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tamaranim1 [39]
3 years ago
13

When a reaction results in an almost complete conversion of reactants to products, chemists say that the reaction goes to comple

tion?
Chemistry
1 answer:
Agata [3.3K]3 years ago
5 0
Yes, that is completion reaction.
<span>This type of reactions do not set up an equilibrium and goes until one of the reactants runs out.
</span>Three major types of completion reactions are:
1. <span>Formation of a slightly ionized product (for example mixture of sodium bromide and potassium chloride).
</span>2. <span>Formation of a gas (for example hydrogen gas from metal and acid).
3.</span> Formation of a precipitate (feor example silver chloride).<span>
</span>
You might be interested in
Which statement about chemical reactions is not true?
igomit [66]

Answer:

1: New atoms are formed as products

Explanation:

matter (atoms) cannot be created or destroyed

unless you are God or Cinderella's fairy Godmother or something

7 0
3 years ago
How many moles of a gas would occupy 22.4 Liters at 273 K and 1 atm?
Molodets [167]

Answer:

1 mole of a gas would occupy 22.4 Liters at 273 K and 1 atm

Explanation:

An ideal gas is a set of atoms or molecules that move freely without interactions. The pressure exerted by the gas is due to the collisions of the molecules with the walls of the container. The ideal gas behavior is at low pressures, that is, at the limit of zero density. At high pressures the molecules interact and intermolecular forces cause the gas to deviate from ideality.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

  • P= 1 atm
  • V= 22.4 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T=273 K

Reemplacing:

1 atm* 22.4 L= n* 0.082 \frac{atm*L}{mol*K} *273 K

Solving:

n=\frac{1 atm* 22.4 L}{0.082 \frac{atm*L}{mol*K} *273 K}

n= 1 mol

Another way to get the same result is by taking the STP conditions into account.

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C (or 273 K) are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<u><em>1 mole of a gas would occupy 22.4 Liters at 273 K and 1 atm</em></u>

4 0
3 years ago
Who discovered sodium?
Free_Kalibri [48]
Humphry Davy First discovered sodium


5 0
3 years ago
Read 2 more answers
Part iii. Is the dissolution reaction of ammonium nitrate enthalpy favored?
lesya [120]

Answer:

No, it is not enthalpy favored since the chemical system gains energy.

Explanation:

The dissolution of ammonium nitrate in water is an endothermic process.

Endothermic process requires the system to gain energy to can dissolve the particles in water.

So, the reaction is not enthalpy favored.

3 0
3 years ago
(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction
ale4655 [162]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2}   \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu}   \\\\m_{Cu}=0.380gCu

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3}   \\\\m_{AgNO_3}=2.03gAgNO_3

Best regards!

5 0
3 years ago
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