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3 years ago

Calculate the mass of water produced when 9.57 g of butane reacts with excess oxygen.

Chemistry

1 answer:

irinina [24]3 years ago

4 0

Answer:

14.9 g

Explanation:

Step 1: Write the balanced equation

C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O

Step 2: Calculate the moles corresponding to 9.57 g of C₄H₁₀

The molar mass of C₄H₁₀ is 58.12 g/mol.

9.57 g × 1 mol/58.12 g = 0.165 mol

Step 3: Calculate the moles of H₂O produced from 0.165 moles of C₄H₁₀

0.165 mol C₄H₁₀ × 5 mol H₂O/1 mol C₄H₁₀ = 0.825 mol H₂O

Step 4: Calculate the mass corresponding to 0.825 mol of H₂O

The molar mass of H₂O is 18.02 g/mol.

0.825 mol × 18.02 g/mol = 14.9 g

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The chemical bonds of carbohydrates and lipids have high potential energy because:

gladu [14]

Answer:

c. Many of their bonds are C-C and C-H

Explanation:

The majority of bonds in carbohydrates and lipids( being an organic compound) are C-C and C-H. Like glucose, fructose or galactose ,etc.

These bonds are strong and do require a lot of energy to break. Thus, a lot of energy are required to break carbs and lipids into simpler compounds.Therefore, carbohydrates and lipids have high potential energy.

The correct answer is c.

4 0

3 years ago

Water changes from a liquid to a gas in the process of evaporation. true or false

Umnica [9.8K]

This it true because in the triangle of the transformation from solid, liquid, and gas
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4 0

3 years ago

Read 2 more answers

when a solution of sodium chloride is added to a solution of copper(ii) nitrate, no precipitate is observed. Write the molexular

zhuklara [117]

Explanation:

Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)

Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)

A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.

Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)

2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)

Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)

2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)

The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.

Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.

6 0

3 years ago

What is the half-life in minutes of a compound if 75.0 percent of a given sample decomposes in 40.0 minutes? Assume first-order

yaroslaw [1]

Answer:

See below

Explanation:

.75 = 1/2^(40/h)

log .75 / ( log 1/2) = 40 / h

h = half life = 96.37683 min

8 0

2 years ago

Which solution below has the highest concentration of hydroxide ions?a. pH= 3.21b. pH= 7.00c. pH= 7.93d. pH= 12.59e. pH= 9.82

Sloan [31]

Answer:

The solution that has the highest concentration of hydroxide ions is d. pH = 12.59.

Explanation:

You can solve this question using just some chemical facts:

pH is a measure of acidity or alkalinity: the higher the pH the lower the acidity and the higher the alkalinity.
The higher the concentration of hydroxide ions the lower the acidity or the higher the alkalinity of the solution, this is the higher the pH.

Hence, since you are asked to state the solution with the highest concentration of hydroxide ions, you just pick the highest pH. This is the option d, pH = 12.59.

These mathematical relations are used to find the exact concentrations of hydroxide ions:

pH + pOH = 14 ⇒ pOH = 14 - pH

pOH = - log [OH⁻] ⇒ $[OH^-]=10^{-pOH}$

Then, you can follow these calculations:

Solution pH pOH [OH⁻]

a. 3.21 14 - 3.21 = 10.79 antilogarithm of 10.79 = 1.6 × 10⁻¹¹

b. 7.00 14 - 7.00 = 7.00 antilogarithm of 7.00 = 10⁻⁷

c. 7.93 14 - 7.93 = 6.07 antilogarithm of 6.07 = 8.5 × 10⁻⁷

d. 12.59 14 - 12.59 = 1.41 antilogarithm of 1.41 = 0.039

e. 9.82 14 - 9.82 = 4.18 antilogarithm of 4.18 = 6.6 × 10⁻⁵

From which you see that the highest concentration of hydroxide ions is for pH = 12.59.

5 0

3 years ago