How many Cl- bond with one Na+ ion?

You have a gas stored in a rigid container (fixed volume). Explain the relationship between a decrease in temperature, collision

alexandr1967 [171]

Answer:

A decrease in temperature would decrease kinetic energy, therefore decreasing collisions possible.

Explanation:

A gas at a fixed volume is going to have collisions automatically. If you decrease the temperature (same thing as decreasing kinetic energy) you are cooling down the molecules in the container which gives them less energy and "relaxes" them. This decrease in energy causes them to move around much slower and causing less collisions, at a much slower rate. In a perfect world, these collisions do not slow down the molecule but we know that they do, just a very very small unmeasurable amount.

6 0

3 years ago

The product of the nuclear reaction in which 31p is subjected to neutron capture followed by alpha emission is ________.

faltersainse [42]

The product of the nuclear reaction in which 31p is subjected to neutron capture followed by alpha emission is ²⁸Al.
Nuclear reaction: ³¹P + n° → ²⁸Al + α (alpha particle).
Alpha decay is radioactive decay in which an atomic nucleus emits an alpha particle (helium nucleus) and transforms into an atom with an atomic number that is reduced by two and mass number that is reduced by four.

6 0

3 years ago

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For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car

Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

Na₂CO₃: 1 mole
Ca(NO₃)₂: 1 mole
CaCO₃: 1 mole
NaNO₃: 2 mole

Being the molar mass of the compounds:

Na₂CO₃: 106 g/mole
Ca(NO₃)₂: 164 g/mole
CaCO₃: 100 g/mole
NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

Na₂CO₃: 1 mole* 106 g/mole= 106 g
Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
CaCO₃: 1 mole* 100 g/mole= 100 g
NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of CaCO₃?

$mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}$