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erastova [34]
3 years ago
9

2 FONS

Chemistry
1 answer:
stiv31 [10]3 years ago
7 0

Answer:

C. It decreases by a factor of 4

Explanation:

F1 = kq1*q2/r²

F2 = kq1*q2/(2r)² = kq1*q2/(4r²) = kq1*q2/(r²*4)  = F1/4

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Consider 5.00 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.50 L and the temperature is increased to 3
pickupchik [31]

First convert celcius to Kelvin.

20 + 273 = 293K

31 + 273 = 304K

Now we can set up an equation based on the information we have.

V1 = 5

P1 = 365

T1 = 293

V2 = 5

P1 = x

T2 = 304

The equation be:  \frac{(5)(365)}{293} = \frac{5x}{304}

Now just solve.

1825/293 = 5x/304

Cross multiply.

554800 = 1465x

Divide both sides by 1465

x = 378.7030717 which can then be rounded to 378.7 mmHg

4 0
3 years ago
Is iron oxide an insulator or conductor?
balu736 [363]

Answer:

It's a conductor

Explanation:

3 0
3 years ago
Find the equation of the line passing through the points (2, 1) and (5, 10).​
Vedmedyk [2.9K]

The equation : y=3x-5

<h3>Further explanation </h3>

Straight-line equations are mathematical equations that are described in the plane of cartesian coordinates

General formula

y-y1 = m(x-x1)

or

y = mx + c

Where

m = straight-line gradient which is the slope of the line

x1, y1 = the Cartesian coordinate that is crossed by the line

c = constant

The formula for a gradient (m) between 2 points in a line

m = Δy / Δx

  • Gradient

\tt \dfrac{10-1}{5-2}=3

  • Equation

\tt y-1=3(x-2)\\\\y-1=3x-6\\\\y=3x-5

3 0
3 years ago
Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,
7nadin3 [17]

Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 )   N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)  ; Kc = 1.54e - 31

2)   N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)  ; Kc = 2.16e - 24

   upon reversing  ( 2 )  equation

     N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)   Kc = 1/2.16e - 24  

    now adding 1 and reversed equation (2)

       N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)

      N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)

   we get ,

                  N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)  Kc = 1.54e-31 × 1/2.61e - 24

       equilibrium constant of equation (3) is -

            Kc = 1.54e - 31 / 2.61e - 24

3 0
3 years ago
A chemist lectures about the following definition for acids and bases. The acid-base behavior is analyzed in terms of how electr
Y_Kistochka [10]

Answer:

the Lewis concept of acids and bases

Explanation:

8 0
3 years ago
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