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Zigmanuir [339]
3 years ago
12

Which tool is most suitable for a chemist to measure the mass of a liquid

Chemistry
2 answers:
I am Lyosha [343]3 years ago
7 0

Chemists use beakers, flasks, burets and pipets to measure the volume of liquids. Plz mark as brainliest

telo118 [61]3 years ago
7 0

Answer:

The most suitable tool is using a graduated cylinder and a calibrated balance.

Explanation:

To measure the mass of a liquid, first you have to define an specific volume of liquid to measure (example 200 milliliters), then using a graduated cylinder you measure this quantity and put it all on the balance. You read the mass value (example 285 grams). To finish you have to measure the mass of the graduated cylinder empty (example 100 grams) and then subtract it to the weight of the liquid with the calibrated cylinder (example 285 g - 100 g = 185 g).

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CH3-CH2-CH=CH-CH=CH2
o-na [289]

Answer:

Hydration (of an alkene)

Mechanism : Electrophilic addition.

8 0
3 years ago
antimony has two naturally occuring isotopes, sb121sb121 and sb123sb123 . sb121sb121 has an atomic mass of 120.9038 u120.9038 u
Luda [366]

Considering the definition of atomic mass, isotopes and atomic mass of an element, sb121 has a percent natural abundance of 0.5726 or 57.26% and sb123 has a percent natural abundance of 0.4284 or 42.84%.

<h3>Definition of atomic mass</h3>

The atomic mass is obtained by adding the number of protons and neutrons in a given nucleus of a chemical element.

<h3>Definition of isotope</h3>

Isotopes are the chemical elements in which atomic numbers are the same, but the number of neutrons is different.

<h3>Definition of atomic mass</h3>

The atomic mass of an element is the weighted average mass of its natural isotopes.

This is, the atomic masses of elements are usually calculated as the weighted average of the masses of the different isotopes of each element, considering the relative abundance of each of them.

<h3>Percent natural abundance of each isotope</h3>

In this case, antimony has two naturally occuring isotopes, sb121 and sb123. You know:

  • sb121 has an atomic mass of 120.9038 u.
  • sb121 has a percent natural abundance of x.
  • sb123 has an atomic mass of 122.9042 u.
  • sb123 has a percent natural abundance of 1 -x (or, what is the same, the abundance is 100% - x%, since both isotopes form 100% of the element.)
  • Antimony has an average atomic mass of 121.7601 u

The average mass of antimony is expressed as:

121.7601 u= 120.9038 u x + 122.9042 u× (1 -x)

Solving:

121.7601 u= 120.9038 u x + 122.9042 u - 122.9042 u x

121.7601 u - 122.9042 u= 120.9038 u x - 122.9042 u x

(-1.1441 u)= (-2.0014) x

(-1.1441 u)÷ (-2.0014)= x

<u><em>0.5726= x or 57.26%</em></u>

So, 1 -x= 1- 0.5716 → <u><em>1-x= 0.4284 or 42.84%</em></u>

<u><em /></u>

Finally, sb121 has a percent natural abundance of 0.5726 or 57.26% and sb123 has a percent natural abundance of 0.4284 or 42.84%.

Learn more about average atomic mass:

brainly.com/question/4923781

brainly.com/question/1826476

brainly.com/question/15230683

brainly.com/question/7955048

#SPJ1

5 0
2 years ago
The earth was created about_____ years ago. 15,000 20,000 6,000 3,000
Sphinxa [80]
It was created about 6,000 years ago
5 0
3 years ago
An acid
tekilochka [14]
The answer is A. Has a low pH in solution.
5 0
3 years ago
Calculate the standard entropy change for the following reactions at 25°C.
Art [367]

Answer:

(a) ΔSº = 216.10 J/K

(b) ΔSº = - 56.4 J/K

(c) ΔSº = 273.8 J/K

Explanation:

We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.

First we need to find in an appropiate reference table the standard  molar entropies entropies, and then do the calculations.

(a)        C2H5OH(l)          +        3 O2(g)         ⇒        2 CO2(g)     +    3 H2O(g)

Sº            159.9                          205.2                         213.8                  188.8

(J/Kmol)

ΔSº = [ 2(213.8) + 3(188.8) ]   - [ 159.9  + 3(205.) ]  J/K

ΔSº = 216.10 J/K

(b)         CS2(l)               +         3 O2(g)               ⇒      CO2(g)      +      2 SO2(g)

Sº          151.0                              205.2                         213.8                 248.2

(J/Kmol)

ΔSº  = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K

(c)        2 C6H6(l)           +        15 O2(g)                     12 CO2(g)     +     6 H2O(g)

Sº           173.3                           205.2                           213.8                    188.8

(J/Kmol)  

ΔSº  = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K

Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4  total mol gas reactants to 3, so the entropy change will be negative.

Note we need to multiply the entropies of each substance by  its coefficient in the balanced chemical equation.

5 0
3 years ago
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