Some chemical changes can be reactive through another chemical change?
The given balanced chemical equation:
![CaCO_{3}(s) +2 HCl (aq) --> CaCl_{2}(aq)+H_{2}O(l) + CO_{2}(g)](https://tex.z-dn.net/?f=%20CaCO_%7B3%7D%28s%29%20%2B2%20HCl%20%28aq%29%20--%3E%20CaCl_%7B2%7D%28aq%29%2BH_%7B2%7DO%28l%29%20%2B%20CO_%7B2%7D%28g%29%20%20%20%20%20)
Calculating the moles of carbon-dioxide:
Given, moles of ![CaCO_{3} = 0.00500 mol CaCO_{3}](https://tex.z-dn.net/?f=%20CaCO_%7B3%7D%20%3D%200.00500%20mol%20CaCO_%7B3%7D%20%20%20)
Moles of ![CO_{2} = 0.00500 mol CaCO_{3} * \frac{1 mol CO_{2}}{1 mol CaCO_{3}}](https://tex.z-dn.net/?f=%20CO_%7B2%7D%20%3D%200.00500%20mol%20CaCO_%7B3%7D%20%2A%20%5Cfrac%7B1%20mol%20CO_%7B2%7D%7D%7B1%20mol%20CaCO_%7B3%7D%7D%20%20%20%20%20)
![= 0.00500 mol CO_{2}](https://tex.z-dn.net/?f=%20%3D%200.00500%20mol%20CO_%7B2%7D%20%20)
Using the ideal gas equation to find out volume from moles, pressure and temperature:
PV = nRT
![P is the pressure = 100 kPa *\frac{1 atm}{101.325 kPa} = 0.987 atm](https://tex.z-dn.net/?f=%20P%20is%20the%20pressure%20%3D%20100%20kPa%20%2A%5Cfrac%7B1%20atm%7D%7B101.325%20kPa%7D%20%3D%20%200.987%20atm%20)
![T is the temperature = 25^{0}C + 273 = 298 K](https://tex.z-dn.net/?f=%20T%20is%20the%20temperature%20%3D%2025%5E%7B0%7DC%20%2B%20273%20%3D%20298%20K%20%20)
![(0.987 atm)(V) = (0.00500 mol)(0.08206 \frac{L.atm}{(mol.K)}(298K)](https://tex.z-dn.net/?f=%20%280.987%20atm%29%28V%29%20%3D%20%280.00500%20mol%29%280.08206%20%5Cfrac%7BL.atm%7D%7B%28mol.K%29%7D%28298K%29%20)
V = 0.124 L or 124 mL
Therefore, 124 mL of carbon dioxide will be collected.
According to Avogadro's law 1 mole contains 6.022 ×10^23 particles
1 mole of carbon = 44.01 g/ mol
Therefore;
44.01 g = 6.022 ×10^23 molecules
Hence, 1.68×10^26 molecules will have a mass of ;
(44.01 × 1.68×0^26) / 6.022×10^23
= 1.228 × 10^4 molecules
Answer:
Sitting at the bottom of the slide will have the most kinetic energy because all the potential energy is transferred to kinetic after he slides
Explanation:
239.8 g so 240g
Explanation:
plz let me know if this is right than I can tell u how I did it :)