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3 years ago

Are the products and balancing correct for this Equation:

Chemistry

1 answer:

goldfiish [28.3K]3 years ago

3 0

Answer:

False

Explanation:

Chemical reaction equation:

C₅H₁₂ + 8O₂ → 2CO₂ + 6H₂O

Most chemical reactions obey the law of conservation of mass. By so doing, the number of chemical elements on both sides of the expression must be balanced.

Reactants Products

C 5 4

H 12 12

O 16 10

We see that for C and O, the number of atoms on both sides of the expression differs and so, it is not balanced.

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What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

4 years ago

Which of the following elements is a transition metal?

GenaCL600 [577]

Answer:

Scandium

Titanium

Vanadium

Chromium

Manganese

Iron

Cobalt

Nickel

Copper

Zinc

Yttrium

Zirconium

Niobium

Molybdenum

Technetium

Ruthenium

Rhodium

Palladium

Silver

Cadmium

Lanthanum

Hafnium

Tantalum

Tungsten

Rhenium

Osmium

Iridium

Platinum

Gold

Mercury

Actinium

Rutherfordium

Dubnium

Seaborgium

Bohrium

Hassium

Meitnerium

Darmstadtium

Roentgenium

Copernicium

Explanation:

all of those are transition metals lol

5 0

4 years ago

Read 2 more answers

When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the ph

Scrat [10]

Answer:

E = 12.92 × 10^(-16) J

Explanation:

Formula for energy is;

E = hc/λ

Where;

h is Planck's constant = 6.63 x 10^(-34) J.s

c is speed of light = 3 × 10^(8) m/s

λ is wavelength = 0.154 nm = 0.154 × 10^(-9) m

E = (6.63 x 10^(-34) × 3 × 10^(8))/(0.154 × 10^(-9))

E = 12.92 × 10^(-16) J

6 0

3 years ago

docker41 [41]

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4 0

3 years ago

When 78.6 g of urea CH4N2O are dissolved in 700. g of a certain mystery liquid X, the freezing point of the solution is 4.9 Â°C

iogann1982 [59]

Answer:

The van't Hoff factor of NaCl in liquid X is 1.69

Explanation:

Step 1: Data given

Mass of urea = 78.6 grams

Molar mass of urea = 60.06 g/mol

Mass of liquid X = 700 grams = 0.700 kg

he freezing point of the solution is 4.9°C lower than the freezing point of pure X

When 78.6 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 8.5°C lower than the freezing point of pure X.

Molar mass of NaCl = 58.44 g/mol

Step 2: Calculate moles

Moles urea = mass / molar mass

Moles urea = 78.6 grams / 60.06 g/mol

Moles urea = 1.31 moles

Moles NaCl = 78.9 grams / 58.44 g/mol

Moles NaCl = 1.35 moles

Step 3: Calculate molality

Molality = moles / mass of liquid

Molality urea = 1.31 moles / 0.700 kg

Molality = 1.87 molal

Molality NaCl = 1.35 moles / 0.700 kg

Molality NaCl = 1.92 molal

Step 4: Calculate the freezing point depression constant of X

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 4.9 °C

⇒with i = the van't hoff factor of urea = 1

⇒with Kf =the freezing point depression consant of X = TO BE DETERMINED

⇒with m = the molality of urea solution = 1.87 molal

4.9 °C = 1 * Kf * 1.87 molal

Kf == 4.9 / 1.87

Kf = 2.62 °C/m

Step 5: Calculate the van't Hoff facotr of NaCl in X

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 8.5 °C

⇒with i = the van't hoff factor of urea = TO BE DETERMINED

⇒with Kf =the freezing point depression consant of X = 2.62 °C/m

⇒with m = the molality of urea solution = 1.92 molal

8.5 °C = i * 2.62 °C/m * 1.92 m

i = 8.5 / (2.62 * 1.92)

i = 1.69

The van't Hoff factor of NaCl in liquid X is 1.69

5 0

3 years ago