Answer:
Explanation:
In cobalt - 65 ,
no of protons is 27 ( p )
no of neutron = 65 - 27 ( n )
= 38
n / p ratio
= 38 / 27
= 1.41
If case of emission of alpha particle
no of proton p = 27 - 2 = 25
no of neutrons = 38 - 2 = 36
n / p ratio = 36 / 25
= 1.44
So it increases
In case of emission of beta particle
No of neutron n = 38 - 1 = 37
No of proton = 27 + 1 = 28
n / p ratio = 37 / 28
= 1.32
Hence ratio decreases.
Hence beta ray decay will result in decrease in n / p ratio.
A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a heat capacity of 9.20 J/K. If the final temperature of the system is 31.10°C, what is the specific heat capacity of the alloy? J g·°C
Answer:
THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K
Explanation:
Mass of alloy = 33 g
Initial temperature of alloy = 93°C
Mass of water = 50 g
Initail temp. of water = 22 °C
Heat capacity of calorimeter = 9.20 J/K
Final temp. = 31.10 °C
specific heat of alloy = unknown
specific heat capacity of water = 4.2 J/g K
Heat = mass * specific heat * change in temperature = m c ΔT
Heat = heat capcity * chage in temperature = Δ H * ΔT
In calorimetry;
Heat lost by the alloy = Heat gained by water + Heat of the calorimeter
mc ΔT = mcΔT + Heat capacity * ΔT
33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)
33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1
2042.7 C = 1911 + 83,72
C = 1911 + 83.72 / 2042.7
C = 1994.72 /2042.7
C =0.9765 J/g K
The specific heat of the alloy is 0.9765 J/ g K
Which of the following is a property of matter?
The major properties of matter are volume, mass, and shape.
All matter however too is made up of tiny particles known as atoms.
Other characteristics properties of matter which can be measured include object's density, color, length, malleability, melting point, hardness, odor, temperature, and others
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The property of matter should be volume, mass, and shape.
The following information should be considered:
The matter should be made up of small & tiny particles that we called as the atoms. It involved the density of an object, length, temperature, melting point, etc.Learn more: https://brainly.com/question/1979431?referrer=searchResults
Which of the following formulas represents an ionic compound?
1.HI 2.HCI 3.LiCI 4.SO2
Answer:
Numbers 4,3
Explanation:
Ionic bond is between nonmental and metals
uses of sodium chloride in daily life
Answer:
sodium chloride can be used as salt
extraction sodium metal by electrolysis
a common chemical in laboratory experiments
Answer:
sodium chloride can be used as preservatives,
in preserving foods.
a gas obeys the equation of state p(v-b)=RT.for the gas b=0.0391L/mol.calculate the fugacity coefficient for the gas at 1000°c and 1000atm
Answer:
The fugacity coefficient is [tex][\frac{f}{p} ] = 1.45[/tex]
Explanation:
From the question we are told that
The gas obeys the equation [tex]p(v-b) = RT[/tex]
The value of b is [tex]b = b = 0.0391 \ L /mol[/tex]
The pressure is [tex]p = 1000 \ atm[/tex]
The temperature is [tex]T= 1000^oC = 1273 K[/tex]
generally
[tex]RT ln[\frac{f}{p} ] = \int\limits^{p}_{o} [ {v_{r} -v_{i}} ]\, dp[/tex]
Where [tex]\frac{f}{p}[/tex] is the fugacity coefficient
[tex]v_r[/tex] is the real volume which is mathematically evaluated from above equation as
[tex]v_r = \frac{RT}{p} + b[/tex]
[tex]v_r = \frac{RT}{p} + 0.0391[/tex]
and [tex]v_{i}[/tex] is the ideal volume which is evaluated from the ideal gas equation (pv = nRT , at n= 1) as
[tex]v_{i} = \frac{RT}{p}[/tex]
So
[tex]RT ln[\frac{f}{p} ] = \int\limits^{1000}_{o} [[ \frac{RT}{p} + 0.0391] - [\frac{RT}{p} ]} ]\, dp[/tex]
=> [tex]RT ln[\frac{f}{p} ] = \int\limits^{1000}_{o} [0.391 ]\, dp[/tex]
=> [tex]RT ln[\frac{f}{p} ] = [0.391p]\left | 1000} \atop {0}} \right.[/tex]
=> [tex]RT ln[\frac{f}{p} ] = 38.1[/tex]
So
[tex]ln[\frac{f}{p} ] = \frac{39.1}{RT}[/tex]
Where R is the gas constant with value [tex]R = 0.082057\ L \cdot atm \cdot mol^{-1}\cdot K^{-1}[/tex]
[tex][\frac{f}{p} ] = \frac{39.1}{ 2.303 *0.082057 * 1273}[/tex]
[tex][\frac{f}{p} ] = 1.45[/tex]
consider an exceptionally weak acid, HA, with Ka= 1 x 10-20. you make 0.1M solution of the salt NA. what is the pH.
