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emmasim [6.3K]
2 years ago
9

250 mL of a solution of calcium oxalate is the evaporated until only a residue of solid calcium

Chemistry
1 answer:
Lana71 [14]2 years ago
4 0

Answer:

2.3 * 10^-5

Explanation:

Recall that the solubility of a solute is the amount of solute that dissolves in 1 dm^3 or 1000cm^3 of solution.

Hence;

Amount of calcium oxalate = 154 * 10^-3/128.097 g/mol = 1.2 * 10^-3 mols

From the question;

1.2 * 10^-3 mols dissolves in 250 mL

x moles dissolves in 1000mL

x = 1.2 * 10^-3 mols * 1000/250

x= 4.8 * 10^-3 moldm^-3

CaC2O4(s) ------->Ca^2+(aq) + C2O4^2-(aq)

Hence Ksp = [Ca^2+] [C2O4^2-]

Where;

[Ca^2+] = [C2O4^2-] = 4.8 * 10^-3 moldm^-3

Ksp = (4.8 * 10^-3)^2

Ksp = 2.3 * 10^-5

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Les étapes nous permettant de sentir une tarte aux pommes
Dmitry [639]

Answer:

L'ordre correct est le suivant;

1. Les cils olfactifs capturent les molécules chimiques odorantes

2. Le signal des molécules chimiques est transformé en impulsions nerveuses

3. L'influx nerveux est transféré au cerveau par le nerf olfactif au bulbe olfactif du cerveau antérieur

4. Les odeurs sont perçues

Explanation:

Pour sentir la pomme, l'air autour de la pomme est inhalé rapidement pour aider les produits chimiques odorants à entrer dans la cavité nasale.

La cavité nasale a des cellules spéciales qui sont réceptives aux produits chimiques odorants, qui envoient un message au cerveau lors de la liaison au produit chimique par le nerf olfactif. Lorsque le message arrive au bulbe olfactif du cerveau antérieur, il est analysé et l'odeur est identifiée et / ou reconnue.

Par conséquent, l'ordre correct est le suivant;

1. 1. Les cils olfactifs captent les molécules chimiques odorantes

2. 2. Le signal des molécules chimiques est transformé en impulsions nerveuses

3. 4. L'influx nerveux est transféré au cerveau par le nerf olfactif au bulbe olfactif du cerveau antérieur

4. 3. Les odeurs sont perçues.

3 0
3 years ago
What mass is in 5 moles of helium?
ArbitrLikvidat [17]

Answer:

Mass = 20 g

Explanation:

Given data:

Number of moles of He = 5 mol

Mass of He = ?

Solution:

Formula:

Number of moles = mass/ molar mass

Molar mass = 4 g/mol

by putting values,

5 mol = Mass / 4 g/mol

Mass = 5 mol × 4 g/mol

Mass = 20 g

3 0
3 years ago
A solution is prepared by mixing 250 mL of 1.00 M CH3COOH with 500 mL of 1.00 M NaCH3COO. What is the pH of this solution? (Ka f
Svetllana [295]

Answer:

A solution is prepared by mixing 250 mL of 1.00 M

CH3COOH with 500 mL of 1.00 M NaCH3COO.

What is the pH of this solution?

(Ka for CH3COOH = 1.8 × 10−5 )

Explanation:

This is a case of a neutralization reaction that takes place between acetic acid,     CH 3 COOH ,   a weak acid, and sodium hydroxide,   NaOH , a strong base.

The resulting solution pH, depends if the neutralization is complete or not.  If not, that is, if the acid is not completely neutralized, a buffer solution containing acetic acid will be gotten, and its conjugate base, the acetate anion.

It's important to note that at complete neutralization, the pH of the solution will not equal  7 . Even if the weak acid is neutralized completely, the solution will be left with its conjugate base, this is the reason why the expectations of its pH is to be over  7 .

So, the balanced chemical equation for this reaction is the ionic equation:

CH 3 COOH (aq]  +  OH − (aq]  →  CH 3 COO − (aq]  +  H 2 O (l]

Notice that:  

1  mole of acetic acid will react with:  1  mole of sodium hydroxide, shown here as hydroxide anions,  OH − , to produce   1   mole of acetate anions:

CH 3 COO −

To determine how many moles of each you're adding , the molarities and volumes of the two solutions are used:

     c  =  n /  V    ⇒     n   =   c  ⋅  V

n  acetic   =   0.20 M   ⋅   25.00   ⋅   10  − 3 L   =   0.0050 moles CH3 COOH

and

n  hydroxide   =   0.10 M   ⋅   40.00   ⋅   10 − 3 L   =   0.0040 moles OH −

There are fewer moles of hydroxide anions, so the added base will be completely consumed by the reaction.

As a result, the number of moles of acetic acid that remain in solution is:

    n  acetic remaining   =   0.0050  −   0.0040   =    0.0010 moles

The reaction will also produce  0.0040   moles of acetate anions.

This is, then a buffer and the Henderson-Hasselbalch equation is applied to find its pH :

pH  =  p K a  +  log  ( [ conjugate base ]  / [ weak acid ] )

Use the total volume of the solution to find the new concentrations of the acid and of its conjugate base .

V total  =  V acetic  +  V hydroxide

V total  =  25.00 mL  +  40.00 mL  =  65.00 mL

Thus the concentrations will be :

[ CH 3 COOH ]  =  0.0010 moles  / 65.00  ⋅  10 − 3 L  =  0.015385 M

and

[ CH 3 COO − ]  =  0.0040 moles  / 65  ⋅  10 − 3 L  =  0.061538 M

The    p K a     of acetic acid is equal to    4.75

Thus the pH of the solution will be:

pH   =   4.75  +  log ( 0.061538 M  /    0.015385 M )

pH   =   5.35

5 0
2 years ago
Read 2 more answers
How many molecules of N204 are in 85.0 g of N2O4?
alukav5142 [94]

Answer:

5.56 × 10^23 molecules

Explanation:

The number of molecules in a molecule can be calculated by multiplying the number of moles in that molecule by Avagadro's number (6.02 × 10^23)

Using mole = mass/molar mass

Molar mass of N2O4 = 14(2) + 16(4)

= 28 + 64

= 92g/mol

mole = 85.0/92

= 0.9239

= 0.924mol

number of molecules of N2O4 (nA) = 0.924 × 6.02 × 10^23

= 5.56 × 10^23 molecules

4 0
3 years ago
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