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kenny6666 [7]
3 years ago
12

Sodium oxide contains Na+ ions and O2- ions. Give the formula of sodium oxide

Chemistry
1 answer:
mixer [17]3 years ago
8 0
To be honest, I learned this in school so I'll tell you XD

The formula of sodium oxide is Na2O
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During what part of the experiment is the hypothesis "judged" as being rejected or supported?
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Answer:

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Explanation:

the hypothesis is a typical statistical theory which suggest that no statistical relationship and significance exists in a set of a given single observed variable between two sets of the observed data and measured phenomena

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The compound Fe2O3 contains how many atoms? <br> A) 2 <br> B) 3 <br> C) 5 <br> D) 7
marishachu [46]
The compound Fe2O3 contains C) 5 total atoms.

This is because there are 2 atoms of Fe (iron) and 3 atoms of O (oxygen) bonded together in this compound, as denoted by the subscripts, and 2+3 = 5.

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4 years ago
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1.A space shuttle's wings are useless as it travels around the exosphere. Explain, in your own words, why the wings are so usele
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Thrust acts on the spacecraft and propels in the space.

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6 0
3 years ago
Will give brainliest, like, 5 stars and 5 points to best answer
Oliga [24]

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6 0
3 years ago
Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °C and 760 T
shtirl [24]

Explanation:

Let us assume that the volume of given aqueous solution is 7.5 L.

Therefore, according to Henry's law, the relation between concentration and pressure is as follows.

                C = \frac{P}{K_{h}}

where,   pressure (P) = 760 torr = 1 atm

According to Henry's law, constants for gases in water at 25^{o}C are as follows.

  p(O_{2}) = 0.21 atm = 0.21 bar

   p(N_{2}) = 0.78 atm = 0.78 bar

   K_{h} for O_{2} = 7.9 \times 10^{2} bar/mol

    K_{h} for N_{2} = 1.6 \times 10^{3} bar /mol

Since, 21% oxygen is present in air so, its mass will be 0.21 g. Similarly, 78% nitrogen means the mass of nitrogen is 0.78 g.

Therefore, concebtrations will be calculated as follows.

      C(O_{2}) = \frac{0.21}{7.9 \times 10^{2}} = 2.66 \times 10^-4 mol/L  

      C(N_2) = \frac{0.78}{1.6 \times 10^3} = 4.875 \times 10^-4 mol/L

Now, we will calculate the number of moles as follows.

         n(O_{2}) = 7.5 \times 2.66 \times 10^-4 = 1.995 \times 10^-3 mol

         n(N_2) = 7.5 \times 4.875 \times 10^-4 = 3.66 \times 10^-3 mol

As the molar mass of O_2 = 32 g/mol

Hence, mass of oxygen will be as follows.

         Mass of O_2 = 32 \times 1.995 \times 10^-3 \times 1000

                           = 63.84 mg

As the molar mass of N_{2} = 28

       Mass of N_{2} = 28 \times 3.66 \times 10^-3 = 102.5 mg

Thus, we can conclude that mass of oxygen is 63.84 mg  and nitrogen is 102.5 mg.

5 0
4 years ago
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