Answer: B
Explanation: adding all the letters together let’s you compare to the other side of the equation. If the have the same number it is right.
The correct answer is option B. 2AB3 + 3C2D – 6CB + A2D3
How do you balance a chemical equation?
The Algebraic Balancing Method. This method of balancing chemical equations involves assigning algebraic variables as stoichiometric coefficients to each species in the unbalanced chemical equation.
What are the 3 rules for balancing equations?
To balance equations on your own, follow these simple rules:
Check that all the formulae in the equation are correct.Deal with only one element at a time.Balancing is adding BIG numbers. You cannot change any of the small numbers in a chemical formula. Check each element again and repeat step 3 again if needed.Learn more about chemical equation balancing here: https://brainly.com/question/11904811
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Helium gas is being pumped into a rigid container at a constant temperature. As a result, the pressure of helium in the container is increasing. Select the one correct statement below.
a) As the pressure increases, helium atoms move faster, on average.
b) As the pressure increases, helium atoms move more slowly, on average.
c) As the pressure increases, helium atoms stay closer to the wall of the container, on average.
d) As the pressure increases, the volume of the container must decrease.
e) As the pressure increases, there are more collisions of helium atoms with the container wall.
Answer: e. As the pressure increases, there are more collisions of helium atoms with the container wall.
Explanation:
We are given the information that helium gas is pumped into a rigid container at a constant temperature.
Due to this reason, there was a rise in the pressure of helium in the container. We should note that as the pressure rises, there are more collisions of helium atoms with the container wall.
The increase in the pressure of He atoms results in the increased collision of atoms with the wall. Thus, option E is correct.
The helium in the container is assumed to be an ideal gas. The ideal gas reaction is given as:
[tex]PV=nRT[/tex]
Where,
P= PressureV=volumen=molesR=Rydberg constantT=temperatureWhat happens to gas with an increase in pressure?The ideal gas equation depicts that the gas is contained in a closed container. Thus, the moles, n of gas are constant. The pressure is pumped at a constant temperature, thereby T is also constant.
The R is a constant in the ideal gas equation. The increase in the pressure of the container results in the increased kinetic energy of the helium atoms.
The increase in the collisions is observed for the molecules with the container wall.
Thus, with the increase in pressure, the collisions in the atoms increase. Hence, option E is correct.
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How many atoms are in 34 g of Cu?
Answer:
Babygirl these are double DD'S
Explanation:
OooOOO I'm hurt oh my neck my neck and my back
What is the balanced equation of:
2Mg+ H2So4
Answer: CH3COO)2Mg + H2SO4 = MgSO4 + CH3COOH
Explanation:
Hope. This helps
How much energy is required to remove a neutron from the nucleus of an atom of carbon-13?
Answer:
uh i think 12?
Explanation:
Some hypothetical alloy is composed of 25 wt% of metal A and 75 wt% of metal B. If the densities of metals A and B are 6.17 and 8.00 g/cm3 , respectively, and their respective atomic weights are 171.3 and 162.0 g/mol, determine whether the crystal structure for this alloy is simple cubic, facecentered cubic, or body-centered cubic. Assume a unit cell edge length of 0.332 nm
Answer:
Simple cubic
Explanation:
The density of metal A (ρa) = 6.17 g/cm³, The density of metal B (ρb) = 8 g/cm³, The atomic weight of metal A (Aa) = 171.3 g/mol, The atomic weight of metal B (Ab) = 162 g/mol, the unit cell edge length (a) = 0.332 nm, concentration of metal A (Ca) = 25%, concentration of metal B (Cb) = 75%
The average density is given by:
[tex]\rho_{ave}=\frac{100}{\frac{C_a}{\rho_a} +\frac{C_b}{\rho_b} } \\\\\rho_{ave}=\frac{100}{\frac{25}{6.17} +\frac{75}{8} } =7.45\ g/cm^3\\\\The\ average\ atomic\ weight\ is:\\\\A_{ave}=\frac{100}{\frac{C_a}{A_a} +\frac{C_b}{A_b} } \\\\A_{ave}=\frac{100}{\frac{25}{171.3} +\frac{75}{162} } =164.23\ g/mol\\\\The\ number\ of\ atoms\ per\ unit(n)\ is:\\\\n=\frac{\rho_{ave}*a^3*N_A}{A_{ave}} \\\\N_A=Avogadro\ constant=6.02*10^{22} \ mol^{-1},a=0.332\ nm=3.32*10^{-8}cm\\\\Substituting:\\\\[/tex]
[tex]n=\frac{\rho_{ave}*a^3*N_A}{A_{ave}} =\frac{7.45*(3.32*10^{-8})^3*6.02*10^{23}}{164.23} \\\\n=0.999\\[/tex]
n≅1
Since n≅1, the crystal structure for this alloy is simple cubic
What are metalloids?
