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3 years ago

The smallest unit that retains the properties of an element is a(n) _____. atom molecule proton compound neutron

Chemistry

1 answer:

Aleks [24]3 years ago

8 0

The answer to your question is ATOM.

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A diesel truck produces 47.5kJ of useful output energy from 125kJ of diesel fuel.what is the truck efficiency?.show the solution

Alex Ar [27]

You have to do a report :
$\frac{E_f}{E_i} = \frac{47,5}{125} = 0,38$

So the truck efficiency is 38%.

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3 years ago

When iodobenzene codistills with water at 760 toor, the boiling point of the mixture is 98.5. the vapor pressure of water is?

katrin [286]

<span>This was one of the old regents. I didn't get this at first glance too so I searched for it and I got this site. lol. I know what to do now. To get a higher boiling point, you have to find which solution has the most moles. Since SO4, PO4, and NO3 are all polyatomic ions, they only count as one mole. </span>

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3 years ago

What is the Equation of Reduction in Mg+F2 gives MgF2, I WILL MARK YOU AS BRAINLIST

skad [1K]

Answer:

Mg+F2= Mgf2

Explanation:

F 2 is an oxidizing agent, Mg is a reducing agent. ; Pale-yellow to greenish gas with a pungent, irritating odor.

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2 years ago

How many grams of solute are needed to order to prepare 100.00 mL of a 0.1000 M solution of a compound with a molecular weight o

Troyanec [42]

The grams of solute are required.

The mass of solute is 3.5 g

c = Molarity = 0.1 M

M = Molar mass = 350 g/mol

V = Volume of solution = 100 mL = 0.1 L

n = Number of moles

m = Mass of solute

Molarity is given by

$c=\dfrac{n}{V}\\\Rightarrow n=cV\\\Rightarrow n=0.1\times 0.1\\\Rightarrow n=0.01\ \text{moles}$

Molar mass is given by

$M=\dfrac{m}{n}\\\Rightarrow m=Mn\\\Rightarrow m=350\times 0.01\\\Rightarrow m=3.5\ \text{g}$

The mass of solute is 3.5 g

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7 0

2 years ago

Read 2 more answers

Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m

erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

$n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3} =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2} =0.06882molBr^-$

Now, we compute the total moles of bromide:

$n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol$

Then, the total volume in liters:

$150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L$

Therefore, the concentration of total bromide is:

$M=\frac{0.12338mol}{0.325L}\\\\M=0.380M$

Best regards!

8 0

2 years ago