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3 years ago

The lecturer compared the humans eye to a camera lens

1 answer:

5 0

An analogy is the closest way to compare something to another that may not be related so that would be my pick. If that’s not it I guess he could maybe be making an observational opinion about it? Good luck hope this helped. Analogy

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Indicate the types of forces that are involved between the solute and solvent when forming a homogeneous solution between ch3ch2

Bess [88]

Answer:
Dispersion Forces are found between n-Pentane (CH₃-CH₂-CH₂-CH₂-CH₃) and n-Hexane (CH₃-CH₂-CH₂-CH₂-CH₂-CH₃).

Explanation:
Dispersion Forces are present and developed by those compounds which are non-polar in nature. In given statement n-Pentane and n-Hexane both are non-polar in nature as the electronegativity difference between Hydrogen atoms and Carbon atoms is less than 0.4.
When non-polar molecules approaches each other, a Dipole is induced in one of them, this step is known as Instantaneous Dipole, This generated Dipole on approaching another non-polar molecule induces dipole in it and the process propagates. Hence, creating intermolecular interactions.

8 0

3 years ago

Read 2 more answers

Given the following unbalanced equation:

antiseptic1488 [7]

<h3>Answer:</h3>

11.84 mol CoF₂

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

<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>

Moles

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[RxN - Balanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[Given] 11.84 moles CoCl₂

[Solve] moles CoF₂

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

[RxN] 1 mol CoCl₂ → 1 mol CoF₂

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 11.84 \ mol \ CoCl_2(\frac{1 \ mol \ CoF_2}{1 \ mol \ CoCl_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 11.84 \ mol \ CoF_2$

7 0

3 years ago

What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature i

Alexxandr [17]

Answer:

a) 48KJ

b) -48KJ

Explanation:

Given that;

ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)

K2= equilibrium constant at T2

K1 = equilibrum constant at T1

R = gas constant

T1 = initial temperature

T2 = final temperature

When we double the equilibrium constant K1; K2 = 2K1

T1 = 310 K

T2 = 310 + 15 = 325 K

ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)

ln2 = -ΔH°/8.314(1/325 - 1/310)

0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)

0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

b) K2 =K1/2

ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)

-0.693 = -ΔH°/8.314 (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

6 0

3 years ago

Sample of 200mls of 0.5 sulphuric acid,was asked to produce 1.2M of the new solution.Calculate the volume of the new solution

AlladinOne [14]

Answer:

$V_2=83.3mL$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the volume of the new solution by using the general formula of dilution:

$V_1M_1=V_2M_2$

In such a way, we solve for the final volume, V2, to obtain:

$V_2=\frac{200mL*0.5M}{1.2M}\\\\V_2=83.3mL$

Regards!

3 0

3 years ago

If element X has 65 protons, how many electrons does it have

Tomtit [17]

In an atom there are the same number of protons as electrons to start with. The answer would be 65 aswell.

8 0

3 years ago

Read 2 more answers