<u>Answer:</u> The temperature of the solution in Kelvins is 422.356 K
<u>Explanation:</u>
Temperature is defined as the measure of coldness or hotness of a body. It also determines the average kinetic energy of the particles in a body.
This term is expressed in degree Celsius, degree Fahrenheit and Kelvins. All these units are interchangeable.
The S.I unit of temperature is Kelvins.
We are given:
Temperature of a solution =
Conversion used to convert degree Celsius and Kelvins is:
Hence, the temperature of the solution in Kelvins is 422.356 K
Answer:
it would be 53
Explanation:
because 4.00 L at 158kPa and 28c would be 53 grams of oxygine
Answer:
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Explanation:
https://courses.lumenlearning.com/boundless-chemistry/chapter/physical-and-chemical-properties-of-matter/
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Answer:
the final temperature is T=305.63 K
Explanation:
using the Stephan-Boltzmann equation for black bodies
q = σ*(T⁴-T₀⁴)
where
q= heat flux = 155 W/m²+150 W/m² = 255 W/m²
σ= Stephan-Boltzmann constant = 5.67*10⁻⁸ W/m²K⁴
T= absolute temperature
T₀= absolute initial temperature = 255 K
solving for T
q = σ*(T⁴-T₀⁴)
T = (q/σ + T₀⁴)^(1/4)
replacing values
T = (q/σ + T₀⁴)^(1/4) = (255 W/m²/(5.67*10⁻⁸ W/m²K⁴) + (255 K)⁴)^(1/4) = 305.63 K
T=305.63 K
thus the final temperature is T=305.63 K
The given choices are as follows
A. add 0.833 mL 12 M HCl stock solution to 99.167 mL water
B.add 0.833 mL 12 M HCl stock solution to 49.167 mL water
C.<span>add 0.833 mL 12 M HCl stock solution to 199.167 mL water
the experiment requires 15.0 mL for each experiment. To conduct at least three experiments the volume needed would be 15.0 x 3 = 45.0 mL
We will need an excess amount of HCl in addition to the 45.0 mL as solutions are required in excess to wash the glassware </span>e.g.: burettes with the solution.
So lets take the required volume of HCl to be prepared as 100.0 mL
now when preparing diluted solutions from stock solutions we can use the dilution equation
c1v1 = c2v2
where c1 = concentration and v1 is volume required from the stock solution
c2 is concentration and v2 is volume of diluted solution to be prepared
substituting the values in the equation
12.0 M x V1 = 0.100 M x 100.0 mL
V1 = 0.833 mL
therefore 0.833 mL should be taken from the 12.0 M stock solution and final volume of 0.100 M HCl to be prepared is 100.0 mL . to make it up to the 100.0 mL mark we need to add (100.0 mL - 0.833 mL ) = 99.167 mL of water.
correct answer is A. add 0.833 mL 12 M HCl stock solution to 99.167 mL water