The answer is H₃C₆H₅O₇ (OPTION D).
In essence, the question is asking which of the options is an acid or a neutral compound; thus not a base. A base is a compound that accepts free hydrogen ions while in aqueous solution and as such they usually lack hydrogen ions that can be released in solution. If you ionize all the compounds/molecules above, the only on that has free hydrogen ions is OPTION D (which is an organic acid and not a base).
You will need the equation PV = nRT
P = Pressure in kPa
V = Volume in L
n = moles
R = 8.314 (constant)
T = Temperature in Kelvin
First convert 2.5 atm into kPa:
2.5 X 101.3 = 253.25 kPa
Convert 125 Celsius into Kelvin:
125 + 273 = 398 K
Convert Gallons to Litres:
1.25 X 3.79 = 4.74 L
Plug your values into the equation to solve for n:
(253.25)(4.74) = n(8.314)(398)
n = (253.25)(4.74)/(8.314)(398)
n = 0.362 moles
Now use M = m/n to solve for the mass of O2
M = Molar Mass
M = mass
n= moles
32 = m/(0.362)
m = (32)(0.362)
m = 11.58g
Answer:
A. Energy is transferred to different forms
.
Explanation:
Hello!
In this case, we need to consider the law of conservation of mass and energy which states that mass and energy cannot be neither created nor destroyed, just modified; it means we can rule out B. and C. so far.
Moreover, since D. is actually true for combustion reactions because they are used to provide energy in industrial operations, this is not the concern here because a combustion reaction is not considered.
Therefore the correct option is A. Energy is transferred to different forms as the energy provided by Rose is transferred to the pendulum system
.
Best regards!
Answer:
b
Explanation:
b/c proton + neutron=mass number and
mass number - proton= neutron
Answer: The correct answer is -297 kJ.
Explanation:
To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:
2SO3 —> O2 + 2SO2 (196 kJ)
Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.
Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:
SO3 —>1/2O2 + SO2 (98 kJ)
S + 3/2O2 —> SO3 (-395 kJ)
Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.
S + O2 —> SO2
Now, we must add the enthalpies together to get our final answer.
-395 kJ + 98 kJ = -297 kJ
Hope this helps!
