Each can hold two electrons
Since the addition of the H2O in the last step of hydroboration is anti-Markovnikov, the starting material is 1-pentyne.
The addition of H2 to C5H8 yields an alkene when a Lindlar catalyst is used. Recall that the Lindlar catalysts poisons the process so that the addition do not go on to produce an alkane.
When hydroboration is carried out on the alkene, we are told that a primary alcohol was obtained. We must note that in the last step of hydroboration, water is added in an anti- Markovnikov manner to yield the primary alcohol. Hence, the starting material must be 1-pentyne as shown in the image attached.
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Answer:- 1337 g
Solution:- For mathematical operations we use PEMDAS. P stands for parenthesis off, E stands for exponents, M stands for multiplication, D stands for division, A stands for addition and S stands for subtraction. All the mathematical operations are done in this sequence.
What we have given in the problem is, 1210.0 g + 157.22 g - 30. g
Here we have addition and subtraction. As per the law, we will do addition first. In addition and subtraction we go with least number of decimal places.
1210.0 has one decimal place where as 157.22 has two decimal places, since the first number has least decimal places(one decimal place), the answer should also have only one decimal place.
1210.0 g + 157.22 g = 1367.2 g
Now, we will do subtraction. 30. does not have any decimal place and so on subtraction the final answer must not have any decimal places.
1367.2 g - 30. g = 1337 g
So, the final answer is 1337 g.
Mg: [Ne] 3s2. The second ionization energy of Mg is larger than the first because it always takes more energy to remove an electron from a positively charged ion than from a neutral atom. The third ionization energy of magnesium is enormous, however, because the Mg2+ ion has a filled-shell electron configuration.
Numa máquina térmica uma parte da energia térmica fornecida ao sistema(Q1) é transformada em trabalho mecânico (τ) e o restante (Q2) é dissipado, perdido para o ambiente.
sendo:
τ: trabalho realizado (J) [Joule]
Q1: energia fornecida (J)
Q2: energia dissipada (J)
temos: τ = Q1 - Q2
O rendimento (η) é a razão do trabalho realizado pela energia fornecida:
η= τ/Q1
Exercícior resolvido:
Uma máquina térmica cíclica recebe 5000 J de calor de uma fonte quente e realiza trabalho de 3500 J. Calcule o rendimento dessa máquina térmica.
solução:
τ=3500 J
Q1=5000J
η= τ/Q1
η= 3500/5000
η= 0,7 ou seja 70%
Energia dissipada será:
τ = Q1 - Q2
Q2 = Q1- τ
Q2=5000-3500
Q2= 1500 J
Exercicio: Qual seria o rendimento se a máquina do exercicio anterior realizasse 4000J de trabalho com a mesma quantidade de calor fornecida ? Quanta energia seria dissipada agora?
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