Temperature change, colour change, releasing gas, bubbles and change in odor
moles NaOH = c · V = 0.2432 mmol/mL · 24.75 mL = 6.0192 mmol
moles H2SO4 = 6.0192 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.0096 mmol
Hence
[H2SO4]= n/V = 3.0096 mmol / 38.94 mL = 0.07729 M
The answer to this question is [H2SO4] = 0.07729 M
a) 56g
<h3>Calculation:</h3>
At STP,
22.4 L of N₂ = 1 mol
We have given 44.8 L of N₂, therefore,
44.8 L of N₂ = 
= 
mol
We know that,
1 mol of N₂ = 28 g
Hence,
2 mol of N₂ = 28 × 2
= 56g
Hence, there are 56 g of N₂ in 44.8 L of nitrogen gas.
Learn more about calculation at STP here:
brainly.com/question/9509278
#SPJ4
<span>The higher the temperature of the gas, the faster the molecules move as they approach evaporating temperature. The lower the temperature of the gas, the slower the molecules move as they approach cooling temperature. The temperature controls how fast the molecules move</span>
