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goblinko [34]
3 years ago
10

Which matter has more attraction between the particles​

Chemistry
1 answer:
julia-pushkina [17]3 years ago
3 0

There are spaces between particles of matter. The average amount of empty space between molecules gets progressively larger as a sample of matter moves from the solid to the liquid and gas phases. There are attractive forces between atoms/molecules, and these become stronger as the particles move closer together.

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CHEMISTRY PLEASE HELP!!!???
Montano1993 [528]

1. Solids

- definite volume & shape

- little energy

-vibrate in place

- very incompressible


2. Liquids

- held together yet can still flow

5 0
2 years ago
Answer soon please ​
erica [24]

Answer:

The sphere on the left as it has more mass.

Explanation:

Inertia is the resistance to changes of motion.

5 0
3 years ago
How many grams of K2Cr207 are in 250. mL of 0.500M of K2Cr207:
goldenfox [79]

<u>Analysing the Question:</u>

We are given a 250 mL solution of 0.5M K₂Cr₂O₇

Which means that we have:

0.5 Mole in 1L of the solution

0.125 moles in 250 mL of the solution      <em>[dividing both the numbers by 4]</em>

<em />

<u>Mass of K₂Cr₂O₇ in the given solution:</u>

Molar mass of K₂Cr₂O₇(Potassium Dichromate) = 194 g/mol

<em>we know that we have 0.125 moles in the 250 mL solution provided</em>

Mass = Number of moles * Molar mass

Mass = 0.125 * 194

Mass = 36.75 grams

7 0
2 years ago
Compare the number of moles calculated in parts a) 0.03 and b). 0.064. Which of the three possible reactions discussed is consis
Snowcat [4.5K]

Answer:don’t click link ur gonna get ur info taken

Explanation:

6 0
3 years ago
When 1000 C of charge is passed through CuSO4 solution, x g of copper is deposited. How much charge should be passed through the
Stels [109]

Number of charge = 5018 C

<h3>Further explanation</h3>

Given

1000 C of charge for x grams of copper

Required

Number of charge

Solution

Faraday's Law :

\tt W=\dfrac{e.i.t}{96500}\\\\W=\dfrac{e.Q}{96500}

For 1000 C, W = x grams

\tt x=\dfrac{1000.e}{96500}=0.0104e

For 5x grams :

\tt 5x=\dfrac{e.Q}{96500}\\\\5\times 0.0104e=\dfrac{e.Q}{96500}\\\\Q=\dfrac{96500\times 5\times 0.0104e}{e}=5018~C

6 0
3 years ago
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