All categories

3 years ago

4 what is the net amount of heat released when two moles of c2h6(g) are formed from its elements at 101.3 kpa and 298 k?

Chemistry

2 answers:

Gennadij [26K]3 years ago

8 0

Answer: Idk

Explanation:

kolbaska11 [484]3 years ago

6 0

Answer:

because reference table I shows that delta h for one mole of c2h6(g) is 84.0 kj so for 2 moles of c2h6(g) is 168.0kj

You might be interested in

Explain how a neutral paper fragment can be attracted to a charged rod

Yuliya22 [10]

Answer:

The particles in the neutral paper can shift, causing the paper to become polarized and attracted to the rod.

Explanation:

The neutral paper has an even distribution of its electrons throughout the paper. If a charged rod is brought near the neutral paper, this can cause the electrons in the paper to shift. If the rod is negative, the electrons will be repelled from the rod and cause the molecules in the paper to have a slight positive charge on the part of the paper closest to the rod. If the rod is positive, the electrons will be attracted to the rod and cause a slight negative charge on the side of the paper closest to the rod.

3 0

3 years ago

To determine the ammonia concentration in a sample of lake water, three samples are prepared. In sample A, 10.0 mL of lake water

Masteriza [31]

Answer:Sample Absorbance (625 nm)

A 0.536

B 0.783

C 0.045

Therefore, I will use these data to solve your question. If you have other absorbances values, just follow my steps and plug in different numbers.

First, we see 1 mole of NH3 gives 1 mole product.

In B moles of NH3 = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA

( mA = moles of NH3 in A) vol of B = 25 = vol of A

now A = el C = eC ( since l = 1cm)

Because, n net absorbance due to complex blank absorbance must be removed.

Here A(A) = 0.536 - 0.045 = 0.491 , A(B) = 0.783 - 0.045 = 0.738

(you can plug in different numbers in this step)

A2/A1 = C2/C1 , A(B)/A(A) = (1.375x10^-6 +mA)/(mA) = 0.738/0.491

So, mA = 2.733 x 10^-6 = moles of NH3 in A (Lake water)

Hence [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M

Lake water vol = 10 ml out of 25,

Concentration of ammonia in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M

Then, A = 0.491 = e x 1 x 1.093 x10^-4

e = 4492 M-1cm-1

Explanation:

4 0

3 years ago

Will give free brainlist !! :)

luda_lava [24]

Answer:

C is the correct answer

Explanation:

7 0

3 years ago

Read 2 more answers

A sample of tin (II) chloride has a mass of 0.49 g. After heating, it has a mass of 0.41 g.

lyudmila [28]

Answer:

16 percent

Explanation:

Just answered the question

5 0

3 years ago

How many moles of ethanol are produced starting with 500.g glucose?

Monica [59]

<h3>Answer:</h3>

5.55 mol C₂H₅OH

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Tables
Moles

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

<u>Step 3: Stoichiometry</u>

[DA] Set up conversion: $\displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})$
[DA} Multiply/Divide [Cancel out units]: $\displaystyle 5.55001 \ mol \ C_2H_5OH$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

8 0

3 years ago