A metalloid is a type of chemical element which has a preponderance of properties in between, or that are a mixture of, those of metals and nonmetals.
Answer:
Metalloid, in chemistry, an imprecise term used to describe a chemical element that forms a simple substance having properties intermediate between those of a typical metal and a typical nonmetal. There is no single property which can be used to unambiguously identify an element as a metalloid.
Select all of the Earth Systems that contain carbon.
Oa Atmosphere
Ob Geosphere
Oc Biosphere
Od Hydrosphere
Answer:
a, c, d
Explanation:
biosphere atmosphere hydrosphere lithosphere
Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . Suppose 14. g of hydrochloric acid is mixed with 6.55 g of sodium hydroxide. Calculate the minimum mass of hydrochloric acid that could be left over by the chemical reaction. Round your answer to significant digits.
Answer:
8.02 g of HCl could be left over by the chemical reaction
Explanation:
We propose the reaction:
HCl(aq) + NaOH (s) → NaCl (aq) + H₂O (l)
Ratio is 1:1. First of all, we determine the moles of reactants:
14 g . 1mol / 36.45g = 0.384 mol of acid
6.55 g. 1mol / 40g = 0.164 mol of base
If a determined mass of HCl, could be left; this means that the acid is the excess reagent.
For 0.164 moles of NaOH, we need 0.164 moles of HCl.
As we have 0.384 moles, (0.384 - 0.164) = 0.220 moles of acid are left over in the reaction. We convert the moles to mass:
0.220 mol . 36.45 g /1mol = 8.02 g
Practice Problem Website: https://www3.nd.edu/~smithgrp/structure/workbook.html
1. Click on Do the problems
2. Click on the number for the practice problem to be completed
3. Click on IR
Example of problem 1 from the website:
Formula: C3H5BrO2
IR Peaks:
1717 nm—Strong peak indicates a carbonyl group C=O
2571-2670 nm- medium peak indicates sp3 hybrid C-H.
3067 nm --- Broad medium peak indicates OH group, specifically in a carboxylic acid (Would not be a carboxylic acid without the carbonyl peak as well)
Complete the following problems: 3, 5, 7, 8, 12, 13, 27, 32, 38, and 40.
Be as accurate as possible. I am looking for the frequency and molecular formula to make sense with the functional group you think is represented.
When 9.00 g C6H12O6 burns with excess oxygen, how many grams
of CO2 will be made?
- 0.44 g
- 6.6 g
- 2.2 g
- 13.2 g
............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
Answer:
Thank you for the free points!!
Explanation:
Choose all true statements about the water molecule: Choose all true statements about the water molecule: Bonding between the oxygen and hydrogen within the molecule happens due to hydrogen bonding. Water is a permanent electric dipole, having permanent charge separation. Water is not naturally polarized, and gets polarized only due to external electric forces.
Answer: Water is a permanent electric dipole, having permanent charge separation.
Explanation:
Hydrogen bonding is an intermolecular force having partial ionic-covalent character.
In [tex]H_2O[/tex], O is a highly electronegative atom attached to a H atom through a covalent bond. The oxygen atoms being more electronegative gets partial negative charge and H atom gets partial positive charge. Thus water is permanent electric dipole.
Hydrogen bonding takes place between a hydrogen atom (attached with an electronegative atom O) and an electronegative atom (O).
The true statement about water molecules are Bonding between the oxygen and hydrogen within the molecule happens due to hydrogen bonding and water is a permanent electric dipole, having permanent charge separation.
What are hydrogen bonds?Hydrogen bonds is a kind of attraction force which is present between the more electronegative atom and hydrogen atom.
In water molecule, 2 hydrogen atoms and one oxygen atom is present and between the more electronegative oxygen atom and hydrogen atom, hydrogen bond is present.And due to electronegative nature of oxygen atom, they withdraw all the electron density of the bond towards itself as a result of which it possesses permanent electric dipole. Because partial negative charge is present on oxygen atom and partial positive charge is present on the hydrogen atom.Hence options (1) and (2) are correct.
