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tino4ka555 [31]
3 years ago
6

Rosa preparo una disolución de agua salada mezclando 120 mg de agua con 5 mg de sal comun ¿ que concentración tiene la disolució

n que preparo ?
Chemistry
1 answer:
enot [183]3 years ago
3 0

Concentration of the solution (% mass) = 4%

<h3>Further explanation</h3>

<em>Rosa prepared a salt water solution by mixing 120 mg of water with 5 mg of common salt, what concentration is the solution she prepare?</em>

<em />

<em />\tt \%mass=\dfrac{mass~solute}{mass~solution}\times 100\%<em />

mass solute=mass of salt=5 mg

mass solution :

mass solute(salt)+mass solvent(water) :

\tt 5~mg+120~mg=125~mg

\tt \%mass=\dfrac{5~mg}{125~mg}\times 100\%\\\\\%mass=4

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2 years ago
A. How many moles of copper equal 8.00 × 109 copper atoms?
BARSIC [14]

Answer:

Explanation:

a ) one mole = 6.02 x 10²³ atoms

no of moles in given no of atoms

= 8 x 10⁹ / 6.02 x 10²³

= 1.329 x 10⁻¹⁴ moles .

b ) one mole of calcium = 40 gram

102 .5 g calcium

= 102 .5 / 40 moles

= 2.5625 moles

c )

no of moles in 5.04 g lead = 5.04 / 207

= 2.4347 x 10⁻² moles

= 2.4347 x 10⁻²x 6.02 x 10²³ no of atoms of lead

= 14.6568 x 10²¹ no of atoms .

d)  

one mole = 6.02 x 10²³ atoms

2.85 mole = 17.157 x 10²³ atoms .

e )

moles of fluorine gas = 1.08 x 10³ / 6.02 x 10²³

= .1794 x 10⁻²⁰ moles

mass in grams =  .1794 x 10⁻²⁰ x 38

= 6.8172 x 10⁻²⁰ grams

f )

no of moles in .584 g of benzene = .584 / 78

= 7.487 x 10⁻³ moles

no of molecules = 6.02 x 10²³ x  7.487 x 10⁻³

= 45.07 x 10²⁰ molecules .

g )

moles of atoms = 5.09 x 10⁹ / 6.02 x 10²³

= .8455 x 10⁻¹⁴ moles

mass in gram = .8455 x 10⁻¹⁴ x 1

=  .8455 x 10⁻¹⁴ g

h )

.45 moles of Ca₃PO₄ = .45 x 6.02 x 10²³ molecules

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7 0
3 years ago
A force of 20 N acts upon 5 kg block caculate the accerleration of the object
Vedmedyk [2.9K]

Answer:

\boxed{\text{4 m $\cdot$ s$^{-2}$}}

Explanation:

F = ma

a = \dfrac{F}{m} = \dfrac{\text{20 N}}{\text{5 kg}} \times \dfrac{\text{1 kg$\cdot$ m $\cdot$ s$^{-2}$}}{\text{1 N}} = \text{4 m $\cdot$ s$^{-2}$}\\\\\text{The acceleration of the object is } \boxed{\textbf{4 m $\cdot$ s$^{\mathbf{-2}}$}}

 

4 0
3 years ago
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