The answer is 0.25 g/cm3
The equation for finding density is mass divided by volume.
Therefore,
12 grams/ 49 cm3= 0.24489795918367 gram/cubic centimeter
you would then round this to the nearest significant figure which is the number of significant numbers you have and the highest amount of sig figs present in this equation is two sig figs. That means that you would round your answer, becasue it has a zero before the decimal which doesn't count as a significant figure, to the nearest hundredth which would be 0.25 and then you add your value after it which is g/cm3
B) Ag because it’s in the middle of the periodic table.
It forms <span>calcium phosphate and potassium nitrate
</span>2 K3PO4 + 3Ca(NO3)2 --> Ca3(PO4)2 + 6KNO3
Answer: R
Explanation: is polarized to the right
Answer:
Molarity = 0.809 M
mole fraction = 0.047
Explanation:
The complete question is
Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.
Solution -
Solution for molarity:
1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.
1)
Mass of 1.09 mole of acetone
= 1.09 mol x 58.0794 g/mol = 63.306 g
Density of acetone = 0.788 g/cm3
Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3
For ethanol
1000 g divided by 0.789 g/cm3 = 1267.427 cm3
Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3 = 1.347 L
a) Molarity:
1.09 mol / 1.347 L = 0.809 M
Mole Fraction
a) moles of ethanol:
1000 g / 46.0684 g/mol = 21.71 mol
b) moles of acetone:
1.09 / (1.09 + 21.71) = 0.047
