Write a complete reaction of succinic acid and acetic anhydride

Which description applies to a physical property? Choose the correct answer

choli [55]

Answer:

the third one, measured or observed without changing the identity and composition of matter. because physical property does not under go any change but can be put back.

6 0

3 years ago

Can someone help me with problem 21.105? Many thanks!

Reika [66]

The correct answer is 1 to the 3rd power

3 0

2 years ago

A chemist measures the enthalpy change ?H during the following reaction: Fe (s) + 2HCl (g) ? FeCl2 (s) + H2 (g) =?H?157.kJ Use t

erik [133]

Correct Question:

A chemist measures the enthalpy change ΔH during the following reaction: Fe(s) + 2HCl(g)-->FeCl2(s) + H2 ΔH=-157.0 kJ. Use this information to complete the table below. Round each of your answers to the nearest kJ/mol

Answer:

-314 kJ

+628 kJ

+157 kJ

Explanation:

The enthalpy change of a reaction measures the amount of heat that is lost or gained by it. If ΔH >0 the heat is gained, and the reaction is called endothermic, if ΔH<0, the heat is lost, and the reaction is called exothermic.

If the reaction is inverted, the value of ΔH is inverted too (the opposite endothermic reaction is exothermic), and if the reaction is multiplied by a constant, ΔH will be multiplied by it too.

1) 2Fe(s) + 4HCl --> 2FeCl2(s) + 2H2(g)

This reaction is the product of the given reaction by 2, so

ΔH = 2*(-157) = -314 kJ

2) 4FeCl2(s) + 4H2(g) --> 4Fe(s) + 8HCl(g)

This reaction is the inverted reaction given multiplied by 4, so

ΔH = 4*(157) = +628 kJ

3) FeCl2(s) + H2(g) --> Fe(s) + 2HCl

This reaction is the inverted reaction given, so

ΔH = +157 kJ

4 0

3 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

3 years ago

How many miles of C are in 32.6 g of C2H6

sammy [17]

32.6 grams divided by the molar mass of C2H6, which is 18.0584g/mol = 1.8 moles of C2H6.

As there are two carbon atoms per C2H6, we must multiply the number of moles of C2H6 by 2 to get the number of moles of Carbon which is 3.6 moles.
The answer is 3.6 moles.

Hope this helps.

(Sorry for previously incorrect answer)

5 0

3 years ago