So mass of sulphur
[tex]\\ \rm\Rrightarrow 0.02\times 757\times 10^{6}[/tex]
[tex]\\ \rm\Rrightarrow 1514\times 10^4[/tex]
S+O_2–≥SO_2Moles =64/32=2
So
So tons of sulphur dioxide
[tex]\\ \rm\Rrightarrow 2(1516)10^4[/tex]
[tex]\\ \rm\Rrightarrow 3032\times 10^4ton[/tex]
Explanation:
2% of 2million tons of sulphur=
2÷100×7.57E9
That is;
[tex] \frac{2}{100} \times 7.57 \times 10 {}^{6} = 0.02 \times 7.57 \times 10 {}^{6} \\ = 2 \times 10 {}^{ - 2} \times7.57 \times 10 {}^{6} = 2 \times 7.57(10 {}^{ - 2 + 6}) \\ = 15.14 \times 10 {}^{4} (for \: sulphur) \\ for \: sulphur \: dioxide = so2 \\ the \: molar \: mass \: of \: so2 = 64g.mol {}^{ -1} \\ if \: 32grams \: of \: sulphur \: weighs \: 15.14 \times 10 {}^{4} tones \\ 64grams \: will \: weigh \: \frac{64}{32} \times 15.14 \times 10 {}^{4} \\ = 2 \times 15.14 \times 10 {}^{4} \\ = 30.28 \times 10 {}^{4} \\ or = 3.028 \times 10 {}^{5} tones[/tex]
1. A balloon contains 100g of oxygen gas (O₂) when the volume is 110 dm³. Find the volume of the balloon when the mass of oxygen gas is decreased to 50g. [take O=16] 1. A balloon contains 100g of oxygen gas ( O₂ ) when the volume is 110 dm³ . Find the volume of the balloon when the mass of oxygen gas is decreased to 50g . [ take O = 16 ]
The volume of the balloon when the mass of oxygen gas is decreased to 50g is 50L.
We will use the ideal gas equation-
PV=nRT
P=Pressure
V=volume
n=no. of moles
T=temperature
In this question temperature and pressure will remains constant then the above equation can be rewritten as-
[tex]V_{1} /n_{1} =V_{2} /n2[/tex]
where V1= initial volume of the balloon
V2= volume of the balloon when the mass of oxygen gas is decreased to 50g
now, substitute the values in the above equation-
V1=100dm3=100Ln1=100/32=3.12n2=50/32=1.56100/3.12=V2/1.56
V2=50 Lhence, the volume of the balloon when the mass of oxygen gas is decreased to 50g is 50L.
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In the laboratory you dissolve 12.2 g of iron(II) iodide in a volumetric flask and add water to a total volume of 500 mL.
What is the molarity of the solution?
M.
What is the concentration of the iron(II) cation?
M.
What is the concentration of the iodide anion?
M.
In the laboratory you dissolve 16.8 g of iron(III) sulfate in a volumetric flask and add water to a total volume of 250 mL.
What is the molarity of the solution?
M.
What is the concentration of the iron(III) cation?
M.
What is the concentration of the sulfate anion?
M.
You wish to make a 0.271 M hydrobromic acid solution from a stock solution of 12.0 M hydrobromic acid. How much concentrated acid must you add to obtain a total volume of 50.0 mL of the dilute solution?
