Answer:
All of these stoichiometry problems are worked alike. Here is the 4 step process.
4Na + O2 → 2Na2O
Step 1. Write and balance the equation. You have that.
Step 2. Convert whatever you have to mols. If a solid then mols = grams/molar mass = ?. If a solution then mols = M x L = ? In this case, the molar mass of Na2O is approx 62 so mols = 2/62 = approx 0.03 but that's just a close guess. You should do it more accurately.
Step 3. Using the coefficients in the balanced equation, convert mols of what you have to mols of what you want. You have mols Na2O and you want to convert that to mols Na.
0.03 mols Na2O x (4 mols Na/2 mols Na2O) = about 0.06. Note that mols Na2O in the numeratorof the first term cancels with mols Na2O in the denominator of the second term which leaves mols Na which is what you want. In practice you ALWAYS know that the coefficient of the unit you are converting to goes on top and the coefficient of mols you have goes on the bottom.
4. Now convert mols of what you want to grams. grams = molsl x molar mass - about 0.06 x 23 = about ?
Copy this. It can come in handy for every chemistry course you take.
Explanation:
Glucose is a simple sugar occurring both in plant and animal tissues. It has a chemical formula of (C6H12O6). Glucose is the byproduct of photosynthesis in plants and a raw material in the metabolism of animals. So, the answer to the above question is Glucose.
The "Nucleus" is the brain of the cell bc it holds the information needed to conduct most of the cell's functions
Colligative properties are usually used in relation to solutions.
Colligative properties are those properties of solutions, which depend on the concentration of the solutes [molecules, ions, etc.] in the solutions and not on the chemical nature of those chemical species. Examples of colligative properties include: vapour pressure depression, boiling point elevation, osmotic pressure, freezing point depression, etc.
For the question given above, the correct option is D. This is because the statement is talking about freezing point elevation, which is not part of colligative properties.
The answer is: all true
<span>A. As the pressure of the gas increased, the volume of the gas decreased.
It is clear that if you compare the data on the left side. When the pressure increased the volume is decreased.
B. For all pairs of data of pressure and volume, P • V was appoximately the same.
The pressure is inversely related to the volume. You can take two data to prove it. Let use the first and second data
V * P= 1.03 * 50= 51.5
</span>V * P= <span>1.08 * 47.5= 51.3
C. For all pairs of data of pressure and volume, P • V mr001-1.jpg k for the same value k.
D. The regression equation was of the form V = kP–1 (which is the same as V = k/P).
The value of k can be expressed as k= P*V. If the equation is turned around, it could be expressed as V= k/P
The value of k is constant on different data, proved by the calculation on the second statement above. The value of k should be around 51.5
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