Answer: 8.12 g NaCl
Explanation: Use Avogadro's number to find the number of m
moles of NaCl:
8.24x10²² molecules NaCl / 1 mole NaCl/ 6.022x10²³ molecules NaCl
= 0.14 mole NaCl
Next convert moles to grams NaCl using its molar mass;
0.14 mole NaCl x 58g NaCl / 1 mole NaCl
= 8.12 g NaCl
Answer:
Pentan_1,5_di-al
Explanation:
OHC-CH₂-CH₂-CH₂-CHO
This is Pentan_1,5_di-al
If we break this compound, we will observe that there is presence aldehyde group and hence the functional group "al". This aldehyde is bonded to carbon 1 and carbon 5 respectively.
Also the pentan is due to presence of 5 carbon atoms.
Therefore, the IUPAC name of this compound (OHC-CH₂-CH₂-CH₂-CHO) is Pentan_1,5_di-al
Cars, aeroplanes, factories etc
Answer:
Enthalpy change for the reaction is -67716 J/mol.
Explanation:
Number of moles of
in 50.0 mL of 0.100 M of 
= Number of moles of HCl in 50.0 mL of 0.100 M of HCl
=
moles
= 0.00500 moles
According to balanced equation, 1 mol of
reacts with 1 mol of HCl to form 1 mol of AgCl.
So, 0.00500 moles of
react with 0.00500 moles of HCl to form 0.00500 moles of AgCl
Total volume of solution = (50.0+50.0) mL = 100.0 mL
So, mass of solution = (
) g = 100 g
Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)
=
= ![\frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}](https://tex.z-dn.net/?f=%5Cfrac%7B-100g%5Ctimes%204.18%5Cfrac%7BJ%7D%7Bg.%5E%7B0%7D%5Ctextrm%7BC%7D%7D%5Ctimes%20%5B24.21-23.40%5D%5E%7B0%7D%5Ctextrm%7BC%7D%7D%7B0.00500mol%7D)
= -67716 J/mol
[m = mass, c = specific heat capacity,
= change in temperature and negative sign is included as it is an exothermic reaction]
Answer:
Red
Explanation:
Violet - shortest wavelength, around 400-420 nanometers with highest frequency. They carry the most energy.
Indigo - 420 - 440 nm
Blue - 440 - 490 nm
Green - 490 - 570 nm
Yellow - 570 - 585 nm
Orange - 585 - 620 nm
Red - longest wavelength, at around 620 - 780 nanometers with lowest frequency and least amount of energy
Therefore, <em>red </em>is the answer you're looking for.
I hope this helps and that you have a great day! :)