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BabaBlast [244]
4 years ago
12

Classify N2 as a substance

Chemistry
1 answer:
ludmilkaskok [199]4 years ago
6 0
N2O is compound becoz it has more than one type of atoms. 

<span>Air is homogeneous mixture becoz it has different gases which are does not bonded to each other</span>
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The maximum allowable concentration of pb2+ ions in drinking water is 0.05 ppm (i.e., 0.05 g of pb2+ in 1 million grams of water
trasher [3.6K]
PbSO₄ partially dissociates in water. the balanced equation is;
                    
                       PbSO₄(s) ⇄  Pb²⁺(aq) + SO₄²⁻(aq)
Initial                                     -                -
Change             -X               +X           +X
Equilibrium                           X              X

Ksp           =    [Pb²⁺(aq)] [SO₄²⁻(aq)]
1.6 x 10⁻⁸  =    X * X
1.6 x 10⁻⁸  =    X²
          X    =   1.3 x 10⁻⁴ M
      
Hence the Pb²⁺ concentration in underground water is 1.3 x 10⁻⁴ M. 
[Pb²⁺]  = 1.3 x 10⁻⁴ M.
           = 1.3 x 10⁻⁴ mol / L x 207 g / mol 
           = 26.91 ppm

8 0
3 years ago
What phase of matter is oxygen at room temperature
Vedmedyk [2.9K]
Gas

good luck mate :)
8 0
3 years ago
In general, ionization energy will?
bija089 [108]
<span>increase while moving left to right within a period and increase while moving upward within a group. </span>
3 0
3 years ago
Read 2 more answers
Is MG an ion, isotope, or neutral atom?
Kitty [74]

Answer:

isotope

isotope

isotope

isotope

isotope

isotope

isotope

isotope

isotope

5 0
3 years ago
El agua del mar contiene aproximadamente un 3,0 % m/v de sal (NaCl, 58,44 g/mol), (asuma que es la única fuente de cloruros) si
Alchen [17]

Answer:

s = 4.41 g/L.

Explanation:

¡Hola!

En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

PbCl_2(s)\rightleftharpoons 2Cl^-(aq)+Pb^{2+}(aq)

Lo cual hace que la expresión de equilibrio se calcule como:

Ksp=[Pb^{2+}][Cl^-]^2

Y que en términos de la solubilidad molar, s, se resuelve como:

1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L

Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):

s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L

¡Saludos!

7 0
3 years ago
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