Answer:
25.2°C
Explanation:
Given parameters:
Energy applied to the water = 1000J
Mass of water = 50g
Final temperature = 30°C
Unknown:
Initial temperature = ?
Solution:
To solve this problem, we use the expression below:
H = m c Ф
H is the energy absorbed
m is the mass
c is the specific heat capacity
Ф is the change in temperature
1000 = 50 x 4.184 x (30 - initial temperature )
1000 = 209.2(30 - initial temperature)
4.78 = 30 - initial temperature
4.78 - 30 = - initial temperature
Initial temperature = 25.2°C
Answer:
c) NOx and VOC
oxides of nitrogen (NOx) and volatile organic compounds (VOC)
Explanation:
-
Answer:
The answer is decreased temperature and increased salinity
Explanation:
It is what is known as the thermohaline circulation
The thermohaline circulation moves the water slowly. This water moves mainly due to differences in its relative density. Much denser water sinks over water that is less dense. Two factors impact the density of seawater: temperature and salinity.
Cold water is denser than hot water:
-Water cools when it loses heat, it occurs at high latitudes.
-Water is heated when it receives energy from the sun, at low latitudes.
Saltier water is much denser than water that has less salt:
-Sea water becomes salty if the evaporation rate increases.
-Sea water becomes less salty if there is a water inlet over the sea.
Answer:
The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol
Explanation:
The chemical reaction of the combustion of methane is given as follows;
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)
Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor
Where:
CH₄ (g): Hf = -74.6 kJ/mol
CO₂ (g): Hf = -393.5 kJ/mol
H₂O (g): Hf = -241.82 kJ/mol
Therefore, the combustion of 1 mole of methane releases;
-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol
Hence the combustion of 2 moles of methae will rellease;
2 × -802.54 kJ/mol or 1,605.08 kJ/mol.
Argon is a noble gas. Argon has a full outer shell. This makes it so that it does not need to react with any of the other elements to be stable.
With Rubidium and Cobalt its a whole different story.
I hope that helps!
