2CO + O2 --> 2002

If 36.2g of Acetic Acid (HC2H302) was dissolved in 300. mL of water, what is the

uysha [10]

Answer:

2.01 M

Explanation:

Step 1: Calculate the moles of acetic acid (HC₂H₃O₂)

The molar mass of acetic acid is 60.05 g/mol. We will use this data to calculate the moles corresponding to 36.2 g of acetic acid.

$36.2g \times \frac{1mol}{60.05g} = 0.603mol$

Step 2: Convert the volume of solution to liters

We will use the relation 1000 mL = 1 L. We assume that the volume of solution is that of water (300 mL)

$300mL \times \frac{1L}{1000mL} = 0.300L$

Step 3: Calculate the molarity of the solution

The molarity is equal to the moles of solute (acetic acid) divided by the liters of solution

$M = \frac{0.603mol}{0.300L} = 2.01 M$

7 0

3 years ago

If an element has a mass number of 222 and an atomic number of 86, how

svlad2 [7]

Answer:

136

Explanation:

The Mass Number is the combination of the amount of Protons and Neutrons in an element, so if the total mass is 222, and the amount of protons is 86, then you can do 86 + x = 222 to find that x is equal to 136

5 0

3 years ago

Which answer tells the reason the earth’s climate is getting warmer?

Soloha48 [4]

Driving cars gives off gases that trap heat in the atmosphere

5 0

3 years ago

Read 2 more answers

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

8 0

4 years ago

A gas with a volume of 4.0 L at a pressure of 2.02 atm is allowed to expand to a volume of 12.0

Maurinko [17]

<h3>Answer:</h3>

P₂ = 0.67 atm

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

Boyle's Law: P₁V₁ = P₂V₂

P₁ is pressure 1
V₁ is volume 1
P₂ is pressure 2
V₂ is volume 2

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] P₁ = 2.02 atm

[Given] V₁ = 4.0 L

[Given] V₂ = 12.0 L

[Solve] P₂

<u>Step 2: Solve</u>

Substitute in variables [Boyle's Law]: (2.02 atm)(4.0 L) = P₂(12.0 L)
[Pressure] Multiply: 8.08 atm · L = P₂(12.0 L)
[Pressure] [Division Property of Equality] Isolate unknown: 0.673333 atm = P₂
[Pressure] Rewrite: P₂ = 0.673333 atm

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>

0.673333 atm ≈ 0.67 atm

4 0

3 years ago