Answer:
Explanation:
You will investigate the breakdown of starch by amylase at different pHs.
The different pHs under investigation will be produced using buffer solutions. Buffer solutions produce a particular pH, and will maintain it if other substances are added.
The amylase will break down the starch.
A series of test tubes containing a mixture of starch and amylase is set up at different pHs.
A sample is removed from the test tubes every 10 seconds to test for the presence of starch. Iodine solution will turn a blue/black colour when starch is present, so when all the starch is broken down, a blue-black colour is no longer produced. The iodine solution will remain orange-brown.
A control experiment must be set up - without the amylase - to make sure that the starch would not break down anyway, in the absence of an enzyme. The result of the control experiment must be negative - the colour must remain blue-black - for results with the enzyme to be valid.
When the starch solution is added:
Start timing immediately.Remove a sample immediately and test it with iodine solution.Sample the starch-amylase mixture continuously, for example every 10 seconds.For each pH investigated, record the time taken for the disappearance of starch, ie when the iodine solution in the spotting tile remains orange-brown.
The time taken for the disappearance of starch is not the rate of reaction.
It will give us an indication of the rate, but is the inverse of the rate - the shorter the time taken, the greater the rate of the reaction.
We can calculate the rate of the reaction by calculating \frac{1}{t}, obtaining a measure of the rate of reaction by dividing one by the time taken for the reaction to occur.
A similar experiment can be carried out to investigate the effect of temperature on amylase activity.
Set up a series of test tubes in the same way and maintain these at different temperatures using a water bath - either electrical or a heated beaker of water.
Depending on the chemical reaction under investigation, you might monitor the reaction in a different way. If investigating the effect of temperature on the breakdown of lipid by lipase, you could monitor pH change - lipids are broken down into fatty acids and glycerol. As the reaction begins, the release of fatty acids will mean that the pH will decrease.
good luck :)
Catalysts are substances that increase the rate of reaction but can be recovered unchanged at the end of the reaction. Catalysts can be classified as either homogeneous (same state as reactants) or heterogeneous (different state than reactants).
Platinum is used to catalyze the hydrogenation of ethylene:
H2(g)+CH2CH2(g)−⟶Pt(s)CH3CH3(g)
Chlorofluorocarbons (CFCs) catalyze the conversion of ozone (O3) to oxygen gas (O2):
2O3(g)−⟶CFC(g)3O2(g)
Magnesium catalyzes the disproportionation of hydrogen peroxide to produce water and oxygen:
2H2O2(aq)−⟶Mg(s)2H2O(l)+O2(g)
What type of catalysts are platinum, CFCs, and magnesium under these conditions?
Answer:
- Platinum acts as a heterogeneous catalyst in the hydrogenation of ethylene.
- CFCs act as homogeneous catalysts in the conversion of ozone to oxygen gas.
- Magnesium acts as a heterogeneous catalyst in the disproportionantion of hydrogen peroxide.
Explanation:
Hello,
For the given reactions, considering the definition of homogeneous and heterogeneous catalyst, we can identify that is each catalyst behave as follows:
- Platinum acts as a heterogeneous catalyst in the hydrogenation of ethylene as all the reactants are gaseous but it remains solid.
- CFCs act as homogeneous catalysts in the conversion of ozone to oxygen gas as it remains gaseous as well as both ozone and oxygen.
- Magnesium acts as a heterogeneous catalyst in the disproportionantion of hydrogen peroxide as it is solid whereas the other species are aqueous, liquid and gaseous
Best regards.
Consider the following reaction:
2NO(g)+O2(g)→2NO2(g)
Estimate ΔG∘ for this reaction at each of the following temperatures and predict whether or not the reaction will be spontaneous. (Assume that ΔH∘ and ΔS∘ do not change too much within the give temperature range.) I need to find the temperature are 298K and 702K. For 298K It is simple because at standard temperature
ΔG∘ = DG(products)- DG(reactants).
Answer:
A. [tex]\mathbf{\Delta G^0 = -72.6 \ kJ/mol}[/tex] ; as such the reaction is said to be spontaneous since the value of [tex]\mathbf{\Delta G^0 }[/tex] is negative.
