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zimovet [89]
3 years ago
13

Water enters a 4.00-m3 tank at a rate of 6.33 kg/s and is withdrawn at a rate of 3.25 kg/s. The tank is initially half full.

Chemistry
1 answer:
olga55 [171]3 years ago
8 0

Answer:

At the start of the process, the volume not occupied by the water is 2 m3

Explanation:

At the start of the process you have a half full tank. It means that also a half is empty (not occupied by water).

Since the volume is 4 m3, 2 m3 are full (occupied by water) and 2 m3 (not occupied by water).

The volume in time will be

V(t)=V_0+(f_i-f_o)*t\\\\V(t) = 2 +(6.33/1000-3.25/1000)*t=2+0.00308*t \, \, [m3]

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How many grams of h2 will be produced if 175g of HCI are allowed to react completely with sodium
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Answer:

4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium

Explanation:

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:

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  • Na: 1 mole
  • NaCl: 2 moles
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You know the following masses of each element:

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So, the molar mass of each compound participating in the reaction is:

  • HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
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Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:

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  • Na: 1 mole* 23 g/mole= 23 g
  • NaCl: 2 moles* 58.45 g/mole= 116.9 g
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So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?

mass of H_{2} =\frac{175 g of HCl*2g ofH_{2} }{72.9 g of HCl}

mass of H₂= 4.8 g

<u><em>4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium</em></u>

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