C2H5OH(aq) + MnO− 4 (aq) → Mn2+(aq) + CH3COOH(aq) of acetic acid from ethanol by the action of permanganate ion in acidic soluti

The quantity of carbon atoms in the pencil mark amounts to 3 x 10^17. Given that the atomic weight of carbon is 12.01 amu, it follows that 12.01 g of carbon contains 6.022 x 10^23 atoms. Thus, we can set up the equation: 12.01 g carbon/ 6.022 x 10^23 atoms (3 x 10^17 atoms) (12.01 g carbon/ 6.022 x 10^23 atoms). By canceling out the atoms, we have (3 x 10^17) (12.01 g carbon/6.022 x 10^23) and then completing the division and multiplication yields 6 x 10^-6 g of carbon. Therefore, the mass of the pencil mark is 6 x 10^-6 g.

Assuming the edge length of the cube is m.
Since the cube's sides are all equal, you can find its volume using the formula:
volume of cube = m^3

Now, using the provided edge length:
volume = (4.33)^3 = 81.1827 cm^3

According to unit conversion standards, 1 cm^3 = 0.001 liters.
So:
volume of cube = 81.1827 cm^3 = 0.0811827 liters

Answer: A= 0.5, B = 0.25, C = 0.125, D = 0.015625 and E = 0.00390625

Explanation:

The half-life of a substance refers to the duration it takes for that substance to decrease to half of its original quantity.

Here, n signifies the count of half-lives.

The remaining quantity of the substance after n half-lives can be determined using the formula provided,

When n = 1,

Fraction of substance remaining = 0.5¹ = 0.5.

This indicates that after the first half-life, the remaining quantity is half of the original amount.

Thus, we find A = 0.5

<pfor n="2," we="" calculate=""><pso after="" two="" half-lives="" the="" remaining="" quantity="" is="" of="" original=""><phence>B = 0.25 <pfor n="3," we="" have=""><pso after="" three="" half-lives="" the="" remaining="" amount="" is="" of="" original.=""><ptherefore we="" find="">C = 0.125<pfor n="6," we="" see="">

Thus, D = 0.015625

<pfor n="8," we="" calculate=""><pconsequently>E = 0.00390625

The calculated values for A, B, C, D, and E are 0.5, 0.25, 0.125, 0.015625, and 0.00390625 respectively.

</pconsequently></pfor></pfor></ptherefore></pso></pfor></phence></pso></pfor>

Response:

Beryllium (Be): 9.01 g/mol

Silicon (Si): 28.09 g/mol

Calcium (Ca): 40.08 g/mol

Rhodium (Rh): 102.91 g/mol

Justification:

Answer:

4702.5 J/g*k

Explanation:

This value varies according to the state of water, whether solid, liquid, or gas. For my calculations, I treated it as a liquid: J=(75g)(4.18 J/g*k)(15K)