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What is the percent enantiomeric excess (ee) of a mixture that has 86% of one enantiomer and 14% of the other
Answer:
86% - 14% = 72%
Explanation:
True or False: In a graph the independent variable will always go on the side.
A.True
B.False
Answer:
False, the independent variable goes on the bottom
The statement “In a graph, the independent variable will always go on the side.” is false. The correct option is B.
What is a graph?A graph is a simple way of representing data. It shows the relation between two or more data with lines, numbers, and charts. The mathematical graphs are of different types. Mainly, a graph has a line and an x-axis and y-axis.
The x-axis contains the independent variable, which is written n the bottom of the graph, horizontally. The y-axis contains the dependent variable, which is written on the left side of the graph vertically.
The independent variable change, but the dependent variable do not change. It is the input. The independent variable changes with the time or what is written on the dependent variables.
Thus, the correct option is B. False.
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The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-1. If the initial pressure of N2O is 3.50 atm at 730°C, calculate the total gas pressure after one half-life. Assume that the volume remains constant. Insert your answer in decimal notation rounded to 3 significant figures.
Answer:
the total gas pressure after one half-life is 4.38 atm
Explanation:
The balanced equation for the decomposition of N2O to N2 and O2 is given as:
2N₂O(gas) ⇒ 2N₂(gas) + O₂(gas)
2 moles of N₂O produce 2 moles of N₂ and 1 mole of O₂
The change in pressure depends on the coefficient (number of moles) of the reactant and product.
N₂O N₂ O₂
number of moles 2 2 1
Initial pressure (atm) 3.50 0 0
change in pressure -2x +2x x
Final pressure (atm) 3.50 - 2x 2x x
The total final pressure is the sum of the individual total pressure. i.e.:
Total final pressure = final pressure of N₂O + final pressure of N₂ + final pressure of O₂
Total final pressure = (3.5 - 2x) + (2x) + x
Total final pressure = 3.5 + x
After one half life, the initial pressure of N₂O would be half its value.
Final pressure of N₂O = half of the initial pressure of N₂O
3.5 - 2x = 0.5(3.5)
3.5 - 2x = 1.75
2x = 1.75
x = 0.875 atm
Therefore, Total final pressure = 3.5 + x = 3.5 + 0.875
Total final pressure = 4.38 atm to 3 significant figures
Strontium chlorate is mixed with ammonium phosphate
Explanation:
Please, if I may ask, what is the exact question?
Which word best describes gamma rays?
A)Weak
B)Destructive
C)Visible
D)Harmless
Answer:
Explanation:
Destructive :D
According to the position of gamma rays in electromagnetic spectrum,the word which best describes gamma rays is destructive as they have high energy.
What is electromagnetic spectrum?The electromagnetic spectrum consists of electromagnetic radiation which consists of waves made up of electromagnetic field which are capable of propogating through space and carry the radiant electromagnetic energy.
The radiation are composed of electromagnetic waves which are synchronized oscillations of electric and magnetic fields . They are created due to change which is periodic in electric as well as magnetic fields.
In vacuum ,all the electromagnetic waves travel at the same speed that is with the speed of air.The position of an electromagnetic wave in an electromagnetic spectrum is characterized by it's frequency or wavelength.They are emitted by electrically charged particles which undergo acceleration and subsequently interact with other charged particles.
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how many molecules of sugar are in 4.67 miles of sugar?
there are 1.204 x 10^24 atoms of oxygen. How many moles is this?
Answer:
7.25 x 10^47
Explanation:
1.204 x 10^24 moles*6.022 x 10^23 avogadro's number= 7.25 x 10^
Two atoms of the element join together at ordinary temperature and pressure to generate dioxygen. Therefore, the correct option is option A.
What is oxygen?The chemical substance with symbol O as well as atomic number 8 is oxygen. It is a component of the periodic table's chalcogen group, a strong oxidizing nonmetal, as well as an oxidizing agent which quickly produces oxides for most elements and other compounds. Oxygen is the most prevalent element on Earth, and it is the third-most plentiful element in the universe after hydrogen and helium.
