Answer:
They blow away from poles to the equator.
Explanation:
Hello,
In this case, we must take into account that global wind systems are formed by the constant increase in the temperature of the Earth’s surface. Thus, they drive the oceans’ surface currents. In such a way, we can say wind is the basic movement of air from an area of higher pressure to an area of lower pressure, for that reason they blow away from the poles to the equator.
Best regards.
The statement that describes the global winds is they travel over short distances.
What is winds?
Wind is a pattern or type of the movement of the natural air or any other composition of gases over to the relative position of the planet's surface.
Global winds are those winds which can travel in a straight path and originated due to global convention currents. Global winds always move from west to east direction and travels short distances only.
Hence, option (2) is correct.
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Determine the half-life of a nuclide that loses 38.0% of its mass in 407 hours. Determine the half-life of a nuclide that loses 38.0% of its mass in 407 hours. 204 hours 568 hour 590 hours 291 hours 281 hours
Answer:
Use 62% - the equation is for the amount present at a given time. 0.62 = (1) e-kt -> ln(0.62)=-kt -> k = -ln(0.62)/t. I get k = .00117 hr-1 t(half) = 0.693/k = 590 hr.
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The half-life is the time at which the substance's concentration is reduced by half of its initial amount. The half-life of a nuclide that lost its 38.0% mass is 590 hr. Thus, option C is correct.
What is half-life?Half-life is the time required by a substance to get reduced to half of its initial concentration. The half-life of the substance can be determined by the rate constant.
Given,
The initial quantity of substance (A₀) = 100
Remaining quantity (At) = 10 - 38 = 62
Time elapse (t) = 407 hours
The rate constant (k) is calculated as:
ln (At ÷ A₀) = - kt
ln (62 ÷ 100) ÷ 407 hour = - k
-0.47803580094 ÷ 407 = - k
k = 0.00117453513
Now, half-life from rate constant (k) is calculated as:
[tex]\rm t ^{\frac{1}{2}}[/tex] = 0.693 ÷ k
[tex]\rm t ^{\frac{1}{2}}[/tex] = 0.693 ÷ 0.00117453513
[tex]\rm t ^{\frac{1}{2}}[/tex] = 590 hours
Therefore, option C. 590 hours is the half-life of the substance.
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At 850 K, the equilibrium constant for the reaction
2SO2(g)+O2(g)↽−−⇀2SO3(g)
is Kc=15. If the given concentrations of the three gases are mixed, predict in which direction the net reaction will proceed toward equilibrium.
Left No net reaction Right
Answer:
Answers are in the explanation.
Explanation:
Given concentrations are:
SO₂ = 0.20M O₂ = 0.60M SO₃ = 0.60MSO₂ = 0.14M O₂ = 0.10M SO₃ = 0.40M And SO₂ = 0.90M O₂ = 0.50M SO₃ = 0.10MIn the reaction:
2SO₂(g) + O₂(g) ⇄ 2SO₃(g)
Kc is defined as:
Kc = 15 = [SO₃]² / [O₂] [SO₂]²
Where concentrations of each species are equilbrium concentrations.
Also, you can define Q (Reaction quotient) as:
Q = [SO₃]² / [O₂] [SO₂]²
Where concentrations of each species are ACTUAL concentrations.
If Q > Kc, the reaction will shift to the left until Q = Kc;
If Q < Kc, the reaction will shift to the right until Q = Kc
If Q = Kc, there is no net reaction because reaction would be en equilibrium.
Replacing with given concentrations:
Q = [0.60M]² / [0.60M] [0.20M]² = 15; Q = Kc → No net reactionQ = [0.40M]² / [0.10M] [0.14M]² = 82; Q > Kc, → Reaction will shift to the leftQ = [0.10M]² / [0.50M] [0.90M]² = 0.015; Q < Kc → Reaction will shift to the right
Ammonia is oxidized with air to form nitric oxide in the first step of the production of nitric acid. Two principal gas-phase reactions occur:
Answer:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O
2NO(g) + O₂(g) → 2 NO₂
Explanation:
First of all, we need to consider the reaction for production of ammonia. In this reaction we have as reactants, nitrogen and hydroge.
