Answer:
The pressure in that cylinder = 1.12atm
Explanation:
We use general gas law to calculate it. General gas law is gotten by combining Boyle's law, Charles' law and Avogadro's law. Thus
P = nRT/V
Where n = number of moles
R = the gas constant
T is the Temperature, V is the volume and P is the pressure.
Given: T = 319K, V = 24L, R = 0.0821 L.atm/K.mol
The first step is to find n using
n = mass of O2/molar mass of O2
=32.7/32
=1.0219
Now, using P =nRT/V
P = 1.0219 ×0.0821×319÷24
Therefore P = 1.12atm
Answer:
36
Explanation:
Since the sample was undiluted the number of colonies is the number that grew on the nutrient agar which is 36 colonies. If it was diluted for example let say 0.1 ml from a dilution in which 1 ml of the sample was added to 9 ml of water, and it grew colonies then 0.1 ml yielded 6 colonies, 1 ml of the diluted sample will yield 60 colonies and 10 ml will have 600 colonies and therefore the 1 ml undiluted sample will have 600 colonies.
If I remember correctly from my science class, it is Protons and Neutrons.
Hope this helps, if not, comment below please!!!
Answer:
The total heat required is 691,026.36 J
Explanation:
Latent heat is the amount of heat that a body receives or gives to produce a phase change. It is calculated as: Q = m. L
Where Q: amount of heat, m: mass and L: latent heat
On the other hand, sensible heat is the amount of heat that a body can receive or give up due to a change in temperature. Its calculation is through the expression:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature (Tfinal - Tinitial).
In this case, the total heat required is calculated as:
- Q for liquid water. This is, raise 248 g of liquid water from O to 100 Celsius. So you calculate the sensible heat of water from temperature 0 °C to 100° C
Q= c*m*ΔT

Q=103,763.2 J
- Q for phase change from liquid to steam. For this, you calculate the latent heat with the heat of vaporization being 40 and being 248 g = 13.78 moles (the molar mass of water being 18 g / mol, then
)
Q= m*L

Q=562.0862 kJ= 562,086.2 J (being 1 kJ=1,000 J)
- Q for temperature change from 100.0
∘
C to 154
∘
C, this is, the sensible heat of steam from 100 °C to 154°C.
Q= c*m*ΔT

Q=25,176.96 J
So, total heat= 103,763.2 J + 562,086.2 J + 25,176.96 J= 691,026.36 J
<u><em>The total heat required is 691,026.36 J</em></u>