2. Which calorimeter lost energy?

What is the theoretical yield of vanadium, in moles, that can be produced by the reaction of 1.0 mole of V2O5 with 4.0 moles of

IRINA_888 [86]

Answer:

Theoretical yield of vanadium = 1.6 moles

Explanation:

Moles of $V_2O_5$ = 1.0 moles

Moles of $Ca$ = 4.0 moles

According to the given reaction:-

$V_2O_5_{(s)} + 5Ca_{(l)}\rightarrow 2V_{(l)} + 5CaO_{(s)}$

1 mole of $V_2O_5$ react with 5 moles of $Ca$

Moles of Ca available = 4.0 moles

Limiting reagent is the one which is present in small amount. Thus, Ca is limiting reagent. (4.0 < 5)

The formation of the product is governed by the limiting reagent. So,

5 moles of Ca on reaction forms 2 moles of V

1 mole of Ca on reaction for 2/5 mole of V

4.0 mole of Ca on reaction for $\frac{2}{5}\times 4$ mole of V

Moles of V = 1.6 moles

<u>Theoretical yield of vanadium = 1.6 moles</u>

4 0

2 years ago

Partner Namets) For each of the following reactions carried out: Write the balanced chemical equation, the full ionic equation,

torisob [31]

Answer : The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

Reactants are lead nitrate and potassium iodide.

Products are lead iodide and potassium nitrate.

The spectator ions are, $K^+,NO_3^-$

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When potassium iodide react with lead nitrate then it gives potassium nitrate and lead iodide as a product.

The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

The ionic equation in separated aqueous solution will be,

$2K^+(aq)+2I^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^{-}(aq)$

In this equation, $K^+\text{ and }NO_3^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Pb^{2+}(aq)+2I^{-}(aq)\rightarrow PbI_2(s)$

4 0

3 years ago

Where can find the neutrons in an atom

il63 [147K]

Answer:

neutrons are found in the nucleus of an atom.

Explanation:

6 0

2 years ago

Read 2 more answers

Why do airbags fill evenly with gases?

Firdavs [7]

Answer:

The signal from the deceleration sensor ignites the gas-generator mixture by an electrical impulse, creating the high-temperature condition necessary for NaN3 to decompose. The nitrogen gas that is generated then fills the airbag.

basically, the nitrogen fills the bag

8 0

2 years ago

Read 2 more answers

A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a

RoseWind [281]

Answer:

$C_3=0.125M$

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

$V_1C_1=V_2C_2$

So we solve for C2:

$C_2=\frac{C_1V_1}{V_2}=\frac{52.0mL*1.20M}{278mL}\\\\C_2=0.224M$

2. Now, since 111 mL of water is added, we compute the final volume, V3:

$V_3=139+111=250mL$

So, the final concentration of the 139 mL portion is:

$C_3=\frac{139 mL*0.224M}{250mL}\\\\C_3=0.125M$

Best regards!

8 0

2 years ago