Answer:
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
Explanation:
Water potential = Pressure potential + solute potential
We have :
C = 0.15 M, T = 273.15 K
i = 1
The water potential of a solution of 0.15 m sucrose= 
(At standard temperature)
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>
</span><span>Where, P is the pressure of the gas (Pa), V
is the volume of the gas (m³), n is the number
of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹)
and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³
n = ?
<span>
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
n = 4.0359 x 10</span>⁻³ mol
<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>
Moles = mass / molar
mass
</span>Mass of the gas = 0.529 g
<span>Molar mass of the gas</span> = mass / number of moles<span>
= </span>0.529 g / 4.0359 x 10⁻³ mol<span>
<span> = </span>131.07 g mol</span>⁻¹<span>
Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹
Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (
), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano
1 gramo de metano aporta 50.125 kilojoules.
Octano
1 gramo de metano aporta 48.246 kilojoules.
Answer:
Weight is the measure of the force of gravity on an object. The mass of an object will never change, but the weight of an item can change based on its location. For example, you may weigh 100 pounds on Earth, but in outer space you would be weightless.
Explanation:
I can't see the options but I hope this helps!
Answer:
2,2,3,3-tetrapropyloxirane
Explanation:
In this case, we have to know first the alkene that will react with the peroxyacid. So:
<u>What do we know about the unknown alkene? </u>
We know the product of the ozonolysis reaction (see figure 1). This reaction is an <u>oxidative rupture reaction</u>. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If 
is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.
<u>What is the product with the peroxyacid?</u>
This compound in the presence of alkenes will produce <u>peroxides.</u> Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product <u>2,2,3,3-tetrapropyloxirane.</u> (see figure 2)
