The answer is <span>B. element.
An element is composed of only one kind of atom and cannot be separated into simpler substances. Oxygen (O) is the element.
A compound is a substance composed </span><span>of two or more <em>different </em>atoms chemically bonded to one another, for example, water (H</span>₂<span>O) consists of 2 atoms of hydrogen (H) and 1 atom of oxygen (O), so it is the compound.
A mixture consists of two or more substances that are not chemically combined. Solutions and colloids are mixtures.</span>
Answer : The correct option is, (D) 3600 kJ
Explanation :
Mass of octane = 75 g
Molar mass of octane = 114.23 g/mole
Enthalpy of combustion = -5500 kJ/mol
First we have to calculate the moles of octane.
Now we have to calculate the heat released in the reaction.
As, 1 mole of octane released heat = -5500 kJ
So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)
= -3608 kJ
≈ -3600 kJ
Therefore, the heat released in the reaction is 3600 kJ
We have that the correct wedge and dash conformation for the following Newman projection is
IV
From the Diagrams above
Two CH_3 groups points on opposite sides in plane
Two Br are on same side of plane
Two H also on same side of the plane so the plausible structure is IV
Therefore
The Correct option is IV
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Answer:
0.85 mole
Explanation:
Step 1:
The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:
When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:
CaCl2 + Na2CO3 -> CaCO3 + 2NaCl
Step 2:
Conversion of 85g of CaCO3 to mole. This is illustrated below:
Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol
Mass of CaCO3 = 85g
Moles of CaCO3 =?
Number of mole = Mass /Molar Mass
Mole of CaCO3 = 85/100
Mole of caco= 0.85 mole
Step 3:
Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.
This is illustrated below :
From the balanced equation above,
1 mole of CaCl2 reacted to produced 1 mole of CaCO3.
Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.
From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3
Answer: all I know it’s not -31.5 for ppl taking the k12 test
Explanation: I took the test
