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PIT_PIT [208]
3 years ago
15

What is the electric field (in N/C) at a point 5.0 cm from the negative charge and along the line between the two charges?

Chemistry
1 answer:
Natali [406]3 years ago
4 0

Answer: E = 2.455 x 10^5 N/C

Explanation:

q1 = 1.2x10^-7C

q2 = 6.2x10^-8C

Electric field, E = kQ/r²

where k = 9.0x10^9

since the location is (27 - 5)cm from q1

hence electric field, E1 = k*q1/r²

E1= (9x10^9 x 1.2x10^-7)/(0.22)² = 22314.05 N/C

for q2:

E1 = k*q2/r²

E2 at 5cm

E2 = (9x10^9 x 6.2x10^-8)/(0.05)² = 223200 N/C

Hence, the total electric field at 5cm position is

E = E1 + E2

E = 22314.05 + 223200 = 245514.05 N/C

E = 2.455 x 10^5 N/C

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The reaction below demonstrates which characteristic of a base?
svlad2 [7]

Answer: option D. the ability of a base to react with a soluble metal salt.


Justification:


NaOH is a strong base, which means that in water it will dissociate according to this reaction:


  • NaOH(aq) → Na⁺ (aq) + OH⁻ (aq)

On the other hand, CuSO₄ is a soluble ionic salt which in water will dissociate into its ions according to this other reaction:

  • CuSO₄(aq) → Cu²⁺ + SO₄²⁻

Hence, in solution, the sodium ion (Na⁺) will  react with the metal salt in a double replacement reaction, where the highly reactive sodium ion (Na⁺) will substitute the Cu²⁺ in the CuSO₄ to form the sodium sulfate salt, Na₂SO₄ (water soluble), and the copper(II) hydroxide, Cu(OH)₂ (insoluble).


That is what the given reaction represents:


  CuSO₄ (aq)     +     2NaOH(aq)    →    Cu(OH)₂(s)       +     Na₂SO₄(aq)

         ↑                                ↑                         ↑                            ↑

soluble metal salt       strong base       insoluble base       solube salt



5 0
3 years ago
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Sulfuryl chloride decomposes at high temperatures to produce sulfur dioxide and chlorine gases:
den301095 [7]

Answer:

[SO2Cl2] = = 0.015 M [SO2] = = 0.0027 M [Cl2] = = 0.0027 M Q = =   = 4.8 × 10−4

No. Q < Kc, so reaction will shift to the right.

Explanation:

7 0
3 years ago
Answer plzzzz im begging
lisabon 2012 [21]

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4 0
2 years ago
A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the
horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution :  Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 Cm^{3}

First, find the Density of given substance.

Formula used :    

Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}

Now,put all the values in this formula, we get

Density=\frac{46.9 g}{3.5 Cm^{3} } = 13.4 g/Cm^{3}

So, we conclude that the density of given substance (13.4 g/Cm^{3}) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/Cm^{3} respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

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3 years ago
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When lewis structures are drawn, do you only show valence electrons??
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