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Annette [7]
3 years ago
7

The image shows a molecular model of a compound using balls and sticks. Each ball is an atom and each stick is a bond. If you we

re to use balls
and sticks to model a molecule of an element, how must your model differ from what's shown?​

Chemistry
1 answer:
jeka943 years ago
7 0
<h2><u>Full Question:</u></h2>

The image shows a molecular model of a compound using balls and sticks. Each ball is an atom and each stick is a bond. If you were to use balls and sticks to model a molecule of an element, how must your model differ from what’s shown?

A. The sticks must be shorter.

B. The balls must be the same color.

C. The balls must be smaller.

D. The balls must change to cubes.

<h2><u>Answer:</u></h2>

The balls must be of same colour.

Option B.

<h3><u>Explanation:</u></h3>

A Molecule is defined as the smallest part of an element or compound which can exist freely in nature. An element is define as the matter which is when divided into smallest particles gives only one type of particle or atom. Where as a compound is defined as the combination of different elements in a particular ratio which has properties totally different from the combining elements. So an element comprises of only one type of atoms.

In this picture we can see that there is a central atom which is marked in blue where as the other atoms are peripheral and marked in red. The bonds between the blue atom and red atoms are marked with sticks. So if only one type of atoms are present in the molecule then only one coloured balls will be seen. Then the the colour of the balls will be only one.

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Answer:

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Explanation:

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Read 2 more answers
A 10 gram sample of water is heated to 105 ℃ and is mixed with a 25 gram sample of water cooled to 25℃ . What is the final tempe
lesya692 [45]

Answer:

The final temperature of the water mixture is 47.85°C

Explanation :

Given,

For Warm Water

mass = 10grams

Temperature = 105°C

For Cold Water

mass = 25grams

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When a sample of warm water is mixed with a sample of cool water,

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<h3>Qlost = Qgain</h3>

However,

Q = (mass) (ΔT) (Cp)

Cp = Specific heat of water = 4.184 J/Kg°C

So,

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

We start by calling the final, ending temperature 'x.' Keep in mind that BOTH water samples will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using is FINAL temperature. This is what we are solving for.  

The warmer water goes down from to 105°C to x, so this means its Δt equals 105°C − x. The colder water goes up in temperature, so its Δt equals x − 25℃

Substituting the values,

(10)( 105°C − x)(4.184) = (25)(x − 25℃)(4.184)

Solving for x, we get

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