Answer : The specific heat of the substance is 0.0936 J/g °C
Explanation :
The amount of heat Q can be calculated using following formula.

Where Q is the amount of heat required = 300 J
m is the mass of the substance = 267 g
ΔT is the change in temperature = 12°C
C is the specific heat of the substance.
We want to solve for C, so the equation for Q is modified as follows.

Let us plug in the values in above equation.


C = 0.0936 J/g °C
The specific heat of the substance is 0.0936 J/g°C
Answer:
B. Concept Formation
Explanation:
All the other answer choices are not equivalent.
Hope this helps!
Brainliest??
Answer:

In which [Ag+] in negligibly small and the concentration of each reactant is 1.0 M
The answer is A) PO43- < NO3- < Na+
Explanation:
Ag+ is removed from the solution just like PO43-, so there are just 2 possible answers at this point: a or b. Then we can notice that Na3PO4 releases 3 moles of Na+ and just 1 mole of NO3-
We have 100mL of each reactant with the same concentration for both (1.0 M) so:
(0.1)(1)(3)= 0.3 mol Na+
(0.1)(1)= 0.1 mol NO3-
so PO43- < NO3- < Na+
The molecular weight of hemoglobin can be calculated using osmotic pressure
Osmotic pressure is a colligative property and it depends on molarity as
πV = nRT
where
π = osmotic pressure
V = volume = 1mL = 0.001 L
n = moles
R = gas constant = 0.0821 L atm / mol K
T = temperature = 25°C = 25 + 273 K = 298 K
Putting values we will get value of moles

we know that

Therefore
