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3 years ago

8

How many units are in 1.54 moles?

1 answer:

Sveta_85 [38]3 years ago

5 0

Answer:

Every mole contains 6.022 x 10^23 atoms regardless of what type of moelcules it is (Gold, Silver). So the anwer is 1.54 times 6.022 x 10^2

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PLEASE HELP!?Consumers must eat other organisms for energy. Which organisms are consumers in this food chain?

Vitek1552 [10]

Answer:

Consumers must consume other organisms to get the food that they need and are known as Heterotrophs as they cannot make their own glucose. These consumers eat producers (plants). Herbivores are considered as first order consumers. These consumers eat consumers and producers (animals and plants).

4 0

2 years ago

A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from

sweet-ann [11.9K]

Answer:

$T_i~=163.1$ ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

$Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT$

$Q_A_u=m_A_u*Cp_A_u*deltaT$

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

$m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT$

Now we can <u>put the values into the equation</u>:

$60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)$

Now we can <u>solve for the initial temperature of gold</u>, so:

$T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20$

$T_i~=163.1$ ºC

I hope it helps!

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3 years ago

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2 years ago

How many total atoms are in the compound:<br> C6H12O6<br><br> 18<br><br> 6<br><br> 12<br><br> 24

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D.) There are 24 atoms in total in that compound.....

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3 years ago

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A runner completes a 400-meter race in 2.5 mins. What is the runner's average speed?

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