1. B
The positive charge in water is provided by hydrogen, and gold provides the same charge. However, gold is not more reactive than hydrogen so it can not replace it in the compound.
2. In order to balance the equation, you must sure there are equal moles of each element on the left and right side of the equation:
2C₂H₆ + 7O₂ → 4CO₂ + ₆H₂O
3. The number of moles of sodium atoms on the left of the equation must be equal to the number of moles of sodium atoms on the right, as per the law of conservation of mass. The answer is B.
4. C.
A synthesis reaction usually results from single displacement because some element or compound is produced in its pure form
5. B.
The gas being produced is being synthesized.
Im pretty sure a cinder-cone volcano
Answer:
His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.
Explanation:
The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.
Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.
Suppose the mass of the solute is m.
Originally, the density is =
![\left(\frac{\text{mass}}{\text{volume}}\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7B%5Ctext%7Bmass%7D%7D%7B%5Ctext%7Bvolume%7D%7D%5Cright%29)
Now after adding extra 10 mL , the density becomes
.
Therefore, ![$\frac{m}{50}>\frac{m}{60}$](https://tex.z-dn.net/?f=%24%5Cfrac%7Bm%7D%7B50%7D%3E%5Cfrac%7Bm%7D%7B60%7D%24)
So the density decreases when we add more solution.
Answer:
2.77 mol N
Explanation:
M(N2O) = 2*14 + 16 = 44 g/mol
61.0 g * 1 mol/44g = (61/44) mol N2O
N2O ---- 2N
1 mol 2 mol
(61/44) mol x mol
x = (61/44)*2/1 = 2.77 mol N
Answer:
Supersaturated solution.
Explanation:
Hello!
In this case, according to the types of solution in terms of the relative amounts of solute and solvent, we can define a point called solubility at which the amount of solute is no longer dissolved in the solvent; thus, a value of solute/solvent less than the solubility is related to unsaturated solutions, equal to the solubility is related to the saturated solutions and more than the solubility to supersaturated solutions.
Thus, since solubility is temperature-dependent, at 30 °C the solubility of sodium chloride is 36.09 g per 100 mL of water; which means that, since the solution has 50 g of sodium chloride, more than 36.09 g, we infer this is a supersaturated solution.
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