Answer:
[tex]m=20kg[/tex]
Explanation:
Hello,
In this case, we define the kinnetic energy as:
[tex]K=\frac{1}{2} m*v^2[/tex]
Thus, for finding the mass we simply solve for it on the previous equation given the kinetic energy and the velocity:
[tex]m=\frac{2*K}{v^2}=\frac{500kg*\frac{m^2}{s^2} }{(5\frac{m}{s})^2} =\frac{500kg*\frac{m^2}{s^2} }{25\frac{m^2}{s^2}}\\\\m=20kg[/tex]
Best regards.
Answer:
The answer is 40 kg
Explanation:
You will this formula below:
m=[tex]\frac{2*\\KE}{v^{2} }[/tex]
Now we know our formula, now we plug in the given numbers:
m=[tex]\frac{2(500J)}{(5m/s)^2}[/tex]
Simplify and we get:
m=40 kg
I hope this was helpful.
An excess of AgNO3 reacts with 185.5 mL of an AlCl3 solution to give 0.325 g of AgCl. What is the concentration, in moles per liter, of the AlCl3 solution? Must show your work on scratch paper to receive credit. AlCl3(aq) + 3 AgNO3(aq) → 3 AgCl(s) + Al(NO3)3(aq)
Answer:
4.07x10⁻³M AlCl₃.
Explanation:
Based on the reaction:
AlCl₃(aq) + 3 AgNO₃(aq) → 3 AgCl(s) + Al(NO₃)₃(aq)
That means 1 mole of AlCl₃ reacts with 3 moles of AgNO₃ to produce 3 moles of AgCl.
As 0.325g of AgCl are produced. Moles of AgCl are (Molar mass AgCl: 143.32g/mol):
0.325g AgCl ₓ ( 1 mol / 143.32g) = 2.27x10⁻³ moles of AgCl
As 3 moles of AgCl are produced from 1 mole of AlCl₃, moles of AlCl₃ that produce 2.27x10⁻³ moles of AgCl are:
2.27x10⁻³ moles of AgCl ₓ (1 mole AlCl₃ / 3 moles AgCl) =
7.56x10⁻⁴ moles AlCl₃
As volume of the AlCl₃ solution that reacts is 185.5mL = 0.1855L, molar concentration of the solution is:
7.56x10⁻⁴ moles AlCl₃ / 0.1855L =
4.07x10⁻³M AlCl₃Consider each pair of compounds listed below and determine whether a fractional distillation would be necessary to separate them or if a simple distillation would be sufficient.
a. Ethyl acetate and hexane
b. Diethyl Ether and 1-butanol
c. Bromobenzene and 1,2-dibromobenzene
What does the atmosphere do for humans?
Which of the following best describes isotopes?
An element with the same number of neutrons, but a different number of protons.
An element with the same number of protons, but a different number of electrons.
An element with the same number of electrons, but a different number of neutrons
An element with the same number of protons, but a different number of neutrons
Answer: An element with the same number of protons, but a different number of neutrons
Explanation:
The # of protons in an atom is what determines what atom it is (hydrogen has 1 proton, helium has 2 protons, etc ...). You cannot change the number of protons in an atom without changing what element the atom is.
The number of electrons in atoms varies greatly because electrons are constantly gained, lost, and shared during chemical reactions.
An isotope is a variation of the same element (so they must have the same # of protons) that have different masses (and therefore a different number of neutrons).
The answer is the fourth choice, "An element with the same number of protons, but a different number of neutrons"
The isotopes refer to an element that consists of a similar number of protons but have a distinct no of neutrons.
What are isotopes:It is considered to be the members of the family with respect to the elements that consist of a similar number of protons but have a distinct no of neutrons. The no of protons in the nucleus measured the atomic number of elements based on the periodic table.
Therefore, the fourth option is correct.
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When 200g of AgNO3 solution mixes with 150 g of NaI solution, 2.93 g of AgI precipitates, and the temperature of the solution rises by 1.34oC. Assume 350 g of solution and a specific heat capacity of 4.184 J/g•oC. Calculate H for the following: Ag+(aq) + I- (aq) → AgI(s)
Answer:
[tex]\Delta H=1962.3J[/tex]
Explanation:
Hello,
In this case, we can compute the change in the solution enthalpy by using the following formula:
[tex]\Delta H=mC\Delta T[/tex]
Whereas the mass of the solution is 350 g, the specific heat capacity is 4.184 J/g °C and the change in the temperature is 1.34 °C, therefore, we obtain:
[tex]\Delta H=350g*4.184\frac{J}{g\°C} *1.34\°C\\\\\Delta H=1962.3J[/tex]
It is important to notice that the mass is just 350 g that is the reacting amount and by means of the law of the conservation of mass, the total mass will remain constant, for that reason we compute the change in the enthalpy as shown above, which is positive due to the temperature raise.
