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Romashka [77]
2 years ago
6

You stay out in the sun too long and get sunburned. What type of energy transfer causes your sunburn?

Chemistry
1 answer:
Allushta [10]2 years ago
3 0

Answer:

Radiation

Explanation:

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What is one way to test whether an unknown solution is acidic or basic?
sertanlavr [38]

Answer: There are several ways. The first that comes to mind is a pH meter. A pH electrode Is lowered into the solution, and (Assuming) the pH Meter has been properly calibrated, and the temperature of the solution is set to the calibration of the Meter, the pH can be read directly from an analogue scale or digital readout. Below 7 is acidic, 7 is Neutral, (like Pure Water), and over 7 is Alkaline, or Basic.

A useful, but less accurate method is the use of any number of “pH Indicator Solutions”, which are essentially a type of various colored dyes that change color within differing pH ranges. Usually, if the pH is unknown, a small amount of solution is removed from the container and tested separately - in a “well plate”, or similar method.

These types of dyes, or Indicator Solutions, can be dried upon strips of “pH indicator Paper”, which, depending upon the type can be very useful when carrying out more precisely arrived at pH tests like Titration.

Just to see if a solution is “Acid” or “Base”, Litmus paper is used; “a Red color shows Acidity, and a Blue color, a Base”; ergo, “An Acid Solution will turn Litmus Paper, Red”.

4 0
3 years ago
Bromophenol blue is the indicator used in detecting the endpoint for the antacid analysis in this experiment. what is the expect
Paul [167]
Hello!

For the antacid analysis, the chemical reactions that occur in the titration are the following ones:

First, the antacid (composed of weak bases and carbonates) is completely neutralized by the H⁺ ions in the HCl

2HCl + CaCO₃ → CO₂ + H₂O + 2CaCl₂

HCl + OH⁻ → H₂O + Cl⁻

The titration reaction consists in titrating the excess H⁺ ions that are left in the solution, by the following reaction:

H⁺ + NaOH → H₂O + Na⁺

So, when the equivalence point is reached, the solution will go from acid to basic. As bromophenol blue is yellow in acidic solution and blue in basic solution, you'll expect the indicator to change from yellow to blue.

Have a nice day!
7 0
3 years ago
Explain how impurities are selected for doping in group 14 semiconductors
aleksklad [387]

Impurities selection for doping in group 14 semiconductors is: based on their ability to add more holes and fewer electrons or to add more electrons and reduce the holes.

<h3>Meaning of Semiconductors</h3>

Semiconductors can be defined as any material that has the ability to exhibit some properties of a conductor and some properties of an insulator.

A semiconductor can be used as either a conductor or an insulator when worked upon.

In conclusion, Impurities selection for doping in group 14 semiconductors is: based on their ability to add more holes and fewer electrons or to add more electrons and reduce the holes.

Learn more about semiconductors: brainly.com/question/1918629

#SPJ1

6 0
2 years ago
Name the type of ion formed by the atom when it loses three electrons. Show by illustration
Nimfa-mama [501]
The atoms of elements can gain or lose electrons and become ions. Ions are charged particles that have gained or lost electrons. The atoms of elements can gain or lose electrons to form monatomic ions (made from a single atom of an element).
5 0
2 years ago
Read 2 more answers
Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne
777dan777 [17]

Answer:

Explanation:

From the information given:

CaF_2 \to Ca^{2+} + 2F^-

Ksp = 3.2 \times 10^{-11}

no of moles of Ca^{2+} = 0.01 L × 0.0010 mol/L

no of moles of Ca^{2+} = 1 \times 10^{-5} \ mol

no of moles of F^- = 0.01 L × 0.00010 mol/L

no of moles of F^- = 1 \times 10^{-6}\ mol

Total volume = 0.02 L

[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\  \\  \[[Ca^{2+}}] = 0.0005 \ mol/L

[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}

[F^{-}] = 5 \times 10^{-5}  \ mol/L

Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}

Since Q<ksp, then there will no be any precipitation of CaF2

3 0
3 years ago
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