mL
Question 1
1) The formula of iron(II) iodide is [tex]\text{FeI}_{2}[/tex] and has a formula mass of 309.654 g/mol. This means that in 12.2 grams, there are
12.2/309.654 = 0.039398812868557 mol.So, the molarity is (0.039398812868557)/(0.5) = 0.0788 M (to 3 sf)
2) In 0.039398812868557 mol of iron(II) iodide, there are 0.039398812868557 moles of iron(II) iodide cations, and thus the molarity is still 0.0788 M (to 3 sf)
3) In 0.039398812868557 mol of iron(II) iodide, there are
2(0.039398812868557) = 0.078797625737114 moles of iodide anionsThus, the molarity is (0.078797625737114)/(0.5) = 0.158 M (to 3 sf)
Question 2
1) The formula of iron(III) sulfate is [tex]\text{Fe}_{2}\text{(SO}_{4})_{3}[/tex] and has a formula mass of 399.9 g/mol. This means that in 16.8 grams, there are
16.8/399.9 = 0.042010502625656 molSo, the molarity is (0.042010502625656)/(0.25) = 0.168 M (to 3 sf)
2) In 0.042010502625656 moles of iron(III) sulfate, there are 2(0.042010502625656)=0.084021005251312 moles of iron(III) cations, so the molarity is:
0.084021005251312/0.25 = 0.336 M (to 3 sf)3) In 0.042010502625656 moles of iron(III) sulfate, there are 3(0.042010502625656)=0.12603150787697 moles of sulfate anions, so the molarity is:
0.12603150787697/0.25 = 0.504 M (to 3 sf)Question 3
Since we need 50.0 mL of solution at 0.271 M,
0.271 = (moles of HBr)/(0.05) moles of HBr = 0.01355 molSo, we need this from 12.0 M HBr so, meaning that:
12.0 = (0.01355)/(liters of 12.0 M HBr stock solution)liters of 12.0 M HBr stock solution = 0.0011291666666667 L = 1.13 M (to 3 sf)How many taste buds are present on the tongue? Name them.
there are five Taste bud present tongue they are -
SWEET SALT SOUR BITTER UMAMI ______________________out of the four type of papillie present on tongue the fungiform circumvallate and folite papillie contain numerous small taste but few such type of Taste but also occur in the mucous membranes of soft palate tonsils and epiglottis all of these days but received taste is normally
each Taste bud is an oval structures formed due to modification of epithelial cells which are situated erect and vertically on the surface long sensory cells and supporting cells are found in this bird is sensory cell is spindle shaped in which a sensory here is found at its free in the century fibre of the facial and vagus cranial nerves are wrapped around the basal and of the gustatory cells each test but open on the epithelial surface through minute taste pore
____________________Identify the products formed in this Brønsted-Lowry reaction.
HCO−3+BrO−↽−−⇀acid+base
Answer:
HCO3- + BrO- → HBrO + CO32-
Explanation:
I think this is the answer hope it helps
How many grams of solid barium sulfate form when 32.0 mL of 0.160 M barium chloride reacts with 70.0 mL of 0.065 M sodium sulfate? Aqueous sodium chloride forms also.
Considering the reaction's stoichiometry and the notion of molarity, 1.167 grams of solid barium sulfate form when 32.0 mL of 0.160 M barium chloride reacts with 70.0 mL of 0.065 M sodium sulfate.
The balanced reaction is:
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2 NaCl(aq)
What is a Limiting Reagent?The reactant that is consumed first in a chemical reaction is the limiting reagent because it stops any more reactions from taking place. The limiting reagent controls how much product is produced during the reaction.
Utilizing the reaction stoichiometry and the estimated number of moles, it is possible to identify the limiting reagent.
According to reaction stoichiometry, which describes the proportions of reagents and products in a chemical reaction, the following numbers of moles of each component are involved in the reaction:
BaCl₂: 1 moleNa₂SO₄: 1 moleBaSO₄: 1 moleNaCl: 2 molesMolarity, on the other hand, refers to how many moles of solute are dissolved in a certain volume.
Molarity is determined by the expression:
[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in units[tex]\frac{moles}{L}[/tex].
In this case, 32.0 mL= 0.032L (being 1000 mL= 1 L) of barium chloride reacts. So, by definition of molarity, the number of moles that participate in the reaction is calculated as:
[tex]0.160 M=\frac{No. of moles of solute}{0.032L}[/tex]
No of moles of barium chloride(solute) = 0.160M × 0.032L
No of moles of barium chloride(solute) = 0.00512moles
On the other side,70.0mL= 0.070 L of 0.065 M sodium sulfate reacts. So, by definition of molarity, the number of moles that participate in the reaction is calculated as:
[tex]0.065 M =\frac{No of moles of solute}{0.070L}[/tex]
No of moles of Sodium sulfate= 0.065M × 0.070L
No of moles of Sodium sulfate = 0.0045moles
if by stoichiometry 1 mole of barium chloride reacts with 1 mole of sodium sulfate, 0.00512moles of barium sulfate reacts with how many moles of sodium sulfate?