Best regards.
Predict the products of the following elimination reaction, and draw the major product formed. Make sure to consider the stereochemistry of the reaction. 3-chloro-3-methylpentane reacts with sodium tertbutoxide in tertbutanol.
Predict the products of the following elimination reaction, and draw the major product formed. Make sure to consider the stereochemistry of the reaction. 3-chloro-3-methylpentane reacts with sodium ethoxide in ethanol.
Answer:
see explanation below
Explanation:
In the first case, we have a reaction where we have the 3-chloro-3-methylpentane reacting with t-butoxide. The t-butoxide is a very voluminous base, so the strength of substracting a hydrogen atom is reduced. Therefore, the reaction taking place here will be an E2 but instead of substracting the hydrogen from the carbons 2 or 4, it will substract it from the methyl group, cause it has less steric hindrance there and the reaction will go faster.
In the second case, the sodium ethoxide is a strong base, so it will rapidly substract an atom of hydrogen from carbon 2 or 4 to form the (Z) - 3 - methyl - 2- pentene and the substitution product.
Look picture for mechanism and products.
The three‑dimensional structure of a generic molecule is given. Identify the axial and equatorial atoms in the three‑dimensional structure. What is the shape of this molecule?
Answer:
Explanation:
CHECK THE ATTACHMENT FOR THE COMPLETE QUESTION AND THE DETAILED EXPLANATION
NOTE:
Equatorial atoms are referred to atoms that are attached to carbons in the cyclohexane ring which is found at the equator of the ring.
Axial atoms are atoms that exist in a bond which is parallel to the axis of the ring in cyclohexane
calculate the moles of 25.2 g Na2S2O8
Answer:
To calculate the moles we must first find the molar mass M
M (Na2S2O8) = (23*2) + (32*2) + (16*8)
= 46 + 64 + 168
= 278g/mol
Molar mass = mass/moles
moles =mass / molar mass
= 25.2/278
= 0.0906mol
Hope this helps.
How could the government enforce ethical standards of scientific
experiments?
A. The government could encourage scientists to make up their own
minds about ethics.
B. The government could take away research funds if ethical
standards are not met.
C. The government could let scientists monitor each other to
encourage ethical behavior.
D. The government could encourage the public to take a stand
against unethical scientists.
Answer: D. The Government could take away research funds if ethical standards are not met
The government enforce ethical standards of scientific experiments
B. The government could take away research funds if ethical
standards are not met.
Ethical standards are a set of principles established by the founders of the organization to communicate its underlying moral values. This code provides a framework that can be used as a reference for decision making processes.
How does the government control scientific research?Politicians and bureaucrats control scientific research and research outcomes by selectively funding projects that look for potential disasters, ideally global disasters.
What are the 8 ethical standards?This analysis focuses on whether and how the statements in these eight codes specify core moral norms (Autonomy, Beneficence, Non-Maleficence, and Justice), core behavioral norms (Veracity, Privacy, Confidentiality, and Fidelity), and other norms that are empirically derived from the code statements.
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You add 5.7 g of iron to 25.20ml of water and observe that the volume of the iron and water together is 25.92ml calculate thw density of the iron
Answer:
7.92gml-1
Explanation:
water=25.20ml
water+iron=25.92ml
iron=5.7g
P=mass/volume (formula of density)
mass=5.7g
volume=25.92-25.20
=0.72ml
p=5.7/0.72
=7.92gml-1
Given:
Initial volume of water = 25.20mL
Volume of water after iron is added = 25.92mL
Mass of iron = 5.7g
So, the volume of iron = 25.92mL - 25.20mL = 0.72mL
∴ Density of iron will be
Density = Mass/Volume
Density = 5.7g / 0.72mL
Density = 7.91 g/mL
What is density in short answer?
The density of a substance is the relationship between its mass and how much space it takes up. Density equals the mass of the substance divided by its volume, D = m/v.
What is the SI unit of density?Though SI unit of density is kg/m³ solids, g/ml for liquids and g/L for gases.
