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son4ous [18]
3 years ago
11

PLZ GIVE ME ANSWERS WITH WORK I WILL GIvE YOU A BRain LIST PLZZZZZ

Chemistry
1 answer:
eduard3 years ago
5 0

Answer:

i only found the 49, hope it still helps, mate

49. In this question (t½) of C-14 is 5730 years, which means that after 5730 years half of the sample would have decayed and half would be left as it is.

After 5730 years ( first half life) 70 /2 = 35 mg decays and 35 g remains left.

After another 5730 years ( two half lives or 11460 years) 35 /2 = 17.5mg decays and 17.5 g remains left .

After another 5730 years ( three half lives or 17190 years) 17.5 /2 = 8.75mg decays and 8.75g remains left.

after three half lives or 17190 years, 8.75 g of C-14 will be left.

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Explanation:

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Assign the oxidation state for all the atoms:<br><br>As2O3<br><br>Show work
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<u>Answer:</u> Oxidation state of Arsenic in the given compound is +3 and that of oxygen is -2.

<u>Explanation:</u>

We are given a compound having formula As_2O_3 and we need to find the oxidation state of all the atoms in this compound. There are 2 different atoms present in this : Arsenic and Oxygen

Oxidation state is the number which is assigned to an element in a compound which represents the number of electrons lost or gained by an atom in a compound.

Oxidation state of oxygen as -2 which is same always. Now, to calculate the oxidation state of Arsenic, we take the oxidation state of it be 'x'

As, there are two arsenic atoms and three oxygen atoms, then:

\Rightarrow 2(x)+3(-2)=0\\\Rightarrow 2x-6=0\\\Rightarrow x=+3

Hence, oxidation state of Arsenic in the given compound is +3 and that of oxygen is -2.

3 0
4 years ago
5. Fill in the boxes below with solid, liquid, or gas.
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Answer:

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Explanation:

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when copper metal is heated it reacts with a gas in the air . what iz the name of the product formed when copper reacts with a g
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Korolek [52]
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
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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?



obs: Entregar foto da resolução ou o cálculo passo a passo na mensagem
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