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2 years ago

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In this experiment, you will use a natural indicator to indirectly determine the H+ concentration of some common household solut

ions to determine a physical characteristic of the solution. In the space below, write a scientific question that you will answer by doing this experiment.

2 answers:

Vlad1618 [11]2 years ago

7 0

Some of the scientific questions that may be answered through the experiment are:

(1) What are the physical changes that may occur in the solution or the indicator when added with acidic/basic solution?

(2) How much of the indicator is needed in order to bring about a significant physical change in the solution to identify its H+ concentration?

Xelga [282]2 years ago

7 0

Answer:

“How is pH used to determine if a solution is acidic or basic?

Explanation:

Sample response i just did it

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A CCD has a greatest possible pixel value of 4095. what is the bit level of this CCD?

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<span>Bit level for a CCD (Charged coupled device) with a greatest possible pixel value of 4095:The relationship between the bit level and pixel value is given as:pixel value = 2^bit level.Most charged coupled devices (CCDs) have 8-bit, 16-bit, 32-bit levels.Using simple mathematics, we can see that 2^12 = 4096.Since the maximum number of pixels is 4095, the bit level is 12., i.e. the CCD has 12-bit level.</span>

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2 years ago

Match each label to the boundary it describes. convergent boundary new crust forms transform boundary crust submerges into the m

Paraphin [41]

The answers would be:

CONVERGENT boundary - Crust submerges into the mantle

TRANSFORM boundary - neither forms nor submerges

DIVERGENT boundary - new crust forms

If you'd like to know more about the different boundaries, read on:

Convergent boundaries occur when two plates move TOWARDS each other. The event where crust submerges into the mantle is called <em><u>subduction</u></em> and this occurs when an oceanic plate and a continental plate collide. The oceanic plate is more dense and thinner than the continental plate, so it slides under it.

Transform boundaries occur when two plates slide against each other. They move slide side by side, so nothing is formed nor do they go under each other. Although, this type of boundaries create strong earthquakes.

Lastly, divergent boundaries occur when two plates move apart. The separation creates a way for magma to come up. New crust is formed when the magma that seeps out is cooled by its cooler surroundings. This is observed in the mid oceanic ridge.

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2 years ago

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Given three capacitors, c1 = 2.0 μf, c2 = 1.5 μf, and c3 = 3.0 μf, what arrangement of parallel and series connections with a 12

Lesechka [4]

Answer:

Connect C₁ to C₃ in parallel; then connect C₂ to C₁ and C₂ in series. The voltage drop across C₁ the 2.0-μF capacitor will be approximately 2.76 volts.

$-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-$ .

Explanation:

Consider four possible cases.

<h3>Case A: 12.0 V.</h3>

$-\begin{array}{c}-{\bf 2.0\;\mu\text{F}-}\\-1.5\;\mu\text{F}- \\-3.0\;\mu\text{F}-\end{array}-$

In case all three capacitors are connected in parallel, the $2.0\;\mu\text{F}$ capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.

<h3>Case B: 5.54 V.</h3>

$-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-$

In case the $2.0\;\mu\text{F}$ capacitor is connected in parallel with the $1.5\;\mu\text{F}$ capacitor, and the two capacitors in parallel is connected to the $3.0\;\mu\text{F}$ capacitor in series.

The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.

The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,

$\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_3}+ \dfrac{1}{C_1+C_2}} = \frac{1}{\dfrac{1}{3.0}+\dfrac{1}{2.0+1.5}} = 1.62\;\mu\text{F}$ .

What will be the voltage across the 2.0 μF capacitor?

The charge stored in two capacitors in series is the same as the charge in each capacitor.

$Q = C(\text{Effective}) \cdot V = 1.62\;\mu\text{F}\times 12\;\text{V} = 19.4\;\mu\text{C}$ .

Voltage is the same across two capacitors in parallel.As a result,

$\displaystyle V_1 = V_2 = \frac{Q}{C_1+C_2} = \frac{19.4\;\mu\text{C}}{3.5\;\mu\text{F}} = 5.54\;\text{V}$ .

<h3>Case C: 2.76 V.</h3>

$-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-$ .

Similarly,

the effective capacitance of the two capacitors in parallel is 5.0 μF;
the effective capacitance of the three capacitors, combined: $\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_2}+ \dfrac{1}{C_1+C_3}} = \frac{1}{\dfrac{1}{1.5}+\dfrac{1}{2.0+3.0}} = 1.15\;\mu\text{F}$ .

Charge stored:

$Q = C(\text{Effective}) \cdot V = 1.15\;\mu\text{F}\times 12\;\text{V} = 13.8\;\mu\text{C}$ .

Voltage:

$\displaystyle V_1 = V_3 = \frac{Q}{C_1+C_3} = \frac{13.8\;\mu\text{C}}{5.0\;\mu\text{F}} = 2.76\;\text{V}$ .

<h3 /><h3>Case D: 4.00 V</h3>

$-2.0\;\mu\text{F}-1.5\;\mu\text{F}-3.0\;\mu\text{F}-$ .

Connect all three capacitors in series.

$\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_1} + \dfrac{1}{C_2}+\dfrac{1}{C_3}} =\frac{1}{\dfrac{1}{2.0} + \dfrac{1}{1.5}+\dfrac{1}{3.0}} =0.667\;\mu\text{F}$ .

For each of the three capacitors:

$Q = C(\text{Effective})\cdot V = 0.667\;\mu\text{F} \times 12\;\text{V} = 8.00\;\mu\text{C}$ .

For the $2.0\;\mu\text{F}$ capacitor:

$\displaystyle V_1=\frac{Q}{C_1} = \frac{8.00\;\mu\text{C}}{2.0\;\mu\text{F}} = 4.0\;\text{V}$ .

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1 year ago

Consider three drinking glasses. All three have the same area base, and all three are filled to the same depth with water. Glass

Kay [80]

The glass which has the greatest liquid pressure at the bottom is all 3 have equal non-zero pressure at the bottom. The correct answer between all the choices given is the first choice or letter A. I am hoping that this answer has satisfied your query about and it will be able to help you.

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1 year ago

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A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

Svet_ta [14]

Answer:

The speed is $\sqrt{2}v_{0}$ .

(a) is correct option.

Explanation:

Given that,

Potential difference $V= V_{0}$

Speed $v = v_{o}$

If it were accelerated instead

Potential difference $V'=2V_{0}$

We need to calculate the speed

Using formula of initial work done on proton

$W = q V$

We know that,

$\Delta W=\Delta K.E$

$q V=\dfrac{1}{2}mv^2$

Put the value into the formula

$q V_{0}=\dfrac{1}{2}mv_{0}^2$

$v_{0}^2=\dfrac{2qV_{0}}{m}$ ....(I)

If it were accelerated instead through a potential difference of $2 V_{0}$ , then it would gain a speed will be given as :

Using an above formula,

$v_{0}'^2=\dfrac{2qV_{0}}{m}$

Put the value of $V_{0}$

$v_{0}'^2=\dfrac{2q\times2V_{0}}{m}$

$v_{0}'=\sqrt{\dfrac{4qV_{0}}{m}}$

$v_{0}'=\sqrt{2}v_{0}$

Hence, The speed is $\sqrt{2}v_{0}$ .

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1 year ago