Answer:
[tex]pH=10.5[/tex]
Explanation:
Hello,
In this case, the dissociation of the given weak acid is:
[tex]HA\rightleftharpoons H^++A^-[/tex]
Therefore, the law of mass action for it turns out:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
That in terms of the change [tex]x[/tex] due to the reaction extent is:
[tex]1x10^{-23}=\frac{x*x}{0.1-x}[/tex]
Thus, by solving with the quadratic equation or solver, we obtain:
[tex]x=31.6x10^{-12}M[/tex]
Which clearly matches with the hydrogen concentration in the solution, therefore, the pH is:
[tex]pH=-log(-31.6x10^{-12})\\pH=10.5[/tex]
Regards.
what is the chemical symbol and name of the third element in the periodic table
Answer: Aluminum symbol Al or aluminum American English
Explanation:
Answer:
Hii
Li( Lithium)
Explanation:
Lithium has the atomic number of three and is the third element in periodic table.
How do the particles in plasmas compare with the particles in solids?
Answer:
Plasmas and solids are both made up of cation-anion pairs. Solids and plasmas are both made up of electrons and cations. Solids are made up of cation-anion pairs, but plasmas are not.
Answer:
Solids are made up of cation-anion pairs, but plasmas are not.
Explanation:
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]
Hcl and 1-isopropylcyclohexane formation
Answer:
Spahgetti
Explanation:
Aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . If of sodium bromide is produced from the reaction of of hydrobromic acid and of sodium hydroxide, calculate the percent yield of sodium bromide.
Answer:
The percentage yield is 50%
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
Determine the limiting reactant in a mixture containing 95.7 g of B2O3, 75.7 g of C, and 369 g of Cl2. Calculate the maximum mass (in grams) of boron trichloride, BCl3, that can be produced in the reaction. The limiting reactant is:
Answer:
[tex]B_2O_3[/tex]
Explanation:
First, we have to find the reaction:
[tex]B_2O_3~+~C~+~Cl_2~->~BCl_3~+~CO[/tex]
The next step is to balance the reaction:
[tex]B_2O_3~+~3C~+~3Cl_2~->~2BCl_3~+~3CO[/tex]
Now, we have to calculate the molar mass for each compound, so:
[tex]B_2O_3=~69.62~g/mol[/tex]
[tex]C=~12~g/mol[/tex]
[tex]Cl_2=~70.96~g/mol[/tex]
With these values, we can calculate the moles of each compound:
[tex]95.7~g~B_2O_3\frac{1~mol~B_2O_3}{69.62~g~B_2O_3}=1.37~mol~B_2O_3[/tex]
[tex]75.7~g~C\frac{1~mol~C}{112~g~C}=6.30~mol~C[/tex]
[tex]369~g~Cl_2\frac{1~mol~Cl_2}{70.96~g~C}=5.20~mol~Cl_2[/tex]
Now we can divide by the coefficient of each compound in the balanced equation:
[tex]\frac{1.37~mol~B_2O_3}{1}=~1.37[/tex]
[tex]\frac{6.30~mol~C}{3}=~2.1[/tex]
[tex]\frac{5.20~mol~Cl_2}{3}=~1.73[/tex]
The smallest values are for [tex]B_2O_3[/tex], so this is our limiting reagent.
I hope it heps!
How many grams of the salt CaF2 (g) are formed when 15.7 mL of 0.612 M KF reacts with an excess of aqueous calcium bicarbonate (Ca(HCO3)2) via a metathesis reaction?