B. [tex]\mathbf{\Delta G^0_{702 \ K} = -13.29 \ kJ/mol.K}}[/tex] and the reaction is spontaneous
Explanation:
The equation for this chemical reaction is :
[tex]2NO_{(g)} +O_{2(g)} \to 2NO_{2(g)}[/tex]
Using the following relation to calculate [tex]\Delta G^0[/tex];
[tex]\Delta G^0 = [2(\Delta G^0_{NO_{2(g)}}] - [1(\Delta G^0_{O_{2(g)}})+ 2(\Delta G^0_{NO_{g}})][/tex]
At 298 K; the standard Gibbs Free Energy for the formation are as follows:
[tex]\Delta G^0_{NO_{2(g)}} = 51.2 \ kJ/mol[/tex]
[tex]\Delta G^0_{O_{2(g)}} = 0[/tex]
[tex]\Delta G^0_{NO_{g}}= 87.6 \ kJ/mol[/tex]
Replacing them into the above equation;
[tex]\Delta G^0 = [2(51.2 \ kJ/mol}] - [1(0)+ 2(87.6 \ kJ/mol})][/tex]
[tex]\Delta G^0 = [102.4 \ kJ/mol}] - [175.2 \ kJ/mol})][/tex]
[tex]\mathbf{\Delta G^0 = -72.6 \ kJ/mol}[/tex]
Thus; [tex]\mathbf{\Delta G^0 = -72.6 \ kJ/mol}[/tex] ; as such the reaction is said to be spontaneous since the value of [tex]\mathbf{\Delta G^0 }[/tex] is negative.
B.
Using the same above chemical equation;
The relation used for calculating [tex]\mathbf{\Delta G^0}[/tex] of the reaction when the temperature is 702 K is:
[tex]\Delta G^0_{702 \ K} = \Delta H^0_{xn} - T \Delta S^0_{rxn}[/tex]
where;
[tex]\Delta G^0_{702 \ K} =[/tex] Gibbs free energy of the reaction at 702 K
[tex]\Delta H^0_{xn}[/tex] = standard enthalpy of the reaction = -116.2 kJ/mol
[tex]\Delta S^0_{rxn}[/tex] = standard entropy of the reaction = -146.6 J/mol/K
Temperature T = 702 K
[tex]\Delta G^0_{702 \ K} = -1162. \ kJ/mol - 702 \ K ( -146.6 \ J/mol. K (\dfrac{1 \ kJ }{1000 \ J})[/tex]
[tex]\Delta G^0_{702 \ K} = -1162. \ kJ/mol - 702 \ K ( 0.1466 \ kJ/mol.K})[/tex]
[tex]\Delta G^0_{702 \ K} = -13.2868 \ kJ/mol.K}[/tex]
[tex]\mathbf{\Delta G^0_{702 \ K} = -13.29 \ kJ/mol.K}}[/tex]
Thus [tex]\mathbf{\Delta G^0_{702 \ K} = -13.29 \ kJ/mol.K}}[/tex] and the reaction is spontaneous
Which of these scientists diagnosed smallpox and measles?
A. Nicolaus Copernicus
B. Al-Razi
C. Archimedes
D. Rosalind Franklin
Answer:
B
Explanation:
A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the final number of microstates is 0.842 times that of the initial number of microstates
Answer: -2.373 x 10^-24J/K(particles
Explanation: Entropy is defined as the degree of randomness of a system which is a function of the state of a system and depends on the number of the random microstates present.
The entropy change for a particle in a system depends on the initial and final states of a system and is given by Boltzmann equation as
S = k ln(W) .
where S =Entropy
K IS Boltzmann constant ==1.38 x 10 ^-23J/K
W is the number of microstates available to the system.
The change in entropy is given as
S2 -S1 = kln W2 - klnW1
dS = k ln (W2/W1)
where w1 and w2 are initial and final microstates
from the question, W2(final) = 0.842 x W1(initial), so:
= 1.38*10-23 ln (0.842)
=1.38*10-23 x -0.1719
= -2.373 x 10^-24J/K(particles)
A chemistry student is given 600. mL of a clear aqueous solution at 37° C. He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 21° C. At that point, the student sees that a precipitate has formed. He pours off the remaining liquid solution, throws away the precipitate, and evaporates the water from the remaining liquid solution under vacuum. More precipitate forms. The student washes, dries and weighs the additional precipitate. It weighs 0.084 kg. Using only the information from above, can you calculate the solubility of X at 21° C?If yes, calculate it. Be sure your answer has a unit symbol and the right number of significant figures.