Two atoms of the element join together at ordinary temperature and pressure to generate dioxygen, a colorless as well as fragrant diatomic gas with both the formula O 2. Diatomic oxygen gas now makes up 20.95% of the Greenhouse effect, however this has fluctuated dramatically throughout time. Oxygen oxides represent more than half of both the Earth's crust.
mole = number of atoms/ 6.022×10²³
= 1.204 x 10²⁴/ 6.022×10²³
= 2 mole
Therefore, the correct option is option A.
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QUESTION 3
How many atoms of Xe does 7.84 moles of Xe contain?
Mole (mol): the amount of material counting 6.02214 × 1023 particles. The value of the mole is equal to the number of atoms in exactly 12 grams of pure ... A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in ... Answers. 1. (a) 2.26 × 10. 24 silver atoms; (b) 1.42 × 10. 24 xenon atoms (c) 3.07 × 10.
Explanation:
54.1miles/gallons how many liters of gas will be consumed traveling 132 km
Answer:
5.75 L.
Explanation:
From the question given above, the following data were obtained:
Rate = 54.1 miles/gallons
Distance = 132 km
Volume (in L) consumed =?
Next, we shall convert 132 km to mile. This can be obtained as follow:
1 km = 0.621 mile
Therefore,
132 km = 132 km × 0.621 mile / 1 km
132 km = 81.972 mile
Next, we shall determine the volume (in gallons) of the gas needed. This can be obtained as follow:
Rate = 54.1 miles/gallons
Distance = 81.972 mile
Volume (in gallon) =?
Rate = Distance / volume
54.1 = 81.972 / volume
Cross multiply
54.1 × volume = 81.972
Divide both side by 54.1
Volume = 81.972 / 54.1
Volume = 1.52 gallon.
Finally, we shall convert 1.52 gallon to litre (L). This can be obtained as follow:
1 gallon = 3.785 L
Therefore,
1.52 gallon = 1.52 gallon × 3.785 L / 1 gallon
1.52 gallon = 5.75 L
Therefore, 5.75 L of the gas will be consumed.
How many significant figures
are in this number?
3 x 10^6
Copper(II) chloride hydrate, also known as copper chloride hydrate, is the inorganic compound with the chemical formula CuCl2 . x(H2O), where x can range from 1 to 7. A 2.060 g sample of copper (II) chloride hydrate was dissolved in 50.0 mL of deionized water. The concentration of copper (II) ion in the solution was determined spectrophotometrically using the equation from the calibration curve, was 0.200 M. Determine the formula of the copper (II) chloride hydrate.
Answer:
CuCl₂·4H₂O
Explanation:
First we calculate the moles of copper (II) ion in the solution, using the given volume and concentration:
50.0 mL ⇒ 50.0/1000 = 0.05 L0.05 L * 0.200 M = 0.01 mol Cu⁺²For each Cu⁺² mol there's also one CuCl₂ mol.
We convert 0.01 CuCl₂ moles into grams, using its molar mass:
0.01 CuCl₂ mol * 134.45 g/mol = 1.344 gThat means that out of the 2.060 g of the sample, 1.344 g are of CuCl₂. This means that there are (2.060 - 1.344) 0.716 g of water.
We convert those 0.716 g of water to moles:
0.716 g ÷ 18 g/mol = 0.04 mol H₂OThere are four times as many H₂O moles as there are CuCl₂ moles. This means the formula is CuCl₂·4H₂O
When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solution has a significantly higher concentration of the given solute (typically 101 to 104 times higher than those of the diluted solutions). The high concentration allows many diluted solutions to be prepared using minimal amounts of the stock solution. What volume of a 6.57 M stock solution do you need to prepare 500. mL of a 0.0490 M solution of HCl
Answer:
3.73 mL
Explanation:
To solve this problem we can use the equation C₁V₁=C₂V₂, where:
C₁ = 6.57 MV₁ = ?C₂ = 0.0490 MV₂ = 500 mLWe input the data:
6.57 M * V₁ = 0.0490 M * 500 mLAnd solve for V₁:
V₁ = 3.73 mLSo 3.73 mL of a 6.57 M stock solution would be required to prepare 500 mL of a 0.0490 M solution.
Does anyone know how to create algae in a water bottle
What are the names of the stable forms of oxygen?