3H₂ (g) + N₂(g) → 2NH₃ (g)
Afterwards, ammonia reacts to oxygen, to produce NO and H₂O
The equation for the process will be:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O
Then, we take the nitric oxide to make it react, to produce NO₂, in order to produce nitric acid, for the final reaction:
2NO(g) + O₂(g) → 2 NO₂
3NO₂(g) + H₂O(g) → 2 HNO₃ (g) + NO(g)
There are parts of a standing wave that do not move at all. These parts are called ___________.
Answer:
They are called nodes.
Explanation:
Answer:
i guess this is the ans nodes
hope this helps
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Phosphorus pentafluoride, PF5, acts as a __________ during the formation of the anion PF−6. Select the correct answer below: A. Lewis acid B. Lewis base C. catalyst D. drying agent
Answer:
Lewis acid
Explanation:
In chemistry, a Lewis acid is any chemical specie that accepts a lone pair of electrons while a Lewis base is any chemical specie that donates a lone pair of electrons.
If we look at the formation of PF6^-, the process is as follows;
PF5 + F^- -----> PF6^-
We can see that PF5 accepted a lone pair of electrons from F^- making PF5 a lewis acid according to our definition above.
Hence in the formation of PF6^-, PF5 acts a Lewis acid.
Zinc bromide is considered which of the following?
A) molecular compound
B) atomic element
C) molecular element
D) ionic compound
Answer:
D
Explanation:
soluble in water and acidicSolid iron(II) oxide reacts with oxygen gas to produce solid iron(III) oxide. Balance the equation for this reaction (in lowest multiple integers). Write the unbalanced equation for this reaction.
Answer
Hello
I think the reaction is like this FeO+OFe²O³
And the balance reaction is 2Fe+OFe²O³
Explanation:
At first we should find sth that has more atoms than the other then for example we realized that we have two atoms of Fe in Fe²O³ then put 2 before FeO and now we have 2 atoms of Fe in right side and 2 atoms of Fe in left then Oxygen in FeO change to 2 atoms of Oxygen and we have an other one in right side that they become 3 atoms of Oxygen and now we have 3 atoms of Oxygen in both right and left side.
Finally our reaction balanced.
Good luck
What is it’s molecular formula for C5H4 if it’s molar mass is 128.17g/mol
✔ C5H4 has a molecular molar mass of :
M(C5H4) = 5 x M(C) + 4 x M(H)
M(C5H4) = 5 x 12 + 4 x 1 M(C5H4) = 60 + 4 M(C5H4) = 64 g/mol✔ The molecular mass of C5H4 is therefore 64 g/mol.
But, 128/64 = 2 This is double the molar mass of C5H4, this molecule has the formula 2C5H4.Answer:
C10H8
Explanation:
I clicked on that answer, and it is correct.
What is the purpose of reacting 2.0mL of HNO3 with 2.0 mL of H2SO4 in a separate test tube, prior to adding it to the solution containing the substrate
The question is incomplete, the complete question is;
What is the purpose of reacting 2.0mL of HNO3 with 2.0 mL of H2SO4 in a separate test tube, prior to adding it to the solution containing the substrate? more than one answer is possible
A) The release of a water molecule that acts as an electrophile in the reaction with methyl benzoate.
B) The formation of nitronium ion, which acts an electrophile in the reaction with methylbenzoate.
C)The formation of bisulfate (hydrogen sulfate), which acts as an electrophile in the reaction with methylbenzoate.
D)The release of a water molecule that acts as a nucleophile in the reaction with methyl benzoate.
Answer:
B) The formation of nitronium ion, which acts an electrophile in the reaction with methylbenzoate.
Explanation:
The benzene ring is known to be stable hence it can only undergo a substitution reaction with the aromatic ring still intact. When the substitution reaction involves an electrophile we refer to the process as electrophillic aromatic substitution. Electrophilic aromatic substitution is a useful synthetic route for many organic compounds.