Answer:
The solubility is [tex]S = 0.0014 \ g[/tex]
Explanation:
From the question we are told that
The volume of the solution is [tex]V = 600 mL[/tex]
The initial temperature is [tex]T_i = 37 ^oC[/tex]
The final temperature is [tex]T_f = 21^oC[/tex]
The additional precipitate is [tex]m = 0.084 \ kg = 84 \ g[/tex]
Yes because the solubility of the substance X is the amount of X needed to saturate a unit volume of the solvent (for solubility of a solute to be calculated the solute must be able to saturate the solvent)
now we see that the substance X saturated the solvent because a precipitate was formed which the student threw away
The solubility at 21 ° C is mathematically represented as
[tex]S = \frac{m}{m_w * 100 g \ of water }[/tex]
Mass of water([tex]m_w[/tex]) in the solution is mathematically represented as
[tex]m_w = V * \rho_w[/tex]
Where [tex]\rho = 1 \frac{g}{mL}[/tex]
So
[tex]m_w =600 * 1[/tex]
[tex]m_w =600g[/tex]
So
[tex]S = \frac{84}{600 * 100 g \ of water }[/tex]
[tex]S = 0.0014 \ g[/tex]
Which diagram represents this molecule?
Answer:
C
Explanation:
The molecule has 8 carbon atoms joined by 7 C-C bonds.
The first two diagrams show 6 carbon atoms, not 8.
The last two diagrams show line segments representing C-C bonds. Only choice C shows 7 such segments.
The appropriate choice is C.
Answer:
C.
Explanation:
One of the radioactive isotopes used in chemical and medical research is sulfur-35, which has a half-life of 87 days. How long would it take for 0.25 g to remain of a 1.00 g sample of sulfur-35
Answer:
Time taken = 174 days
Explanation:
Half life is the time take for a subsrtance taken to decay to half of it's origial or initial concentration.
In this probel, the haf life is 87 days, this means that after evry 87 days, the concentration or mass of sulfur-35 decreases by half.
If the starting mass is 1.00g, then we have;
1.00g --> 0.5g (First Half life)
0.5g --> 0.25g (Second half life)
This means that sulphur-335 would undergo two half lives for 0.25g to remain.
Total time taken = Number of half lives * Half life
Time taken = 2 * 87
Time taken = 174 days
what is the equation for "acid dissociation constant" of "carbonic acid"
Answer:
H2CO3 = 2H+ + CO3-
Explanation:
It is simply what carbonic acid breaks down into when placed in water. Since carbonic acid is made up of H and CO3, these are the products.
3. Crystalline structural unit of barium metal is a body-centered cubic cell. The edge length of the unit cell is 5.02x10-8 cm. The density of the metal is 5.30 g/cm3. Assume that 68% of the unit cell is occupied by Ba atoms. The molar mass of barium is 137.3 g/mol. Using this information, calculate Avogadro’s number. Show your calculation procedure that allows you to derive Avogadro’s number. Your answer must show six digits after the decimal point (i.e., 6.pppx1023) that is not necessarily the same as the known value. By showing your calculation-result down to six digits after the decimal point, you showcase that you did calculate the number, instead of simply adopting the known Avogadro’s number available in open resources.
Answer:
The Avogadro's number is [tex]N_A = 6.02289 *10^{23}[/tex]
Explanation:
From the question we are told that
The edge length is [tex]L = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100} = 5.02 * 10^{-10}[/tex]
The density of the metal is [tex]\rho = 5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3} * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3[/tex]
The molar mass of Ba is [tex]Z = 137.3 \ g/mol = \frac{137.3}{1000} = 0.1373 \ kg / mol[/tex]
Generally the volume of a unit cell is
[tex]V = L^3[/tex]
substituting value
[tex]V = [5.02 *10^{-10}]^3[/tex]
[tex]V = 1.265*10^{-28}\ m^3[/tex]
From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be (BCC),
The volume of barium atom is
[tex]V_a = \frac{V}{2} * 0.68[/tex]
substituting value
[tex]V_a = \frac{ 1.265*10^{-28}}{2} * 0.68[/tex]
[tex]V_a = 4.301 *10^{-29} \ m^3[/tex]
The Molar mass of barium is mathematically represented as
[tex]Z = N_A V_a * \rho[/tex]
Where [tex]N_A[/tex] is the Avogadro's number
So
[tex]N_A = \frac{ Z}{ V_a * \rho}[/tex]
substituting value
[tex]N_A = \frac{ 0.1373}{ 4.301*10^{-29} * 5.3*10^{3}}[/tex]
[tex]N_A = 6.02289 *10^{23}[/tex]
Why is tape attracted to my skin? Give explanation
Answer:
Since the tape has extra electrons, it has a negative charge. When you move your finger close to the tape, electrons in your skin are repelled and move away. This makes the skin on your finger tip have a slight positive charge. Since positive and negative attract, the tape moves toward your finger.
