Answer:
I know that the 100-mL graduated cylinders are always read to 1 decimal place.
I think for 50 mL graduated cylinders, it lets you measure volumes up to 50.0 mL to the nearest 0.1 or 0.2 mL, depending on your exact cylinder.
Answer:
Q = 1455.12 Joules.
Explanation:
Given the following data;
Mass = 300 grams
Initial temperature = 22.3
Final temperature = 59.9°C
Specific heat capacity = 0.129 J/gºC.
To find the quantity of energy;
Where,
Q represents the heat capacity.
m represents the mass of an object.
c represents the specific heat capacity of water.
dt represents the change in temperature.
dt represents the change in temperature.
dt = T2 - T1
dt = 59.9 - 22.3
dt = 37.6°C
Substituting the values into the equation, we have;
Q = 1455.12 Joules.
213034 torr is the osmotic pressure.
Explanation:
osmotic pressure is calculated by the formula:
osmotic pressure= iCrT
where i= no. of solute
c= concentration in mol/litre
R= Universal Gas constant
T = temp
It is given that solution is 3% which is 3gms in 100 ml.
let us calculate the concentration in moles/litre
3gm/100ml*1000ml/1L*1mol NaCl/55.84g NaCl
= 5.372 gm/litre
Putting the values in the formula, Temp in Kelvin 318.5K
osmotic pressure= 2*5.372*0.083 * 318.5 Gas constant 0.083
= 284.023 bar or 213018 torr. c= 5.372 moles/L
i=2 for NaCl
Assuming that the contents of the chamber ar ideal gases. We can use the relation PV=nRT. At a constant
temperature and number of moles of the gas the product of PV is equal to some
constant. At another set of condition of temperature, the constant is still the
same. Calculations are as follows:
P1V1 =P2V2
P2 = (1)(450)/ 48
P2 = 9.375 atm