No. of moles of sodium sulfate=
[tex]\frac{ 0.00512moles of barium sulfate * 1 mole of sodium sulfate}{1 mole of barium chloride}[/tex]
amount of moles of sodium sulfate= 0.00512 moles
But 0.00512moles of sodium sulfate are not available, 0.0045moles moles are available. Since you have fewer moles than you need to react with 0.00512moles of barium chloride, sodium sulfate will be the limiting reagent.
Then, it is possible to determine the number of moles of barium sulfate produced by another rule of three: if by stoichiometry 1 mole of sodium sulfate produces 1 mole of barium sulfate, 0.00512 moles of sodium sulfate produces how many moles of barium sulfate?
[tex]Amt of mole of barium sulfate=\frac{0.00512 moles of sodium sulfate* 1mole of barium sulfate}{1 mole of sodium sulfate}[/tex]
amount of moles of barium sulfate= 0
Being the mass molar of barium sulfate is 233.34 g/mole, the mass-produced of the compound is calculated as:
= [tex]\frac{0.00512 moles * 233.34 gm}{1 mole}[/tex]
=1.167 gm
Hence, 1.167 gm of solid barium sulfate is formed.
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When developing an experimental design, which action would improve the
quality of the results?
O A. Have a different researcher make the measurements.
OB. Ensure that it answers a question about cause and effect.
C. Include as many responding variables as possible.
OD. Keep the sample size of test subjects to a minimum.
SUBMIT
Answer:
A
Explanation:
I'm right I took the test
Use the chemical equation to answer the question.
2H₂(g) + O₂(g) → 2H₂O(1)
Which statement describes the breaking and forming of bonds in the reaction?
The reaction requires breaking one H-H bond and two O=O bonds, and then forming four O-H bonds.
The reaction requires breaking one O=O bond and two H-H bonds, and then forming four O-H bonds.
The reaction requires breaking four O-H bonds, and then forming one H-H bond and two O=O bonds.
The reaction requires breaking four O-H bonds, and then forming one O=O bond and two H-H bonds
Answer: The reaction requires breaking one O=O bond and two H-H bonds, and then forming four O-H bonds.
How would you draw the ionic compound Beryllium Arsenide lewis structure?
Answer:
[Na]^+ [Cl]^-
Explanation:
Lets say its sodium, its number of electrons is 11, but when its stable (an ion), it is 10. and chloride, number of electrons is 17, but when its stable (an ion) it is 18. So the lewis structure for that is, remember with the straight brackets (not sure what it's called, but you know what I mean I guess) its this one: [ ]
Sodium will be + because it has more protons (11-10 = +1), and chloride will be - because it gained an electron, so has more electrons than protons (17-18 = -1)
So the lewis structure would be:
[Na]^+ for sodium
and
[Cl]^- for chlorine
Sodium chloride:
[Na]^+ [Cl]^-
Also just to add, only 1 of each atom (Na and Cl) was needed for the bonding, but if let's say example; 2 Cl was needed to bond with sodium, there would be 2 Cl (same) and 1 Na.
How many atom are present in a piece of iron that has a mass of 65.
0g
Iron has an atomic mass of 55.845 g/mol, so the amount of iron in 65.0 g is 65.0/55.845 = mol.
Avogadro's law states that in every mole of a substance, there are [tex]6.022 \times 10^{23}[/tex] particles per mole, so in this case, that means there are [tex](1.16393589)(6.022 \times 10^{23})=\boxed{7.01 \times 10^{23}} \text{ atoms (to 3 sf)}}[/tex]
Which of the following alkali metal is expected to melt if the room temperature
rise to 30oc (a) Na (b) K (c) Rb (d) Cs
A piece of metal with a mass of 23.2 g at 120.1oC is placed in a styrofoam cup containing 35.0 g of water at 22.2oC. Once the system has reached equilibrium, the final temperature of the water is 34.0oC. What is the specific heat of the metal to four significant digits? The specific heat of water is 4.18 J/g-oC.
