Answer:
-2.79 × 10³ cal
Explanation:
Step 1: Given data
- Mass of water (m): 35.0 g
- Latent heat of fusion of water (L): -79.7 cal/g
Step 2: Calculate the heat required to freeze 35.0 g of water
We have 35.0 g of liquid water and we want to freeze it, that is, to convert it in 35.0 g of ice (solid water), at 0 °C (melting point). We can calculate the heat (Q) that must be released using the following expression.
Q = L × m
Q = -79.7 cal/g × 35.0 g
Q = -2.79 × 10³ cal
Answer:
Explanation:
From the given study,
the molecular formula of the given aromatic compound = C9H12
the first process we suppose to carry out is to calculate the double bond equivalence.
So, the double bond equivalence 

DBE = 4
Hence, the aromatic compound possesses 4 double bonds, this signifies that the might be a presence of benzene ring.
The substitutional products of the aromatic hydrocarbon C9H12 can be seen in the attached file below.
So that the mircrobes present could get help to perform anariobïc repiration (respiration without oxygen).....as it helpfull for decomposing the sewage waste ...also to prepare sideproduct called metane gas
Answer:
The distance the truck traveled is 190
Explanation:
I added 30, 45, 50, 65 together to get the total distance
Answer:
Explanation:
Use Dalton's law and the vapor pressure of water at 23.0 o C to correct the pressure to units of atmoshperes.
PT = Poxygen +Pwater
At 23.0 o C the vapor pressure of water is 21.1 mmHg. (This can be found on a vapor pressure table.)
762 mmHg = Poxygen + 21.1 mmHg
Poxygen = 762 mmHg - 21.1 mmHg
Poxygen =741 mmHg
Convert the corrected pressure to atmospheres.
(741 mmHg) (1 atm / 760 mmHg) = 0.975 atm
Use the ideal gas law to find out how many moles of gas were produced:
PV = nRT (remember to put volume in liters and temperature in Kelvin)
(0.975 atm) (.193 L) = n (.0821 L atm / mol K) (298 K)
n = (0.975 atm) (.193 L) / (.0821 L atm / mol K) (298 K)
n = 7.69 X 10-4 mol
Use the number of moles and the molecular weight of oxygen to find out how many grams of oxygen were collected.
(7.69 X 10-4 mol) (32.0 g / 1 mol) = 2.46 X 10-2 g