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3 years ago

¿cuántas partículas elementales hay en 20g de Ca?

Chemistry

1 answer:

snow_lady [41]3 years ago

5 0

Answer:

la esperanza ayuda

Explanation:

Cuando se utiliza el mol deben especificarse las entidades elementales de que se trata. ... ¿Cuántas partículas (moléculas) de oxígeno (O2) hay en una mol de dicho gas? ... 20 g de Fe; 20 g de S; 20 g de oxígeno; 20 g de Ca; 20 g de CaCO3.

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While under constant pressure, the temperature of 892 mL of a gas is changed from 193 °C to 144 °C. What is the new gas volume?

galina1969 [7]

Answer:

please mark my answer brainliest

Explanation:

nice question...

ok solution goes right the way through...

in temp 193°C the volume of gas was 892ml

so for...144°C the volume of gas will be 892/193x144=665.533 or 665.534...(rounded off)...or 665°...(approx)....so got ur answer...

I hope it helps....

please comment if you have any further questions...i will be solving it for u...

please mark my answer brainliest

7 0

4 years ago

How many molecules of water are in a 45 g sample of H2O?

Ymorist [56]

Answer:

3. 0.75 mole. or (C)

Explanation:

1.5 moles. 9.03X1023 . 3. 0.75 mole. 4581023. 4. 15 moles.

4 0

3 years ago

Calculate the equilibrium constant for the following reaction: Co2+ (aq) + Zn(s> CO (s) + Zn2+ (aq)

Simora [160]

<u>Answer:</u> The $K_{eq}$ of the reaction is $1.73\times 10^{16}$

<u>Explanation:</u>

For the given half reactions:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Co^{2+}+2e^-\rightarrow Co(s);E^o_{Co^{2+}/Co}=-0.28V$

Net reaction: $Zn(s)+Co^{2+}\rightarrow Zn^{2+}+Co(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

$E^o_{cell}=-0.28-(-0.76)=0.48V$

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Equating these two equations, we get:

$nfE^o_{cell}=RT\ln K_{eq}$

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard electrode potential of the cell = 0.48 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = $25^oC=[273+25]=298K$

$K_{eq}$ = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

$2\times 96500\times 0.48=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=1.73\times 10^{16}$

Hence, the $K_{eq}$ of the reaction is $1.73\times 10^{16}$

8 0

3 years ago

How many atoms of titanium are in 0.085 mol of titanium

aleksley [76]

Answer:

5.12x10^23 atoms

Explanation:

To find atoms, you need to use Avogadro's number. Here is the equation:

0.085 mol x 6.022x10^23 atoms = 5.12x10^22 atoms of titanium

Keep in mind that units need to cancel and make sure to double check those sig figs!

7 0

3 years ago

Select the objects that hold the same amount of liquid as a 160−fluid-ounce jug. Select all that apply. A four 1−quart bottles B

Finger [1]

B, D and E

Explanation:

conversion factors

1c = 8oz

1pt = 2c

1qt = 2pt

For A and B

ounces to cup = 160/8 = 20c

cup to pints = 20c / 2c = 10pt

pint to quarts = 10pt/2pt = 5qt

B applies as four 1-quart and two 1-pt = 5 1-quart

Considering C

4 8oz = 32oz

160-32 = 128oz /8 = 16c/2 = 8pints C does not apply

considering D

8 * 8 = 64oz

160 - 64 = 96oz/8 = 12c/2 = 6pt/2 = 3qt

So D applies

E applies

5 0

3 years ago

¿cuántas partículas elementales hay en 20g de Ca?​

¿cuántas partículas elementales hay en 20g de Ca?