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Paraphin [41]
3 years ago
10

How many days is 1.6 x 10^7 minutes?

Chemistry
1 answer:
Over [174]3 years ago
8 0

(1.6 \times  {10}^{7}) \: min \times  \frac{1 \: hour}{60 \: min} \times  \frac{1 \: day}{24 \: hour}

= 11111.11111 \: days

(round as you wish)

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What volume of chlorine gas at 27 °C, 812 mmHg, is required to react with an excess of carbon disulfide so that 5.00kJ of heat i
Kamila [148]

Answer:

The correct answer is 1.21 L.

Explanation:

Based on the given information, the reaction will be,

CS2 (l) + 3Cl2 (g) ⇒ CCl4 (l) + S2Cl2 (l)

By using the standard values of the substances, the standard enthalpy of the reaction is,

ΔH° = [(-139.5) + (-58.5) – 0 – (87.3)] kJ/mol

= -285.3 kJ/mol

The amount of heat evolved for 3 moles of chlorine reacted us 285.3 kJ.

Now the number of moles of chlorine needed to react to produce 5.00 kJ is,

= 5.00 kJ × 3 mol Cl2/285.3 kJ

= 0.0526 mol Cl2

Now the volume of chlorine gas at 27degree C and 812 mmHg will be,

Volume = 0.0526 mol Cl2 × 0.0821 Latm/mol K × 300 K/ 1.07 atm

= 1.21 L

4 0
3 years ago
Describe the redox reaction. Check all that apply 2AI(s) + 3ZnCI2(g) → 3Zn(s) + 2AICI3(aq).
Inga [223]

Answer:

Zinc is reduced.

The oxidation number of chlorine does not change.

Aluminum is oxidized.

Explanation:

  • For the reaction:

<em>2AI(s) + 3ZnCI₂(g) → 3Zn(s) + 2AICI₃(aq). </em>

<em></em>

  • For the elements in the reactants side:

Al has (0) oxidation state and converted to (+3) oxidation state in the products (AlCl₃), which means Al is oxidized and it is the reducing agent.

So, we can check that: Aluminum is oxidized.

Zn has the oxidation state (+2) in the reactants side (ZnCl₂) and converted to (0) in the products side (Zn), which means that Zn is reduced and it is the oxidizing agent.

So, we can check that: Zinc is reduced.

<em>and can not check: Zinc is the reducing agent.  </em>

<em />

The oxidation state of Cl does not change, it is the same in both sides (-1).

So, we can check: The oxidation number of chlorine does not change.

<em>and can not check: Aluminum atoms transfer electrons to chlorine atoms. </em>

<em />

  • So, the choices will be:

<em>Zinc is reduced.</em>

<em>The oxidation number of chlorine does not change.</em>

<em>Aluminum is oxidized. </em>

4 0
4 years ago
Read 2 more answers
For scientific notation, when i move the decimal to the left the the exponent get bigger or smaller?
xeze [42]
I'm not sure what you mean. Besides, I feel like you're talking math.
But anyways, if you have 120 let's say
The scientific notation is 1.20 × 10^2
if you have 125000
the scientific notation is 1.25 × 10^ 5
The number of times you go left the decimal, I guess exponent increases
So yea
3 0
3 years ago
Can u pls help me with this question ​
ziro4ka [17]
I think it is A as solid atoms are close and gases are far away
8 0
3 years ago
Copper oxide, CuO, reacts with hydrochloric acid, HCI, to produce copper chloride, CuCL2 and water
spayn [35]

Explanation:

El óxido de cobre (II), también llamado antiguamente óxido cúprico ({\displaystyle {\ce {CuO}}}{\displaystyle {\ce {CuO}}}), es el óxido de cobre con mayor número de oxidación. Como mineral se conoce como tenorita.

{\displaystyle {\ce {2Cu + O2 = 2CuO}}}{\displaystyle {\ce {2Cu + O2 = 2CuO}}}

Aquí, se forma junto con algo de óxido de cobre (I) como un producto lateral, por lo que es mejor prepararlo por calentamiento de nitrato de cobre (II), hidróxido de cobre (II) o carbonato de cobre (II):

{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}

{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}

{\displaystyle {\ce {CuCO3 = CuO + CO2}}}{\displaystyle {\ce {CuCO3 = CuO + CO2}}}

El óxido de cobre (II) es un óxido básico, así se disuelve en ácidos minerales tales como el ácido clorhídrico, el ácido sulfúrico o el ácido nítrico para dar las correspondientes sales de cobre (II):

{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}

{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}

{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}

Reacciona con álcali concentrado para formar las correspondientes sales cuprato.

{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}

Puede reducirse a cobre metálico usando hidrógeno o monóxido de carbono:

{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}

{\displaystyle {\ce {CuO + CO = Cu + CO2}}}{\displaystyle {\ce {CuO + CO = Cu + CO2}}}

6 0
3 years ago
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