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HACTEHA [7]
3 years ago
9

In the Energy and Specific Heat lab, how should the water bath be stabilized over the heat source? Select one:

Chemistry
2 answers:
Reika [66]3 years ago
5 0

Answer:

c

Explanation:

butalik [34]3 years ago
3 0

Answer:

C

Explanation:

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Consider the equation below. Upper C a upper C upper O subscript 3 (s) double-headed arrow upper C a upper O (s) plus upper C up
stiks02 [169]

Answer:K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction

Explanation: the answer has it's root in Law of mass action which states that; the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants raised to their respective stoichiometric coefficients.

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3 years ago
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Is backbonding is possible in BF4- ? If no then why?
Rudik [331]

Answer:

No, because Flourine can only form 1 bond, thus backbonding is not obtainable

4 0
2 years ago
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
Even though the neighborhood shown by the green star was not next to any of these fires, how was the neighborhood affected by th
iogann1982 [59]
Im pretty sure the answers b

7 0
3 years ago
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Does an astronaut have more mass on earth then space? why or why not?
katovenus [111]

Answer:

No

Explanation:

No, his mass remains the same no matter where he is in the universe.

But then again the moon has less gravitational pull, therefore your weight and mass will be smaller in space and on the moon than on earth

I hope this was helpful! ;)

3 0
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