! mole of CO2 at STP has a volume of 22.4 liters
88 grams = 2 moles
so the required volume = 2*22.4 = 44.8 liters
The atomic number of an element is based on the number of protons in the atomic nuclei of its atoms.
Answer:
Mass = 114.26 g
Explanation:
Given data:
Number of gold atoms = 3.47×10²³ atoms
Mass in gram = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
3.47×10²³ atoms × 1 mol /6.022 × 10²³ atoms
0.58 mol
Mass of gold:
Mass = number of moles × molar mass
Mass = 0.58 mol × 197 g/mol
Mass = 114.26 g
★ « <em><u>what is oxidation number of S in H2SO5??</u></em><em><u> </u></em><em><u>»</u></em><em><u> </u></em><em><u>★</u></em>
- <em><u>it's </u></em><em><u> </u></em><em><u>6</u></em><em><u>!</u></em><em><u>!</u></em>
Explanation:
- <em>Oxidation number of S in H2SO5 is 6 .</em>
Answer:
the activation energy Ea = 179.176 kJ/mol
it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.
Explanation:
From the given information



Thus; 
Because at 113.0°C; the rate is 7 time higher than at 100°C
Hence:

1.9459 = 



Ea = 179.176 kJ/mol
Thus; the activation energy Ea = 179.176 kJ/mol
b)
here;






where ;


Now;

t = 7.0245 mins
Therefore; it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.