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3 years ago

How many grams of CaCl2 are need to prepare 250 mL of a 0.15M solution?

Chemistry

1 answer:

Tanzania [10]3 years ago

4 0

Answer:

Here is what I got from reading this question...

Explanation:

So if you dissolve 7.5625- g of CaCl2 in water, dissolve well and then add distilled in a 250- ml standard flask upto the mark and shake well for uniform Concentration, you will get 0.25- M CaCl2 solution.

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Without doing any calculations, determine which mass is closest to the atomic mass of carbon. without doing any calculations, de

yarga [219]

The mass closest to the mass of carbon is 12.00 amu

This is because carbon is used as a standard compound in chemistry, and its mass is defined as being 12.00 amu per mole of carbon. Its definition as a standard is such that the mole is defined as, "the amount of substance containing the same number of atoms as 12 grams of carbon-12".

7 0

3 years ago

Formic acid (HCO2H) has a dissociation constant of 1.8 x 10^-4 M. The acid dissociates 1:1. What is the [H+] if 0.35 mole of for

Tju [1.3M]

Initial [ HCO2H] = moles * volume
=0.35 moles * 1 L = 0.35 M

by using ICE table:

HCO2H ↔ H+ + HCO2-
initial 0.35 M 0 0
change - X +X +X
Equ (0.35 - X) X X

∴ Ka = [H+][HCO2-] / [HCO2H]

by substitution:

1.8 x 10^-4 = X^2 / (0.35-X) by solving for X

∴ X = 0.0079 or 7.9 x 10^-3

∴ [H+] = X = 7.9 x 10^-3 M

7 0

3 years ago

T he thermochemical equation for the reaction between hydrazine, N2H4, and dinitrogen tetroxide is given as: 2N2H4(2) N204 (1) 3

SpyIntel [72]

Answer:

4N2H4(2) + 2N204 (1) --> 6N2 (g) + 8H20 (g), AHO-2098 kJ

Explanation:

2N2H4(2) + N204 (1) --> 3N2 (g) + 4H20 (g),AHO-1049 kJ

when 6 mol of nitrogen are formed?

The ratio of Nitrogen in the previous therrmochemical equation to when 6 mol f nitrogen are formed is 2; 6/3 = 2.

So to get the new thermoochemical equation, multiply all parameters in the given equation by 2.

We have;

4N2H4(2) + 2N204 (1) --> 6N2 (g) + 8H20 (g), AHO-2098 kJ

6 0

3 years ago

What is the voltage of the voltaic cell Zn|Zn2+||Cu2+|Cu at 298 K if [Zn2+] = 0.2 M and [Cu2+] = 4.0 M? Cu2+ + 2e- → Cu Eo = +0.

OLga [1]

Answer : The voltage of the voltaic cell is 1.14 V

Explanation :

From the given cell representation, we conclude that

The copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

To calculate the $E^o{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}$

Putting values in above equation, we get:

$E^o_{cell}=(+0.34V)-(-0.76V)$

$E^o_{cell}=1.1V$

Now we have to calculate the emf or voltage of the cell.

Using Nernest equation at 298 K :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

Now put all the given values in the above equation, we get:

$E_{cell}=1.1-\frac{0.0592}{2}\log \frac{0.2}{4.0}$

$E_{cell}=1.14V$

Therefore, the voltage of the voltaic cell is 1.14 V

5 0

3 years ago

Which of the following is not true about limiting and excess reagents?

Bogdan [553]

Answer : The incorrect option is, (d) The reactant that was the smallest given mass is the limiting reagent.

Explanation :

Limiting reagent : It is the reagent that is completely consumed in the chemical reaction when the chemical reaction is complete. No amount is left after the reaction is complete. The amount of product obtained is determined by the limiting reagent. A balanced equation is necessary to determine which reactant is limiting reagent.

Excess reagent : It is the reagent that are not completely consumed in the chemical reaction. That means the reagent is in excess amount. Some amount of the excess reagent is left over after the reaction is complete.

From this we conclude that the options, A, B and C are correct. While the option D is incorrect.

Option D is incorrect because it is not necessary the reactant that was the smallest given mass is the limiting reagent but it is judge by the number of moles present in the reaction.

Hence, the incorrect option is, (d)

4 0

3 years ago