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3 years ago

What principle do Tesla coils use to generate electricity?

Chemistry

2 answers:

Zepler [3.9K]3 years ago

8 0

Electromagnetic
Hope this helps!

zaharov [31]3 years ago

7 0

When u give supply to a coil then a current will be passed through it .
naturally a magnetic field is produced around the c.c.c(current carrying conductor). when u make more turns in the same direction the whole magnetic field will be stronger than that of a wire has. and 
by lenz's law there will be an opposition to the changing current in a coil due to the change in magnetic field produced

You might be interested in

1. Except for special cases such as peroxides, oxygen has an oxidation number of _______ in compounds. A. +2 B. –1 C. +1 D. –2

egoroff_w [7]

Answer:

The Typical Oxidation number of Oxygen is –2.

And as the question said... its Oxidation changes to –1 when it is in Peroxides and –½ when its in SuperOxides.

Correct Answer : Option D.

7 0

2 years ago

Given the initial rate data for the reaction A + B –––> C, determine the rate expression for the reaction.

umka21 [38]

Answer:

B) Δ[C]/Δt = 3,60x10⁻² M⁻¹s⁻¹ [A] [B]

Explanation:

For the reaction A + B → C

The formula for rate of reaction is:

Δ[C]/Δt = k [A] [B]

As you have [A], [B] and Δ[C]/Δt information you can multiply [A] times [B] and take this value as X and Δ[C]/Δt as Y. The slope of this lineal regression will be k.

Thus, you must obtain:

y = 3,60x10⁻² X

Thus, rate of reaction is:

B) Δ[C]/Δt = 3,60x10⁻² M⁻¹s⁻¹ [A] [B]

I hope it helps!

3 0

3 years ago

Read 2 more answers

What is the rate constant of a first-order reaction that takes 5.50 minutes for the reactant concentration to drop to half of it

bulgar [2K]

For a first order reaction, the half life is inversely proportional to the rate constant.
The formula is
half life = ln(2)/k = 0.693/k
where k is the rate constant

t = 5.50 minutes

k = ln(2)/5.50 = 0.126 min^-1

Your rate constant is 0.126 min^-1.

8 0

3 years ago

Perform the following conversions:

ololo11 [35]

Answer:

a) $20^0C$ and $293K$

b) $437K$ and $327.2^0F$

c) $-273^0C$ and $459.44^0F$

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like $^0C$ , $^0F$ and $K$

These units of temperature are inter convertible.

$t^0C=(t+273)K$

$t^oC=\frac{5}{9}\times (t^oF-32)$

$K-273=\frac{5}{9}\times (^oF-32)$

a) 68°F (a pleasant spring day) to °C and K.

Converting this unit of temperature into $^0C$ and $K$ by using conversion factor:

$t^oC=\frac{5}{9}\times (68^oF-32)$

$t=20^0C$

$20^0C=(20+273)K=293K$

b) 164°C (the boiling point of methane, the main component of natural gas) to K and °F

Conversion from degree Celsius to Kelvins and Fahrenheit

$164C=\frac{5}{9}\times (t^oF-32)$

$t^0F=327.2$

$164°C=(164+273)K=437 K$

c) 0K (absolute zero, theoretically the coldest possible temperature) to °C and °F.

$K-273=\frac{5}{9}\times (t^oF-32)$

$0-273=\frac{5}{9}\times (t^oF-32)$

$t=-459.4^0F$

$t^K=(0-273)^0C=-273^0C$

8 0

3 years ago

An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle a

Lisa [10]

Answer:

1. 0.0637 moles of nitrogen.

2. The partial pressure of oxygen is 0.21 atm.

Explanation:

1. If we assume ideal behaviour, we can use the Law of ideal gases to find the moles of nitrogen, considering that air composition is mainly nitrogen (78%), oxygen (21%) and argon (1%):

$V_{N_2}=V_{T}\times 0.78=2L \times 0.78 =1.56 L\\PV=nRT\\n_{N_2}=\frac{PV}{RT}=\frac{1 atm\times 1.56 L}{0.0821\frac{atmL}{molK}\times 298 K}\\n_{N_2}= 0.0637 mol$

2. Now, in order to find he partial pressure of oxygen we need to find the total moles of air, and then the moles of oxygen. Then, we use these results to determine the molar fraction of oxygen, to multiply it with total pressure and get the partial pressure of oxygen as follows:

$n_{total}=\frac{1 atm \times 2L}{0.0821 \frac{atmL}{molK}298K}=0.0817 mol$

$V_{O_2}=2L \times 0.21 = 0.42 L\\n_{O_2}=\frac {1atm \times 0.42 L}{0.0821 \frac{atm L}{mol K}298 K}=0.0172 mol\\X_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.0172 mol}{0.0817 mol}= 0.21$

$P_{O_2}=X_{O_2} \times P = 1 atm \times 0.21 = 0.21 atm$

As you see, the molar fraction and volume fraction are the same because of the assumption of ideal behaviour.

3 0

3 years ago