Answer:
[tex]m_{CaF_2}0.375gCaF_2[/tex]
Explanation:
Hello,
In this case, for the studied reaction:
[tex]2KF+Ca(HCO_3)_2\rightarrow CaF_2+2KHCO_3[/tex]
Thus, the first step is to compute the reacting moles of potassium fluoride by using its volume and molarity:
[tex]n_{KF}=0.0157L*0.612\frac{mol}{L} =9.61x10^{-3}molKF[/tex]
Then, we apply the 2:1 molar ratio between potassium fluoride and calcium fluoride to compute the produced moles of calcium fluoride:
[tex]n_{CaF_2}=9.61x10^{-3}molKF*\frac{1molCaF_2}{2molKF} =4.80x10^{-3}molCaF_2[/tex]
Finally, by using the molar mass of calcium fluoride (78.07 g/mol) we can compute its produced grams:
[tex]m_{CaF_2}=4.80x10^{-3}molCaF_2*\frac{78.07gCaF_2}{1molCaF_2} \\\\m_{CaF_2}0.375gCaF_2[/tex]
Best regards.
Given the density of iron (Fe) is 7.87 g/cm3, determine the mass of iron (in grams) in a rectangle block with the dimensions of 12.5 in long, 3.50 in wide, and 2.50 in high. (1in = 2.54 cm).
Answer:
[tex]m=14,105.71 g Fe[/tex]
Explanation:
Hello,
In this case, the first step is to compute the volume of the block considering the length, height and width:
[tex]V=L \times W \times H =12.5 in\times 3.50 in \times 2.50 in =109.375 in^3[/tex]
Then, we compute the volume in cubic centimetres:
[tex]V=109.375in^3\times \frac{16.3871 cm^3}{1in^3} =1792.34cm^3[/tex]
Finally, as the density is given by:
[tex]\rho =\frac{m}{V}[/tex]
We solve for the mass:
[tex]m=\rho \times V= 7.87\frac{g}{cm^3} \times 1792.34 cm^3\\\\m=14,105.71 g Fe[/tex]
Best regards.
The pyruvate dehydrogenase complex is subject to allosteric control, especially inhibition by reaction products. The main regulatory process controlling pyruvate dehydrogenase's activity in eukaryotes is
a. exchange of ADP and ATP on the pyruvate dehydrogenase complex.
b. phosphorylation by ATP, which turns the complex on, and dephosphorylation, which turns the complex off.
c. AMP binding to and activating the enzyme.
d. phosphorylation by ATP, which turns the complex off, and dephosphorylation, which turns the
Answer:
D. Phosphorylation by ATP, which turns the complex off, and dephosphorylation, which turns the complex on.
Explanation:
The pyruvate dehydrogenase complex (PDH) is responsible for the conversion of pyruvate to acetylCoA, the fuel for the citric acid cycle.
The regulation of the activity of PDH is allosterically by the products of the reaction which it catalyses. These products are ATP, acetylCoA and NADH. When their is sufficient fuel available for the needs of the cells in the form of ATP, the complex is turned off by phosphorylation of one of the two subunits of E1 (pyruvate dehydrogenase). This phosphorylation inactivates E1. When the concentration of ATP declines, a specific phosphatase removes the phosphoryl group from E1, thereby activating the complex again.
Which of the following is a conjugate acid-base pair in the reaction represented by the
equation below?
H2PO4 + H20 H3PO, + OH
H2PO, and H2O
b) H,PO, and OH
c) H2PO, and H3PO,
None of the above
Answer: [tex]H_2PO_4[/tex] and [tex]H_3PO_4[/tex]
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
For the given reaction:
[tex]H_2PO_4^-+H_2O\rightleftharpoons H_3PO_4+OH^-[/tex]
Here, [tex]H_2O[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]OH^-[/tex] which is a conjugate base.
Similarly , [tex]H_2PO_4^-[/tex] is gaining a proton, thus it is considered as an base and after gaining a proton, it forms [tex]H_3PO_4[/tex] which is a conjugate acid.
Thus [tex]H_2PO_4[/tex] and [tex]H_3PO_4[/tex] is a conjugate acid-base pair in the reaction represented by the equation below
The Bronsted-Lowry conjugate acid-base hypothesis defines an acid as a substance that loses protons and donates them to another chemical to produce conjugate base, and a base as a substance that takes protons to generate conjugate acid.
Thus, a proton is being lost, making it an acid, and once a proton is lost, a conjugate base is formed. Similar to that, is gaining a proton, making it a base, and then it produces a conjugate acid after gaining a proton.
The Brnsted-Lowry hypothesis, often known as the proton theory of acids and bases, is an independent theory of acid-base reactions that was put forth in 1923 by Johannes Nicolaus Brnsted and Thomas Martin Lowry.
This theory's central idea is that when an acid and a base interact, the acid creates its conjugate base and the base creates its conjugate acid by exchanging a proton (the hydrogen cation, or H+).
