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Nitella [24]
3 years ago
10

Is it living or nonliving?

Chemistry
1 answer:
Fittoniya [83]3 years ago
8 0

Answer:

One kind is called living things. Living things eat, breathe, grow, move, reproduce and have senses. The other kind is called nonliving things. Nonliving things do not eat, breathe, grow, move and reproduce.

You might be interested in
5. Hydrogen & oxygen react chemically to form water. How much water
stepladder [879]

39.25 g of water (H₂O)

Explanation:

We have the following chemical reaction:

2 H₂ + O₂ → 2 H₂O

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of H₂ = 14.8 / 2 = 7.4 moles

number of moles of O₂ = 34.8 / 32 = 1.09 moles

We see from the chemical reaction that 2 moles of H₂ will react with 1 mole of O₂ so 7.4 moles of H₂ will react with 3.7 moles of O₂ but we only have 1.09 moles of O₂ available. The O₂ will be the limiting reactant. Knowing this we devise the following reasoning:

if        1 moles of O₂ produces 2 moles of H₂O

then  1.09 moles of O₂ produces X moles of H₂O

X = (1.09 × 2) / 1 = 2.18 moles of H₂O

mass = number of moles × molar weight

mass of H₂O = 2.18 × 18 = 39.25 g

Learn more about:

limiting reactant

brainly.com/question/7144022

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6 0
3 years ago
The pk1, pk2, and pkr for the amino acid glutamate are 2.1, 9.5, and 4.1, respectively. at ph 11.0, glutamate would be charged p
vfiekz [6]
There are no answer choices
8 0
3 years ago
What kind of bond is the result of the transfer of an electron?
MrRa [10]
An Ionic bond is the result of transfer of electrons between atoms
5 0
3 years ago
Read 2 more answers
Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks
charle [14.2K]

Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

  = 3*2*10^-15 * 10^15

  = 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

                Zn(OH)₂ + 2OH⁻(aq)  --> Zn(OH)₄²⁻(aq)

Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

                                    = 0.3 -(2*0.089)

                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

     = 13.09

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:

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#SPJ4

3 0
1 year ago
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