Answer:16.38085
Explanation:Aline the decimals then start multiplying.
Answer:
CaCl₂
Step-by-step explanation:
The <em>empirical formula</em> is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the molar ratio of Ca to Cl.
Data:
Mass of Ca = 3.611 g
Mass of Cl = 6.389 g
Calculations
Step 1. <em>Calculate the moles of each element
</em>
Moles of Ca = 3.611 g Ca × (1 mol Ca/(40.08 g Ca)= 0.090 10 mol Ca
Moles of Cl = 6.389 g Cl
Step 2. <em>Calculate the molar ratio of the elements
</em>
Divide each number by the smallest number of moles
Ca:Cl = 0.090 10:0.1802 = 1:2.000
Step 3. Round the molar ratios to the nearest integer
Ca:Cl = 1:2.000 ≈ 1:2
Step 4: <em>Write the empirical formula
</em>
EF = CaCl₂
From the statement of Hess' law, the enthalpy of the reaction A---> C is +90 kJ
<h3>What is Hess' law?</h3>
Hess' law of constant heat summation states that for a multistep reaction, the standard enthalpy of reaction is always constant and is independent of the pathway or intermediate routes taken.
From Hess' law, the enthalpy change for the reaction A ----> C is calculated as follows:
A---> C = A ---> B + B ---> C
ΔH of A---> C = 30 kJ + 60 kJ
ΔH = 90 kJ
Therefore, the enthalpy of the reaction A---> C is +90 kJ
The above reaction A---> C can be shown in the enthalpy diagram below:
A -------------------> C (ΔH = +90 kJ)
\ /
\ / (ΔH = +60 kJ)
(ΔH = +30 J) \ /
> B
Learn more about enthalpy and Hess law at: brainly.com/question/9328637
94.0 L if the reaction takes place under STP.
<h3>Explanation</h3>
The molar mass of glucose C₆H₁₂O₆ is
12.01 × 6 + 1.008 × 12 + 16.00 × 16.00 = 180.16 g / mol.
126 grams of glucose will contain 126 / 180.16 = 0.69939 mol of C₆H₁₂O₆. (To avoid rounding errors, keep a couple more digits than necessary.)
6 moles of CO₂ will be produced when 1 mole of C₆H₁₂O₆ is consumed. 0.69939 moles of C₆H₁₂O₆ will give rise to 4.196 mol of CO₂.
Assuming that the reaction takes place under STP, where T = 0 °C = 273 K and P = 1 atm. Each mole of any ideal gas will occupy a volume of 22.4 liters. The 4.196 moles of CO₂ will occupy 4.196 × 22.4 = 94.0 L. (The least significant number given is 126 g, the mass of glucose. This number has three significant figures. Thus, round the result to three significant figures.)
The volume of CO₂ can be found using the ideal gas law if the condition isn't STP. For example, T = 25 °C = 297 K and P = 1.00 × 10⁵ will lead to a different volume. By the ideal gas law,
V = (n · R · T) / (P)
where
- V is the volume of the gas,
- n is the number of moles of gas particles,
- R is the ideal gas constant<em>,</em>
- P is the pressure on the gas,
- T is the absolute temperature of the gas (in degrees Kelvin.)
R = 8.314 × 10³ L · Pa / (K · mol)
Taking T = 297 K and P = 1.00 × 10⁵ Pa,
V = (4.196 × 8.314 × 10³ × 297 ) / (1.00 × 10⁵ ) = 104 L.
