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Effectus [21]
3 years ago
8

Which compound has the highest melting point?

Chemistry
1 answer:
motikmotik3 years ago
3 0
I think b carbon dioxide
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what are we/what has someone did to prevent/lessen the negative side effects of aluminium extraction?
kogti [31]

Aluminum is one of the main factors that reduce plant growth in acid soils. Although it is generally harmful to plants in soils with a neutral medium, the concentration of positive aluminum ions in acid soils increases and malfunctions in root and function growth.

Most acid soils are saturated with aluminum rather than hydrogen ions. Soil acidity is the result of hydrolysis of aluminum compounds. This principle (lime correction) to determine the degree of base saturation in the soil has become the basis of the methods used in soil testing laboratories to determine the lime requirements for soil. Application of lime to soil reduces the toxicity of aluminum to plants. Note This connector loads slowly.

Adaptation of wheat to allow aluminum to be carried out is due to the fact that aluminum releases organic compounds that in turn combine with harmful aluminum cations. It is believed that sorghum has the same endurance. The first genes found to withstand aluminum were found in wheat. Aluminum sulphide bearing has been found to be governed by an individual gene, such as in wheat. This is not the case in all plants.

7 0
3 years ago
WILL OFFER BRAINLIEST IF YOU CAN ANSWER THIS
AleksAgata [21]

Answer:

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Explanation:

Thats the correct anwers

5 0
3 years ago
Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t
Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

A=\epsilon \times C\times l

A=\log \frac{I_o}{I}

\log \frac{I_o}{I}=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution = 3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}

l = path length = 2.5 mm = 0.25 cm

I_o = incident light

I = transmitted light

\epsilon = molar absorptivity coefficient = 855dm^3mol^{-1}cm^{-1}

Now put all the given values in the above formula, we get:

\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)

\log \frac{I_o}{I}=0.6947

\frac{I_o}{I}=10^{0.6947}=4.951

If we consider I_o = 100

then, I=\frac{100}{4.951}=20.198

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light I_A = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100

\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%

Therefore, the percentage reduction in intensity is 79.80 %

3 0
3 years ago
What is the gcf of 16 and 72? <br> a. 2 <br> b. 4 <br> c. 16 <br> d. 8
nadya68 [22]
Your answer is D. 8

16 = 2^4
72 = 2^3*3^2

So you'll choose 2^3 = 8
7 0
3 years ago
Read 2 more answers
Which student is responses are correct?​
irga5000 [103]

Answer:

Neutrons: Electron cloud

Explanation:

7 0
3 years ago
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