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3 years ago

6. How are Fisher projections used to represent three-dimensional perspective?

Chemistry

2 answers:

Levart [38]3 years ago

8 0

Answer:lol study more

Explanation:

Nitella [24]3 years ago

5 0

It is to get more stuff

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It would be carbon because its relative atomic mass is 12

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3 years ago

Which BEST explains why the ice cube sank in the liquid?

ValentinkaMS [17]

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What element is added to toothpaste and city water to help prevent tooth decay?

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4 years ago

3. The sum of gravitational potential energy and kinetic energy is called mechanical

SCORPION-xisa [38]

Answer:

Explanation:

Mechanical energy = Kinetic energy + Potential energy

We shall calculate mechanical energy at different points of time .

Time Kinetic energy Potential energy mechanical energy

0 0 4 4

2 2 2 4

4 4 0 4

6 2 2 4

8 0 4 4

10 2 2 4

and so on ........

Hence we see that total mechanical energy is constant at the value of 4

at all time .

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3 years ago

You placed 43.1 g of an unknown metal at 100 °C into a coffee cup calorimeter that contained 50.0 g of water that was initially

nika2105 [10]

Answer :

(a) The heat released by the metal is -312.48 J

(b) The specific heat of the metal is $0.0944J/g^oC$

Explanation :

Heat released by the metal = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the metal

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $51.5J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 50.0 g

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=23.2-22.0=1.2^oC$

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

$q=[(51.5J/^oC\times 1.2^oC)+(50.0g\times 4.184J/g^oC\times 1.2^oC)]$

$q=312.48J$

Thus, the heat released by the metal is -312.48 J

$q=m\times c\times \Delta T$

q = heat released by the metal = -312.48 J

m = mass of metal = 43.1 g

c = specific heat of metal = ?

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=23.2-100=76.8^oC$

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

$-312.48J=43.1g\times c\times 76.8^oC$

$c=0.0944J/g^oC$

Thus, the specific heat of the metal is $0.0944J/g^oC$

8 0

4 years ago