Most seeds germinate and start to grow in the dark. Why don’t they need light right away?

Please answer truthfully:))

makvit [3.9K]

Answer:

Fast, Direction, Time

Explanation:

4 0

2 years ago

a chemist wishes to mix some pure acid with some water to produce 16L of a solution that is 30% acid how much pure acid and how

Aloiza [94]

<u>Answer:</u> The volume of acid and water that must be mixed will be 4.8 L and 11.2 L

<u>Explanation:</u>

We are given:

Volume of mixture = 16 L

Percent of acid present = 30 %

Calculating the percentage of acid present in the mixture:

$\Rightarrow 16\times \frac{30}{100}=4.8L$

The mixture is made entirely of acid and water.

Volume of acid in the mixture = 4.8 L

Volume of water in the mixture = 16 - 4.8 = 11.2 L

Hence, the volume of acid and water that must be mixed will be 4.8 L and 11.2 L

4 0

3 years ago

A compound containing chromium, Cr; chlorine, Cl; and oxygen, O, is analyzed and found to be 33.6% chromium, 45.8% chlorine, and

Assoli18 [71]

Answer

The empirical formula is CrO₂Cl₂

Explanation:

Empirical formula is the simplest whole number ratio of an atom present in a compound.

The compound contain, Chromium=33.6%

Chlorine=45.8%

Oxygen=20.6%

And the molar mass of Chromium(Cr)=51.996 g mol.

Chlorine containing molar mass (Cl)= 35.45 g mol.

Oxygen containing molar mass (O)=15.999 g mol.

Step-1

Then,we will get,

Cr= $\frac{1}{51.996} \times33.6=0.64$ mol

Cl= $\frac{1}{35.45} \times45.8=1.29$ mol.

O= $\frac{1}{15.99} \times=1.28$ mol.

Step-2

Divide the mole value with the smallest number of mole, we will get,

Cr= $\frac{0.64}{0.64} =1$

Cl= $\frac{1.29}{0.64} =2$

O= $\frac{1.28}{0.64} =2$

Then, the empirical formula of the compound is CrO₂Cl₂ (Chromyl chloride)

4 0

3 years ago

Read 2 more answers

A student has an unknown sample. How can spectroscopy be used to identify the sample?

iogann1982 [59]

Answer: A. It can identify the elements in the sample.

Explanation: on edge

4 0

3 years ago

identify the reagents you would use to convert each of the following compounds into pentanoic acid: (a) 1-pentene (b) 1-bromobut

Morgarella [4.7K]

a)BH3.THF is used to convert 1-pentane to pentanoic acid and b)NaCN is used to convert Bromobutane to pentanoic acid.

a) The conversion of 1-pentane to pentanoic acid using BH3, also known as hydroboration-oxidation, is a two-step reaction involving the reaction of 1-pentane with borane (BH3), followed by oxidation of the resulting 1-pentylborane with hydrogen peroxide or other oxidizing agents.

In the first step, 1-pentane reacts with borane (BH3) to form 1-pentylborane, through a process known as hydroboration. This reaction is catalyzed by a Lewis acid, such as aluminum chloride, and proceeds via a hydride transfer from the borane to the 1-pentane.

In the second step, the 1-pentylborane is oxidized to pentanoic acid using hydrogen peroxide (H₂O₂) or other suitable oxidizing agents. The oxidation is catalyzed by an acid, such as hydrochloric acid (HCl), and proceeds via a proton transfer from the 1-pentylborane to the hydrogen peroxide. The end result is the conversion of 1-pentane to pentanoic acid.

The overall chemical reaction for the conversion of 1-pentane to pentanoic acid using borane (BH₃) and hydrogen peroxide (H₂O₂) is as follows:

1-pentane + BH₃ + H₂O₂ → pentanoic acid + H₂O + BH₂

b)The conversion of 1-Bromo butane to pentanoic acid using sodium cyanide (NaCN) proceeds via a nucleophilic substitution reaction. The reaction mechanism involves the following steps:

1. Attack of the nucleophile, NaCN, on the carbon atom of 1-Bromo butane to form a tetrahedral intermediate.

2. Loss of a proton from the tetrahedral intermediate to form a carbanion.

3. Protonation of the carbanion by water (or another proton source) to form pentanoic acid.

The overall reaction can be represented as follows:

1-Bromo butane + NaCN → Pentanoic Acid + NaBr

To know more about reagents, click below:

brainly.com/question/26283409

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3 0

1 year ago