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Alexus [3.1K]
2 years ago
12

Which is a pure substance (A) Mixture (B) Solution (C) Molecule

Chemistry
2 answers:
d1i1m1o1n [39]2 years ago
8 0
C. Molecule is a pure substance. It may be 2 or more atoms binding together via covalent bonding (I.e. the sharing of electrons). The bonding of two or more atoms from the same element forms stable molecules.
The bonding of two different atoms or more will result in a completely new substance as compared to the original substances of atoms.

Mixtures, from the name itself, are being mixed with several solutions, where one part of it is soluble, and the other part is insoluble.
●For e.g., sand and seawater being put together forms a mixture, where sea water is water and salt (soluble together) and sand (insoluble with water and salt).

Solutions may be 2 or more substances being mixed together which are soluble.
● For the sea water example above, sea water is the solution, since salt can be desolved in water (i.e. the solute).

Hope this helps! :)
raketka [301]2 years ago
4 0
It's C, I believe


good luck
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5 0
2 years ago
Kc = 3.07 x 10-4 at 24°C for 2NOBr(g) ↔ 2NO(g) + Br2(g). If the initial concentration of NOBr = 0.878 M, what is the equilibrium
pav-90 [236]

Answer:

The equilibrium concentration of NO is 0.02124 M.

Explanation:

Given that,

Initial concentration of NOBr = 0.878 M

k_{c}=3.07\times10^{-4}

Temperature = 24°C

We know that,

The balance equation is

2NOBr\Rightarrow 2NO+Br_{2}

Initial concentration is,

0.878\Rightarrow 0+0

Concentration is,

-2x\Rightarrow 2x+x

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Using formula of concentration

k_{c}=\dfrac{[NO][Br_{2}]}{[NOBr]^2}

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3.07\times10^{-4}=\dfrac{[2x][x]}{[0.878-2x]^2}

2x^2=3.07\times10^{-4}\times(0.878)^2+3.07\times10^{-4}\times4x^2-2\times2x\times0.878\times3\times10^{-4}

2x^2=0.0002367+0.001228x^2-0.0010536x

2x^2-0.001228x^2+0.0010536x-0.0002367=0

1.998772x^2+0.0010536x-0.0002367=0

x=0, 0.01062

We need to calculate the equilibrium concentration of NO

Using formula of concentration of NO

concentration\ of\ NO=2x

Put the value of x

concentration\ of\ NO=2\times0.01062

concentration\ of\ NO=0.02124

Hence, The equilibrium concentration of NO is 0.02124 M.

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