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One mole of C= 12 grams
two moles of O =32
so one mole of CO^2 is44 grams
.1 moles or 1/10 moles of 44 grams is 4.4 grams
D. you are not making any motion
Answer:
75603.86473 K
Explanation:
Given that:
The 1st excited electronic energy level of He atom = 3.13 × 10⁻¹⁸ J
The objective of this question is to estimate the temperature at which the ratio of the population will be 5.0 between the first excited state to the ground state.
The formula for estimating the ratio of population in 1st excited state to the ground state can be computed as:

From the above equation:
Δ E = energy difference = 3.13 × 10⁻¹⁸ J
k = Boltzmann constant = 1.38 × 10⁻²³ J/K

Thus:





T = 75603.86473 K
Answer:
.926 moles
Explanation:
Rounding :
H2 0 = 18 gm/mole
50 gm would then be 50 / 18 = 2.7777 moles of water
every two moles of 2 C6H6 produces 6 moles of water
2.7777/6 * 2 = .926 moles