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Neporo4naja [7]
4 years ago
5

How has the Internet changed communication in the environmental sciences?

Chemistry
1 answer:
Naily [24]4 years ago
3 0
A. It has made the spread of information faster and easier
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What does it mean that something is a conservative ion?
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The major ionic constituents whose concentrations can be determined from the salinity are known as conservative substances. Their constant relative concentrations are due to the large amounts of these species in the oceans in comparison t

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Which statement best explains why earth and the other planets of our solar system became layered as they were being formed
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The top of the plate where food is placed is called:
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Design Shape

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Give the balanced chemical reaction for the preparation of black oxide from blue vitriol​
sasho [114]

Answer:

Blue vitriol is crystal copper sulphate

Explanation:

The black oxide is the higher oxide of Copper i.e Cu. It is also called as tenorite. It is solid black in color. It is formed by heating copper with air.

$2Cu +O_2 \rightarrow 2CuO$

Blue vitriol is the crystalline form of copper sulphate. It is also known as copper (II) sulphate.

When heating, the blue vitriol undergoes various steps of changes to the black oxide of copper.

The chemical reaction is --

$CuSO_4 \cdot5H_2O \rightarrow CuSO_4\cdot 3H_2O {\rightarrow} CuSO_4\cdot H_2O \rightarrow CuSO_3 \rightarrow CuO +SO_2+O_2

3 0
3 years ago
Beryllium-11 is a radioactive isotope of the alkaline metal Beryllium. It decays at a rate of 4.9% every second. Assuming you st
rewona [7]

Answer:

11.02 % of an isotope will be left after 45 seconds.

Explanation:

N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}

where,

N_o = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

t_{\frac{1}{2}} = half life of the isotope

\lambda = rate constant

We have :

Mass of Beryllium-11 radioactive isotope= N_o=100

Mass of Beryllium-11 radioactive isotope after 45 seconds = N=?

t = 45 s

\lambda = rate constant  = 4.9 \% s^{-1}=0.049 s^{-1}

N=N_o\times e^{-(\lambda \times t}

Now put all the given values in this formula, we get

N=100\times e^{-0.049 s^{-1}\ties 45 s}

N=11.02 g

Percentage of isotope left :

\frac{N}{N_o}\times 100=\frac{11.02 g}{100 g}\times 100=11.02\%

11.02 % of an isotope will be left after 45 seconds.

7 0
4 years ago
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