Based on this information what is the most likely reason for refrigerating most foods reduce the rate at which they spoil
Answer: The lower temperature reduces molecule speeds, reducing the number of effective collisions.
Explanation:
At 25 oC, the rate constant for the first-order decomposition of a pesticide solution is 6.40 x 10-3 min-1. If the starting concentration of pesticide is 0.0314 M, what concentration will remain after 62.0 minutes at 25 oC? 3.12 x 10-2 M 47.4 M 2.11 x 10-2 M 4.67 x 10-2 M 8.72 M
Answer:
[tex]2.11\ * 10^{-2}[/tex] is the correct answer to the given question.
Explanation:
Given k=6.40 x 10-3 min-1.
According to the first order reaction .
The concentration of time can be written as
[tex][\ A\ ]\ = \ [\ A_{0}\ ] * e \ ^\ {-kt}[/tex]
Here [tex][\ A\ ]_{0}[/tex] = Initial concentration.
So [tex][\ A\ ]_{0}= 0.0314 M[/tex]
Putting this value into the above equation.
[tex]0.0314 \ *\ e^{6.40 x 10^{-3} \ * \ 62.0 }[/tex]
=0.211 M
This can be written as
[tex]=\ 2.11 *\ 10^{-2}[/tex]
Ethanol, , boils at 78.29 °C. How much energy, in joules, is required to raise the temperature of 2.00 kg of ethanol from 26.0 °C to the boiling point and then to change the liquid to vapor at that temperature? (The specific heat capacity of liquid ethanol is 2.44 J/g ∙ K, and its enthalpy of vaporization is 855 J/g.)
Answer:
THE HEAT REQUIRED TO CHANGE 2 KG OF ETHANOL FROM 26 °C TO THE BOILING POINT AND TO VAPOR AT THAT TEMPERATURE IS 1965.175 KJ.
Explanation:
Boiling point of ethanol = 78.29 °C = 78.29 + 273 K = 351.29 K
Mass = 2 kg = 2000 g
Final temp. = 26.0 °C = 26 + 273 K= 299 K
Change in temperature = (78.29 - 26) °C = 52.29 °C
1. Heat required to raise the temperature from 26 °C to the boiling point?
Heat = mass * specific heat * change in temperature
Heat = 2000 * 2.44 * 52.29
Heat = 255 175.2 J
2. Heat required to change the liquid to vapor at that temperature?
Heat = mass * enthalphy of vaporization
Heat = 2000 * 855
Heat =1 710000 J
The total heat required to raise the temperature of 2 kg of ethanol from 26 °C to the boiling point and then to change the liquid to vapor at that temperature will be:
Heat = mcT + m Lv
Heat = 255 175.2 J + 1710000 J
Heat = 1965175.2 J
Heat = 1965.175 kJ of heat.
Calculate the percent saturated fat in the total fat in butter
Give two examples (i.e. list 2 elements that are examples) of: a. an atom with a half-filled subshell b. an atom with a completely filled outer shell c. an atom with its outer electrons occupying a half-filled subshell and a filled subshell
Answer:
an atom with a half-filled subshell - hydrogen
an atom with a completely filled outer shell - argon
an atom with its outer electrons occupying a half-filled subshell and a filled subshell- copper
Explanation:
The outermost shell or the valence shell of the atom is the last shell in the atom. Chemical reactions occur at this outer most shell. The number of electrons on the outermost shell of an atom determines the group to which it belongs in the periodic table as well as its chemical properties.
Hydrogen has a half filled 1s sublevel. Only one electron is present in this sublevel.
Let us consider argon
1s2 2s2 2p6 3s2 3p6
The outermost ns and np levels are completely filled. Thus the outermost shell is completely filled.
In the last case; let us look at the electronic configuration of nitrogen;
1s2 2s2 2p3
The outermost 2p subshell is exactly half filled while the 2s sublevel is fully filled. The outermost shell of nitrogen is made up of 2s2 and 2p3 sublevels.
