Answer:
The heat that was used to melt the 15.0 grams of ice at 0°C is 4,950 Joules
Explanation:
The mass of ice in the beaker = 15.0 grams
The initial temperature of the ice = 0°C
The final temperature of the ice = 0°C
The latent heat of fusion of ice = 330 J/g
The heat required to melt a given mass of ice = The mass of the ice to be melted × The latent heat of fusion of ice
Therefore, the heat, Q, required to melt 15.0 g of ice = 15.0 g × 330 J/g = 4,950 J
The heat that was used to melt the 15.0 grams of ice = 4,950 Joules.
C is the answer hope the answer is right
Answer:
The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M
Explanation:
As HCl is a strong acid and hence a strong electrolyte, it will dissociate as
HCl ⟶ H⁺ + Cl⁻
So, The concentration of H⁺ will be 2.5 M (same as HCl)
Thus, The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M
<u>-TheUnknownScientist</u><u> 72</u>
For this question, assume that you have 1 compound. This compound is divided in half once, so you are left with 0.5. That 0.5 that remains is divided in half again, this is the second half-life, and you are left with 0.25. The final half life involves dividing 0.25 in half, which means you are left with 0.125. For the answer to make sense, you need to know your conversions between decimals and fractions. To make it simple, if you have 0.125 and you times it by 8, you are left with your initial value of 1. Therefore, after three half-lives, you are left with 1/8th of the compound.
Answer:
See the explanation
Explanation:
In this case, in order to get an <u>elimination reaction</u> we need to have a <u>strong base</u>. In this case, the base is the phenoxide ion produced the phenol (see figure 1).
Due to the resonance, we will have a more stable anion therefore we will have a less strong base because the negative charge is moving around the molecule (see figure 2).
Finally, the phenoxide will attack the <u>primary carbon</u> attached to the Cl. The C-Cl bond would be broken and the C-O would be produced <u>at the same time</u> to get a substitution (see figure 1).
