The Great Pacific Garbage Patch is created by deep oceans. The Marine Debris is litter that ends up in oceans or other large bodies of water.
Answer:
Explanation:
Hello!
In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):
In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:
Best regards!
Answer:
Explanation:
A 12.48 g sample of an unknown metal, heated to 99.0 °C was then plunged into 50.0 mL of 25.0 °C water. The temperature of the water rose to 28.1 Go to calculating final temperature when mixing two samples of water ... Problem #1: A 610. g piece of copper tubing is heated to 95.3 °C and placed in an ... The two rings are heated to 65.4 °C and dropped into 12.4 mL of water at 22.3 °C. ... Problem #4: A 5.00 g sample of aluminum (specific heat capacity = 0.89 J g¯1
Answer:
506.912 L
Explanation:
From the question given above, the following data were obtained:
Number of mole of O₂ = 22.63 moles
Volume of O₂ =?
Recall:
1 mole of a gas occupy 22.4 L at STP.
With the above information, we obtained the volume occupied by 22.63 moles of O₂ as follow:
1 mole of O₂ occupied 22.4 L at STP.
Therefore, 22.63 moles of O₂ will occupy = 22.63 × 22.4 = 506.912 L at STP.
Thus, 22.63 moles of O₂ is equivalent to 506.912 L.
Answer: Final temperature = 206∘C
Explanation:
Heat Energy is given as
q= mCΔT
ehere
q= Heat energy = 87.4J
m= mass=1.25g
C=specific heat c= 0.386Jg∘C) ,
ΔT = Change in temperate of which the final temperature= 25.0∘C
q= mCΔT
ΔT = q/mC
ΔT = 87.4/ 1.25 X 0.386=181.14∘C
But,
T final- T initial = ΔT
T final = T initial + ΔT
T final = 25.0∘C +181.14∘C=206.14∘C rounded to 206∘C
