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julia-pushkina [17]
3 years ago
8

What is the most abundant gas in our atmosphere? oxygen hydrogen nitrogen water

Chemistry
2 answers:
Artyom0805 [142]3 years ago
8 0

The answer is NItrogen

Inessa [10]3 years ago
7 0
It seems that nitrogen is, but its odd tbh.
You might be interested in
Complete and balance the following:(c) FeS(s) + HCl (aq) →
Scrat [10]

balance the following:(c) FeS(s) + HCl (aq) = FeS + 2 HCl → FeCl2 + H2S.

FeS Names: Iron(II) sulfide source: wikipedia, accessed: 2019-09-27, Iron sulfide source: wikipedia, accessed: 2019-09-27, Ferrous sulfide source: wikipedia, accessed: 2019-09-27source: wikidata, accessed: 2019-09-02

Appearance: Gray, sometimes in lumps or powder source: wikipedia, accessed: 2019-09-27

HCl – Chlorane source: wikipedia, accessed: 2019-09-27, Hydrogen chloride source: wikipedia, accessed: 2019-09-27source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02

Other names: Hydrochloric acid source: balance, accessed: 2019-09-27source: NIOSH NPG, accessed: 2019-09-02, M source: wikipedia, accessed: 2019-09-27, Uriatic acid source: wikipedia, accessed: 2019-09-27

Appearance: Colorless, transparent liquid, fumes in air if concentrated source: wikipedia, accessed: 2019-09-27; Colorless gas source: wikipedia, accessed: 2019-09-27; Colourless compressed liquefied gas with pungent odour source: ICSC, accessed: 2019-09-04; Colorless to slightly yellow gas with a pungent, irritating odor. [Note: Shipped as a liquefied compressed gas.] source: NIOSH NPG, accessed: 2019-09-02.

Learn more about FeS on:
brainly.com/question/1119048

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5 0
2 years ago
How many miles are there if you have 70.5g of Nacl
GenaCL600 [577]
70.5g x 1 mol / 58.5g = 1.2mol of nacl
4 0
3 years ago
A chemistry student needs of acetic acid for an experiment. He has available of a w/w solution of acetic acid in acetone. Calcul
Volgvan

Answer:

49.4 g Solution

Explanation:

There is some info missing. I think this is the original question.

<em>A chemistry student needs 20.0g of acetic acid for an experiment. He has 400.g  available of a 40.5 %  w/w solution of acetic acid in acetone.  </em>

<em> Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to 3 significant digits.</em>

<em />

We have 400 g of solution and there are 40.5 g of solute (acetic acid) per 100 grams of solution. We can use this info to find the mass of acetic acid in the solution.

400gSolution \times \frac{40.5gSolute}{100gSolution} = 162 g Solute

Since we only need 20.0 g of acetic acid, there is enough of it in the solution. The mass of solution that contains 20.0 g of solute is:

20.0gSolute \times \frac{100gSolution}{40.5gSolute} = 49.4g Solution

4 0
3 years ago
Please help <br><br> 19.0 kg/0.021 m^3
irina1246 [14]

Answer:

904.8 kg / m3

Explanation:

19.0 kg/0.021 m^3 = 904.761905 kg / m3

6 0
2 years ago
Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
Gnom [1K]

Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10^{-2.62}

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{2.81}

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

5 0
3 years ago
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