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trapecia [35]
3 years ago
12

A student has four resistors. Each resistor has a resistance of 100 ohms.

Chemistry
1 answer:
ValentinkaMS [17]3 years ago
7 0

Answer:

The minimum resistance is 25 ohms.

Explanation:

Resistance of each resistor is 100 ohms. When resistors are connected in parallel, the equivalent resistance is lowest. For parallel combination, the equivalent resistance is given by :

\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+....

Here, all resistors are 100 ohms. So,

\dfrac{1}{R}=\dfrac{1}{100}+\dfrac{1}{100}+\dfrac{1}{100}+\dfrac{1}{100}\\\\R=25\ \Omega

So, the minimum resistance is 25 ohms.

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Answer:

B. 1-Butene rightarrow (1) BH3: THF (2)H202, OH-

Explanation:

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the reagent BH₃:THF is the way that borane is used in organic reactions. The BH₃ adds to the double bond of an alkene to form an alkyl borane. Peroxide hydrogen in basic medium oxidizes the alkyl borane to form an alcohol. Indeed, hydroboration-oxidation converts alkenes to alcohols by adding water through the double bond, with anti-Markovnikov orientation.

8 0
3 years ago
What are metals that have been mixed together called
d1i1m1o1n [39]

Answer:

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3 years ago
What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4
dsp73

<u>Given information:</u>

Concentration of NaF = 0.10 M

Ka of HF = 6.8*10⁻⁴

<u>To determine:</u>

pH of 0.1 M NaF

<u>Explanation:</u>

NaF (aq) ↔ Na+ (aq) + F-(aq)

[Na+] = [F-] = 0.10 M

F- will then react with water in the solution as follows:

F- + H2O ↔ HF + OH-

Kb = [OH-][HF]/[F-]

Kw/Ka = [OH-][HF]/[F-]

At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x

10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x

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pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92

pH = 14 - pOH = 14-5.92 = 8.08

Ans: (b)

pH of 0.10 M NaF is 8.08

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