Thus, The Bronsted-Lowry conjugate acid-base hypothesis defines an acid as a substance that loses protons and donates them to another chemical to produce conjugate base, and a base as a substance that takes protons to generate conjugate acid.
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What is it called when a gas changes into a liquid?
Answer:
Explanation:
Condensation is the word you seek.
Answer:
Condensation
Explanation:
When a gas is subjected to decrease in temperature it is condensed
Real images can be projected onto a screen.
A. True
B. False
Answer:
True
Explanation:
A real image can be projected or seen on a screen but a virtual image cannot because a real image is formed when light rays coming from an object actually meet at a point after refraction through a lens while a virtual image is formed when light rays coming from an object only appear to meet at a point when produced ...
Hope this helps you, and Good luck!
Without doing any calculations, arrange the elements in CF2Cl2 in order of decreasing mass percent composition. Rank from highest percent to lowest.
a. C > F > Cl
b. F < Cl > C
c. Cl > C > F
d. Cl > F > C
Answer:
a. C > F > Cl
Explanation:
We know that atomic mass of Chlorine is greater than of Florine than that of carbon. Moreover, in CF2Cl2, therefore, there are two atoms of Cl, F and one atom of C. Therefore, in CF2Cl2 in order of decreasing mass percent composition C > F > Cl. Therefore, the correct option is a.
7.Which one of the following statements is not true?
1 point
O The molecules in a solid vibrate about a fixed position
O The molecules in a liquid are arranged in a regular pattern
The molecules in a gas exert negligibly small forces on each other, except during
collisions
The molecules of a gas occupy all the space available
Answer:
B. the molecules in liquid are loosely packed and scattered thus, they cannot be arranged
Empirical formula for compound of 2.17 mol N and 4.35 mol O
Answer:
Explanation:
ratio of moles of N and O in molecule =
N / O = 2.17 / 4.35
1/2
empirical formula = NO₂
Suppose the amount of a certain radioactive substance in a sample decays from to over a period of days. Calculate the half life of the substance. Round your answer to significant digit.
The given question is incomplete, the complete question is:
Suppose the amount of a certain radioactive substance in a sample decays from 1.30 mg to 100. ug over a period of 29.5 days. Calculate the half life of the substance Round your answer to 2 significant digits.
Answer:
The correct answer is 7.974 days.
Explanation:
Based on the given question, the concentration of a radioactive substance present in a sample get decays to 100 micro grams from 1.30 milligrams in 29.5 days. There is a need to find the half-life of the substance.
Radioactive decay is an illustration of first order reaction.
K = (2.303 / t) log [a/(a-x)]
Here a is 1.30 mg and (a-x) is 100 micrograms = 100 * 10^-3 mg or 0.1 mg, and t is 29.5 days. Now putting the values we get,
K = (2.303 /29.5)log (1.30/0.1)
= 2.303/29.5 log13
= 2.303/29.5 * 1.1139
K = 0.0869
The half-life or t1/2 is calculated by using the formula, 0.693 / K
= 0.693 / 0.0869
= 7.974 days.
Two hypothetical ionic compounds are discovered with the chemical formulas XCl2 and YCl2, where X and Y represent symbols of the imaginary elements. Chemical analysis of the two compounds reveals that 0.25 mol XCl2 has a mass of 100.0 g and 0.50 mol YCl2 has a mass of 125.0 g. (a) What are the molar masses of XCl2 and YCl2
Answer:
THE MOLAR MASS OF XCL2 IS 400 g/mol
THE MOLAR MASS OF YCL2 IS 250 g/mol.
Explanation:
We calculate the molar mass of XCL2 and YCL2 by bringing to mind the formula for molar mass when mass and amount or number of moles of the substance is given.
Number of moles = mass / molar mass
Molar mass = mass / number of moles.
For XCL2,
mass = 100 g
number of mole = 0.25 mol
So therefore, molar mass = mass / number of moles
Molar mass = 100 g / 0.25 mol
Molar mass = 400 g/mol.
For YCL2,
mass = 125 g
number of mole = 0.50 mol
Molar mass = 125 g / 0.50 mol
Molar mass = 250 g/mol.
So therefore, the molar mass of XCL2 and YCL2 IS 400 g/mol and 250 g/mol respectively.
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]
The temperature program for a separation starts at a temperature of 50 °C and ramps the temperature up to 270 °C at a rate of 10 °C/minute. Which statement is NOT true for this separation?
