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agasfer [191]
3 years ago
15

The greater the molality of solute, the less (easily) a solution evaporates.

Chemistry
1 answer:
Alina [70]3 years ago
7 0

Answer:

The answer is "False".

Explanation:

The given statement is false, which can be defined as follows:

  • Molarity is focused on solution liters, which is focussed on kg of solvents when analyzing the property of solutions related to vapor temperature.
  • Its pressure variations, quantities represented in molar concentration can be used.  
  • It is used when its meaning doesn't change with temperature changes. It can be calculated as the M1 x V1 = M2 x V2.

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What is science about?
Olin [163]

Answer:

Science is the study of the nature and behavior of natural things and the knowledge that we obtain about them. ... A science is a particular branch of science such as physics, chemistry, or biology. Physics is the best example of a science which has developed strong, abstract theories.

Explanation:

8 0
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What did Thomson’s model of the atom include that Dalton’s model did not have?
7nadin3 [17]
Thomson's model included Protons and Electrons. His model is referred to as 'Plum Pudding' because of it.
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Calculate the number of moles found in 3.045x1024 atoms of helium.<br><br> PLS HELP
Kisachek [45]

Explanation:

so for this u have to use this equation where

Moles = number of particle/6.02×10^23

= 3.045 × 10^24/6.02×10^23

= 5.0581

write it to 3 S.F so 5.06 moles

4 0
2 years ago
a 25.0-ml volume of a sodium hydroxide solution requires 19.6 ml of a 0.189 m hydrochloric acid for neutralization. a 10.0- ml v
Rashid [163]

<u>Concentration of NaOH = 0.148 molar, M</u>

<u>Concentration of H3PO4 = 0.172 molar, M</u>

<u></u>

Concentration x Volume  will give the number of moles of solute in that volume.  C*V = moles

Concentration  has a unit of (moles/liter).  When multiplied by the liters of solution used, the result is the number of moles.

Original HCl solution:  (0.189 moles/L)*(0.0196 L)= 0.00370 moles of HCl

The neutralization of 25.0 ml of sodium hydroxide, NaOH, requires 0.00370 moles of HCl.  The reaction is:

  NaOH + HCl > NaCl and H2O

This balanced equation tells us that neutralization of NaOH with HCl requires the same number of moles of each.  We just determined that the  moles of HCl used was 0.00370 moles.  Therefore, the 25.0 ml solution of NaOH had the same number of moles:  0.00370 moles NaOH.

The 0.00370 moles of NaOH was contained in 25.0 ml (0.025 liters).  The concentration of NaOH is therefore:  

    <u>(0.00370 moles of NaOH)/(0.025 L) = 0.148 moles/liter or Molar, M</u>

====

The phosphoric acid problem is handled the same way, but with an added twist.  Phosphoric acid is H3PO4.  We learn the 34.9 ml of the same NaOH solution (0.148M) is needed to neutralize the H3PO4.  But now the acid has three hydrogens that will react.  The balanced equation for this reaction is:

  H3PO4 + 3NaOH = Na3PO4 + 3H2O

Now we need <u><em>three times</em></u> the moles of NaOH to neutralize 1 mole of H3PO4.

The moles of NaOH that were used is:

  (0.148M)*(0.0349 liters) = 0.00517 moles of NaOH

Since the molar ratio of NaOH to H3PO4 is 3 for neutralization, the NaOH only neutralized (0.00517)*(1/3)moles of H3PO4 = 0.00172 moles of H3PO4.

The 0.00172 moles of H3PO4 was contained in 10.0 ml.  The concentration is therefore:

     (0.00172 moles H3PO4)/(0.010 liters H3PO4)

<u>Concentration of H3PO4 = 0.172 molar, M</u>

 

5 0
1 year ago
You need to prepare 1 L of the citric acid/citrate buffer. You have chosen to use Method 1 (see lab presentation). Calculate the
prisoha [69]

Answer:

3.11 is the pH of the buffer

Explanation:

The pH of a buffer is obtained using H-H equation:

pH = pKa + log [Conjugate base] / [Weak acid]

<em>Where pH is the pH of the buffer, pKa = -log Ka = 3.14 for the citric buffer and [] could be taken as the moles of each species.</em>

The citric acid,HX (Weak acid), reacts with NaOH to produce sodium citrate, NaX (weak base) and water:

HX + NaOH → H2O + NaX

That means the moles of NaOH added = Moles of sodium citrate produced

And the resulitng moles of HX = Initial moles - Moles NaOH added

<em>Moles HX and NaX:</em>

Moles NaOH = 0.100L * (0.65mol / L) = 0.065 moles NaOH = Moles NaX

Moles HX = 0.300L * (0.45mol / L) = 0.135 moles HX - 0.065 moles NaOH = 0.070 moles HX

Replacing in H-H equation:

pH = 3.14 + log [0.065mol] / [0.070mol]

pH = 3.11 is the pH of the buffer

8 0
3 years ago
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