permanganate (KMnO4).