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DedPeter [7]
3 years ago
8

What happened as the catalyst CuO was added to the thermal decomposition of potassium chlorate?

Chemistry
1 answer:
Ber [7]3 years ago
8 0
"O2 was produced at a faster rate.(C)" 
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It is desired to produce 2.25 grams of dichloromethane (CH2Cl2) by the following reaction. If the percent yield of dichlorometha
Allushta [10]

Answer:- 3.12 g carbon tetrachloride are needed.

Solution:- The balanced equation is:

CH_4+CCl_4\rightarrow 2CH_2Cl_2

From given actual yield and percent yield we will calculate the theoretical yield that would be further used to calculate the grams of carbon tetrachloride.

percent yield formula is:

percent yield = (\frac{actual}{theoretical})100

65.5=(\frac{2.25}{theoretical})100

theoretical=(\frac{2.25(100)}{65.5})

theoretical = 3.44 g

From balanced equation, there is 2:1 mol ratio between dichloethane and carbon tetrachloride.

Molar mass of dichloroethane is 84.93 gram per mol and molar mass of carbon tetrachloride is 153.82 gram per mol.

3.44gCH_2Cl_2(\frac{1molCH_2Cl_2}{84.93gCH_2Cl_2})(\frac{1molCCl_4}{2molCH_2Cl_2})(\frac{153.82gCCl_4}{1molCCl_4})

= 3.12gCCl_4

So, 3.12 grams of carbon tetrachloride are needed to be reacted.

8 0
3 years ago
How much heat is needed warm 65.34 g of water from 18.43 to 21.75
BaLLatris [955]

Given that  

Mass of water = 65.34 g

Amount of heat = mass of water * specific heat (temperature change

)

= 65.34 g * 4.184 J / g-C ( 21.75-18.43 )C

= 907.63  J  

= 0.908 KJ

And  

1 cal = 4.186798 J

907.63 J * 1 cal / 4.186798 J =216.78  cal

Or0.218 kcal  


5 0
3 years ago
The equilibrium constant K changes with changes in<br>the temperature.​
chubhunter [2.5K]
<h3>Answer:</h3><h2>Equilibrium constants are changed if you change the temperature of the system. Kc or Kp are constant at constant temperature, but they vary as the temperature changes. You can see that as the temperature increases, the value of Kp falls.</h2>

4 0
2 years ago
What is a ribosome?
Cloud [144]
A good lucklolololol
3 0
2 years ago
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How much nano3 is needed to prepare 225 ml of a 1.55 m solution of nano3?
andrezito [222]

Answer:

= 29.64 g  NaNO3

Explanation:

Molarity is given by the formula;

Molarity = Moles/Volume in liters

Therefore;

Number of moles = Molarity × Volume in liters

                             = 1.55 M × 0.225 L

                             = 0.34875 moles NaNO3

Thus; 0.34875 moles of NaNO3 is needed equivalent to;

   = 0.34875 moles × 84.99 g/mol

   = 29.64 g

5 0
3 years ago
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