In the electrophilic substitution of methyl benzoate using the 1:1 volume ratio mixture of H2SO4/HNO3, the nitronium ion (NO2+) is the electrophile generated in the test tube. It is this NO2+ that now reacts with the methyl benzoate to yield the reaction product.
A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. what volume will the dry hydrogen gas occupy at 20°C and 760 torr?
Answer:
V2 = 17371.43ml
Explanation:
We use Boyles laws
since temperature is constant
P1V1=P2V2
760 x 400 = 17.5 x V2
304000 = 17.5 x V2
V2 = 304000/17.5
V2 = 17371.43ml
The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at 760 torrs will be 18 ml.
What is vapor pressure?
The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.
The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -
P1 V1 / T1 = P2 V2 / T1
here, P = pressure
T = temperature
V = volume
substituting the value in the equation,
400 ×760 / 20 = 17.5× V / 20
V = 400× 760 / 20 × 17.5 / 20
V = 18 ml
Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.
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What's the difference between velocity time graph and distance time graph
Explanation:
Hi there!
I attached a photo of a unit summary that states the difference between s-t and v-t graph.
Hope this helps ;) ❤❤❤
If a 275 mL gas container had pressure of 732.6 mm Hg at -28°C and the gas was condensed into a liquid with a mass of 1.95 g, what is the molar mass of the gas?
Answer:
THE MOLAR MASS OF THE GAS IS 147.78 G/MOLE
Explanation:
Using PV = nRT
n = Mass / molar mass
P = 732.6 mmHg = 1 atm = 760 mmHg
So therefore 732.6 mmHg will be equal to 732.6 / 760 = 0.964 atm
P = 0.964 atm
V = 275 mL = 275 *10 ^-3 L
R = 0.082 Latm/ mol K
T = -28 C = 273 - 28 K = 245 K
mass = 1.95 g
molar mass = unknown
Having known the other variables in the formula, the molar mass of the gas can be obtained.
PV = m R T/ molar mass
Molar mass = m RT / PV
Molar mass = 1.95 * 0.082 * 245 / 0.964 * 275 *10^-3
Molar mass = 39.1755 / 265.1 *10^-3
Molar mass = 39.1755 / 0.2651
Molar mass = 147.78 g/mol
The molar mass of the gas is 147.78 g/mol
1. The following thermochemical equation is for the reaction of water(l) to form hydrogen(g) and oxygen(g). 2H2O(l)2H2(g) + O2(g) H = 572 kJ How many grams of H2O(l) would be made to react if 110 kJ of energy were provided? _____ grams
2. The following thermochemical equation is for the reaction of carbon monoxide(g) with hydrogen(g) to form methane(g) and water(g). CO(g) + 3H2(g) CH4(g) + H2O(g) H = -206 kJ When 6.27 grams of carbon monoxide(g) react with excess hydrogen(g),_____ kJ of energy are ____ a.evolved b.absorbed
Answer:
1. 6.92 g of H2O
2i. - 46 KJ of energy.
ii. Option A. Evolved.
Explanation:
1. Determination of the mass of H2O that would be made to react if 110 kJ of energy were provided.
This can be obtained as follow:
The equation for the reaction is given below
2H2O(l) —> 2H2(g) + O2(g) H = 572 kJ
Next, we shall determine the mass of H2O required to produce 572 kJ from the balanced equation.
Molar mass of H2O = (2x1) + 16 = 18 g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36 g
From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.
Finally, we shall determine the mass of water (H2O) needed to produce 110 kJ of energy.
This is illustrated below:
From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.
Therefore, Xg of H2O will react to 110 kJ of energy i.e
Xg of H2O = (36 x 110)/572
Xg of H2O = 6.92 g
Therefore, 6.92 g of H2O is needed to react in order to produce 110 KJ of energy.
2i. Determination of the energy.
The balanced equation for the reaction is given below:
CO(g) + 3H2(g) —> CH4(g) + H2O(g) H = -206 kJ
Next, we shall determine the mass of CO that reacted to produce -206 kJ of energy from the balanced equation.
