# Deres utrum wotassium common ials solubilj gunds in water common exceptions contain none ionics barium calcium sodium magnesium

###### Question:

utrum

wotassium

common

ials

solubilj

gunds in water

common exceptions

contain

none

ionics

barium

calcium

sodium

magnesium

aluminum

manganese

zinc

chromium

none

ne

office depot

19 activity

4. how many moles of iron (1) oxide, feo, will be produced if two moles of iron are reacted with

an excess of oxygen?

2fe + 02 - 2 feo mule

a. 1 mole of feo

b. 2 moles of feo

c. 3 moles of feo

d. 4 moles of feo

## Answers

Answer is in a photo. I can't attach it here, but I uploaded it to a file hosting. link below! Good Luck!

tinylnk.cf/8oKd

Explanation:

2. First we have to calculate the heat gained by the solution.

[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]

where,

q = heat gained = ?

c = specific heat of solution = [tex]5.20kJ/^oC[/tex]

m = Mass of the solution = 100.0 g

[tex]T_{final}[/tex] = final temperature

[tex]T_{initial}[/tex] = initial temperature

ΔT = [tex]T_{final}-T_{initial}[/tex] = 5.6°C

Now put all the given values in the above formula, we get:

[tex]q_{solution}=100.0 g\times 4.18 J/^oC\times (5.6)^oC[/tex]

[tex]q_{solution}=2,340.8 J[/tex]

The amount of heat energy gained by the solution 2,340.8 J.

3.

Heat gained by solution = Heat released on neutralization reaction

[tex]q_{solution}=q_{recation}=2,340.8 J[/tex]

2,340.8 Joules is the value [tex]q_{reaction }[/tex] of reaction for the neutralization reaction described in number 2.

4. [tex]Molarity=\frac{Moles}{Volume(L)}[/tex]

Moles of phosphoric acid = n

Volume of the phosphoric acid solution = 50.0 mL = 0.050 L

Molarity of the phosphoric acid solution = 0.60 M

[tex]n=0.60 M\times 0.050 L= 0.03 mol[/tex]

0.03 moles of phosphoric acid are contained in 50.0 mL of 0.60 M phosphoric acid solution.

5. To calculate the enthalpy change during the reaction.

[tex]\Delta H=-\frac{q}{n}[/tex]

where,

n = moles of phosphoric acid = 0.03 moles

[tex]\Delta H[/tex] = enthalpy change = ?

q = heat gained = 2,340.8 J = 2.3408 kJ

[tex]\Delta H=-\frac{ 2.3408 kJ}{0.03 mol}=-78.027 kJ/mole[/tex]

Therefore, the enthalpy change during the reaction is -78.027 kJ/mole

honeslty, i need help with this subject too. Can someone pls share their answer and the explanation? thx

Explanation:

I uрlоаdеd thе аnswеr tо а filе hоsting. Hеrе's link:

xtiny.cf/H5ct