To find the density you must divide the mass by the density.
180kg ÷ 90m³ = 2kg/m<span>³
The density is </span>2kg/m³
Explanation:
The given data is as follows.
= 286 kJ = 
= 286000 J
, 
Hence, formula to calculate entropy change of the reaction is as follows.
= ![[(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20S_%7BO_%7B2%7D%7D%29%20-%20%281%20%5Ctimes%20S_%7BH_%7B2%7D%7D%29%5D%20-%20%5B1%20%5Ctimes%20S_%7BH_%7B2%7DO%7D%5D)
= ![[(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20205%29%20%2B%20%281%20%5Ctimes%20131%29%5D%20-%20%5B%281%20%5Ctimes%2070%29%5D)
= 163.5 J/K
Therefore, formula to calculate electric work energy required is as follows.
= 
= 237.277 kJ
Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.
Answer:
D
Explanation:
Miss Girl. Snow isn't in Nevada-
Answer:
See the answer below
Explanation:
<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>
1. The first one is to allow the molten substance to solidify again and then repeat the experiment. This time around, a critical attention should be paid to be able to notice the melting point temperature once the temperature gets to 132 C.
2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.
