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denis23 [38]
3 years ago
8

What are the characteristics of a mixture?

Chemistry
1 answer:
lora16 [44]3 years ago
4 0
Properties of mixtures:
1. Parts of a mixture are physically combines.
2. Parts of mixtures keep their own properties.
3.Can be mixed together in different amounts.
4. Can be physically separated.
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I have to covert 4.4×10^4 Pa in to atm
Kitty [74]
The answer is 4.40000
5 0
2 years ago
Which of the following compounds has the most deshielded protons?A) CH3ClB) CH3IC) CH3BrD) CH4
Korolek [52]

Given :

Some compounds :

A)\ CH_3Cl\ B)\ CH_3I\ C)\ CH_3Br\ D)\ CH_4 .

To Find :

Which of the following compounds has the most deshielded protons .

Solution :

Deshielded means nucleus whose chemical shift has been increased due to removal of electron density, magnetic induction, or other effects .

In simple words deshielding means the ability to shift protons .

Now , among Cl , I , Br and H . Cl is the most electron negative .

Therefore , deshielding will be more in CH_3Cl .

Hence , this is the required solution .

6 0
3 years ago
Can you guys help me answer question 5 on homogeneous mixture tysm
hjlf

Answer:

Hope this helped :) good luck! ❤️

Explanation:

A <em>coolant solution</em> is a <u><em>homogeneous </em></u>mixture because the coolant particles are not chemically combined with the water (keep their properties) and they are evenly distributed throughout the water.

7 0
3 years ago
If the oxidation number of nitrogen in a certain molecule changes from +3 to -2 during a reaction, is the nitrogen oxidized or r
aliina [53]
The nitrogen has been reduced or has undergone reduction and it has gained one electron
7 0
3 years ago
A gas has a pressure of 5.7 atm at 100.0°C. What is its pressure at20.0°C (Assume volume is unchanged)
son4ous [18]

Answer:

\large \boxed{\text{4.5 atm}}

Explanation:

The volume and amount of gas are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}

Data:

p₁ =5.7 atm; T₁ = 100.0 °C

p₂ = ?;          T₂ =  20.0 °C

Calculations:

1. Convert the temperatures to kelvins

T₁ = (100.0 + 273.15) K = 373.15

T₂ =  (20.0 + 273.15) K = 293.15

2. Calculate the new pressure

\begin{array}{rcl}\dfrac{5.7}{373.15} & = & \dfrac{p_{2}}{293.15}\\\\0.0153 & = & \dfrac{p_{2}}{293.15}\\\\0.0153\times 293.15 &=&p_{2}\\p_{2} & = & \textbf{4.5 atm}\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.5 atm}}$}

6 0
3 years ago
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