This is illustrated below:
Molar mass of CO = 12 + 16 = 28 g/mol
Mass of CO from the balanced equation = 1 x 28 = 28 g
From the balanced equation above,
28 g of CO reacted to produce -206 kJ of energy.
Finally, we shall determine the amount of energy produced by reacting 6.27 g of CO. This is illustrated below:
From the balanced equation above,
28 g of CO reacted to produce -206 kJ of energy.
Therefore, 6.27 g of CO will react to produce = (6.27 x -206)/28 = - 46 KJ of energy.
Therefore, - 46 KJ of energy were produced from the reaction.
2ii. Since the energy obtained is negative, it means heat has been given off to the surroundings.
Therefore, the heat is evolved.
Need help please! Theres three parts to this that I don't understand at all
Answer:
Chemical reaction B governs the process
Explanation:
The first part of the question asks to convert the mass of the calcium carbonate given to number of moles.
Mathematically;
Number of moles = mass/molar mass
Molar mass of CaCO3 = 100 g/mol
So the number of moles of CaCO3 will be 2.49/100 = 0.0249 moles
The second part of the question asks to convert the mass of carbon iv oxide to moles of carbon iv oxide
Mathematically;
That is same as ;
Number of moles = mass/molar mass
molar mass of CO2 is 44 g/mol
Number of moles of CO2 = 1.13/44 = 0.0256 moles
Now, if we compare the values of these number of moles, we can see that there are almost equal.
What this means is that the number of moles of calcium carbonate reacted is equal to the number of moles of carbon iv oxide produced.
So what we conclude here is that we have an equal mole ratio between the two compounds.
So the reaction that would be the correct answer will present equal number of moles of carbon iv oxide and calcium carbonate
Thus, we can see that reaction B is the one that governs this process as it is the only reaction out of the three options that present the two compounds with equal number of moles.
Determine the volume occupied by 10 mol of helium at
27 ° C and 82 atm
Answer:
3.00 L
Explanation:
PV = nRT
(82 atm × 101325 Pa/atm) V = (10 mol) (8.314 J/mol/K) (27 + 273) K
V = 0.00300 m³
V = 3.00 L
Considering that catalysts are not consumed in a reaction, how do you think increasing the amount of catalyst would affect the reaction rate for the decomposition of hydrogen peroxide?
a. increase
b. decrease
c. no effect
Answer:
a. increase
Explanation:
Catalysis is the process of increasing the rate of a chemical reaction by adding a substance known as a catalyst, which is not consumed in the catalyzed reaction.
By default, catalysts exists to speed up the rate of reactions. Increasing the amount of catalysts means that there would be an increase in the rate of reaction. The correct option is A.
Given a fixed amount of gas help at a constant pressure, calculate the temperature to which the gas would have to be changed if a 1.75 L sample at 23.0*C were to have a final volume of 3.50 L.
A. 46.0*C
B. 89.5*C
C. 169*C
D. 319*C
E. 592*C
Answer:
592 K or 319° C
Explanation:
From the statement of Charles law we know that the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Thus;
V1/T1= V2/T2
Initial volume V1 = 1.75 L
Initial temperature T1= 23.0 +273 = 296 K
Final volume V2= 3.50 L
Final temperature T2 = the unknown
T2= V2T1/V1= 3.50 × 296 / 1.75
T2 = 592 K or 319° C
Draw the major product(s) obtained when the following compounds are treated with bromine in the presence of iron tribromide.
a. Bromobenzene
b. ortho-Xylene
c. Benzene sulfonic acid
d. Benzaldehyde
e. meta-Nitrotoluene
f, para-Dibromobenzene
g. Nitrobenzene tert-Butylbenzene
h. Benzoic acid
i. Dibromobenzene
Answer:
The halogens are the ortho and para directing groups. Whenever they react with other benzene compounds they will attach to the ortho or para positions of the benzene ring.
Major products which are obtained by reacting these given compounds are given in attached pictures with complete reactions.
HBr will always be the side product of the bromine reactions along with the major compound.
