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3 years ago

Based on the chemical bonds present in this molecule, we would classify it as a(n)

Chemistry

2 answers:

MrMuchimi3 years ago

8 0

Answer is C – Unsaturated fat.

Fats are large molecules basically made from a combination of glycerol and three chains of fatty acids. Usually fats consist of C, H and O. According to the C – C bond, there are two types of fats as saturated fats and unsaturated fats. Saturated fats have only C – C single bonds and all C atoms are sp³ hybridized atoms. But in unsaturated fats other than C – C single bonds there are C – C double bonds. Hence, both sp³ and sp² hybridized C atoms can be seen.

Blizzard [7]3 years ago

4 0

The answer is C. Unsaturated Fat<span />

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How many grams are in 0.39 moles of water

Shtirlitz [24]

Answer: 7.025959200000001

Explanation:

5 0

2 years ago

The activation energy for the isomerization ol cyclopropane to propene is 274 kJ/mol. By what factor does the rate of this react

Bond [772]

Answer:

The rate of the reaction increased by a factor of 1012.32

Explanation:

Applying Arrhenius equation

ln(k₂/k₁) = Ea/R(1/T₁ - 1/T₂)

where;

k₂/k₁ is the ratio of the rates which is the factor

Ea is the activation energy = 274 kJ/mol.

T₁ is the initial temperature = 231⁰C = 504 k

T₂ is the final temperature = 293⁰C = 566 k

R is gas constant = 8.314 J/Kmol

Substituting this values into the equation above;

ln(k₂/k₁) = 274000/8.314(1/504 - 1/566)

ln(k₂/k₁) = 32956.4589 (0.00198-0.00177)

ln(k₂/k₁) = 6.92

k₂/k₁ = exp(6.92)

k₂/k₁ = 1012.32

The rate of the reaction increased by 1012.32

3 0

3 years ago

Read 2 more answers

a sample of ammonia contains 9g hydrogen and 42g nitrogen. another sample contains 5g hydrogen .calculate the amount of nitrogen

balandron [24]

9 g of hydrogen - 42 g of nitrogen
5 g of hydrogen - x g of nitrogen

$9x=42 \cdot 5 \\
9x=210 \\
x \approx 23.33$

The mass of nitrogen in the second sample is 23.33 g.

4 0

3 years ago

Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stresse

amm1812

Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

6 0

3 years ago

Read 2 more answers

How many formula units are there in 50.3 moles potassium chloride?

Alona [7]

<h3>Answer:</h3>

3.03 × 10²⁵ formula units KCl

<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>Chemistry</u>

Reading a Periodic Table
Writing Compounds

<u>Atomic Structure</u>

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

<u>Step 1: Define</u>

50.3 mol KCl (Potassium chloride)

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

Avogadro's Number

<u>Step 3: Convert</u>

<u /> $\displaystyle 50.3 \ mol \ KCl(\frac{6.022 \cdot 10^{23} \ formula \ units \ KCl}{1 \ mol \ KCl} )$ = 3.02907 × 10²⁵ formula units KCl

<u>Step 4: Check</u>

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

3.02907 × 10²⁵ formula units KCl ≈ 3.03 × 10²⁵ formula units KCl

6 0

3 years ago