Answer : 1.12 grams
Yo find the mass of the sample, you take the increased mass and subtract the original mass.
6.130 - 5.010 = 1.12
<span>E = mCdT
E = energy, m = mass, C = specific heat capacity, dT = change in temperature.
526 = 0.074C x 17
E = 0.074C x 55
Divide the equations
E/526 = (0.074C x 55)/(0.074C x 17) = 55/17
E = (55 x 526)/17 = 1702 J</span>
<u>Answer:</u> The mass of sucrose required is 69.08 g
<u>Explanation:</u>
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

Or,

where,
= osmotic pressure of the solution = 8.80 atm
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of solute (sucrose) = ?
Molar mass of sucrose = 342.3 g/mol
Volume of solution = 564 mL (Density of water = 1 g/mL)
R = Gas constant = 
T = Temperature of the solution = 290 K
Putting values in above equation, we get:

Hence, the mass of sucrose required is 69.08 g
Answer:
yes, it is a tenth of a centimeter
Explanation: