Answer:
alkanes
Alkynes have the strongest bond. Followed by alkenes and then alkanes. This is because in alkynes there are 3 bonds between at least two carbons
How much Ba(OH)2, in grams, is needed to make a 1.00m aqueous solution?
Answer:
We can say that we need 171 g barium hydroxide needed to make one model equals solution.
Explanation:
I hope this answered your question.
I will give brainlist
How much Ba(OH)2, in grams, is needed to make a 1.00m aqueous solution?
Answer:
We can say that we need 171 g barium hydroxide needed to make one model equals solution.
Explanation:
I hope this answered your question.
171 gram of Ba(OH)2, in grams, is needed to make a 1.00 m aqueous solution.
What is mass?Mass is the quantity of matter. The mass is calculated by the density divide by the volume.
In 1.00 solution, the mass of Ba(OH)2 is 171 gram.
Thus, the mass of Ba(OH)2 is 171 gram.
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what is another extraction that uses gravity filtration and describe it
Answer -ˋˏ ༻༺ ˎˊ-
A common use for gravity filtration is for separating anhydrous magnesium sulfate (MgSO4) from an organic solution that it has dried (Figure 1.68b). Anhydrous magnesium sulfate is powdery, and with swirling in an organic solvent creates a fine dispersal of particles like a snow globe.
Which diagram best illustrates the difference between distance and
displacement?
Answer:
It is the one that has start and finish and a compass on it...I am sorry I can share images for some reason. And underneath the compass, it says 1.0 Mile. It also has a compass in the bottom right-hand corner.
Explanation:
I just took the quiz and got it correct!
How are mass and weight alike?
A. Both change with gravitational force.
B. Both are measured in pounds.
C. Both depend on how much matter an object has.
D. Both are the same everywhere in the universe.
Answer:
C. Both depend on how much matter an object has.
Explanation:
weight is the gravitational force acting on an object, so it does vary depending on gravity.
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How many atoms can be found in a sample of 75g of silicon
Answer:
[tex]\boxed {\boxed {\sf 1.6 \times 10^{24} \ atoms \ Si}}[/tex]
Explanation:
We are asked to find how many atoms are in a sample of 75 grams of silicon.
1. Convert Grams to MolesFirst, we must convert grams to moles using the molar mass. This is the mass of 1 mole of a substance. The molar mass is found on the Periodic Table because it is equal to the atomic mass, but the units are grams per mole instead of atomic mass units.
Look up the molar mass of silicon.
Si: 28.085 g/molWe convert using dimensional analysis, so we must set up a conversion factor.
[tex]\frac {28.085 \ g \ Si}{1 \ mol \ Si}[/tex]
We are converting 75 grams of silicon to moles, so we multiply by this value.
[tex]75 \ g \ Si*\frac {28.085 \ g \ Si}{1 \ mol \ Si}[/tex]
Flip the conversion factor so the units of grams of silicon cancel.
[tex]75 \ g \ Si*\frac {1 \ mol \ Si}{28.085 \ g \ Si}[/tex]
[tex]75 *\frac {1 \ mol \ Si}{28.085 }[/tex]
[tex]\frac {75}{28.085} \ mol \ Si[/tex]
[tex]2.670464661 \ mol \ Si[/tex]
2. Convert Moles to AtomsNext, we convert moles to atoms using Avogadro's Number, or 6.022 ×10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are atoms of silicon.
Set up another conversion factor, this time with Avogadro's Number.
[tex]\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}[/tex]
Multiply by the number of moles we found.
[tex]2.670464661 \ mol \ Si*\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}[/tex]
The units of moles of silicon cancel.
[tex]2.670464661 *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 }[/tex]
[tex]2.670464661 * {6.022 \times 10^{23} \ atoms \ Si}[/tex]
[tex]1.60815382 \times 10^{24} \ atoms \ Si[/tex]
The original value of grams (75) has 2 significant figures, so our answer must have the same. For the number we found, that is the tenths place. The 0 in the hundredths place tells us to leave the 6 in the tenths place.
