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Softa [21]
3 years ago
8

Your computer is an example of which type of energy? Question options: Nuclear Kinetic Light All of the above

Chemistry
1 answer:
xz_007 [3.2K]3 years ago
4 0

<u>Answer</u>:  Light

<em>Computer is an example of light energy which is the third option out of the given four choices. </em>

<u>Explanation:</u>

We know how a computer works it takes in <em>the electrical energy</em> and does a <em>lot of mathematical mechanical work</em> and for giving answers. It uses a screen on which light blinks in pattern such that it represents letters or mathematical numbers or expressions.

Hence by using this statement we can say <em>computer converts electrical energy into light energy. </em>

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How would you make a 2.00L of 0. 500M sodium chloride solution. (assume you have a fully equipped lab with water); sketch, calc
Gwar [14]

Answer:

Sodium chloride solution:

First you need to calculate the mass of salt needed (done in the explanation), which is 58.44g. Then it have to be weighted in an analytical balance in a weighting boat and then transferred into a 2L volumetric flask that is going to be filled until the mark with distilled water.

Sulfuric acid dilution:

First you need to calculate the volume needed (done in the explanation), it is 16.6 mL. Using a graduated pipette one measures this volume and transfer it into a 2L volumetric flask that is already half filled with distilled water, and then one fills it until its mark.

Explanation:

Sodium chloride solution:

Each liter of a 0.500M solution has half mol, so 2L of said solution has 1 mol of salt. Sodium chloride molar mass is 58.44g/mol, so in 2L of solution there is 58.44g of salt. That`s the mass that`s going to be weighted and transferred to a 2L volumetric flask.

Sulfuric acid dilution:

This is the equation for dilution of solutions:

c_{1} v_{1} =c_{2} v_{2}

Where "c1" stands for the initial concentration (stock solution concentration), "v1" for the initial volume (volume of stock solution used), "c2" for the desired concentration and "v2" for the desired volume.

When we are diluting from a stock solution we want to know how much do we have to pipette from the stock solution into our volumetric flask. We do so by isolating the "v1" term from the dilution equation:

v_{1} =\frac{c_{2} v_{2} }{c_{1} }

in this case that would be:

v_{1} =\frac{0.100 x2.0 }{12.0 }=0.0166L=16.6mL

5 0
3 years ago
Which one of the following will change the value of an equilibrium constant?
Ad libitum [116K]

Answer:

(E) changing temperature

Explanation:

Consider the following reversible balanced reaction:

aA+bB⇋cC+dD

If we know the molar concentrations of each of the reaction species, we can find the value of Kc using the relationship:

Kc = ([C]^c * [D]^d) / ([A]^a * [B]^b)

where:

[C] and [D] are the concentrations of the products in the equilibrium; [A] and [B] reagent concentrations in equilibrium; already; b; c and d are the stoichiometric coefficients of the balanced equation. Concentrations are commonly expressed in molarity, which has units of moles / 1

There are some important things to remember when calculating Kc:

-  <em>Kc is a constant for a specific reaction at a specific temperature</em>. If you change the reaction temperature, then Kc also changes

- Pure solids and liquids, including solvents, are not considered for equilibrium expression.

- The reaction must be balanced with the written coefficients as the minimum possible integer value in order to obtain the correct value of Kc

8 0
3 years ago
In Part B the given conditions were 1.00 mol of argon in a 0.500-L container at 18.0°C. You identified that the ideal pressure (
Natasha2012 [34]

Answer:

4,38%

small molecular volumes

Decrease

Explanation:

The percent difference between the ideal and real gas is:

(47,8atm - 45,7 atm) / 47,8 atm × 100 = 4,39% ≈ <em>4,38%</em>

This difference is considered significant, and is best explained because argon atoms have relatively <em>small molecular volumes. </em>That produce an increasing in intermolecular forces deviating the system of ideal gas behavior.

Therefore, an increasing in volume will produce an ideal gas behavior. Thus:

If the volume of the container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to <em>decrease</em>

<em />

I hope it helps!

5 0
3 years ago
Give me a statement regarding the process of science discovery and development of atomic theory
matrenka [14]
Can you say please? Just kidding!
The process of science discovery depends upon changing your theories based upons new evidence from new experiments.  New technology allows for new experiments, leading to new theories.  
5 0
3 years ago
Calculate ΔG o for the following reaction at 25°C: 3Mg(s) + 2Al3+(aq) ⇌ 3Mg2+(aq) + 2Al(s) Enter your answer in scientific notat
larisa [96]

Answer:

-3.7771 × 10² kJ/mol

Explanation:

Let's consider the following equation.

3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)

We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)

where,

n: moles

ΔG°f(): standard Gibbs free energy of formation

p: products

r: reactants

ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))

ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)

ΔG° = -377.71 kJ = -3.7771 × 10² kJ

This is the standard Gibbs free energy per mole of reaction.

5 0
3 years ago
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