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IceJOKER [234]
3 years ago
8

One difference between mixtures and pure substances is that

Chemistry
2 answers:
horsena [70]3 years ago
8 0

Answer: B) mixtures are made of one type of atom.

pure substances have no chemical bonds

devlian [24]3 years ago
4 0

Answer:

I think b but I could be wrong

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I think so since in the universe there are other universes as well so maybe the might be even more space in space it is kind of confusing but it makes sense so yes the is space in space. Hope it helped pls give brainlyest
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A body moving with a velocity of 20 m/s begins to accelerate at 3 m/s2. How far does the body move in 5 seconds?
Norma-Jean [14]
The distance covered will be:

d = v_0\cdot t + \frac{1}{2}a\cdot t^2 = 20\frac{m}{s}\cdot 5\ s + \frac{5}{2}\frac{m}{s^2}\cdot 5^2\ s^2 = \bf 162,5\ m

The correct answer is D).
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What portion of your salt solution in ice water system is: a. A mixture? b. A solution?
ahrayia [7]

The salt and water are a homogeneous mixture but when salt dissolves in the water system is called a solution of salt and water.

<h3>What is a mixture? </h3>

A mixture is defined as the combination of two or more substances that are not chemically bonded together.

There are two types of mixture which include:

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  • Heterogeneous mixtures

When salt is added to the ice water system, it lowers the freezing point of the ice water thereby forming a homogenous mixture of water and salt.

The dissolution of salt in ice water leads to the formation of salt and water solution.

Learn more about mixture here:

brainly.com/question/10677519

8 0
2 years ago
Do you think “doing the wave” is a wave? Why or why not?
zmey [24]

Answer:

i mean kinda they should throw people in the air tho lol

Explanation:

3 0
4 years ago
A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
Lilit [14]

Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

3 0
3 years ago
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