Answer:
Explanation:
Hello!
In this case, according to the given description of how the temperature changes for aluminum in agreement to the loss of heat of 6120.0 J, we can use the following equation:
Thus, by knowing Q, m, C and the initial temperature, we are able to obtain:
Regards!
Answer: Option (B) is the correct answer.
Explanation:
According to the given reaction equation, formula to calculate is as follows.
= coefficients of gaseous products - gaseous reactants
= 1 - 0
= 1
Also we know that,
For the equation,
Activity of solid and liquid = 1
As,
Hence, = 0.0056 atm
Thus, we can conclude that partial pressure of oxygen gas at equilibrium is 0.0056 atm.
HI BRAINLY USER!!
<h3>JUST CLICK THE ATTACHMENT PHOTO </h3>
⊱─━━━━━━━━━⊱༻●
༺⊰━━━━━━━━━─⊰
- 》Mass = 5.90 x 114 = <u>672.6g.</u>
<u>This can be rounded to 673g</u>
⊱─━━━━━━━━━⊱༻
●༺⊰━━━━━━━━━─⊰
<h2>#CARRY ON LEARNING </h2>
Answer:
B
Explanation:
Statement A
Vrms (root-mean-square velocity) is defined as follows.
Vrms = √(2RT/Mm), where R is the ideal gas constant, T is the temperature, and Mm is the molar mass of the gas. R and T are the same for both gases, but the molar mass is different, so Vrms for the two gases is different
Statement B
Kinetic energy KE of a molecule is related to temperature as follows:
KE = 3/2KbT, where Kb is the Boltzmann's constant and T is the temperature. The temperature is the same for both gases, so KE is the same for both.
Statement C
The rate depends on the number of molecules present. Since there are twice as many molecules of methane as propane, the rate is different for both gases.
Statement D
Pressure is directly proportional to the number of moles, based on the ideal gas law (pV = nRT). The volume and temperature are the same for both gases, so since there are twice as many molecules of methane as propane, the pressure is twice as great in the cylinder of methane.
Answer:
ddndnndndndkzkaoaoowowjebd
Explanation:
nzndbdnxkskksksjsjsjsjsiisisjs