A student checks the air in her bicycle tires early in the morning when it is cool outside. If she measures it again later in th

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To achieve the cancellation of electrons, the oxidation half-reaction needs to be multiplied by 4 while the reduction half-reaction must be multiplied by 3. Explanation: The oxidation reaction accounts for the loss of electrons, increasing the oxidation state, while the reduction implies gaining electrons, leading to a decrease in oxidation state. The respective half-reactions illustrate this, confirming that multiplying the oxidation by 4 and the reduction by 3 achieves the desired effect.

Result:

I believe it’s called Trinitrogen Pentaseleniumide

Explanation:

Tri means three

Penta means five

The second element concludes with -ide

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.

To tackle this problem, one must first determine the specific heat of water, which is the energy required to raise the temperature of 1 g of water by 1 degree C. The relationship is given by the formula q = c X m X delta T, where q indicates the specific heat of water, m signifies the mass, and delta T denotes the temperature change. The specific heat of water is 4.184 J/(g X degree C). The temperature of the water increased by 20 degrees, therefore: 4.184 x 713 x 20.0 = 59700 J, rounded to 3 significant digits, equals 59.7 kJ. This value indicates the energy required to produce B2O3 from 1 gram of boron. To convert this to kJ/mole, additional calculations are required. The gram atomic mass of Boron is 10.811, so dividing 1 gram of boron by 10.811 results in.0925 moles of boron. Given that 2 moles of boron are needed for the formation of 1 mole of B2O3, dividing the moles of boron by two yields.0925/2 =.0462 moles. Consequently, dividing the energy in KJ by the number of moles provides KJ/mole: 59.7/.0462 = 1290 KJ/mole.