Question 13
No of moles of glucose = 0.5 x 0.4 = 0.2 Moles
Hence 0.2 Moles of glucose present at 0.5 L of 0.4 M glucose.
Question 14
Volume of NaI = 0.405 x 1000 / 0.724 = 559..39 ml
Hence 559.32 ml of 0.724 M NaI contains 0.405 mol.
Question 15
Mass of Mg(NO3)2 = 45 gm
Molar mass of Mg(NO3)2 = 148.30 g/mol
Moels of Mg(NO3)2 = 45 gm / 148.30 g/mol = 0.30343 Moles
Volume of Mg(NO3)2 = 0.30343 x 1000 / 0.32 = 948.25 ml
948.25 ml of 0.32 M of Mg(NO3)2 contains 45 gm of Mg(NO3)2
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6:24 < Question 5 of 10 Submit What volume in L of a 0.724 M Nal solution contains 0.405 mol of Nal STARTING AMOUNT ADD FACTOR ANSWER RESET 22.99 1.79 L 559 0.293 g Nal 148.89 mol Nal 0.405 0.724 g Nal/mol 0.559 M Nal 126.90 ml 6.022 x 1023 Tap here or pull up for additional resources
Question 8 of 14 mol of Nal M al solution contains 0.405 What volume in Loa 0.724 () -0 o 6.022 x 100.7240.405 0.559 1.790.293 Nol mol Na ML 2299 148.89 126.90 559 M Nalg Nal/mol
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