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krek1111 [17]
2 years ago
9

certain conductor with a resistance of 10 Ohms is crossed by an electric current of intensity 100 mA. What is the potential diff

erence at the terminals of that conductor?
Chemistry
1 answer:
Morgarella [4.7K]2 years ago
8 0

Answer:

1.0 V

Explanation:

Step 1: Given data

  • Resistance of certain conductor (R): 10 Ω
  • Current's intensity (I): 100 mA
  • Potential difference at the terminals of that conductor (V): ?

Step 2: Convert "I" to Ampere

We will use the conversion factor 1 A = 1000 mA.

100 mA × 1 A/1000 mA = 0.100 A

Step 3: Calculate the potential difference

We will use Ohm's law.

V = I × R

V = 0.100 A × 10 Ω = 1.0 V

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During lab, a student used a Mohr pipet to add the following solutions into a 25 mL volumetric flask. They calculated the final
kompoz [17]

Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) =  0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

                       = 10.63 + 1.42

                       = 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

                     = 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ =  2.126 mmol -  0.00284 mmol

                  =2.123 mmol

For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

                    =0.00109

V(standard 2) = 4.63 + 5.17

                       = 9.8 mL

M (FeSCN⁺²)  = 0.00109/9.8

                      = 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

7 0
2 years ago
How to calculate delta S surroundings? calculate Delta S(surr) at the indicated temperature for a reaction having each of the fo
astraxan [27]
You can use the equation ΔS(surr)=q(surr)/T or ΔS(surr)=-q(rxn)/T.
the two equations are equal since we know that the energy the system (reactoin) puts out just goes into the surroundings.  
(In other words q(surr)=-q(rxn))

Using the equation, <span>ΔS(surr)=-(-283kJ/298K)=0.9497kJ/K or 949.7J/K

This answer makes sense since the reaction is exothermic which means it released energy into the system which usually causes the entropy to increase.

I hope that helps.</span>
8 0
3 years ago
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