All categories

2 years ago

Write the correct balanced equation for

1 answer:

7 0

The given reaction is between solid iron and aqueous copper (II) bromide to form solid copper and aqueous iron (II) bromide. Iron being more reactive than copper can displace copper ion from its aqueous solution. Therefore, this reaction is an example of single replacement reaction in which Iron displaces copper from an aqueous solution of copper (II) bromide.

Word equation :

Iron (solid)+Copper(II) bromide(aqueous)--->Copper (solid)+Iron (II) bromide (aqueous)

Chemical equation:

$Fe(s) +CuBr_{2}(aq)-->Cu(s)+FeBr_{2}(aq)$

You might be interested in

Help me, I don’t understand

marysya [2.9K]

Answer:

option c

Explanation:

ffuunkie. tfnhc dhrhrjebrjeieie

7 0

3 years ago

Read 2 more answers

Balanced chemical equations show:

lara31 [8.8K]

the products formed from the reaction

all of the above

4 0

3 years ago

The equilibrium constant for the dissolution of silver chloride (AgCl(s) Ag+(aq) + Cl–(aq)) has a value of 1.79 × 10–10. Which s

ryzh [129]

"Silver chloride is essentially insoluble in water" this statement is true for the equilibrium constant for the dissolution of silver chloride.

Option: b

<u>Explanation</u>:

As silver chloride is essentially insoluble in water but also show sparing solubility, its reason is explained through Fajan's rule. Therefore when AgCl added in water, equilibrium take place between undissolved and dissolved ions. While solubility product constant $\left(\boldsymbol{K}_{s p}\right)$ for silver chloride is determined by equilibrium concentrations of dissolved ions. But solubility may vary also at different temperatures. Complete solubility is possible in ammonia solution as it form stable complex as water is not good ligand for Ag+.

To calculate $\left(\boldsymbol{K}_{s p}\right)$ firstly molarity of ions are needed to be found with formula: $\text { Molarity of ions }=\frac{\text { number of moles of solute }}{\text { Volume of solution in litres }}$

Then at equilibrium cations and anions concentration is considered same hence:

$\left[\mathbf{A} \mathbf{g}^{+}\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }$

Hence from above data $\left(\boldsymbol{K}_{s p}\right)$ can be calculated by: $\left(\boldsymbol{K}_{s p}\right)$ = $\left[\mathbf{A} \mathbf{g}^{+}\right] \cdot[\mathbf{C} \mathbf{I}]$

6 0

3 years ago

A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The

V125BC [204]

Taking into account the definition of calorimetry, the specific heat of metal is 0.165 $\frac{cal}{gC}$ .

<h3>Definition of calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

Q is the heat exchanged by a body of mass m.
C is the specific heat substance.
ΔT is the temperature variation.

<h3>Specific heat capacity of the metal</h3>

In this case, you know:

For metal:

Mass of metal = 50 g
Initial temperature of metal= 45 °C
Final temperature of metal= 11.08 ºC
Specific heat of metal= ?

For water:

Mass of water = 250 g
Initial temperature of water= 10 ºC
Final temperature of water= 11.08 ºC
Specific heat of water = 1.035 $\frac{cal}{gC}$

Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater= 1.035 $\frac{cal}{gC}$ × 250 g× (11.08 C - 10 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 $\frac{cal}{gC}$ × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 $\frac{cal}{gC}$ × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= $\frac{279.45 cal}{1696 gC}$