The Symbol below is used to represent what component in a circuit?

Chemistry

2 answers:

FrozenT [24]3 years ago

8 0

Actually, I strongly believe it is a switch.

sineoko [7]3 years ago

8 0

Answer: Option (C) is the correct answer.

Explanation:

A device which is used to operate an electrical appliance is known as switch.

When a switch is closed then circuit gets complete and current completely flows into the circuit. Whereas when a switch is open then circuit is incomplete and current does not flow into the circuit.

Therefore, we can conclude that the symbol below is used to represent switch component in a circuit.

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PLEASE HELP ME!!!

kondor19780726 [428]

<h3>Answer:</h3>

0.424 J/g °C

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Thermochemistry</u>

Specific Heat Formula: q = mcΔT

q is heat (in Joules)
m is mass (in grams)
c is specific heat (in J/g °C)
ΔT is change in temperature

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] m = 38.8 g

[Given] q = 181 J

[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C

[Solve] c

<u>Step 2: Solve for Specific Heat</u>

Substitute in variables [Specific Heat Formula]: 181 J = (38.8 g)c(11.0 °C)
Multiply: 181 J = (426.8 g °C)c
[Division Property of Equality] Isolate <em>c</em>: 0.424086 J/g °C = c
Rewrite: c = 0.424086 J/g °C

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.424086 J/g °C ≈ 0.424 J/g °C

6 0

2 years ago

Predict the initial and isolated products for the reaction. The starting material is a 6 carbon chain where there is a triple bo

satela [25.4K]

Answer:

See explanation and image attached

Explanation:

This reaction is known as mercuric ion catalyzed hydration of alkynes.

The first step in the reaction is attack of the mercuric ion on the carbon-carbon triple bond, a bridged intermediate is formed. This bridged intermediate is attacked by water molecule to give an organomercury enol. This undergoes keto-enol tautomerism, proton transfer to the keto group yields an oxonium ion, loss of the mercuric ion now gives equilibrium keto and enol forms of the compound. The keto form is favoured over the enol form.