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Len [333]
3 years ago
13

Please help Asap………………

Chemistry
2 answers:
vladimir2022 [97]3 years ago
6 0

Answer:

3rd

Explanation:

Gemiola [76]3 years ago
3 0

The coeffecients should be 2, 1, 2.  

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at what height is an object that has a mass of 16kg, if it’s gravitational potential energy is 7500J?
dolphi86 [110]

Answer:

46.875

Explanation:

P.E=MGH

7500=16*10*h

7500=160h

h=7500/160

h=46.875

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3 years ago
All of the following are uses of a lake/pond except:
umka2103 [35]
Well give me the options lol
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Who was the first to propose the idea of atoms?
VMariaS [17]

Answer::Democritus

Explanation

The idea that all matter is made up of tiny, indivisible particles, or atoms, is believed to have originated with the Greek philosopher Leucippus of Miletus and his student Democritus of Abdera in the 5th century B.C. (The word atom comes from the Greek word atomos, which means “indivisible.”) These thinkers held that,

5 0
3 years ago
Read 2 more answers
A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin
lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, 3.0\times 10^1g

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron metal = 0.45J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of iron metal = 32.3 g

m_2 = mass of water = ?

T_f = final temperature of mixture = 59.2^oC

T_1 = initial temperature of iron metal = 21.9^oC

T_2 = initial temperature of water = 63.5^oC

Now put all the given values in the above formula, we get

32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC

m_2=30.16g\approx 3.0\times 10^1g

Therefore, the mass of the water in two significant figures is, 3.0\times 10^1g

3 0
3 years ago
Anton van Leeuwenhoek used a microscope to observe
Goshia [24]
B. He was the first person to observe and identify living cells.
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