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Monica [59]
3 years ago
6

? + HCl -> FeCl2 + ? CaO + ? -> Ca(NO3)2 + ? ? + Cu(OH)2 -> CuCl2 + ?

Chemistry
2 answers:
koban [17]3 years ago
7 0

Answer:

1. Fe / H2

2. HNO3 / H2O

3. HCI / H2O

Oksana_A [137]3 years ago
4 0

Answer:

1. Fe / H2

2. HNO3 / H2O

3. HCI / H2O

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Which of the following statements explains which trial has a lower concentration of the reactant? (5 points) Trial 1, because th
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A)Trial 1 because the average rate of reaction is lower.

Explanation:

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The process in which an organic acid and an alcohol react to form an ester and water is known as esterification. Ethyl butanoate
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Answer:

697 g

Explanation:

Ethanol (C₂H₅OH) and butanoic acid (C₃H₇COOH) react to form ethyl butanoate (C₃H₇COOC₂H₅) and water (H₂O).

C₂H₅OH + C₃H₇COOH → C₃H₇COOC₂H₅ + H₂O

The molar ratio of C₂H₅OH to C₃H₇COOC₂H₅ is 1:1. The moles of C₃H₇COOC₂H₅ produced from 6.00 moles of C₂H₅OH are:

6.00 mol C₂H₅OH × (1 mol C₃H₇COOC₂H₅/1 mol C₂H₅OH) = 6.00 mol C₃H₇COOC₂H₅

The molar mass of C₃H₇COOC₂H₅ is 116.16 g/mol. The mass corresponding to 6.00 mol is:

6.00 mol × (116.16 g/mol) = 697 g

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3 years ago
How do radiography and sonography compare?
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Radiography techs are trained to operate medical equipment that uses X-rays and other types of radiation. Ultrasound techs, referred to professionally as diagnostic medical sonographers, operate equipment that uses high-frequency sound waves to create images.

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What common item can detect the presence of radiation? A. Bleach B. A roll of photographic film C. LED watch D. Baking soda
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Instead of using ratios for back titrations we can also use molarities, if our solutions are standardized. A 0.196 g sample of a
kvv77 [185]

Answer:

The mass percent of Al(OH)₃ is 15.3%

Explanation:

The reaction is:

Al(OH)₃ + 3HCl = AlCl₃ + 3H₂O

The excess acid is neutralized with a solution of sodium hidroxide, in the reaction:

NaOH + HCl = NaCl + H₂O

The total moles of HCl is:

n_{HCl,total} =M_{HCl} *V_{HCl} =0.111*0.025=2.78x10^{-3} moles

From the second titration, the moles of excess of HCl is:

n_{HCl,excess} =n_{NaOH} =M_{NaOH} *V_{NaOH} =0.132*0.01105=1.46x10^{-3} moles

The difference between the total and excess of HCl, it can be know the moles that reacts with the aluminum hydroxide, is:

n_{HCl,reacts} =n_{HCl,total}-n_{HCl,excess} =2.78x10^{-3} moles-1.46x10^{-3} moles=1.32x10^{-3} moles

The ratio between HCl and Al(OH)₃ is 3:1. The MW for aluminum hydroxide is 78 g/mol, thus:

m_{Al(OH)3} =1.32x10^{-3} molesHCl*\frac{1molAl(OH)3}{3molesHCl} *\frac{78gAl(OH)3}{1molAl(OH)3} =0.03g

The percentage of Al(OH)₃ is:

Percentage-Al(OH)3=\frac{m_{Al(OH)3} }{m_{antiacid} } *100=\frac{0.03}{0.196} =15.3%

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3 years ago
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