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2 years ago

What is the molar mass of 12?

Chemistry

1 answer:

andrew-mc [135]2 years ago

3 0

Answer:

I assume your talking about carbon when you say 12 so it'd be 12 grams if you are

Explanation:

The molar mass of any substance in grams per mole is numerically equal to the mass of that substance expressed in atomic mass units.

Hope this helps you some

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Why phosphorus forms 3 bonds in molecule such as PH3 and PCl3-

gregori [183]

Phosphorous has three lone electrons that need pairing. Similar to how carbon has 4 lone electrons, and forms CH4

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3 years ago

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Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

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3 years ago

Ionic formula for sodium oxide

Bas_tet [7]

The ionic formula of sodium oxide would be Na20

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2 years ago

Propiedades químicas del óxido

tangare [24]

Óxidos básicos: Son formados por metales. El metal presente en su fórmula puede presentar carga eléctrica +1 y +2, o sea, poseer carácter iónico. Ejemplos: Na2O (óxido de sodio), BaO (óxido de bario).

Óxidos neutros: Son compuestos por no metales. No reaccionan con agua, ácido o base, en razón del enlace covalente que une sus componentes; de ahí el por qué de ser llamados óxidos inertes. Ejemplos: monóxido de dinitrógeno (N2O) y monóxido de carbono (CO).

Óxidos ácidos: También conocidos como anhídridos de ácidos, son formados por no metales y presentan carácter covalente. En la presencia de agua, producen ácidos y en la presencia de bases, origina sal y agua. Ejemplo: CO2 (dióxido de carbono o gas carbono) y el SO2 (dióxido de azufre)

Óxidos dobles o mixtos: La combinación de dos óxidos de un mismo elemento, da origen a este tipo de óxidos. Ejemplo: magnetita (Fe2O4), unión de los óxidos de hierro (Fe) y oxígeno (O).

Óxidos anfóteros: Presentan ambigüedad, en la presencia de un ácido se comportan como óxidos básicos y en la presencia de una base, como óxidos ácidos. Ejemplos: óxido de aluminio (Al2O3 ) y el óxido de zinc (ZnO).

Peróxidos: Compuestos que poseen en su fórmula el grupo (O2)2- . Los peróxidos más comunes son formados por hidrógeno, metales alcalinos y metales alcalinos térreos. Ejemplos: agua oxigenada (H2O) y peróxido de sodio (Na2O2).

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3 years ago