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3 years ago

The mass of 2.50 moles of calcium fluoride is ____ grams

Chemistry

2 answers:

lapo4ka [179]3 years ago

8 0

Answer:

195.187016

Explanation:

Vesna [10]3 years ago

3 0

Answer:

195.187016

Explanation:

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Calculate the cell potential for the reaction as written at 25.00 °C 25.00 °C , given that

Taya2010 [7]

Answer:

E = 2.02 V

Explanation:

In order to do this, we need to apply the Nernst equation which is:

E = E° - RT/nF lnQ

The value of RT/F can be simplified to just 0.059 because we are doing this experiment at 25 °C, and R and F are constants. so we need the value of Q which in this case is:

Q = [Mg²⁺] / [Ni²⁺]

We already have the concentrations, so, all we have left is the standard reduction potential, which are:

E° Mg = -2.38 V

E° Ni = -0.25 V

According to the overall reaction:

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s)

we can see that one element is reducting and the other is oxidizing, so we need to write the semi equation of reduction for each element:

Mg(s) ---------> Mg²⁺ + 2e⁻ E° = 2.38 V oxidizing (Value of E° inverted)

Ni²⁺ + 2e⁻ -----------> Ni(s) E° = -0.25 V reducting

------------------------------------------------------------

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s) E° = 2.13 V

We have the value of the standard potential, now we need to replace all given data into the nernst equation to solve for the cell potential:

E = 2.13 - 0.059/2 ln(0.757/0.0160)

E = 2.13 - 0.0295 ln(47.3125)

E = 2.13 - 0.11

E = 2.02 V

This is the cell potential

3 0

3 years ago

In a Dam, If we doubled the depth of the dam the hydrostatic force will be ?

AnnZ [28]

Answer: Option (A) is the correct answer.

Explanation:

Force acting on a dam is as follows.

F = $\frac{1}{2}\rho g\omega H^{2}$ .......... (1)

Now, when we double the depth then it means H is increasing 2 times and then the above relation will be as follows.

F' = $\frac{1}{2}\rho g\omega (2)^{2}$

F' = $\frac{1}{2}\rho g\omega 4$ ........... (2)

Now, dividing equation (1) by equation (2) as follows.

$\frac{F}{F'}$ = $\frac{\frac{1}{2}\rho g\omega H^{2}}{\frac{1}{2}\rho g\omega 4}$

Cancelling the common terms we get the following.

$\frac{F}{F'}$ = $\frac{1}{4}$

4F = F'

Thus, we can conclude that if doubled the depth of the dam the hydrostatic force will be 4F.

4 0

3 years ago

A pump contains 0.5 L of air at 203kPa. You draw back on the piston of the pump until the pressure reads 25.4kPa. What is the vo

densk [106]

<h3> Answer;</h3>

= 4.0 L

<h3>Explanation;</h3>

Boyle's law states that the volume of a fixed mass of a gas is inversely proportional to pressure at a constant temperature.

Therefore; Volume α 1/pressure

Mathematically; V α 1/P

V = kP, where k is a constant;

P1V1 = P2V2

V1 = 0.5 l, P1 =203 kPa, P2 = 25.4 kPa

V2 = (0.5 × 203 )/25.4 

 = 3.996 

 ≈ 4.0 L

6 0

3 years ago

Choose the correct coefficients to balance the following equation:

umka21 [38]

B) 1 Zn+ 2HCl--> 1 ZnCl2+1H2 This is balanced

5 0

3 years ago

Read 2 more answers

2.40 kJ of heat is added to a slug of platinum and a separate 2.40 kJ of heat is added to a slug of zinc. The heat capacity of t

leonid [27]

Final temperature of platinum :- 36.5°C

Final temperature of Zinc :- 29.17°C

3 0

3 years ago