Answer:
0.8797 J/g oC
Explanation:
use mcAT=mcaT
37) Which chemical equation is correctly balanced?
a) Al(OH)3(s) → Al₂O3(s) + 2 H₂O(g)
b) KI(aq) + Pb(NO3)2(aq) → KNO3(aq) + 4 Pbl₂(aq)
c) 2 Na3PO4(aq) + 3 CaCl₂(aq) →6 NaCl(aq) + Ca3(PO4)2(aq) to S
d) CsH12) +6 O₂(g)→ 5 CO₂(g) + 6 H₂O(g)
0 (0
Answer: c) [tex]2\text{Na}_{3}\text{PO}_{4}+3\text{CaCl}_{2} \longrightarrow 6\text{NaCl}+\text{Ca}_{3}\text{(PO}_{4})_{2}[/tex]
Which of these reactions are acid-base reactions?
Answer:
The fifth answer choice (NH3 + H2O = NH4+ + OH-)
Explanation:
An acid-base reaction occurs when an acid and base combine, and the acid, (which is H2O in this case) "donates" a proton, or an H+ to the base. In this reaction, H2O gave one of its H+ to NH3, to make NH4+ and OH-. NH4+ now has a positive charge because it gained a positive ion, while OH- now has a negative charge because it gained a negative ion.
For the reaction below determine the ▲H for the reaction and state whether the reaction was endothermic or exothermic. Show your work.
C2H5OH + 3 O2 → 2 CO2 + 3H2O
C-C = 83 kcal; C-H = 99 kcal; C-O = 84 kcal; O-H = 111 kcal; C=O 192 kcal; O=O = 119 kcal
I got stuck on this and need help please, thank you so much!
Answer:exothermice
Explanation: it is relaeing heat not keeping it in
explain how magic numbers are related to nuclear stabilize
Answer:
A magic number is a number of nucleons (either protons or neutrons, individually) that form complete bands within the atomic nucleus. As a result, atomic nuclei with a magic number of protons or neutrons are significantly more persistent than other nuclei.
Use Table B in your Student Guide to answer the questions about ion concentrations.
A solution with a pH = 13 has approximately how many moles of OH ions per liter?
How many moles of H* would this same solution have per liter?
(Use the decimal form of your answer.)
A different solution with an H+ concentration of 1.0 x 10-4 would have a pH =
Question 1
[tex]\text{pH}=-\log(\text{H}^{+})\\\\13=-\log(\text{H}^{+})\\ \\ \text{[H}^{+}]}=1 \times \boxed{10^{-13} \text{ moles per liter}} \\ \\ \text{[H}^{+}] \times \text{[OH}^{-}] =1 \times 10^{-14} \text{ M}\\\\\text{[OH}^{-}]=\boxed{1 \times 10^{-1} \text{ moles per liter}}[/tex]
Question 2
[tex]\text{pH}=-\log(\text{H}^{+})\\\\\text{pH}=-\log(1.0 \times 10^{-4})=\boxed{4}[/tex]
The average atomic mass of carbon is 12.01 amu. Based on the atomic
masses of the two isotopes of carbon, how do the relative abundances of the
isotopes compare?
Isotope
Atomic mass (amu)
C-12
12.000
C-13
13.003
OA. There is a slightly larger percentage of C-12 than C-13.
OB. There is a very large percentage of C-12.
OC. There is a very small percentage of C-12.
OD. They are about the same.
Answer: There is a very large percentage of C-12.
Explanation:
The atomic mass is must closer to that of C-12 than of C-13, meaning that there is a much larger percentage of C-12.
For the following reaction, 22.0 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas . nitrogen monoxide ( g ) + hydrogen ( g ) nitrogen ( g ) + water ( l ) What is the maximum amount of nitrogen gas that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams
NO is the limiting reagent and 4.34 g is the amount of the excess reagent that remains after the reaction is complete
What is a limiting reagent?The reactant that is entirely used up in a reaction is called as limiting reagent.