Answer:
18 O, 17 O, and 16 O
Explanation:
three naturally stable isotopes
The stable forms of oxygen are molecular oxygen ([tex]O_{2}[/tex]) and ozone ([tex]O_{3}[/tex]). Molecular oxygen is the most common form in the Earth's atmosphere, while ozone is found in the ozone layer of the atmosphere and has a different molecular structure than [tex]O_{2}[/tex].
The most stable type of oxygen in the Earth's atmosphere is [tex]O_{2}[/tex], sometimes referred to as molecular oxygen or dioxygen. It is made up of two oxygen atoms joined by a link.
Another stable form of oxygen is [tex]O_{3}[/tex], also referred to as ozone. It is a molecule made up of three linked oxygen atoms. The ozone layer in the stratosphere of the Earth contains ozone, which is essential for protecting life on the planet by absorbing damaging ultraviolet (UV) radiation from the Sun.
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Can someone help me with this? And provide an explanation on how they found their answer? Using conversion factors. I'm confused :(
You have used 3.0×102 L of distilled water for a dialysis patient. How many gallons of water is that?
Answer:
This would be 63 gallons :)
Explanation:
The Volume of water = 2.4 × 10² L
The Volume of water in gal = ?
The solution: We know that one gal is equal to 3.785 litter.
So in conclusion, 2.4 × 10²L × 1 gal / 3.785 L
2.4 × 10²L × 0.264 gal. L⁻¹
0.634 × 10² gal
Hopefully this helps :3
63 gal
A careless student forgets to label one of their reagent containers. They know it contains one of the following solutions: 0.1 M NaCl, 0.1 M Ca(NO3)2, 0.1 M KCH3COO The student decides to add some 0.1 M Pb(NO3)2 to a sample of the reagent in question. Upon stirring, a white precipitate formed. Which of the possible solutions was present in the unlabeled container based on this observation
Answer:
He kinda lazy ngl
Explanation:
Rgheeghfgufugf
Name this molecule c4h7bro
Answer:
I believe its : "2-Bromobutanal"
Phosphorus is commercially prepared by heating a mixture of calcium phosphate, sand, and coke in an electric furnace. The process involves two reactions. 2 Ca3(PO4)2(s) 6 SiO2(s) 6 CaSiO3(l) P4O10(g) P4O10(g) 10 C(s) P4(g) 10 CO(g) The P4O10 produced in the first reaction reacts with an excess of coke (C) in the second reaction. Determine the theoretical yield of P4 if 293.5 g Ca3(PO4)2 and 378.5 g SiO2 are heated. (No Response) g If the actual yield of P4 is 44.9 g, determine the percent yield of P4.
Answer:
76.6% is percent yield of P₄
Explanation:
Percent yield is defined as 100 times the ratio of actual yield and theoretical yield. To solve this quesiton we need to find the theoretical yield of the reaction. Using:
2Ca₃(PO₄)₂(s) + 6SiO₂(s) → 6CaSiO₃(l) + P₄O₁₀(g)
P₄O₁₀(g) + 10C(s) → P₄(g) + 10CO(g)
We need to find the moles of Ca₃(PO₄)₂ and SiO₂ to find limiting reactant. With limiting reactant we can find moles of P₄O₁₀ = Moles of P₄. We must convert the moles of P₄ to mass using Molar mass (P₄ = 123.895g/mol):
Moles Ca₃(PO₄)₂ -Molar mass: 310.1767g/mol-
293.5g * (1mol / 310.1767g) = 0.9462moles
Moles SiO₂ -Molar mass: 60.08g/mol-:
378.5g * (1mol / 60.08g) = 6.30 moles
For a complete reaction of 6.30 moles of SiO₂ there are required:
6.30 moles SiO₂ * (2 moles Ca₃(PO₄)₂ / 6 moles SiO₂) =
2.10 moles Ca₃(PO₄)₂. As there are just 0.9462 moles, Ca₃(PO₄)₂ is limiting reactant
Moles P₄O₁₀ = Moles P₄:
0.9462moles Ca₃(PO₄)₂ * (1mol P₄O₁₀ / 2 mol Ca₃(PO₄)₂) = 0.4731 moles P₄O₁₀ = Moles P₄.
The mass is:
0.4731 moles P₄ * (123.895g / 1mol) = 58.6g = Theoretical yield.
Percent yield is:
44.9g / 58.6g * 100 =
76.6% is percent yield of P₄