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Calculate the percentage of the void space out of the total volume occupied by 1 mole of water molecules. The density of water is assumed to be 1.0 g/mL that is 1.0 g/cm3. The molar mass of water is 18.0 g/mol. The atomic radius of hydrogen is 37 pm and of oxygen is 73 pm. The formula for the volume of a sphere is 4/3(r3
Answer:
The percentage of the void space out of the total volume occupied is 93.11%
Explanation:
A mole of water contains 2 atoms of hydrogen and 1 atom of oxygen.
To calculate the volume of a mole of water, we calculate 2 times the volume of the hydrogen atom and 1 times the volume of the oxygen atom
Let's calculate this one after the other.
For the hydrogen, formula for the volume will be
[tex]V_{hydrogen[/tex] = 2 × 4/3 × π × [tex]r_{H}^{3}[/tex]
where [tex]r_{H}^{3}[/tex] = 37 pm which is read as 37 picometer (1 picometer = 10^-12 m) = 37 × [tex]10^{-12}[/tex] meters
Volume of the hydrogen = 8/3 × (37 × 10^-12)^3 = 4.05 * 10^-31
we multiply this by the avogadro's number = 6.02 * 10^23
= 4.05 * 10^-31 * 6.02 * 10^23 = 2.6 * 10^-8 m^3
We do same for thr oxygen, but this time we do not multiply the volume of the oxygen by 2 as we have only one atom of oxygen
Volume of oxygen = 4/3 * π * (73 * 10^-12) ^3 * avogadro's number = 9.81 * 10^-7 m^3
adding both volumes together, we have 1.24 * 10^-6 m^3 or simply 1.24 ml ( 0.01 m = 1 ml)
Dividing the molar mass of one mole of water by its density, we can get the volume of 1 mole of water
= (18g/mol)/(1 g/ml) = 18 ml/mol
Now we proceed to calculate the volume of void = Total volume - volume of molecule = 18 - 1.24 = 16.76 ml
Now, the percentage of void = volume of void/total volume * 100%
= 16.76/18 * 100% = 93.11%
50.0 g N204 (92.02 g/mol) react with 45.0 g N2H4 (32.05 g/mol) forming nitrogen gas, N2
(28.01 g/mol) and water, H20 (18.02 g/mol). What mass in grams of excess-reactant is
left over?
Answer:
The excess reactant is N2H4 and the leftover mass is 10.17g.
Explanation:
Step 1:
The balanced equation for the reaction.
N2O4 + 2N2H4 —> 3N2 + 4H2O
Step 2
Determination of the masses of N2O4 and N2H4 that reacted from the balanced equation:
Molar mass of N2O4 = 92.02 g/mol
Mass of N2O4 from the balanced equation = 1 x 92.02 = 92.02g
Molar mass of N2H4 = 32.05 g/mol
Mass of N2H4 from the balanced equation = 2 x 32.05 = 64.1g
From the balanced equation above, 92.02g of N2O4 reacted with 64.1g of N2H4.
Step 3:
Determination of the excess reactant. This is illustrated below:
From the balanced equation above, 92.02g of N2O4 reacted with 64.1g of N2H4.
Therefore, 50g of N2O4 will react with = (50 x 64.1)/92.02 = 34.83g of N2H4.
From the calculations made above, we can see that only 34.83g of N2H4 reacted out of 45g that was given. Therefore, N2H4 is the excess reactant.
Step 4:
Determination of the mass of excess reactant that is leftover.
The excess reactant is N2H4 and the leftover mass can be obtained as follow:
Mass of N2H4 given = 45g
Mass of N2H4 that reacted = 34.83g
Leftover mass of N2H4 =..?
Leftover mass of N2H4 = (Mass of N2H4 given) – (Mass of N2H4 that reacted)
Leftover mass of N2H4 = 45 – 34.83
Leftover mass of N2H4 = = 10.17g.
If 25.8 mL of an AgNO3 solution is needed to precipitate all Cl- ions in a 1570 mg of KCl (forming AgCl), what is the molarity of the AgNO3nsolution?
