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Gennadij [26K]
2 years ago
15

In a laboratory, Nakisha mixes a sodium hydroxide solution with an indicator called phenolphthalein. When combined, they create

a pink solution. Nakisha wonders if mixing other solutions with phenolphthalein will also create this pink color. How could Nakisha use the scientific inquiry process to determine whether mixing other solutions with phenolphthalein will also create a pink color? Check all that apply. by making qualitative observations about the sodium hydroxide and phenolphthalein in solution by asking a classmate if sodium hydroxide turns pink when mixed with other solutions by comparing the given solution to other familiar solutions containing phenolphthalein by designing an experiment to test phenolphthalein with other solutions by repeating the experiment at different temperatures
Physics
2 answers:
Gnesinka [82]2 years ago
8 0

by making qualitative observations about the sodium hydroxide and phenolphthalein in solution

by comparing the given solution to other familiar solutions containing phenolphthalein

by designing an experiment to test phenolphthalein with other solutions

Lapatulllka [165]2 years ago
8 0

Answer:

1. by making qualitative observations about the sodium hydroxide and phenolphthalein in solution

3. by comparing the given solution to other familiar solutions containing phenolphthalein

4. by designing an experiment to test phenolphthalein with other solutions

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fomenos

Answer:

Its traveling in the +x direction

Explanation:

The E-field is in the +y-direction, and the B-field is in the +z-direction, so it must be moving along the +x-direction, since the E-field, B-field and the direction of moving are all at right angles to each other.

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2 years ago
What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?
kkurt [141]

Answer:

U = 12,205.5 J

Explanation:

In order to calculate the internal energy of an ideal gas, you take into account the following formula:

U=\frac{3}{2}nRT        (1)

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n: number of moles = 10 mol

T: temperature of the gas = 100K

You replace the values of the parameters in the equation (1):

U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J

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6 0
2 years ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
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Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 So

       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

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Answer:

Explanation:

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c )

When resistance is added in series , they are added up to obtain equivalent resistance . So equivalent resistance R₁₂ will be more than R₁ OR R₂.

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Explanation:

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