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LekaFEV [45]
3 years ago
13

50 POINTS!!!!!!!

Chemistry
1 answer:
KATRIN_1 [288]3 years ago
7 0
We could reduce soil erosion and recycle phosphorus from farm and human waste so that we could help make food production sustainable and prevent algae blooms. We can also do land reclamation as well to help solve this problem. With the land, we would have to design a system to where the land could be functional again in order to plant crops, trees, also to help the wildlife that was once a part of the island. Therefore if the design is done before the mining then afterward we can do the reclamation of the land which would help the people to be able to function after the mining. It would also help the future generations that come along after the previous generations. Everyone must work together in the process in order for everyone to survive. If all this is done then the people of the island would not have to import their food. The reclamation process is the most important thing that has to be designed first whether it is land, soil, water, lakes, and clay then after plant trees, vegetation, and other forms of plants to help replenish the land after the mining is done.

I hope I helped :3
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What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a
Vadim26 [7]

Answer:

\boxed {\boxed {\sf 333 \ grams}}

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

Q= mc \Delta T

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

  • ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

  • ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

  • Q= 4500.0 J
  • c= 0.4494 J/g°C
  • ΔT = 30.1 °C

Substitute these values into the formula.

4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)

Multiply on the right side of the equation. The units of degrees Celsius cancel.

4500.0 \ J = m (13.52694 J/g)

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}

The units of Joules cancel.

\frac {4500.0 \ J }{13.52694 J/g}= m

332.6694729 \ g =m

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

333 \ g \approx m

The mass of the sample of metal is approximately <u>333 grams.</u>

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