Explanation:
Find the percentage composition of each element in the compound having 9.8 grams of nitrogen,0.7 grams of hydrogen and 33.6 grams of oxygen
Answer: The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.
Explanation:
Percentage composition is defined as the ratio of mass of substance to the total mass in terms of percentage.
Percentage composition=[tex]\frac{\text {mass of the element}}{\text {Total mass of the substance}}\times 100\%[/tex]
a) [tex]{\text {percentage composition of nitrogen}}=\frac{\text {mass of nitrogen}}{\text {Total mass}}\times 100\%[/tex]
[tex]{\text {percentage composition of nitrogen}}=\frac{9.8g}{9.8+0.7+33.6}\times 100\%=22.2\%[/tex]
b) [tex]{\text {percentage composition of hydrogen}}=\frac{\text {mass of hydrogen}}{\text {Total mass}}\times 100\%[/tex]
[tex]{\text {percentage composition of hydrogen}}=\frac{0.7}{9.8+0.7+33.6}\times 100\%=1.59\%[/tex]
c) [tex]{\text {percentage composition of oxygen}}=\frac{\text {mass of oxygen}}{\text {Total mass}}\times 100\%[/tex]
[tex]{\text {percentage composition of oxygen}}=\frac{33.6}{9.8+0.7+33.6}\times 100\%=76.2\%[/tex]
The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.
Describe the process of scientific inquiry ?
Answer:
It usually consists of six steps: question, observation or investigation, hypothesis, experiment, analysis of data (reviewing what happened during the experiment), and conclusion. Scientific inquiry, on the other hand, is non-linear, which means it does not follow a consistent step-by-step process.
Explanation:
Hope it helps
The ionization constant of lactic acid ch3ch(oh) co2h am acid found in the blood after strenuous exercise is 1.36×10^-4 If 20.0g of latic acid is used to make a solution with a volume of 1.00l what is the concentration of hydronium ion in the solution
Answer:
Explanation:
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
ionisation constant = 1.36 x 10⁻⁴ .
molecular weight of lactic acid = 90 g
moles of acid used = 20 / 90
= .2222
it is dissolved in one litre so molar concentration of lactic acid formed
C = .2222M
Let n be the fraction of moles ionised
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
C - nC nC nC
By definition of ionisation constant Ka
Ka = nC x nC / C - nC
= n²C ( neglecting n in the denominator )
n² x .2222 = 1.36 x 10⁻⁴
n = 2.47 x 10⁻²
nC = 2.47 x 10⁻² x .2222
= 5.5 x 10⁻³
So concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per litre .
The concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per liter .
Ionization of lactic acid can be represented as:
CH₃CHOHCOOH⇄ CH₃CHOHCOO⁻ + H⁺
Given:
ionization constant = 1.36 x 10⁻⁴
mass= 20.0 g
Now, Molecular weight of lactic acid = 90 g
[tex]\text{Number of moles}=\frac{20}{90} =0.22mol[/tex]
It is dissolved in 1.00L so molar concentration of lactic acid formed will be
C = 0.22M
Consider "n" to be the fraction of moles ionized
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
C - nC nC nC
By definition of ionization constant Ka
[tex]K_a =\frac{nC*nC}{C-nC}[/tex]
[tex]K_a= n^2C[/tex] ( neglecting n in the denominator )
On substituting the values we will get:
[tex]n^2 *0.22 = 1.36 *10^{-4}\\\\n = 2.47 * 10^{-2}[/tex]
To find the concentration of hydronium ion in the solution,
[tex]nC = 2.47 *10^{-2} *0.22\\\\nC= 5.5 * 10^{-3}[/tex]
So, concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per liter.
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What may be expected when K < 1.0? Choose the THREE correct statements. The concentration of one or more of the reactants is small. The concentration of one or more of the products is small. The reaction will not proceed very far to the right. The reaction will generally form more reactants than products.
Answer:
The concentration of one or more of the products is small.
The reaction will not proceed very far to the right.