The reaction:
[tex]2NO(g) +2H_2(g)[/tex] → [tex]N_2 +2H_2O[/tex]
Moles of nitrogen monoxide
Molecular weight: [tex]M_(_N_O_)[/tex]=30g/mol
[tex]n_(_N_O_) =\frac{mass}{molar \;mass}[/tex]
[tex]n_(_N_O_) =\frac{22.0}{30g/mol}[/tex]
[tex]n_(_N_O_) = 0.73 mol[/tex]
Moles of hydrogen
Molecular weight: [tex]M_(_H_2_)[/tex]=30g/mol
[tex]n_(_H_2_) =\frac{mass}{molar \;mass}[/tex]
[tex]n_(_H_2_) =\frac{5.80g}{2g/mol}[/tex]
[tex]n_(_H_2_) = 2.9 mol[/tex]
Hydrogen gas is in excess.
NO is the limiting reagent.
The amount of the excess reagent remains after the reaction is complete.
[tex]n_(_N_2_) =[/tex] (2.9 mol- 0.73 mol NO x [tex]\frac{1 \;mol \;of \;H_2}{2 \;mole \;of \;NO}[/tex]) x [tex]\frac{2g \;of \;H_2}{mole \;of \;H_2}[/tex]
[tex]n_(_N_2_) =[/tex]4.34 g
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Which buffer can be suitable to prepare a solution that needs to be buffered at ph 4?
Answer:
For pH= 4.00 :
Add 0.1 ml of 0.1 molar NaOH to 50 ml of 0.1 molar potassium hydrogen phthalate .
Hazmat Poison gas is in what class?
HAZMAT Class 6 Toxic and infectious substances.
Answer:
CLASS 2 Gases
Explanation:
Hope this helps
Determine the trends in radii for the following ions and atoms: Mg2+, F- , Ne, and Na+ .explain your choice
The trend for the radii is Mg²⁺ < Na+ < Ne <F⁻
What does Ionic size varies with number of protons ?With the increase in number of protons in the nucleus the electrons are more attracted towards nucleus and so the ionic radius decreases.
Ionic size increases with the addition of electrons and ionic size decreases with the loss of electrons.
Higher is the atomic number, higher will be the effective nuclear charge, lower will be the size.
F⁻
no. of proton=9
no of electron = 10
Na+
no. of proton=11
no of electron=10
Mg²⁺
no. of proton=12
no of electron=10
Ne
no. of proton=10
no of electron=10
Mg²⁺ < Na+ < Ne <F⁻
Therefore the trend for the radii is Mg²⁺ < Na+ < Ne <F⁻
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An unknown compound contains only C , H , and O . Combustion of 6.80 g of this compound produced 16.0 g CO2 and 4.37 g H2O . What is the empirical formula of the unknown compound?
Answer:
I am a small palm-sized pointing device
What is the percent yield for the reaction below when
705.0 g SO2 and 80.0 g 0₂ produce 586.0 g SO3?
2SO2(g) + O₂(g) → 2SO3(g)
The percent yield for the chemical reaction producing sulfur trioxide is 66.51%.
What is percent yield?Percent yield is defined as the ratio of actual yield to the theoretical yield multiplied by 100. If the actual and theoretical yield are same then the percent yield is 100%.If actual yield is less than the theoretical yield then the percent yield is less than 100%.Reason of this condition arising is the incompletion of reaction or loss of sample during recovery process.
In cases where percent yield is over 100% it indicates that more sample is recovered than the predicted amount.
For the given reaction, theoretical yield is calculated as,
128.132 g of sulfur dioxide gives 160.12 g of sulfur trioxide
∴705 g of sulfur dioxide gives 705×160.12/128.132= 881 g
Percent yield is, 586/881×100=66.51%
Therefore, the percent yield for the chemical reaction is 66.51%.
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A doctor prescribed a tablet to a patient suffering from iron deficiency. The tablet does not look like iron. Explain.
Answer:
Tablets are made from salts of iron hence it does not look like iron.
Explanation:
Write and balance molecular equations for the following reactions between aqueous solutions. You will need to decide on the formulas and phases of the products in each of the cases.
An aqueous solution of barium hydroxide is mixed with an aqueous solution of sulfuric acid.
Answer:
Ba(OH)2(aq) + H2SO4(aq) ----> BaSO4(s) + 2H2O(l)
Question 3
What part of a water molecule is the negatively charged side?
Answer:
The oxygen atom in water has a negative charge.
