Answer:- 64015 J
Solution: There is 4250 mL of water in the calorimeter at 22.55 degree C.
density of water is 1 g per mL.
So, the mass of water = 
= 4250 g
Final temperature of water after adding the hot copper bar to it is 26.15 degree C.
So, 
for water = 26.15 - 22.55 = 3.60 degree C
Specific heat for water is 4.184 
The heat gained by water is calculated by using the formula:
where, q is the heat energy, m is mass and c is specific heat.
Let's plug in the values in the formula and do the calculations:
q = 64015 J
So, 64015 J of heat is gained by the water.
1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = <span>0.0135 mol Cr
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
</span>0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr /0.0135= 1 mol Cr
0.0538 mol O/0.035= 4 mol Cr
K2CrO4
The answer:
<span>the overall balanced equation for the precipitation reaction occurring between silver nitrate and calcium bromide:
</span>CaBr2 + AgNO3-----------> AgBr2 + CaNO3, so among the given choices, the
true answe is
<span>agno3(aq)+cabr2(aq) (as reactants)</span>
To calculate the new pressure, we can use Boyle’s law to relate these two scenarios (Boyle’s law is used because the temperature is assumed to remain constant). Boyle’s law is:
P1V1 = P2V2,
Where “P” is pressure and “V” is volume. The pressure and volume of the first scenario is 215 torr and 51 mL, respectively, and the second scenario has a volume of 18.5 L (18,500 mL) and the unknown pressure - let’s call that “x”. Plugging these into the equation:
(215 torr)(51 mL) =(“x” torr)(18,500 mL)
x = 0.593 torr
The final pressure exerted by the gas would be 0.593 torr.
Hope this helps!
Depression is freezing point is a colligative property. It is mathematically expressed as ΔTf = Kf X m
where Kf = <span>freezing point depression constant = 1.86°c kg /mol (for water)
m = molality of solution = 1.40 m
</span>∴ ΔTf = Kf X m = 1.86 X 1.40 = 2.604 oC
Now, for water freezing point = 0 oC
∴Freezing point of solution = -2.604 oC