The reaction will generally form more reactants than products
Explanation:
We often write
K =[Products]/[Reactants]
Thus, if K is small
We have fewer products than reactants We have more reactants than products The position of equilibrium lies to the leftA. is wrong. Usually, if K < 1, the concentration of reactants is greater than that of the products.
A sample of an unknown gas effuses in 11.1 min. An equal volume of H2 in the same apparatus at the same temperature and pressure effuses in 2.42 min. What is the molar mass of the unknown gas
Answer:
Molar mass of the gas is 0.0961 g/mol
Explanation:
The effusion rate of an unknown gas = 11.1 min
rate of [tex]H_{2}[/tex] effusion = 2.42 min
molar mass of hydrogen = 1 x 2 = 2 g/m
molar mas of unknown gas = ?
From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.
from
[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]
where
[tex]R_{h}[/tex] = rate of effusion of hydrogen gas
[tex]R_{g}[/tex] = rate of effusion of unknown gas
[tex]M_{h}[/tex] = molar mass of H2 gas
[tex]M_{g}[/tex] = molar mass of unknown gas
substituting values, we have
[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587
[tex]\sqrt{M_{g} }[/tex] = 0.31
[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol
The molar mass of the unknown gas will be "0.0961 g/mol".
Given:
Effusion rate of unknown gas,
[tex]R_g = 11.1 \ min[/tex]Effusion rate of [tex]H_2[/tex],
[tex]R_h = 2.42 \ min[/tex]Molar mass of hydrogen,
[tex]M_h = 1\times 2[/tex][tex]= 2 \ g/m[/tex]
According to the Graham's law, we get
→ [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]
By substituting the values, we get
→ [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]
→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]
→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]
[tex]\sqrt{M_g} = 0.31[/tex]
[tex]M_g = 0.0961 \ g/mol[/tex]
Thus the above solution is right.
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Which of these groups of elements show the least electronegativity?
Explanation:
On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group. As a result, the most electronegative elements are found on the top right of the periodic table, while the least electronegative elements are found on the bottom left. The answer is alkali metals.Name the following alkanes, please need answer for f,g,h?!
Answer:
f is =2,2-dimethyl butane
g is = 2,2-dimethyl propane
h is = 3,3-diethyl pentane
Explanation:
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Which of the following reagents should be used to convert to Question 26 options: A) H2, Lindlar's catalyst B) Na, liquid NH3 C) H2 / Nickel D) H2SO4, H2O
Answer: The question is is not complete...here is the complete question.
Which of the following reagents should be used to convert 3-Hexyne to E-3-hexene
Option B.
Na, liquid NH3.
Explanation:
3-Hexyne to E-3-hexene can be converted with by using the reagent of Na, NH3 (birch reduction) and this can be done by hdrogenation of H2.
The reagents NaNH3 convert 3-Hexyne to E-3-hexene because it is a reducing agents that convert or has the ability to reduce alkynes to trans alkenes.
3 Hexyne is an alkynes and it is converted to E- 3 hexene Na and NH3.
The heat of vaporization delta Hv of dichloromethane (Ch2CL2) is 28.0 kJ/mol . Calculate the change in entropy delta S when 473 g of dichloromethane boils at 39.8 degree.
Answer:
16 J/K.mol
Explanation:
From the question,
ΔS = ΔH/T............... Equation 1
Where ΔH = Heat change, T = Temperature
But,
ΔH = n(Hv).................. Equation 2
Where n = number of mole, Hv = heat of vaporization.
Given: Hv = 28.0 kJ/mol, n = 473/85 = 5.59 mole.
Substitute these values into equation 2
ΔH = 28/5.59
ΔH = 5.01 kJ.
