Temperatures in the central pacific ocean are warmer than usual and there is less rainfall. hope this helps! :)
Answer:
See explanation.
Explanation:
Hello!
In this case, we consider the questions:
a. Ideal gas at:
i. 273.15 K and 22.414 L.
ii. 500 K and 100 cm³.
b. Van der Waals gas at:
i. 273.15 K and 22.414 L.
ii. 500 K and 100 cm³.
Thus, we define the ideal gas equation and the van der Waals one as shown below:
Whereas b and a for hydrogen sulfide are 0.0434 L/mol and 4.484 L²*atm / mol² respectively, therefore, we proceed as follows:
a.
i. 273.15 K and 22.414 L.
ii. 500 K and 100 cm³ (0.1 L).
b.
i. 273.15 K and 22.414 L: in this case, v = 22.414 L / 1.00 mol = 22.414 L/mol
ii. 500 K and 100 cm³: in this case, v = 0.1 L / 1.00 mol = 0.100 L/mol
Whereas we can see a significant difference when the gas is at 500 K and occupy a volume of 0.100 L.
Best regards!
The volume of the radiator solution formed with the 197 mL of antifreeze is 788 mL.
<h3>What is a percentage?</h3>
A percentage is given as the respective ratio of the quantity present with respect to others.
The solution of 25% v/v can be depicting the presence of a 25% volume of antifreeze in the radiator solution.
The volume of the radiator solution has 197 mL of antifreeze. The total volume of the solution suppose is x.
25% of x = 197 mL
25/100 * x = 197
x = 788 mL
The volume of the radiator solution formed with 25% v/v antifreeze is 788 mL.
Learn more about percentage dilution, here:
brainly.com/question/14528274
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Explanation:
Let us assume that 5.49 g of Iron chloride is dissolved in 350 ml of 66.0 mM aqueous solution of silver nitrate
Hence, the balanced chemical equation for this reaction is as follows.
So, moles of Iron chloride = 5.49 g of molar mass of Iron chloride
No. of moles =
=
= 0.0338 moles
And, moles of silver nitrate = molarity × volume in L
= 66.0 Mm
or, = 0.066 M × 350 ml
= 0.0231 moles
Therefore, here silver nitrate is the limiting agent
.
Now, we will calculate the moles of which are reacted as follows:.
= 0.01155 moles
Remaining moles of = 0.0338 moles - 0.01155 moles
= 0.02225 mole
Or, 0.0445 moles ions.
Now, we will calculate the molarity as follows.
Molarity =
=
= 0.127 M
Thus, we can conclude that final molarity of chloride anion in the solution is 0.127 M.
C. Weak acid is the correct answer