Calculate the kilojoules needed to heat 231 g of gold from 18 ∘C to 195 ∘C.
1 answer:
5.27 kJ of heat are required to heat 231 g of gold from 18 °C to 195 °C.
We have 231 g of gold at 18 °C and supply it with heat to increase its temperature to 195 °C. We can calculate the amount of heat required using the following expression.
where,
- Q: heat
- <em>c: specific heat capacity of gold</em> (0.129 J/g.°C)
- m: mass
- ΔT: change in the temperature
5.27 kJ of heat are required to heat 231 g of gold from 18 °C to 195 °C.
You can learn more about heating here: brainly.com/question/1105305
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Answer:
30, 50
Explanation:
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In this since an element's atomic number is equal to the number of protons in its atom, we can infer that selenium's atomic number is 30. Moreover, due to the fact the the neutrons are equal to the atomic mass minus the atomic number or the number of protons, by knowing the number of neutrons we compute the atomic as follows:
Thus, answer is 30, 50.
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Orbital shell notation of fluorine is 2. 7 while that of oxygen s 2. 6. This means that these elements (that follow each other in the periodic table) will have high electronegativity in molecules due to their high atomic number (which causes them to strongly attract electron orbital shell closer to their nucleus). NB: Atomic number of a peroid increased from left to right of the periodic table.
Therefore, in the first molecule, the negative dipole would most likely be located between the F atoms In the second molecule the negative molecule would be most likely located in the between the O and F atoms.
Answer:
336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.
Explanation:
In this case, the balanced reaction is:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:
- C₃H₈: 1 mole
- O₂: 5 moles
- CO₂: 3 moles
- H₂O: 4 moles
Being the molar mass of each compound:
- C₃H₈: 44 g/mole
- O₂: 16 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
Then, by stoichiometry, the following quantities of mass participate in the reaction:
- C₃H₈: 1 mole* 44 g/mole= 44 grams
- O₂: 5 moles* 16 g/mole= 80 grams
- CO₂: 3 moles* 44 g/mole= 132 grams
- H₂O: 4 moles* 18 g/mole= 72 grams
So you can apply the following rules of three:
- If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?
mass of CO₂= 336.6 grams
- If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?
mass of H₂O= 183.6 grams
<u><em>336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.</em></u>