Methods in electrochemistry that can be used for the separation of proteins and enzymes?
Answer:
Some of the methods in Methods in electrochemistry that can be used for the separation of proteins and enzymes are as follows:
Redox transformations: In this method enzymes or proteins would be adsorbed on the electrode surface and facilitates direct electron transfer that causes denaturation and loss of their electrochemical activities and bioactivities. It is widely used in biosensors and biofuel cells.
Protein electrophoresis: In this process proteins are seperated by placing them in a gel matrix in the presence of an electrical field. In this method a negative charge is applied so that proteins move towards a positive charge.
The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Determine how much heat (in kJ) is produced by the decomposition of 1.71 mol of H2O2 under standard conditions.
Answer:
[tex]Q=-361.56kJ[/tex]
Explanation:
Hello,
In this case, the decomposition of hydrogen peroxide is given by:
[tex]2H_2O_2\rightarrow 2H_2O+O_2[/tex]
Which occurs in gaseous phase, therefore the enthalpy of reaction is:
[tex]\Delta _rH=2\Delta _fH_{H_2O}-2\Delta _fH_{H_2O_2}[/tex]
Oxygen is not included as it is a pure element. The enthalpies of formation for both hydrogen peroxide and water are -136.11 and -241.83 kJ/mol respectively, so we compute the enthalpy of reaction:
[tex]\Delta _rH=2(-241.83kJ/mol)-2(-136.11kJ/mol)=-211.44kJ/mol[/tex]
Then, the total heat that is released for 1.71 mol of hydrogen peroxide is:
[tex]Q=n*\Delta _rH=1.71mol*-211.44kJ/mol\\\\Q=-361.56kJ[/tex]
Whose sign means a released heat.
Regards.
When 1.550 gg of liquid hexane (C6H14)(C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 ∘C∘C to 38.13 ∘C∘C. Find ΔErxnΔErxn for the reaction in kJ/molkJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.73 kJ/∘CkJ/∘C.
Answer:
ΔErxn[tex]= -3.90*10^3KJ[/tex]
Explanation:
Given from the question
T1 = 25.87∘C
T2= 38.13∘C.
C= 5.73Kj/C
CHECK THE ATTACHMENT FOR DETAILED EXPLATION
At that volume is measured to be 755 mm of Hg. If the lungs are compressed to a newA healthy male adult has a lung capacity around 6.00 liters. The pressure in the lungs volume of 3.81 liters, what would be the new pressure in the lungs? What would happen to the air in the lungs?
Answer:
1188.976 mmHg
Explanation:
Initial pressure P1= 755 mmHg
Initial volume V1 = 6.00 litres
Final volume V2 = 3.81 litres
Final pressure P2= the unknown
Now applying Boyle's Law,we have;
P1V1 = P2V2
Since P2 is the unknown then it has to be made the subject of the formula.
P2=P1V1/ V2
P2= 755 × 6.00/ 3.81
P2= 1188.976 mmHg
Therefore, the new pressure is; 1188.976 mmHg
What is the probability that an offspring will have a
heterozygous genotype? |
Answer:
25,50,25
Explanation:
Consider the following system at equilibrium: P(aq)+Q(aq)⇌3R(aq) Classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the net reaction. Drag the appropriate items to their respective bins.
Items:1) Increase [P]2) Increase [Q]3) Increase [R]4) Decrease [P]5) Decrease [Q]6) Decrease [R]7) Triple [P] and reduce [Q] to one third8) Triple both [Q] and [R]
Explanation:
P(aq)+Q(aq)⇌3R(aq)
This problem involves applying LeChatelier's principle.
LeChatelier's principle states that whenever a system in equilibrium is disturbed, the equilibrium position would change in order to annul that change.
1) Increase [P]
This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.
2) Increase [Q]
This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.
3) Increase [R]
This would cause the equlibrium position to shift to the left. This is because more products have been formed, to annul that change more reactants have to be formed.
4) Decrease [P]
This would cause the equlibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.
5) Decrease [Q]
This would cause the equilibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.
6) Decrease [R]
This would cause the equilibrium position to shift to the right. This is because there are now less products, to annul that change more products have to be formed.
7) Triple [P] and reduce [Q] to one third
No shift in the direction of the net reaction because both changes cancels each other.
8) Triple both [Q] and [R]
No shift in the direction of the net reaction because both changes cancels each other.
which statements describe how chemical formulas, such as H2O, represent compounds?