[tex]1.6 \times 10^{24} \ atoms \ Si[/tex]
Stems are important plant structures because they _________.
A.
are the main sites of photosynthesis for plants
B.
are the only site of water and nutrient transport
C.
allow plants to absorb water and nutrients from the soil
D.
give the plant structure and support
Answer:
c
Explanation:
allow plants to absorb water and nutrients from the soil
Can anyone help me with this question?
sure i can help
so u have to ask someone else
Air is a mixture of (mostly) oxygen (molecular mass 16) and nitrogen (molecular mass 14) gases. At room temperature, which molecules in this room have more kinetic energy (on average)
The molecules of the two gases will have the same kinetic energy at room temperature.
The average kinetic energy of gaseous molecules can be calculated using the formula:
Kinetic Energy = 3/2RT, where R = constant and T = temperature in Kelvin
This means that the kinetic energy of a gaseous molecule is dependent on the temperature of the molecule only.
In other words, the molecular mass of molecules of gases has no bearing on the kinetic energy of each molecule.
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Someone help pls this is due today
Answer:
arrow down is gravity
Explanation:
Answer: The person above me is right
Explanation:
Have a Great day!
A city carrier resigns and trains as an office receptionist this is a person’s metabolism is likely to
leopard has canine teeth because
Explanation:
Leopards have 32 teeth, 4 of which are long, pointed canine teeth. Canine teeth are used to kill prey. Other teeth are for cutting flesh and grinding bone
Under an O2(g) pressure of 1.00 atm, 28.31 mL of O2(g) dissolves in 1.00 L H2O at 25 °C. What will be the molarity of O2 in the saturated solution at 25 °C when the O2 pressure is 4.47 atm? Assume that the solution volume remains at 1.00 L.
In the first case, the molarity of oxygen gas is 0.0012 mol/L. In the second case, the molarity of oxygen gas is 0.0052 mol/L .
We have to obtain the number of moles of oxygen as follows;
P = 1.00 atm
V = 28.31 mL or 0.02831 L
n = ?
R = 0.082 atm LK-1mol-1
T = 25 °C + 273 = 298 K
n = PV/RT
n= 1.00 atm × 0.02831 L/ 0.082 atm LK-1mol-1 × 298 K
n = 0.0012 moles
Molarity = Number of moles/Volume = 0.0012 moles/1.00 L = 0.0012 mol/L
For the second case;
P = 4.47 atm
V = 28.31 mL or 0.02831 L
n = ?
R = 0.082 atm LK-1mol-1
T = 25 °C + 273 = 298 K
n = PV/RT
n = 4.47 atm × 0.02831 L/0.082 atm LK-1mol-1 × 298 K
n = 0.1265/24.436
n = 0.0052 moles
Molarity of oxygen = 0.0052 moles/1.00 L = 0.0052 mol/L
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(100 POINTS!!!!) In one to two sentences, describe the process by which the ionic compound, LiCl, would dissolve in the polar solvent, CH3COCH3
Answer:
LICL as an Ionic compounds has covalent ability and they are soluble in polar solvent such as CH3COCH3 but they are insoluble in non-polar solvents. Due to their polarity, CH3COCH3 will decreases the electrostatic forces of attraction thereby resulting in free ions in aqueous solution.
Water is the known as a polar solvent that can dissolve an ionic compounds very easy and as a polar solvent, the arrangement of oxygen and hydrogen atoms in water is in bent shape.
Ionic compounds like LICL that is very polar is soluble in the polar solvent water.
Polar solvents like CH3COCH3 will dissolve polar and ionic solutes because of the attraction of the opposite charges on the solvent and solute particles.
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Explanation:
Balancing Chemical Equations
Based on the chemical equation, use the drop down menu to choose the coefficients that will balance the chemical
equation
(vbo,- (v]),
Will make Brainlyest plzzzz help!!!
Answer:
3O2→ 2O3
Explanation:
by multiplying 2 by 3 we get 6 on both sides
Calculate the freezing point of a solution of 3.46 g of a compound, X, in 160 g of benzene. When a separate sample of X was vaporised, its density was found to be 3.27 g/Lat 116°C and 773 torr. The freezing point of pure benzene is 5.45°C, and Kis 5.12°C kg/mol.
Answer:
Calculate the freezing point of a solution of 3.46 g of a compound,X,in 160 g of benzene. When separate sample of X was vaporised, its density was found 3.27 g/L at 116 c and 773 torr. The freezing point of pure benzen is 5.45 c, and kf is 5 .12 c/m.
Explanation:
First determine molar mass from
P*molar mass = density*R*T
Solve for molar mass. Then
moles = grams/molar mass
Solve for moles
molality = moles/kg solvent
Solve for molality
delta T = Kf*m
Solve for delta T, then convert to freezing point
What is the reaction between CCl4 and H2O?
Answer:
At temperatures > 400 °C, CCl4 reacts with H2O over a MgO catalyst to yield HCl and CO2.
ethylene glycol, an antifreeze boils at 197 ⁰C. Convert 197 ⁰C to:
⁰F =
K =
[tex]\boxed{\sf °F=\dfrac{9}{5}°C+32}[/tex]
[tex]\\ \sf\longmapsto °F=\dfrac{9}{5}(197)+32[/tex]
[tex]\\ \sf\longmapsto °F=\dfrac{1773}{5}+32[/tex]
[tex]\\ \sf\longmapsto °F=354.6+32[/tex]
[tex]\\ \sf\longmapsto °F=386.6°F[/tex]
[tex]\rule{200pt}{5pt}[/tex]
[tex]\boxed{\sf K=°C+273}[/tex]
[tex]\\ \sf\longmapsto K=197+273[/tex]
[tex]\\ \sf\longmapsto K=470K[/tex]
To prepare 250.0 mL of 2.5 M KCl you will need to dilute _____ mL of 8.0 M KCl solution to a volume of 250.0 mL.
Answer:
78.125ml
Explanation:
Number of moles in 250ml of 2.5M KCl is (250÷1000)litres×2.5M so we divide these moles by 8M. The answer gotten will be in litres so multiply by 1000 to get it in ml
The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO2 dissolved in water is 1.65 x 103 atm, and the density of water is 1.0 g/cm3.
If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.
The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.
[tex]C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 4.60 atm = 7.59 \times 10^{-3} M[/tex]
We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.
[tex]\frac{7.59 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.367 g[/tex]
Now, we will repeat the same procedure for a partial pressure of 1.28 atm.
[tex]C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 1.28 atm = 2.11 \times 10^{-3} M[/tex]
[tex]\frac{2.11 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.102 g[/tex]
The mass of CO₂ released will be equal to the difference in the masses at the different pressures.
[tex]m = 0.367 g - 0.102 g = 0.265 g[/tex]
If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.
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The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.
Elements that form diatomic molecules include
a oxygen
b hydrogen
C sodium
d two of the above
Answer:
D, two of the above, oxygen and hydrogen.
Explanation:
To solve this problem we can either use orbital stuff or common knowledge. Oxygen and hydrogen are known to form O2 and H2, while sodium forms metallic bonds between the atoms.
Suppose you are performing a precipitation titration to study the K sp of zinc(II) iodate, Z n ( I O 3 ) 2 . The literature value of K sp is 3.9 × 10 − 6 . If you use 0.200 M K I O 3 as the titrant, what concentration of I O 3 − (in M) do you predict will be needed to start precipitation in a 0.229 M solution of Z n ( N O 3 ) 2 ?