Answer:
M=0.816M
Explanation:
Hello,
In this case, we should consider the following reaction:
[tex]AgNO_3+KCl\rightarrow KNO_3+AgCl[/tex]
Thus, by knowing the 1:1 molar ratio of silver nitrate and potassium chloride, we can easily compute the moles of silver nitrate precipitating the 1570 mg of potassium chloride considering its molar mass of 74.5513 g/mol:
[tex]n_{AgNO_3}=1570mgKCl*\frac{1gKCl}{1000mgKCl} *\frac{1molKCl}{74.5513gKCl}*\frac{1molAgNO_3}{1molKCl} \\\\n_{AgNO_3}=0.021molAgNO_3[/tex]
Then, by using the volume of silver nitrate in liters (0.0258 L), we can directly compute the molarity:
[tex]M=\frac{0.021molAgNO_3}{0.0258L}\\ \\M=0.816M[/tex]
Regards.
When you look at an ant up close, using a convex lens, what do you see?
Answer:
You would be able to see the ants clearly with the unique body parts.
Explanation:
Convex lens is also known as Converging lens. It helps in converging rays of light to become a principal focus which is usually very clear and visible to the eyes. The converging lens when used to view the ant makes the ant appear very visible and the individual is able to see all the unique body parts of the ant. This type of lens is used in individuals who are longsighted.
Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the amount of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.850 A that flows for 60.0 min.
Answer:
Mass of Ga = 0.73694 gram
Explanation:
Given:
Current = 0.850 A
Time = 60 minutes
Find:
Amount of gas deposit.
Computation:
Total charge = Current × Time in second
Total charge = 0.850 × 60 × 60
Total charge = 3,060 C
Mole of electron = Total charge / Faraday constant [Faraday constant = 96,485.3329]
Mole of electron = 3,060 / 96,485.3329
Mole of electron = 0.0317146
Moles of Ga = 1/3 [Mole of electron]
Moles of Ga = 1/3 [0.0317146]
Moles of Ga = 0.01057
Mass of Ga = molar mass × Moles of Ga
Mass of Ga = 69.72 × 0.01057
Mass of Ga = 0.73694 gram
Suppose a reaction mixture, when diluted with water, afforded 300 mL of an aqueous solution of 30 g of the reaction product malononitrile [CH2(CN)2], which is to be isolated by extraction with ether. The solubility of malononitrile in ether at room temperature is 20.0 g/100 mL, and in water is 13.3 g/100 mL. What weight of malononitrile would be recovered by extraction with (a) three 100-mL portions of ether and (b) one 300-mL portion of ether
The given question is not complete, the complete question is:
Suppose a reaction mixture, when diluted with water, afforded 300 mL of an aqueous solution of 30 g of the reaction product malononitrile [CH2(CN)2], which is to be isolated by extraction with ether. The solubility of malononitrile in ether at room temperature is 20.0 g/100 mL, and in water is 13.3 g/100mL. The ratio of these quantities is equal to the partition coefficient, k, which equals What weight of malononitrile would be recovered by extraction of (a) three 100-mL portions of ether and (b) one 300-mL portion of ether? SHOW WORK (Can be written in pen and attached to report). Suggestion: For each extraction, let x equal the weight extracted into the ether layer. In part (a), for the first of the three extractions, the concentration of malononitrile in the ether layer is x/100 and in the water layer is (30-x)/100.
Answer:
The correct answer is 10 grams and 18 grams.
Explanation:
Based on the given question, 20 gram per 100 ml is the solubility of malononitrile in ether, and 13.3 gram per 100 ml is the solubility of malononitrile in water.
Thus, the ration of the solubility is,
Solubility in water/solubility in ether = 20/13.3 = 1.50
a) Let w be the weight of malononitrile extracted into water in every extraction. Then the concentration of the ether layer will be w/100. The concentration in the water layer will be 30-w/300. Now the ratio will be,
Ratio = w/100 / (30-w)/300
1.50 = w/100 * 300 (30-w)
w = 10
Hence, the weight of malononitrile recovered by extraction is 10 grams.
b) The concentration in the ether layer will be w/300. The concentration in the water layer will be (30-w) / 300. Now the ratio will be,
Ratio = w/300 / (30-w) / 300
1.50 = w/300 * 300 (30-w)
w = 18
Hence, 18 grams is the weight of malononitrile recovered by extraction.
What are extensive properties of Oxygen?
Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 4.3 g of hexane is mixed with 7.14 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
We can produce 6.20 grams of CO2
Explanation:
Step 1: Data given
Mass of hexane = 4.3 grams
Molar mass of hexane = 86.18 g/mol
Mass of oxygen = 7.14 grams
Molar mass of oxygen = 32.0 g/mol
Step 2: The balanced equation
2C6H14 + 19O2 → 12CO2 + 14H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles hexane = 4.3 grams / 86.18 g/mol
Moles hexane = 0.0499 moles
Moles oxygen = 7.14 grams / 32.0 g/mol
Moles oxygen = 0.2231 moles
Step 4: Calculate the limiting reactant
For 2 moles hexane we need 19 moles O2 to produce 12 moles CO2 and 14 moles H2O
Oxygen is the limiting reactant. It will completely be consumed ( 0.2231 moles). Hexane is in excess. There will react 2/19 * 0.2231 = 0.02348 moles
There will be porduced 12/19 * 0.2231 = 0.1409 moles CO2
Step 5: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 0.1409 moles * 44.01 g/mol
Mass CO2 = 6.20 grams
We can produce 6.20 grams of CO2
For the following reaction, draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.
When drawing hydrogen atoms on a carbon atom, either include all hydrogen atoms or none on that carbon atom, or your structure may be incorrect.
Select the correct IUPAC name for the organic reactant:
a) 2-methylbutene
b) 2-methyl-1-butene
c) 3-methyl-3-butened) 3-methylbutene
Answer:
The correct IUPAC name for the organic reactant is :
d) 3-methylbutene
Explanation:
Firstly the missing diagram is attached in the diagram below.
The objective of this question is to draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.
From the image attached below; we would see the reaction that occurs between the alkene and the HBr (hydrobromic acid). What really occur in the reaction is that; in the presence of HBr with an alkene compound a secondary 2° carbocation is usually formed. This secondary 2° carbocation formed is usually unstable, so what we called an hydride shift occurs (Markovnikov's product) here to form a stable tertiary 3° carbocation.
The correct IUPAC name for the organic reactant is : 3-methylbutene
What would cause a balloon to expand if taken to the top of a mountain?
O A. Increased molecular collision
O B. Increased amount of molecules
O C. Lowered temperature
D. Lowered pressure
Answer:
D. Lowered pressure
Explanation:
As you go to more altitude or height, the atmospheric pressure significantly lowers so the gas molecules are free to expand and take up as room as possible.
This is best explained by Boyle's law where pressure and volume are inversely related, where if one thing goes up another goes down. Here the pressure goes down, so volume increases and ballon expands.
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
How has human action affected the
population?
Answer:
Human action has affected the population in quite a negative way. Technological innovations and urbanization gave rise to a high degree of pollution on the land, air and water.
Emission of hydrocarbons from automobiles and factories are known to pollute and cause sicknesses related to the respiratory system thereby shortening the lifespan of humans. Pollution of water bodies also cause death of sea animals and thereby reducing food availability of humans.
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]
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.
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How many moles of h2 can be formed if a 3.25g sample of Mg reacts with excess HCl
Answer:
0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl
Explanation:
The balanced reaction is:
Mg + 2 HCl → MgCl₂ + H₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles react:
Mg: 1 moleHCl: 2 molesMgCl₂: 1 moleH₂: 1 moleBeing:
Mg: 24. 31 g/moleH: 1 g/moleCl: 35.45 g/molethe molar mass of the compounds participating in the reaction is:
Mg: 24.31 g/moleHCl: 1 g/mole + 35.45 g/mole= 36.45 g/moleMgCl₂: 24.31 g/mole + 2*35.45 g/mole= 95.21 g/moleH₂: 2*1 g/mole= 2 g/moleThen, by stoichiometry of the reaction, the following quantities of mass participate in the reaction:
Mg: 1 mole* 24.31 g/mole= 24.31 gHCl: 2 moles* 36.45 g/mole= 72.9 gMgCl₂: 1 mole* 95.21 g/mole= 95.21 gH₂: 1 mole* 2 g/mole= 2 gThen you can apply the following rule of three: if by stoichiometry 24.31 grams of Mg form 1 mole of H₂, 3.25 grams of Mg how many moles of H₂ will they form?
[tex]moles of H_{2} =\frac{3.25 grams of Mg*1 mole of H_{2} }{24.31 grams of Mg}[/tex]
moles of H₂= 0.134
0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl
0.134 moles of H₂ are formed by the reaction of 3.25 g of Mg with excess HCl.
Let's consider the balanced equation between Mg and HCl.