Answer:
2 Hydrogen One oxygen
Explanation:
For some hypothetical metal the equilibrium number of vacancies at 750°C is 2.8 × 1024 m−3. If the density and atomic weight of this metal are 5.60 g/cm3 and 65.6 g/mol, respectively, calculate the fraction of vacancies for this metal at 750°C.
Answer:
The correct answer is 5.447 × 10⁻⁵ vacancies per atom.
Explanation:
Based on the given question, the at 750 degree C the number of vacancies or Nv is 2.8 × 10²⁴ m⁻³. The density of the metal is 5.60 g/cm³ or 5.60 × 10⁶ g/m³. The atomic weight of the metal given is 65.6 gram per mole. In order to determine the fraction of vacancies, the formula to be used is,
Fv = Nv/N------ (i)
Here Nv is the number of vacancies and N is the number of atomic sites per unit volume. To find N, the formula to be used is,
N = NA×P/A, here NA is the Avogadro's number, which is equivalent to 6.022 × 10²³ atoms per mol, P is the density and A is the atomic weight. Now putting the values we get,
N = 6.022 × 10²³ atoms/mol × 5.60 × 10⁶ g/m³ / 65.6 g/mol
N = 5.14073 × 10²⁸ atoms/m³
Now putting the values of Nv and N in the equation (i) we get,
Fv = 2.8 × 10²⁴ m⁻³ / 5.14073 × 10²⁸ atoms/m^3
Fv = 5.44669 × 10⁻⁵ vacancies per atom or 5.447 × 10⁻⁵ vacancies/atom.
Acetonitrile, CH3CN, is a polar organic solvent that dissolves many solutes, including many salts. The density of a 1.80 M acetonitrile solution of LiBr is 0.826 g/mL. Calculate the concentration of the solution in units of (a) molality; (b) mole fraction of LiBr; (c) mass percentage of CH3CN.
Answer:
(a) [tex]m=2.69m[/tex]
(b) [tex]x_{LiBr}=0.099[/tex]
(c) [tex]\% LiBr=18.9\%[/tex]
Explanation:
Hello,
In this case, given the molality in mol/L, we can compute the required units of concentration assuming a 1-L solution of acetonitrile and lithium bromide that has 1.80 moles of lithium bromide:
(a) For the molality, we first compute the grams of lithium bromide in 1.80 moles by using its molar mass:
[tex]m_{LiBr}=1.80mol*\frac{86.845 g}{1mol}=156.32g[/tex]
Next, we compute the mass of the solution:
[tex]m_{solution}=1L*0.826\frac{g}{mL}*\frac{1000mL}{1L}=826g[/tex]
Then, the mass of the solvent (acetonitrile) in kg:
[tex]m_{solvent}=(826g-156.32g)*\frac{1kg}{1000g}=0.670kg[/tex]
Finally, the molality:
[tex]m=\frac{1.80mol}{0.670kg} \\\\m=2.69m[/tex]
(b) For the mole fraction, we first compute the moles of solvent (acetonitrile):
[tex]n_{solvent}=669.68g*\frac{1mol}{41.05 g} =16.31mol[/tex]
Then, the mole fraction of lithium bromide:
[tex]x_{LiBr}=\frac{1.80mol}{1.80mol+16.31mol}\\ \\x_{LiBr}=0.099[/tex]
(c) Finally, the mass percentage with the previously computed masses:
[tex]\% LiBr=\frac{156.32g}{826g}*100\%\\ \\\% LiBr=18.9\%[/tex]
Regards.
A Carbon-10 nucleus has 6 protons and 4 neutrons. Through radioactive beta decay, it turns into a Boron-10 nucleus, with 5 protons and 5 neutrons, plus another particle. What kind of additional particle, if any, is produced during this decay
Answer:
No additional particle was produced during the decay.
Explanation:
The equation of decay is given as;
¹⁰₆C + ⁰₋₁ e → ¹⁰₅B + x
To identify x, we have to calculate its atomic and mass number.
In the reactants side;
Atomic Number = 6 + (-1) = 5
Mass number = 10 + 0 = 10
In the products side;
Atomic Number = 5 + x
Mass Number = 10 + x
Generally, reactant = product
Atomic Number;
5 = 5 + x
x = 5 - 5 = 0
Mass Number;
10 = 10 + x
x = 10 - 10 = 0
This means no additional particle was produced during the decay.
iron oxide + oxygen equals to ?
