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4 years ago

In a solution the solute is the substance present in the greatest amount true or false

Chemistry

2 answers:

emmasim [6.3K]4 years ago

7 0

Answer: False

Explanation:

Binary Solution is a homogeneous mixture of two components called as solute and solvent.

Solute is the component which is present in smaller proportion and is solid for solid in liquid solution and solvent is the component which is present in larger proportion and is liquid for solid in liquid solution.

For example: NaCl in water will have NaCl as solute and will be lesser in amount and water will be solvent and will be greater in amount.

Thus the statement that in a solution the solute is the substance present in the greatest amount is false.

kati45 [8]4 years ago

3 0

Solute is something that is being dissolved { ex : sugar , salt}

Solvent is something that has ability to dissolve things { ex : water}

False because the solvent is present in larger amounts...

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3 years ago

You are given a small bar of an unknown metal. You find the density of the metal to be 10.5 g/cm3. An X-ray diffraction experime

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4 years ago

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4) 10.00 mL of sulfuric acid is neutralized in a titration using 18.54 mL of 0.100 M NaOH.

sdas [7]

Answer:

a) $H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

b) $M_{H_2SO_4}=0.0927M$

Explanation:

Hello!

In this case, for this acid-base reaction which also known as a neutralization because sulfuric acid is neutralized with sodium hydroxide, we can write the undergoing chemical reaction as shown below:

$H_2SO_4+NaOH\rightarrow Na_2SO_4+H_2O$

However, it needs to be balanced as two sodium atoms are yielded:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

Next, since there is a 1:2 mole ratio between the acid and the base, at the equivalence point, at which the moles of acid and base are consumed, we write:

$2n_{H_2SO_4}=n_{NaOH}$

Which can also be written in terms of the given volumes and concentration of the base:

$2M_{H_2SO_4}V_{H_2SO_4}=M_{NaOH}V_{NaOH}$

In such a way, we solve for the concentration of sulfuric acid as shown below:

$M_{H_2SO_4}=\frac{M_{NaOH}V_{NaOH}}{2V_{H_2SO_4}} =\frac{18.54mL*0.100M}{2*10.00mL}\\\\ M_{H_2SO_4}=0.0927M$

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3 years ago

Ba(oh)2 is a strong electrolyte. determine the concentration of each of the individual ions in a 0.750 m ba(oh)2 solution.

Sindrei [870]

The concentration of each of the individual ions in a 0750 M Ba(OH)2 solution is

[Ba2+] = 0.750 M

[OH-]= 1.50 M

<h3>calculation</h3>

write the equation for dissociation

that is Ba(OH)2 (s)→ Ba2+(aq) + 2OH-(aq)

by use of mole ratio of Ba(OH)2 : Ba2+ which is 1: 1 the concentration of Ba2+ is therefore= 0.750M

by use of mole ratio of Ba(OH)2 : OH- which is 1:2 the concentration of OH- =0.750 M x2/1=1.50 M

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3 years ago