<u>Answer:</u> The new water level of the cylinder is 24.16 mL
<u>Explanation:</u>
To calculate the volume of water displaced by silver, we use the equation:
Density of silver = 10.49 g/mL
Mass of silver = 35.2 g
Putting values in above equation, we get:
We are given:
Volume of graduated cylinder = 20.8 mL
New water level of the cylinder = Volume of graduated cylinder + Volume of water displaced by silver
New water level of the cylinder = (20.8 + 3.36) mL = 24.16 mL
Hence, the new water level of the cylinder is 24.16 mL
To calculate the number of atoms of Cr, we first find the number of moles per unit of cubic centimeter of Cr. Then, use avogadros number for the number of atoms. Calculations are as follows:
1 cm^3 (7.15 g/cm^3) (1 mol / 51.996 g Cr) = 0.14 mol Cr
0.14 mol Cr ( 6.022 x 10^23 atoms Cr / 1 mol Cr ) = 8.28 x 10^22 atoms Cr
The reaction, 2 C4H10 (g) + 13 O2 (g) = 8 CO2 (g) + 5 H2O (g), is the combustion of butane. A combustion reaction involves the reaction of a hydrocarbon with oxygen producing carbon dioxide and water. This reaction is exothermic which means it releases energy in the form of heat. Therefore, as the reaction proceeds,a heat energy is being given off by the reaction. This happens because the total kinetic energy of the reactants is greater than the total kinetic energy of the products. So, the excess energy should be given off somewhere which in this case is released as heat.
Explanation:
96.485 columbs=1 faraday will
deposit 64/2g= 32 g cu ion
therfore it will require
96,485 ×2/32 =? coulombs or 1/16 of
Faraday= 1 / 16 mole of electrons .
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62
