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3 years ago

5

Iron has a density of 7.87mL. How much mass would a 12.3mL sample contain?

1 answer:

algol [13]3 years ago

4 0

The correct answer is 96.80 grams

Explanation:

The density of a substance shows the total mass the substance contains in 1 mL or 1 cm3 as density is calculated by using the formula D= M (mass) / V (volume). This implies, Iron contains 7.87 grams per milliliter. Moreover, this value and formula can be used to calculate the mass or volume of any other sample. The process to calculate the mass of an iron sample with a volume of 12.3 mL is shown below.

D = M ÷ V

7.87 mL = x ÷ 12.3 mL - x represents the missing value. Now find the value of x by solving the equation

x = 7.87 · 12.3

x = 96.80

This means a sample of 12.3 mL contains a mass of 96.80 grams. Also, you can know this value is correct because if you divide the mass by the value the density is the same (96.80 grams ÷ 12.3 mL = 7.87 g/mL)

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Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

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

The formation reaction of $C_2H_4$ will be,

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_1=-1411kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.8kJ$

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) $2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H_1=+1411kJ$

(2) $2C(s)+2O_2(g)\rightarrow 2CO_2(g)$ $\Delta H_2=2\times (-393.5kJ)=-787kJ$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.8kJ)=-571.6kJ$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)$

$\Delta H=52.4kJ$

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

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How to tell what side is the reactant and what side is the products?

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Reactant at the left product at the rig

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A common fuel additive that is composed of C, H, and O enhanced the performance of gasoline began being phased out in 1999 becau

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Explanation:

44 g of CO₂ contains 12 g of C

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