The predicted concentration of IO₃⁻ needed to start the precipitation titration is 1.703 × 10⁻⁷ M
The dissociation of zinc (II) iodate can be expressed as:
[tex]\mathbf{Zn (IO_3)_2 \to Zn^{2+} + 2IO_3^-}[/tex]
Given that the solubility product constant Ksp value = 3.9 × 10⁻⁶
For the above dissociation,
[tex]\mathbf{Ksp = [Zn^{2+ }] [IO_3^-]^2}[/tex]
Since [tex]\mathbf{ [Zn^{2+ }] = [Zn(NO_3)_2] = 0.229 \ M}[/tex]
∴
[tex]\mathbf{3.9\times 10^{-8} =(0.229) \times [IO_3^-]^2}[/tex]
[tex]\mathbf{ [IO_3^-]^2 = \dfrac{3.9\times 10^{-8} }{(0.229)}}[/tex]
[tex]\mathbf{ [IO_3^-]^2 = 1.703 \times 10^{-7} \ M}[/tex]
Therefore, we can conclude that the initial predicted concentration of IO₃⁻ needed to start the precipitation titration is 1.703 × 10⁻⁷ M
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Usain Bolt went from 0.0 to
11.0m/s in 3.4s. What was his
acceleration?
Answer:
Answer: Change in speed = 11 m/sTime taken = 3.4sHence, acceleration = (11/3.4) m/s² = 3.235 m/s²
Explanation:
(winks and runs off)
Acceleration is 3.235 m/s² when Usain Bolt went from 0.0 to 11.0 m/s in 3.4 seconds.
What is acceleration?
Acceleration is defined as a rate at which the velocity changes with time with respect to both speed and direction.An object is said to be accelerated if it's speeds up r slows down.
Motion along a circle is always accelerated even when speed is constant it's direction is continuously changing.Acceleration is a vector quantity as it has both magnitude as well as direction.The magnitude of acceleration is described by Newton's second law of motion.
Formula of acceleration is change in speed /time.Substituting the given values in the given formula,(11.0-0.0)/3.4=3.235 m/s².
Thus, the acceleration is 3.235 m/s² when Usain Bolt went from 0.0 to 11.0 m/s in 3.4 seconds.
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what is chemical reaction ?
How many atoms are in mercury phosphate
Answer:
There are 95 atoms in mercury phosphate
Explanation:
You just add the atomic number for both elements
Write the formula of the conjugate base of C6H5SH.
Answer:
C6H5S^-
Explanation:
This is because you are removing the hydrogen to make it a conjugate base.
The formula for conjugate base of C6H5SH is C6H5S⁻.
What is conjugate base?Conjugate base is defined as the substance which is formed when an acid liberate its protons.
The acid donate protons and the acid changes into base.
Conjugate acid is defined as the pair of compound that differs by its protons by gaining protons.
Conjugate acid and base are based in acid base theory.
Thus, the formula for conjugate base of C6H5SH is C6H5S⁻.
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13500 mL convert to milliliters of mercury
Answer:
5.627 × 1019 milliliters
Explanation:
is this what you wanted?
A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the butane completelycondenses and the pressure drops to 1.0 atm. Calculate the mole fraction of nitrogenin the original gaseous mixture.
A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.
A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.
We can calculate the total number of moles using the ideal gas equation.
[tex]P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V[/tex]
At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.
We can calculate the moles of nitrogen using the ideal gas equation.
[tex]P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V[/tex]
The mole fraction of nitrogen in the mixture is:
[tex]X(N_2) = \frac{0.030 mol/L \times V}{0.091 mol/L \times V} = 0.33[/tex]
A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.
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How do the ecological pyramid shapes compare
Answer:
it allow us to compare the amount of energy passing through each tropical level over a period of time
Which of the four is least like the other three? A)Platinum B)Ivory C)Silver D)Gold
Answer:
B) Ivory
Explanation:
Ivory is a hard, white material from the tusks and teeth of animals, that consists mainly of dentine, one of the physical structures of teeth and tusks.
Platinum, Silver, and Gold are precious metals, meaning they are rare metallic chemical elements of high economic value, shiny, hard, strong with high melting points
Ivory is regarded as the least of the options. Ivory consists of dentine and
are usually found in the tusks and teeth of animals such as elephants. They
however don't have much value as the other metals in the options.
Precious metals include the following:
PlatinumSilverGoldThey are used for ornamental purposes. They are usually mined and refined
to produce a variety of objects.
Their use are however legal and they are expensive which makes the Ivory
the least out of the options provided.
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