Also: T = 273+39.8 = 312.8 J
Substitute into equation 1
ΔS = 5.01/312.8
ΔS = 0.016 kJ/K
ΔS = 16 J/K.mol
What amounts of sodium benzoate would be required to prepare 2.5L of 0.35M benzoic buffer solution with a pH of 6.10? Ka of benzoic acid = 6.5 x 10-5 MW benzoic acid, HC7H5O2, is 122.01 MW sodium benzoate, NaC7H5O2, is 144.01
Answer:
Benzoic acid: 1.288g
Sodium benzoate: 124.48g
Explanation:
Benzoic acid, HC7H5O2 is in equilibrium with its conjugate base, C7H5O2⁻ producing a buffer. The pH of the buffer can be determined following H-H equation:
pH = pKa + log [C7H5O2⁻] / [HC7H5O2] (1)
Where pH is desire pH = 6.10 pKa is -log Ka = 4.187 and [] are molar concentrations of the buffer.
As you want to prepare 2.5L of a 0.35M of buffer, moles of buffer are:
2.5L ₓ (0.35mol / L) = 0.875moles of buffer.
And you can write:
0.875 moles = [C7H5O2⁻] + [HC7H5O2] (2)
Replacing (2) in (1)
pH = pKa + log [C7H5O2⁻] / [HC7H5O2]
6.10 = 4.187 + log [C7H5O2⁻] / [HC7H5O2]
1.913 = log [C7H5O2⁻] / [HC7H5O2]
81.846 = 0.875mol - [HC7H5O2] / [HC7H5O2]
81.846 [HC7H5O2] = 0.875mol - [HC7H5O2]
82.846 [HC7H5O2] = 0.875mol
[HC7H5O2] = 0.01056 molesAnd moles of the benzoate, [C7H5O2⁻]:
[C7H5O2⁻] = 0.875mol - 0.01056mol =
[C7H5O2⁻] = 0.8644molUsing molar mass of benzoic acid and sodium benzoate, amount of each compound you must add to prepare 2.5L of the buffer are:
Benzoic acid: 0.01056mol ₓ (122.01g/mol) = 1.288g
Sodium benzoate: 0.8644mol ₓ (144.01g/mol) = 124.482g
Determine the pH during the titration of 25.5 mL of 0.276 M hydroiodic acid by 0.105 M barium hydroxide at the following points: (1) Before the addition of any barium hydroxide .55 (2) After the addition of 16.7 mL of barium hydroxide (3) At the equivalence point (4) After adding 40.7 mL of barium hydroxide
Answer:
1) before the addition of barium hydroxide
pH = -log[H⁺] = -log (0.276) = 0.559≈0.56
2)after the addition of barium hydroxide
pH = -log [H⁺] = -log(0.0857) = 1.067
3)at equivalent point, the solution will be neutral
pH = 7.0
4) after adding 40.7mL barium hydroxide
Explanation:
equation of reaction
2HCl(aq) + Ba(OH)₂(aq) ------->BaCl₂(aq) + 2H₂O(l)
1) Before the addition of barium hydroxide
concentration of HBr = 0.276M
[H⁺] = 0.276M
pH = -log[H⁺] = -log (0.276) = 0.559≈0.56
2) After adding 16.7mL barium hydroxide
moles of [OH⁻] = 16.7mL × 0.105 × 2
=3.507m mol = 3.507 × 10³mol
moles of [H⁺] = 25.5mL × 0.276M
=7.038m mol = 7.038 × 10³mol
moles of [H⁺] remaining = (7.038 - 3.421)m mol
= 3.617m mol = 3.617 × 10³mol
[H⁺]= [tex]\frac{3.617}{25.5 + 16.7}[/tex] = 0.0857
pH = -log [H⁺] = -log(0.0857) = 1.067
3) At equivalent point, the solution will be neutral
pH = 7.0
4) After adding 40.7mL barium hydroxide
moles of [OH⁻] = 40.7mL × 0.105M × 2
=8.547
moles of [OH⁻] remaining = 8.547 - 7.038
= 1.509m mol = 1.509 × 10³mol
pOH= -log[OH⁻]= 2.82
pH = 14 - 2.82 = 11.18
Daniel has a sample of pure copper.its mass is 89.6 grams (g),and its volume is 10 cubic centimeters (cm3). whats the destiny of the sample.
Answer:
8.96g \ cm3
Explanation:
(89.6\ 10) (g\ cm3) = 8.96g\cm3