Definitions:Polar molecule: A molecule in which the centroid of the positive charges is different from the centroid of the negative charges.
Oxygen: A colorless, odorless, gaseous element constituting about one-fifth of the volume of the atmosphere and present in a combined state in nature. It is the supporter of combustion in air and was the standard of atomic, combining, and molecular weights until 1961, when carbon 12 became the new standard. Symbol: O; atomic weight: 15.9994; atomic number: 8; density: 1.4290 g/l at 0°C and 760mm pressure.
Water: A transparent, odorless, tasteless liquid, a compound of hydrogen and oxygen, H2O, freezing at 32°F or 0°C and boiling at 212°F or 100°C. that in more or less impure state constitutes rain, oceans, lakes, rivers, etc.: it contains 11.188 percent hydrogen and 88.812 percent oxygen, by weight.
Hydrogen: A colorless, odorless, flammable gas that combines chemically with oxygen to form water: the lightest of the known element. Symbol: H; atomic weight: 1.00797; atomic number: 1; density: 0.0899 g/l at 0°C and 760 mm pressure.
Atom: Am atom is the smallest constituent particle of a chemical element which has the properties of that element. They re comprised of at least an electron and a portion, as is the case for Hydrogen. Atoms of all other elements however, contain at least one neutron.
Proton: A positively charged elementary particle that is a fundamental constituent of all atomic nuclei. It is the lightest and most stable baryon, having a charge equal in magnitude to that of the electron, a spin of 1/2, and a mass of 1.673 × 10-27kg. Symbol: P.
Electron: An elementary particle that is a fundamental constituent of matter, having a negative charge of 1.602 × 10-19 coulombs, ha mass of 9.108 × 10-31 kilograms, and spin of 1/23, and existing independently or as the component outside the nucleus of an atom.
Neutron: An elementary particle having no charge, mass slightly greater than that of a proton, and spin of 1/2: a constituent of the nuclei of all atoms except those of hydrogen. Symbol: n.
What is a negative charge?A object has a negative charge when it consists of more electrons than protons.
With a partial positive charge on 2 Hs and a partial negative charge on oxygen, water molecules are polar molecules.
Actually, two hydrogen atoms and the core oxygen atom of water are covalently connected. Due to its greater electronegative nature than hydrogen, oxygen pulls the bound electron pair in the middle of the atom toward it.
When illustrating higher electron densities over oxygen, partial negative charge is used, whereas partial positive charge is used to illustrate lower densities over hydrogen atoms.
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Chromium (III) oxide reacts with hydrogen sulfide gas to form chromium (III) sulfide and water. To produce 421 g of cr2s3, how many moles of cr2o3 and grams of cr2o3 are required?
Cr₂O₃ ( s ) + 3H₂S ( g ) → Cr₂S₃ ( s ) + 3H₂O ( l )
mol Cr₂S₃ = 421 : 200.19 g/mol = 2.103
mol Cr₂O₃ ≈ mol Cr₂S₃ = 2.103 ( equivalent coefficient)
mass Cr₂O₃ = 2.103 x 151.99 g/mol = 319.63 gr
Question 26 of 30
For a reaction, AH = -75 kJ/mol and SO = -0.081 kJ/(K.mol). At what
A
temperatures is this reaction spontaneous?
O A. T<930 K
OB. At all temperatures
C. T< 100 K
OD. T> 930 K
Answer: B
Explanation:
The heat capacity of nickel is 0.444 J/(g · °C). Calculate the amount of heat needed to raise the temperature of 12 g of nickel from 23 °C to 69 °C. Now imagine those same joules were used instead to accelerate the same mass of nickel from rest. What would be the final speed, in m/s
Answer:245.088
Explanation: you subtract final temperature with initial temperature = 46 then multiply by grams = 552 then multiplied by J = 245.088
Which data are shown on topographic maps?
Answer:
Topographic maps generally shows Natural features such as Rivers and Lakes.
sometimes they also show some features made by humans like brodges, elevations etc.
Answer:
Topographic maps show contours, elevation, forest cover, marsh, pipelines, power transmission lines, buildings, and various types of boundary lines such as international, provincial, and administrative, and many others.
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