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zubka84 [21]
3 years ago
8

I NEED HELP PLEASE !!!!

Chemistry
2 answers:
iogann1982 [59]3 years ago
6 0

Answer:

the mean i believe sorry if im wrong

Explanation:

hope this helps ;)

inysia [295]3 years ago
3 0

Answer:

over 10 weeks the plant will grow 24 time longer than the the same hieght of the plant and the plant age will become 45 and so the plant is very much give hey man is my answer corecct

Explanation:

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What are air pollutants
KIM [24]

Answer:

things that pollute the air

Explanation:

nuclear energy, fossil fuels, gas from cars and other transportation, and these all contribute to greenhouse gases that are harmful to the environment

hope this helped!

5 0
2 years ago
N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?
just olya [345]

Answer:

\large \boxed{\text{197.4 g}}

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:     28.01               17.03

            N₂ + 3H₂ ⟶ 2NH₃

m/g:                          240.0

(a) Moles of NH₃

\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}

(b) Moles of N₂

\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}

(c) Mass of N₂

\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}

7 0
3 years ago
Read 2 more answers
Gases and particles which are put into the air or emitted by various sources are called __________. photochemical smog emissions
Leto [7]
I would say the answer is emissions. These are the particles that are not supposed to be present in air but due to the production of different substances from humans daily activities these substances go with the air we breath. Hope this helped.
7 0
3 years ago
How do cells reproduce? ∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵
lora16 [44]

Answer:

Cells divide to reproduce. There are two primary methods used, one for somatic cells, which compose the organism’s body, and one for reproductive cells, or gametes.

Scientists call the process of somatic cell division mitosis. Mitosis has six distinct steps in which the cell organizes and copies the DNA in the nucleus. Once copied, each new cell has its own copy of the DNA. The six steps of mitosis are prophase, prometaphase, metaphase, anaphase, telophase and cytokinesis. Some authorities consider the non-dividing portion of the cell’s lifecycle, known as interphase, to be one of seven steps involved in mitosis; however, interphase cells are not actively dividing.

Sex cells, such as sperm or eggs, must divide differently. Sex cells only possess one-half of the DNA that makes up a new animal. That way, when they combine, the resulting organism gets half of its DNA from its mother and half from its father. Because sex cells only want half of the DNA in each cell, they go through a different division process called meiosis. In meiosis, the cells split a second time, which yields four daughter cells rather than two as with mitosis; however, this provides each daughter cell with 23 chromosomes in contrast to the 46 chromosomes in somatic cells.

8 0
4 years ago
From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp
Nina [5.8K]

<u>Answer:</u> The concentration of NH_3 required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of NiC_2O_4, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of NiC_2O_4 = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M

For the given chemical equations:

NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}

Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9

Net equation: NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?

To calculate the equilibrium constant, K for above equation, we get:

K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48

The expression for equilibrium constant of above equation is:

K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}

As, NiC_2O_4 is a solid, so its activity is taken as 1 and so for C_2O_4^{2-}

We are given:

[[Ni(NH_3)_6]^{2+}]=0.016M

Putting values in above equations, we get:

0.48=\frac{0.016}{[NH_3]^6}}

[NH_3]=0.285M

Hence, the concentration of NH_3 required will be 0.285 M.

7 0
3 years ago
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