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monitta
3 years ago
5

When does phenolphthalein turn pink?

Chemistry
2 answers:
lesya692 [45]3 years ago
5 0
A.) <span>Phenolphthalein is colorless in an acid, but it turns bright pink in the presence of a base.

Hope I helped you!!</span>
Svetach [21]3 years ago
3 0

Answer is: a) in the presence of a base.

Phenolphthalein is colorless in acidic solutions and pink in basic solutions.

Acid-base indicators are usually weak acids or bases and they are chemical detectors for hydrogen or hydronium cations.

Example for acid-base indicator is phenolphthalein (molecular formula C₂₀H₁₄O₄).

When solution turns phenolphthalein pink, it means it is basic (pH>7).

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Explain the two parts of a scientific name?
enot [183]

Answer:The binomial nomenclature system combines two names into one to give all species unique scientific names. The first part of a scientific name is called the genus. The second part of a species name is the specific epithet. Species are also organized into higher levels of classification.

6 0
3 years ago
How did Henry Moseley revise Mendeleev's periodic table? (Please answer A.S.A.P.)
CaHeK987 [17]
<span>If you give it a good search, the most used answer would probably be as follows,

</span><span>In 1914 Henry Moseley found a relationship between an element's X-ray wavelength and its atomic number (Z), and therefore rearranged the table by nuclear charge / atomic number rather than atomic weight. Before this discovery, atomic numbers were just sequential numbers based on an element's atomic weight. Moseley's discovery showed that atomic numbers had an experimentally measurable basis.
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5 0
3 years ago
Read 2 more answers
Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C
umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}

Given:

\left[I_2_{Equilibrium} \right] = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}

So,

\left[I_{Equilibrium} \right]^2=0.011\times 0.10

\left[I_{Equilibrium} \right]^2=0.0011

\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M

<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>

3 0
3 years ago
WILL MARK BRAINLIEST
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The answer is C. Life
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7 0
3 years ago
Which atom would have the largest radius?
ryzh [129]

Answer:

D, Rubidium

Explanation

Rubidium- 290 pm

Cobalt- 200 pm

Beryllium- 153 pm

Helium- 140 pm

5 0
3 years ago
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