Answer:
It's ferric oxide Fe2O3
Explanation:
I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me plz...
WILL GIVE BRAINLIEST!!!!! will give brainliest!!!!! ---------Write the molecular equation, complete ionic equation and net ionic equation of Barium nitrate reacting with potassium carbonate.
Answer:
Molecular:
Ba(NO3)2 + K2CO3 -> BaCO3 + 2KNO3
Complete ionic:
Ba2+ + 2NO3- + 2K+ + CO3 2- -> BaCO3 + 2K+ + 2NO3-
Net ionic:
Ba2+ + CO3 2- - > BaCO3
Explanation:
Molecular consists of all species reacting.
Complete ionic consists of all ionic species (ex. K+, NO3-) separated, as well as any compounds that didn't dissociate into ions (BaCO3 doesn't dissolve).
Net ionic doesn't include spectator ions (in this case, nitrate and potassium) and only species that aren't present on both sides of the arrow (barium and carbonate become a solid precipitate, so the ions aren't present as products, making them appear in the net ionic equation).
1.Draw the born-Haber lattice energy cycle for sodium chloride. Explain the concept of resonance using the nitrate ion structure.
Answer:
1. Born Haber cycle is used to calculate enthalpy of formation of an ionic solid
2. Resonance structures are used to represent the bonding in some chemical species.
Explanation:
The Born–Haber cycle is a method popularly known in chemistry used in computing enthalpy. The enthalpy of formation of an ionic solid cannot be measured directly. The lattice enthalpy refers to the enthalpy change involved in the formation of an ionic compound from gaseous ions the process is exothermic process. A Born–Haber cycle works on the principle of Hess's law. It can be used to calculate the lattice enthalpy by comparing the standard enthalpy change of formation of the ionic compound from the elements to the enthalpy required to make gaseous ions from the elements.
Resonance is an idea introduced by Linus Pauling to explain chemical bonding from the valence bond perspective. The idea of resonance affords us the opportunity to describe the bonding in certain molecules by combining several structures called chemical or canonical structures. The real structure of the specie lie somewhere between the structures indicated by the resonance structures. The resonance structures of the nitrate ion are shown in the image attached.
(a) Titration curve for the titration of 5.00 mL 0.010 M NaOH(aq) with 0.005 M HCl(aq), indicating the pH of the initial and final solutions and the pH at the stoichiometric point.
What volume of HCl has been added at
(b) the stoichiometric point
(c) the halfway point of the titration?
Answer:
AT STOICHIOMETRIC POINT, THE VOLUME OF ACID ADDED IS 0.01 L
AT HALF-WAY POINT, THE VOLUME OF ACID IS 0.0050 L
Explanation:
In solving titration problems, you must remember this formula;
MaVa = MbVb
Since M a= 0.005 M
Mb = 0.010 M
Vb = 5 mL = 5 /1000 = 0.005 L
Va = unknown.
Solving for Va, we have:
Va = MbVb / Ma
Va = 0.010 * 0.005 / 0.005
Va = 0.01 L
So therefore, the volume of acid added at:
1. the stoichiometric point is 0.01 L
2. half-way point of titration is 0.01 /2 = 0.0050 L
For the pH:
Since HCl is a strong acid, it dissociate into {H30}+ ion.
First calculate the number of moles of hydronium ion
number of mole = concentration of hydronium ion {H30}+ * Volume
n = 0.005 * 0.01 = 0.00005 moles
A. At initial point of the titration, the volume of base added is 0 L
{H30]+ = n(H+)/ V = 0.00005 / 0.01 = 0.005 M
pH = - log {0.005}
pH = 2.3
B. At the final point, since the volumes and concentrations of acid and base are the same, the pH is equal to 7.
n(H+) = n(OH^-)
pH = 7
Which of the following is named using the unmodified element name and adding the word "ion"? Select the correct answer below:
a. simple cations (monatomic cations of elements of only one possible charge)
b. simple anions (monatomic anions of elements of only one possible charge)
c. simple protons
d. simple neutrons
Answer:
simple cations (monatomic cations of elements of only one possible charge)
Explanation:
Simple cations (monatomic cations of elements of only one possible charge) are named using the unmodified element name and adding the word "ion"
For example, the Na+ is named the sodium ion.
An atom or molecule with a net electric charge as a result of the loss or gain of one or more electrons is known as an ion.
