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lions [1.4K]
3 years ago
12

Please help me! U will get a 30 points!

Chemistry
1 answer:
Rudiy273 years ago
4 0

Answer:

C

Explanation:

The stem and leaves need sunlight to undergo photosynthesis, they will grow in such a way until they receive some sunlight.

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What is the concentration (in M) of a sample of the unknown dye with an absorbance of 0.25 at 542 nm?
Sladkaya [172]

Answer:

see explanation below

Explanation:

Question is incomplete, so in picture 1, you have a sample of this question with the missing data.

Now, in general terms, the absorbance of a substance can be calculated using the beer's law which is the following:

A = εlc

Where:

ε: molar absortivity

l: distance of the light in solution

c: concentration of solution

However, in this case, we have a plot line and a equation for this plot, so all we have to do is replace the given data into the equation and solve for x, which is the concentration.

the equation according to the plot is:

A = 15200c - 0.018

So solving for C for an absorbance of 0.25 is:

0.25 = 15200c - 0.018

0.25 + 0.018 = 15200c

0.268 = 15200c

c = 0.268/15200

c = 1.76x10⁻⁵ M

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3 years ago
Assuming equal concentrations, rank these solutions by ph.? assuming equal concentrations, rank these solutions by ph. koh (aq h
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I remember coming across this question and the options were:
KOH, HCN, NH₃, HI, Sr(OH)₂

Now, a substance with a low pH is one that dissociates completely in water to release hydrogen ions, while basic substances dissociate completely to release hydroxide ions. Therefore, in the order of increasing pH:
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Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y
olya-2409 [2.1K]

Answer:

0.9715 Fraction of Pu-239 will be remain after 1000 years.

Explanation:

\lambda =\frac{0.693}{t_{\frac{1}{2}}}

A=A_o\times e^{-\lambda t}

Where:

\lambda= decay constant

A_o =concentration left after time t

t_{\frac{1}{2}} = Half life of the sample

Half life of Pu-239 = t_{\frac{1}{2}}=24,000 y[

\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]

Let us say amount present of  Pu-239 today = A_o=x

A = ?

A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}

A=0.9715\times x

\frac{A}{A_0}=\frac{A}{x}=0.9715

0.9715 Fraction of Pu-239 will be remain after 1000 years.

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