Mg + 2 HCl ⇒ MgCl₂ + H₂
The molar mass of Mg is 24.3 g/mol. The moles corresponding to 3.25 g of Mg are:
[tex]3.25 g \times \frac{1mol}{24.3g} = 0.134 mol[/tex]
The molar ratio of Mg to H₂ is 1:1. The moles of H₂ formed by 0.134 moles of Mg are:
[tex]0.134 mol Mg \times \frac{1molH_2}{1molMg} = 0.134molH_2[/tex]
0.134 moles of H₂ are formed by the reaction of 3.25 g of Mg with excess HCl.
Learn more: https://brainly.com/question/9743981
During chemical reaction 7.55gKI and 9.06g were allowed to react. How many grams of excess reagent are left over after the reaction is complete. Reaction: Pb(NO3)2(s) + 2KCI(s) > 2KNO3(s) + PbI(s)
Answer: 7.45 g of [tex]Pb(NO_3)_2[/tex] excess reagent are left over after the reaction is complete.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
a) [tex]{\text{Number of moles of} KI}=\frac{7.55g}{166g/mol}=0.045moles[/tex]
b) [tex]{\text{Number of moles of} Pb(NO_3)_2}=\frac{9.06g}{331.2g/mol}=0.027moles[/tex]
The balanced chemical reaction is :
[tex]Pb(NO_3)_2(s)+2KI(s)\rightarrow 2KNO_3(s)+PbI(s)[/tex]
According to stoichiometry :
2 moles of [tex]KI[/tex] require = 1 mole of [tex]Pb(NO_3)_2[/tex]
Thus 0.045 moles of [tex]KI[/tex] will require=[tex]\frac{1}{2}\times 0.045=0.0225moles[/tex] of [tex]Pb(NO_3)_2[/tex]
Thus [tex]KI[/tex] is the limiting reagent as it limits the formation of product and [tex]Pb(NO_3)_2[/tex] is the excess reagent as (0.045-0.0225) = 0.0225 moles are left
Mass of [tex]Pb(NO_3)_2=moles\times {\text {Molar mass}}=0.0225moles\times 331.2g/mol=7.45g[/tex]
Thus 7.45 g of [tex]Pb(NO_3)_2[/tex] of excess reagent are left over after the reaction is complete.
To determine the concentration of a sample of calcium hydroxide, 1.45M HCl is added drop-wise using a burst. Write the balanced net ionic equation for the acid-base reaction.
Answer:
H^+(aq) + OH^-(aq) —> H2O(l)
Explanation:
We'll begin by writing the balanced equation for the reaction.
2HCl(aq) + Ca(OH)2(aq) —> CaCl2(aq) + 2H2O(l)
Ca(OH)2 is a strong base and will dissociates as follow:
Ca(OH)2(aq) —> Ca^2+(aq) + 2OH^-(aq)
HCl is a strong acid and will dissociates as follow:
HCl(aq) —> H^+(aq) + Cl^-(aq)
Thus, In solution a double displacement reaction occurs as shown below:
2H^+(aq) + 2Cl^-(aq) + Ca^2+(aq) + 2OH^-(aq) —> Ca^2+(aq) + 2Cl^-(aq) + 2H2O(l)
To get the net ionic equation, cancel out Ca^2+ and 2Cl^-
2H^+(aq) + 2OH^-(aq) —> 2H2O(l)
H^+(aq) + OH^-(aq) —> H2O(l)
Heat is added to a 1.0-kg block of ice at OC. Determine if the process is
endothermic or exothermic. Explain your answer. *
Answer:
endothermic
Explanation:
Heat is added to make the process possible.
9. Predict the major products formed when: (a) Toluene is sulfonated. (c) Nitrobenzene is brominated. (b) Benzoic acid is nitrated. (d) Isopropylbenzene reacts with acetyl chloride and AlCl3. If the major products would be a mixture of ortho and para isomers, you should so state.
Answer:
a) ortho-para isomers predominates
b) 3-nitrobenzoic acid ( meta isomer predominates)
c) 3-bromo nitrobenzene ( meta isomer predominates)
d) the ortho- para isomers predominates
Explanation:
a) Toluene contains -CH3 which is an ortho- para- director hence the major product of the sulphonation of toluene should be the ortho- para isomers.
b) The major product of the nitration of benzoic acid is 3-nitrobenzoic acid. This is an electrophilic substitution in which the meta isomer predominates.
c) The meta isomer predominates giving 3-bromo nitrobenzene as the major product.
d) The isopropyl group is an ortho- para director hence the ortho- para isomers predominates .
