Acid more H3O+ than OH-
Base less H3O+ than OH-
Answer: As mass is constant during free-weight resistance training, a greater impulse will result in a greater velocity. Therefore, as generation of force greater than the weight of the resistance increases higher movement velocities and decreased movement times result. (from google)
Explanation: hope this helps! :)
With reference to present question, following things may be noted
1) Oxidation is process of lose of electron. Atom//ion undergoing oxidation is refereed as reducing agent
2) Reduction is process of gain of electron. Atom/ion undergoing reduction is refereed as oxidizing agent.
in present system, Ag+ gain an electron to reduced to Ag. Therefore, Ag+ is an oxidizing agent.
of octane had been converted to carbon dioxide CO₂.
<h3>Explanation</h3>
Octane has a molar mass of
1.000 gallon of this fuel would have a mass of 2.650 kilograms or 
, which corresponds to 
of octane.
Octane undergoes complete combustion to produce carbon dioxide and water by the following equation:
An incomplete combustion of octane that gives rise to carbon monoxide and water but no carbon dioxide would consume not as much oxygen:
The mass of the product mixture is 
heavier than that of the octane supplied. Thus 
of oxygen were consumed in the combustion. There are 
of oxygen molecules in 
of oxygen.
Let the number of moles of octane that had undergone complete combustion as seen in the first equation be 
(
). The number of moles of octane that had undergone incomplete combustion through the second equation would thus equal 
.
25 moles of oxygen gas is consumed for every two moles of octane that had undergone complete combustion and 17 moles if the combustion is incomplete.
Therefore 
out of the 23.2 moles of octane had undergone complete combustion to produce carbon dioxide.
Correct Question :
Mass of water = 50.003g
Temperature of water= 24.95C
Specific heat capacity for water = 4.184J/g C
Mass of metal = 63.546 g
Temperature of metal 99.95°C
Specific heat capacity for metal ?
Final temperature = 32.80°C
In an experiment to determine the specific heat of a metal student transferred a sample of the metal that was heated in boiling water into room temperature water in an insulated cup. The student recorded the temperature of the water after thermal equilibrium was reached. The data we shown in the table above. Based on the data, what is the calculated heat absorbed by the water reported with the appropriate number of significant figures?
Answer:
1642 J
Explanation:
Given:
Mass of water = 50.003g
Temperature of water= 24.95C
Specific heat capacity for water = 4.184J/g C
Mass of metal = 63.546 g
Temperature of metal 99.95°C
Specific heat capacity for metal ?
Final temperature = 32.80° C
To calculate the heat absorbed by water, Q, let's use the formula :
Q = ∆T * mass of water * specific heat
Where ∆T = 32.80°C - 24.95°C = 7.85°C
Therefore,
Q= 7.85 * 50.003 * 4.184
Q = 1642.32 J
≈ 1642 J
