Matter that has a definite shape and value

Identify whether each reaction represents a chemical reaction or a nuclear reaction by filling in the blank with the word ‘chemi

valkas [14]

A chemical reaction is defined as the reaction between two or more substances that produces new substances/compounds.

On the other hand, a nuclear reaction is a reaction that alters the nucleus of a certain atom. This can be done either by splitting the nucleus or by joining this nucleus with another one.

Reaction 1:
Cl₂ + H₂ ............> 2 HCl
This is the combination of two elements to produce a new compound.
Therefore, this reaction is a chemical reaction

Reaction 2:
²₁H + ³₁H .........> ⁴₂H + ¹₀n
This is an alteration in the nucleus of a hydrogen atom by bonding it with another nucleus.
Therefore, this reaction is a nuclear reaction

Hope this helps :)

5 0

3 years ago

Read 2 more answers

How many liters of H2O gas are produced when

anyanavicka [17]

1 mole C3H8 produces 4 moles H2O. So, first we convert 32 grams of propane to moles and then find moles of H2O. Then convert moles of H2O to grams of H2O
Moles of H2O produced = 32 g C3H8 x 1 mole/44 g x 4 moles H2O/mole C3H8 = 2.909 moles H2O
Grams H2O produced = 2.909 moles H2O x 18 g/mole = 52.36 g = 52 g H2O

8 0

2 years ago

Which response has both answers correct? Will a precipitate form when 250 mL of 0.33 M Na 2CrO 4 are added to 250 mL of 0.12 M A

olchik [2.2K]

Answer:

A precipitate will form.

[Ag⁺] = 2.8x10⁻⁵M

Explanation:

When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:

Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)

Ksp is defined as:

Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are in equilibrium

Reaction quotient, Q, is defined as:

Q = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are the actual concentrations

If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,

The actual concentrations are -Where 500mL is the total volume of the solution-:

[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M

[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M

And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴

As Q > Ksp; a precipitate will form

In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:

[Ag⁺] = 0.06M - 2X

[CrO₄²⁻] = 0.165M - X

Where X is defined as the reaction coordinate

Replacing in Ksp expression:

1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]

Solving for X:

X = 0.165M → False solution. Produce negative concentrations.

X = 0.0299986M

Replacing, equilibrium concentrations are:

[Ag⁺] = 0.06M - 2(0.0299986M)

[CrO₄²⁻] = 0.165M - 0.0299986M

[CrO₄²⁻] = 0.135M

6 0

3 years ago

The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)

cestrela7 [59]

Answer:

a. The limiting reactant is $NaHCO_{3}$

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

a)

You know the following reaction:

$3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}$ ⇒ $3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}$

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

Na: 23
H: 1
C: 12
O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

$NaHCO_{3}$ :23+1+12+16*3=84 g/mol
$H_{3} C_{6} HO_{7}$ :1*3+12*6+1*5+16*7= 192 g/mol
$CO_{2}$ :12+16*2= 44 g/mol
$H_{2} O$ :1*2+16= 18 g/mol
$Na_{3} C_{6} H_{5} O_{7}$ : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of $NaHCO_{3}$ react with 1 mole of $H_{3} C_{6} HO_{7}$ Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

$grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}$

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So the limiting reagent is sodium bicarbonate.

b)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of $NaHCO_{3}$ react with 3 mole of $CO_{2}$ . Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

$grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}$

grams of carbon dioxide= 0.73 g

Then, 0.73 g of carbon dioxide are formed.

c)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first. This means that citric acid will not react everything, leaving an excess.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

$grams of citric acid=\frac{1.4 g * 192 g}{252 g}$

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

So the grams of excess reactant that do not participate in the reaction are 0333 g.

3 0

3 years ago

Explain how you find the instantaneous speed or velocity of an object

puteri [66]

Answer:

divide the distance by the average

Explanation:

ash QC ok

3 0

3 years ago