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Alenkinab [10]
3 years ago
10

Gram formula for carbom tetrachloride

Chemistry
2 answers:
astraxan [27]3 years ago
4 0
CCl4

(1x12.011 g/mol)+(4x35.45 g/mol) = 153.811 g
suter [353]3 years ago
4 0
I agree with the answer as well
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What is the role of a catalyst in a chemical reaction?
Zinaida [17]

A catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy without itself being consumed by the reaction.

4 0
2 years ago
The chemical reaction in which two molecules combine to form a larger molecule, and water is released is called
Semmy [17]

Answer: condenstation.


Justification:


The polymerization by condensation is a well know chemical reaction in which two monomers ("small" molecules), each with (at least) two functional groups, combine and relase water as by-product. Actually, even if the by-product released is not water, yet the reaction is called condenstation, since the mechanism is basically the same.


An example of such reaction is the manufacturing of nylon 6,6, which is produced from adipic acid and 1,6-diamine hexane:


HOOC - [CH₂]₄ - COOH + nH₂N - [CH₂]₆ - NH₂ → - nylon - + nH₂O


I omitted the formula of nylon because it is large, and that is not the core of the question but the fact the kind of reaction: two molecules combine to form is a larger molecule, and water is released

7 0
3 years ago
Read 2 more answers
For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.
Nikolay [14]

Explanation:

Molar mass of KClO_{3} = 39.1 + 35.5 + 3(16.0) = 122.6 g

Molar mass of KCl = 39.1 + 35.5 = 74.6 g

Molar mass of O_{2} = 32.0 g

According to the equation, 2 moles of KClO_{3} reacts to give 3 moles of oxygen.

Therefore, 2 (122.6) = 245.2 g of KClO_{3} will give 3 (32.0) = 96.0 g of oxygen. Thus, 245.2 g of KClO_{3} gives 96.0 g of oxygen.

(a) Calculate the amount of oxygen given by 2.72 g of KClO_{3} as follows.

       \frac {96g}{245.2g} \times 2.72 g = 1.06 g of O_{2}


(b) Calculate the amount of oxygen given by 0.361 g of KClO_{3} as follows.

    \frac {96g}{245.2g} \times 0.361 g = 0.141 g of O_{2}


c) Calculate the amount of oxygen given by 83.6 kg KClO_{3} as follows.  

    \frac {96g}{245.2g} \times 83.6 g = 32.731 kg of O_{2}

Convert kg into grams as follows.

    \frac{32.731 kg}{1 kg} \times 1000 g = 32731 g of O_{2}


(d) Calculate the amount of oxygen given by 22.5 mg of KClO_{3} as follows.  

    \frac {96g}{245.2g} \times 22.5 mg = 8.79 mg

Convert mg into grams as follows.

  \frac{8.79 mg}{1 mg}\times 10^{-3} g = 8.79 \times 10^{-3} g of O_{2}

8 0
3 years ago
Argon-40 undergoes positron emission as shown:
Rudik [331]

Answer:

\frac{40}{18}Ar => \frac{40}{17}Cl + \frac{0}{1}e\\

Explanation:

3 0
4 years ago
N2O5 decomposes to form NO2 and O2 with first-order kinetics. The initial concentration of N2O5 is 3.0 M and the reaction runs f
kow [346]

Answer : The final concentration of N_2O_5 is, 2.9 M

Explanation :

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 5.89\times 10^{-3}\text{ min}^{-1}

t = time passed by the sample  = 3.5 min

a = initial concentration of the reactant  = 3.0 M

a - x = concentration left after decay process = ?

Now put all the given values in above equation, we get

3.5=\frac{2.303}{5.89\times 10^{-3}}\log\frac{3.0}{a-x}

a-x=2.9M

Thus, the final concentration of N_2O_5 is, 2.9 M

3 0
3 years ago
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