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3 years ago

Calculate the molar mass, in g/mol, for the compound strontium nitrate, Sr(NO3)2.

Chemistry

1 answer:

nikdorinn [45]3 years ago

8 0

By definition, the molar mass for the compound strontium nitrate Sr(NO₃)₂ is 211.62 g/mole.

<h3>Definitio of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

The molar mass of a compound (also called Mass or Molecular Weight) is the sum of the molar mass of the elements that form it (whose value is found in the periodic table) multiplied by the number of times they appear in the compound.

<h3>Molar mass of strontium nirate</h3>

In this case, you know the molar mass of the elements is:

Sr= 87.62 g/mole
N= 14 g/mole
O= 16 g/mole

So, the molar mass of the compound Sr(NO₃)₂ is calculated as:

Sr(NO₃)₂= 87.62 g/mole + 2× (14 g/mole + 3×16 g/mole)

Solving:

Sr(NO₃)₂= 211.62 g/mole

Finally, the molar mass for the compound strontium nitrate Sr(NO₃)₂ is 211.62 g/mole.

Learn more about molar mass:

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Carbon, hydrogen and ethane each burn exothermically in an excess of air. AHⓇ =-393.7 kJ mol. C(s) + O2(g) → CO2(g) H2(g) + % O2

Salsk061 [2.6K]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is 51.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the reaction of carbon and water follows:

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_1=-393.7kJ$ ( × 2)

(2) $H_2+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_2=-285.9kJ$ ( × 2)

(3) $2C_2H_4(s)+2O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_3=-1411kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[2\times \Delta H_1]+[2\times \Delta H_2]+[1\times (-\Delta H_3)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(2\times (-393.7))+(2\times (-285.9))+(1\times -(-1411))]=51.8kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is 51.8 kJ.

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