A) At 10 °C/minute, a total of 22 minutes is needed to reach 270 oC.
B) Strongly retained solutes will remain at the head of the column while the temperature is low.
C) Weakly retained solutes will separate and elute early in the separation.
D) The vapor pressure of strongly retained solutes will increase as temperature increases.
E) Strongly retained analytes will give broad peaks.
Answer:
The correct answer to the question is Option E (Strongly retained analytes will give broad peaks).
Explanation:
The other options are true because:
A. Initial temp = 50 °C
Final temp = 270 °C
Differences in temp = 270 - 50 = 220°C
Rate = 10 °C/minute.
So, at 10 °C/minute,
total of 220°C /10 °C = number of minutes required to reach the final temp.
220/10 = 22 minutes
B. A column has a minimum and maximum use temperature. Solutes that are already retained would remain stationary while temperatures are low. This would only change if there is an increase in temperature. Heat transfers more energy to the liquid which would make the solute interact with the column phase.
C. Weakly retained solutes may contain larger molecules, will separate by absorbing into the solvent early in separation making the mobile phase separates out into its components on the stationary phase.
D. Retained solute's vapor pressure is higher at higher temperatures making it possible for particle to escape more from the solute when the temperature is high than when it is low.
The wolf gets enegry from____
The rabbit gets energy from____
The plant gets energy from___
The mushoom gets energy from___
Answer:
The wolf gets energy from other Animals through Cellular respiration. it's a carnivore
The rabbit gets energy from Carbohydrates,Fats.... obtained through different sources. A common example is the grass. It's an herbivore
The plant gets energy from the sun during photosynthesis. It's Autotrophic.
The mushroom gets energy from the decomposition of other organic matter. It's heterotrophic.
Explanation:
In a food chain; The Wolf eats the rabbit, when the Wolf dies, decomposers such as mushrooms breaks down its body returning it to the soil, where it provides nutrients for plants
One proposed mechanism of the reaction of HBr with O2 is given here. HBr + O2 → HOOBr (slow) HOOBr + HBr → 2HOBr (fast) HOBr + HBr → H2O + Br2 (fast) What is the equation for the overall reaction?
Answer:
4 HBr + O2 → + 2H2O + 2Br2
Explanation:
Based on the following reaction mechanism:
HBr + O2 → HOOBr (slow)
HOOBr + HBr → 2HOBr (fast)
HOBr + HBr → H2O + Br2 (fast)
The equation for the overall reaction is the sum of the three reactions in which intermediaries of reaction (HOBr and HOOBr are canceled). That is 1 + 2 + 2*(3):
HBr + O2 + HOOBr + HBr + 2HOBr + 2HBr → HOOBr + 2HOBr + 2H2O + 2Br2
4 HBr + O2 → + 2H2O + 2Br2Beeing this reaction the equation of the overall reaction.
Which of these statements gives a correct reason as to why our body needs water?
(1 Point)
1. It provides us with energy.
2. It helps us to eliminate waste.
3. It helps regulate our body temperature.
Answer:
2. It helps us to eliminate waste
3. It helps regulate our body temperature
Explanation:
In addition to the function of bringing nutrients to the cells, water provides the elimination of substances out of the body. This occurs, for example, through urine, which is basically formed by water and toxic or excess substances dissolved.
Water also helps in regulating body temperature. This occurs when the heat becomes exaggerated, sweat is released, which has water in its composition. When in contact with the medium, the sweat evaporates on the surface of the skin, causing the body to cool.
An ideal gaseous reaction occurs at a constant pressure of 35.0 atm and releases 66.8 kJ of heat. Before the reaction, the volume of the system was 8.20 L. After the reaction, the volume of the system was 2.21 L. Calculate the total change in internal energy for the system. Enter your answer numerically in units of kJ.
Answer:
U = -45.557kj
Explanation:
Before we can calculate the totally internal energy change in kilojoules firstly we need to calculate W
U=q + w .
We know that
w = PΔ V
where P is the pressure of
and V is the volume
then we can calculate the work
w = 35 atm * ( 8.20L - 2.21L)
W=35atm* 5.99L
W=209.65atmJ
But 1 atm = 101.325J
then ,
w = 209.65* 101.325 J = 21242.79 J
let us convert it to Kj
But we know that 1kJ = 10^3 J .
Then w = 21.243 kJ .
Then we can now calculate the internal energy as
U = 21.243- 66.8 kJ = -45.557kj
But we know that heat was released. Theeefore, the total internal energy change was -45.557kj