Answer:
Given info : 500 cc of 2N Na2CO3 are mixed with 400 cc of 3N H2SO4 and volume was diluted to one litre. To find : will the resulting solution is acidic , basic or neutral ? Calculate the molarity of the dilute solution. solution : no of moles of Na2CO3 = normality/n %3D - factor x volume 2/2 x 500/1000 = 0.5 mol %D no of moles of H2SO4 = 3/2 x 400/100O = 0.6 mol %3D We see, Na2CO3 + H2S04 => Na2S04 + CO2 + H2O Here one mol of Na2C03 reacts with one mole of H2SO4. So, 0.5 mol of Na2CO3 reacts with 0.5 mol of H2SO4. so, remaining 0.1 mol of H2SO4 makes solution acidic. Now molarity of solution = remaining no of moles of H2SO4/volume of solution= 0.1/1 = %3D 0.1M
Answer:
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Explanation:
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Answer:
Molar mass = 94972.745 g/mol
Explanation:
Given data:
Density = 2.25 g/ml
Pressure = 700 mmHg
Temperature = 200°C
Molar mass = ?
Solution:
Density = 2.25 g/ml (2.25×1000 = 2250 g/L)
Pressure = 700 mmHg (700/760 = 0.92 atm)
Temperature = 200°C (200+273 = 473K)
Formula:
d = PM/RT
M = dRT/P
M = 2250 g/L × 0.0821 atm.L /mol.K × 473K / 0.92 atm
M = 87374.93 g/mol / 0.92
M = 94972.745 g/mol
Answer:
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Explanation:
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Answer:
0.464 L
Explanation:
Molarity (M) = number moles (n) ÷ volume (V)
According to the information given in this question:
number of moles (n) = 4.36 moles
Molarity = 9.4M
Volume = ?
Using M = n/V
9.4 = 4.36/V
9.4V = 4.36
V = 4.36/9.4
V = 0.464 L
Hence, 0.464L of water are needed the volume of water.
