What is the name for temperature changes which involve no heat transfer?

Chemistry

2 answers:

Nina [5.8K]3 years ago

5 0

Adiabatic.

In thermodynamic chemistry, an adiabatic process is defined as the process in which there is no transfer of heat and mass takes place between system and surroundings. The energy will transfer only in the form of work. The condition is expressed as q=0, which means no heat is transferred. It has various applications in the study of thermodynamics reactions.

Ganezh [65]3 years ago

4 0

The name for temperature changes which involve no heat transfer is Adiabatic. Adiabatic is a greek word, that means, “unable to cross”.And it means change in temperature where no heat is transferred.

Adiatic process in thermodynamics means , change in q=o, that is no heat is transferred.

So the answer here is adiabatic that is The name for temperature changes which involve no heat transfer is Adiabatic.

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A certain isotope has 53 protons 78 neutrons and 53 electrons. What is its atomic number? What is the mass number of this atom?

statuscvo [17]

Answer:

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3 years ago

If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W

olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

$CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2$

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

$m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl$

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

$m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3$