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3 years ago

10

PLEASE HELP

2 answers:

Oksana_A [137]3 years ago

5 0

Answer:

The answer is A

Explanation: hope this helps :)

noname [10]3 years ago

3 0

Answer:

D

Explanation:

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Relation between Electric and magnetic fields. I've read that both the electric and magnetic field vectors are perpendicular to each other in an electromagnetic wave. Passing steady current through a straight conductor shows some magnetic flux (because most of the energy is wasted into outer space as magnetic lines).

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Why might the current in a short circuit be higher than the current in the original circuit?

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Answer:

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Explanation:

The current in a short circuit may be very high because the resistance in the short circuit is probably less than the resistance in the original circuit.

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Automobile air bags use the decomposition of sodium azide as their sources of gas for rapid inflation, represented in the reacti

NeTakaya

Answer : The mass of $NaN_3$ required is 71.175 grams.

Explanation :

To calculate the moles of nitrogen gas, we use the equation given by ideal gas :

PV = nRT

where,

P = Pressure of nitrogen gas = 763 torr

V = Volume of the nitrogen gas = 40.0 L

n = number of moles of gas = ?

R = Gas constant = $62.364\text{ L.torr }mol^{-1}K^{-1}$

T = Temperature of helium gas = $25^oC=273+25=298K$

Putting values in above equation, we get:

$763torr\times 40.0L=n\times 62.364\text{ L.torr }mol^{-1}K^{-1}\times 298K\\\\n=1.642mol$

Now we have to calculate the moles of $NaN_3$ .

The balanced chemical reaction is:

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced chemical reaction, we conclude that

As, 3 moles of $N_2$ produced from 2 moles $NaN_3$

So, 1.642 moles of $N_2$ produced from $\frac{2}{3}\times 1.642=1.095$ moles $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

Molar mass of $NaN_3$ = 65 g/mol

$\text{Mass of }NaN_3=\text{Moles of }NaN_3\times \text{Molar mass of }NaN_3$

$\text{Mass of }NaN_3=1.095mole\times 65g/mole=71.175g$

Therefore, the mass of $NaN_3$ required is 71.175 grams.

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4 years ago

Three liquids are at temperatures of 6 ◦C, 23◦C, and 38◦C, respectively. Equal masses of the first two liquids are mixed, and th

kupik [55]

The equilibrium temperature is T13=3.12 ◦C

<u>Explanation:</u>

<u>Given </u>

The temperature of liquids: T1=6◦C, T2=23◦C, T3=38◦C

The temperature of 1+2 liquids mix: T12= 13◦C.

The temperature of 2+3 liquids mix: T23=26.8 ◦C.

The temperature of 1+3 liquids mix: T13= ??

<u>1.When the first two liquids are mixed:</u>

mC1(T1-T12)+mC2(T2-T12)=0
C1(6-13)=C2(23-13)=0
7C1=10C2
C1=1.42C2

<u>2.When the second and third liquids are mixed</u><u>:</u>

mC2(T2-T23)+mC3(T3-T23)=0
C2(23-26.8)=C3(38-26.8)=0
3.8C2=12.8C3
C2=3.36C3

<u>3.When the first and third liquids are mixed:</u>

mC1(T1-T13)+mC3(T3-T13)=0
C1(6-T13)+C3(38-T13)=0
C1=1.42C2 C2=3.36C3
C1=1.42C2(3.36C3)
C1=4.77C3
C1(6-T13)+C3(38-T13)=0
4.77C3(6-T13)+C3(38-T13)=0
By solving the equation we get,
T13=3.12 ◦C
The equilibrium temperature is T13=3.12 ◦C

<u></u>

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