Answer:The mass numbr is 22
Explanation:
Mass number=number of protons+ number of neutrons....which is 10+12=22
Use the Heat formula for both problems.
q=m*c*∆t
Where
q= heat in Joules
m= mass in grams
c= specific heat which is a constant 4.18
∆t= change in temperature
The metabolic process that takes place in an organism's cells is called cellular respiration. Both photosynthesis and cellular respiration contribute carbon dioxide to the biogeochemical cycle.
<u>What is biogeochemical cycle?</u>
The mechanism that distributes and circulates abiotic chemical elements among the various realms of the planet is known as the biogeochemical cycle. The carbon, water, nitrogen, phosphorus cycle, etc. are all included.
Carbon dioxide is released during cellular respiration as a waste gas into the atmosphere, where it is then taken by plants to produce energy, which is then utilized by the organism, which then releases more carbon dioxide. The ecological chain's carbon cycle continues from the producer to the consumer.
Therefore, the carbon cycle is influenced by photosynthesis and cellular respiration.
Learn more about the biogeochemical cycle here:
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Answer:
There is a production of 11.6 moles of CO₂
Explanation:
The reaction is this:
2C₂H₆(g) + 7O₂(g) ⟶ 4CO₂(g) + 6H₂O(g)
2 moles of ethane reacts with 7 moles of oxygen, to make 4 mol of dioxide and 6 moles of water vapor.
If the oxygen is in excess, we make the calculate with the ethane (limiting reactant)
2 moles of ethane produce 4 moles of dioxide
5.8 moles of ethane produce (5.8 .4)/2 = 11.6 moles
Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol
