77.78 kJ of energy is released when 1 mol of glucose is burned, 2802.5 kJ of energy is released.
<h3>What are moles?</h3>
A mole is defined as 6.02214076 ×
of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
Calculate the moles of 5.00g of glucose.
Given mass = 5.00g
The molar mass of glucose = 180.156 g/mol
Moles =0.02775372455
The quantity of energy released to a person by eating 5.00g of glucose in a candy.
0.02775372455 x 2802.5 kJ
77.77981305 kJ =77.78 kJ
Hence, 777.78 kJ of energy is released.
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Answer:
1.1 × 10⁻⁴ M
Explanation:
Let's consider the following double displacement reaction.
CuCl₂(aq) + 2 AgNO₃(aq) → 2 AgCl(s)+ Cu(NO₃)₂(aq)
We can establish the following relations:
- The molar mass of AgCl is 143.32 g/mol.
- The molar ratio of AgCl to CuCl₂ is 2:1
The moles of CuCl₂ that reacted to produce 7.7 mg of AgCl are:
The molarity of CuCl₂ is:
Answer:
CaCO3 is the limiting reactant
55 g of CO2 is made
Explanation:
First we must put down the reaction equation;
CaCO3(s) + 2HCl(aq) ---------> CaCl2(s) + H2O(l) + CO2(g)
Number of mole of CaCO3 = 125g/100gmol-1 = 1.25 moles
From the reaction equation;
1 mole of CaCO3 yields 1 mole of CO2
Hence 1.25 moles of CaCO3 yields 1.25 moles of CO2
For HCl;
number of moles of HCl = 125g/36.5 g mol-1 = 3.42 moles
From the reaction equation;
2 moles of HCl yields 1 mole of CO2
3.42 moles of HCl yields 3.42 * 1/2 = 1.71 moles of CO2
Hence CaCO3 is the limiting reactant.
Mass of CO2 produced = 1.25g * 44 gmol-1 = 55 g of CO2
Since Alai is hitting each with a hammer, the physical property which he must be comparing on the two materials must be hardness. Solubility can be tested when you put them in a solvent. Odor can be tested with smell. Conductivity can be tested with electricity. Hardness can be tested with its resistance to the force that hits it.
37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2
<span>4.60X10^-3 mol CH2Br2 X (2 mol Br / 1 mol CH2Br2) X 6.02X10^23 atoms/mol = 5.54X10^21 bromine atoms</span>
