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kenny6666 [7]
3 years ago
6

HOW MANY GRAMS OF NAOH ARE NEEDED TO MAKE 500ML OF 2.5M NAOH

Chemistry
1 answer:
azamat3 years ago
3 0
Mass = volume  ×  concentration

Mass of NaOH needed  =  500 mL  ×  2.5 M
                                       =  1250 g
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Using the standard enthalpies of formation found in the textbook, determine the enthalpy change for the combustion of ethanol c2
ArbitrLikvidat [17]
Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)

Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol

Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol

<em>Answer: -1367.5 kJ/mol</em>
6 0
3 years ago
Read 2 more answers
What is the percent by mass of oxygen fe2o3 (formula mass= 160)?
GarryVolchara [31]
30% should be the percentage of oxygen if the total mass of fe2o3 is 160.
5 0
3 years ago
A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o
Andreas93 [3]
<h3>Answer:</h3>

2.809 L of H₂SO₄

<h3>Explanation:</h3>

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

<h3>First: We need to write the balanced equation for the reaction.</h3>
  • The reaction between NaOH and H₂SO₄ is a neutralization reaction.
  • The balanced equation for the reaction is;

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

<h3>Second: We calculate the umber of moles of NaOH used </h3>
  • Number of moles = Mass ÷ Molar mass
  • Molar mass of NaOH is 40.0 g/mol
  • Therefore;

Moles of NaOH = 13.20 g ÷ 40.0 g/mol

                          = 0.33 moles

<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>
  • From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
  • Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
  • Thus, Moles of H₂SO₄ = moles of NaOH × 2

                                    = 0.33 moles × 2

                                   = 0.66 moles of H₂SO₄

<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>
  • When given the molarity of an acid and the number of moles we can calculate the volume of the acid.
  • That is; Volume = Number of moles ÷ Molarity

In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

                                = 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

7 0
3 years ago
When iron(III) oxide (Fe2O3) is exposed to carbon monoxide (CO) under high pressure, metallic iron (Fe) and carbon dioxide (CO2)
DedPeter [7]

Answer:

The equation is Fe₂O₃ + CO ⇒ Fe + CO₂.

The balanced reaction equation is Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂.

Explanation:

First, we have to write our equation. It's actually pretty straightforward - first we look for our reactants (looks like it's Fe₂O₃ and CO), then we look for our products (Fe and CO₂). Then, we have to balance it so that both sides have the same number of both element.

Currently, we have the equation Fe₂O₃ + CO ⇒ Fe + CO₂. There are 2 Fe atoms, 4 O atoms, and 1 C atom on the left side. There is 1 Fe atom, 2 O atoms, and 1 C atom on the right side.

First thing we can do is give our Fe on the right side a coefficient of 2. This will make it equivalent to the 2 Fe atoms on the left side:

Fe₂O₃ + CO ⇒ 2Fe + CO₂

Next, we need to make sure that we have the same number of C and O atoms on each side. This takes a little bit of thinking, but what we have to do is give CO a coefficient of 3 and CO₂ a coefficient of 3. This gives us 6 O atoms on the left side (when we include the O₃) and 6 O atoms on the right side (since there are 3 O₂ atoms and 3 times 2 is 6). Here's what that looks like:

Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂

And that's how I balanced the equation. It can be confusing, but with enough practice, it will get easier and easier. :)

8 0
3 years ago
Convert 72g of O2 to moles
Volgvan
Moles = mass / molar mass
molar mass of O2 is 32
therefore moles = 72/32
= 2.25 moles
4 0
3 years ago
Read 2 more answers
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