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Komok [63]
3 years ago
15

Measuring the level of lead in drinking water is the job of?

Chemistry
2 answers:
yulyashka [42]3 years ago
7 0

Answer:

Analytical chemistry is the field of chemistry which deals with the application of methods used for separation, identification and quantification of the elements present in the sample or specimen under concern.

The lead can be identified, separated and quantified by the methods used in the analytic chemistry present inside the water. Therefore, measuring the level of lead in drinking water is the job of analytical chemistry.

tamaranim1 [39]3 years ago
3 0
Analytical chemistry -- A task that would fall into this area of chemistry is measuring the level of lead in drinking water.

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Un estudiante de noveno año decide realizar varias mezclas de compuestos quimicos y tiene un frasco rotulado con la formula quim
expeople1 [14]

Answer:

- Oxido de magnesio

Explanation:

Mg → magnesio

O → oxígeno

La unión del oxígeno (no metal) con un metal, forma el determinado óxido que se nombra como óxido de .... y el no metal.

En este caso, podemos deducir que el contenido del frasco es de óxido de magnesio, aunque viendo que hay un sólo átomo de oxígeno podemos llamarlo como monóxido de magnesio, similar al CO (monóxido de carbono).

A partir de que el magnesio sólido entre en contacto con el aire, se produce MgO de acuerdo a la siguiente reacción:

2Mg  +  O₂ →  2MgO

7 0
3 years ago
Sulfuric acid (H2SO4) is prepared commercially from elemental sulfur using the contact process. In a typical sequence of reactio
Elan Coil [88]
I believe the end result is still 83 moles since there is never an amount of sulfur atoms added to the initial amount, but rather oxygen and water is repeatedly added to it. To find it's weight, first find the molar mass of H2SO4:

H2 + S + O4 = 2.00 + 32.1 + 64.0 = 98.1 g/mol

and mass = (98.1 g/mol)(83 mol) = 8142.3 g

rounded to 8.1 x 10^3 g assuming 100% yield?
8 0
3 years ago
Read 2 more answers
How many atoms of phosphorus are in 2.50 million of copper (II) phosphate
Dmitry [639]
Are you sure it is 2.50 million and not 2.50 mol?
7 0
4 years ago
Colorea los recuadros que representan la acciones viables para usar racionalmente los recursos y contaminar menos
In-s [12.5K]

Algunas acciones viables para utilizar los recursos ambientales de forma racional y con el objetivo de reducir la contaminación son:

  • Ahorra agua y energía
  • Reducir, reciclar y reutilizar los residuos generados en el hogar.
  • Reducir el uso de vehículos de combustión interna.
  • Usa energías renovables
  • No quemes basura al aire libre
  • No utilices aerosoles ni frigoríficos que dañen la capa de ozono.
<h3 /><h3>¿Qué importancia tiene la sostenibilidad?</h3>

Es fundamental para la preservación y mantenimiento del medio ambiente y la calidad de vida de las generaciones actuales y futuras. El desarrollo de acciones sostenibles promueve la protección del medio ambiente, los ecosistemas y los recursos naturales.

Por lo tanto, todas las personas e instituciones económicas pueden promover la sostenibilidad a través de prácticas fáciles de ejecutar que ayuden a preservar y proteger el medio ambiente.

Encuentre más información sobre sostenibilidad aquí:

brainly.com/question/23701037

7 0
3 years ago
Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =
MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{c}

For a general chemical reaction:

aA+bB\rightleftharpoons cC+dD

The expression for K_{eq} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)

The expression of K_c for above equation is:

K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}

We are given:

[H_2S]_{eq}=0.671M

[O_2]_{eq}=0.587M

K_c=1.35

Putting values in above expression, we get:

1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}

[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M

Hence, the equilibrium concentration of water is 0.597 M

8 0
4 years ago
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