Answer:
Noble gas
Explanation:
With the exception of helium, all elements in Group 18 contain 8 electrons in the outermost shell; these are the noble gases.
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Translation
(2) 0.05 g (4) 5 g
Answer: The correct answer is (1).
Explanation:
Concentration of the solute in 1000 g of solution: 5 parts per million
1 part per million = 1 milligram/Liter
Then , 5 parts per million = 5 mg/liter
This means that 5 milligrams of solute is present in 1 liter solution.
1 gram = 1000 milligrams
Then 5 milligram = = 0.005 grams
Hence, 0.005 grams is the total mass of solute in 1000 grams of a solution having a concentration of 5 parts per million.
Parts per million (ppm) is how many parts a certain molecule or compound makes up within the one million parts of the whole solution.
Concentration of the solute in 1000 g of solution: 5 parts per million
1 part per million = 1 milligram/Liter
Then , 5 parts per million = 5 mg/liter
This means that 5 milligrams of solute is present in 1 liter solution.
1 gram = 1000 milligrams
Then 5 milligram = 5 X = 0.005 grams.
Hence, 0.005 grams is the total mass of solute in 1000 grams of a solution having a concentration of 5 parts per million.
Learn more about the parts per million here:
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Answer:
C
Explanation:
I just took the test twice and it wasn't D the first time
When dynamite explodes, it releases a large amount of light and heat. This is an example of _____.
slow combustion
rapid combustion
an exothermic reaction
an endothermic reaction
a chemical change
a physical change
Answer
Rapid Combustion
Explanation:
An explosion is a combustion and a quick reaction, That makes it rapid
Given that the molar mass of CO2 is 44.01 g/mol, how many liters of oxygen is required at STP to produce 88.0 g of CO2 from this reaction?
44.8 L
45.00 L
89.55 L
89.6 L
89.6 L of O₂
The balanced chemical equation is as,
CH₄ + 2 O₂ → CO₂ + 2 H₂O
As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,
44 g ( 1 mol) CO₂ is produced by = 44.8 L (2 mol) of O₂
So,
88 g CO₂ will be produced by = X L of O₂
Solving for X,
X = (88 g × 44.8 L) ÷ 44 g
X = 89.6 L of O₂