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3 years ago

NEED HELP NOW which materials allow heat to pass through them easily?

2 answers:

4 0

The materials which allow heat to pass through them easily are conductors of heat. For examples,aluminum, iron, and copper.

Hope this helps! ;)

coldgirl [10]3 years ago

3 0

The materials which allow heat to pass through them easily are. conductors of heat. For examples,aluminum, iron and copper.

Hope this helps! :)

You might be interested in

Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit

luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

$K~=~10^{-pK}$

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

$K~=~10^{4.9}$

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

$K`=~10^{1.12]$

Reaction can take place

2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

$K~=~10^{ -3.56}$

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

$K`=~10 ^{-12.93}$

Reaction does not happen

3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

$K~=~10^{-3.26}$

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

$K~=~ 10^{-12.63}$

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

the lowest ph

brainly.com/question/9875355

the concentrations at equilibrium.

brainly.com/question/8918040

the ph of a solution

brainly.com/question/9560687

Keywords : acid base reaction, the equilibrium constant

5 0

3 years ago

Read 2 more answers

6. What is the equilibrium constant for the following reaction?<br> C+02 = CO2

Orlov [11]

<h3> $\tt Kc=\dfrac{[CO_2]}{[C][O_2]}$ </h3><h3>Further explanation</h3>

Given

Reaction

C+02 = CO2

Required

The equilibrium constant

Solution

The equilibrium constant is the ratio of concentration or pressure between the product and the reactant with each reaction coefficient raised

The equilibrium constant is based on the concentration (Kc) in a reaction

pA + qB -----> mC + nD

$\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}$

So for the reaction :

C+O₂ ⇔ CO₂

$\tt Kc=\dfrac{[CO_2]}{[C][O_2]}$

8 0

3 years ago

If the neutral atom of an element has only 4 valence electrons it must be in which group? 1. IVB 2. VA 3. VIIA 4. VIA 5. IVA 6.

lisabon 2012 [21]

Answer:option

Option 5..IVA

Explanation:

5 0

3 years ago

Breathing equipment used by rescue workers needs to capture the CO2 the humans breath out and produce O2 for them to breath in,

const2013 [10]

Answer: $4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2$

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced chemical equation for reaction of potassium superoxide with carbon dioxide to produce oxygen and potassium carbonate will be:

$4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2$

8 0

3 years ago

B. The following reaction takes place in a basic solution. (7 points)

Natali5045456 [20]

<u>Answer:</u> The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

<u>Explanation:</u>

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

The given redox reaction follows:

$MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)$

To balance the given redox reaction in basic medium, there are few steps to be followed:

Writing the given oxidation and reduction half-reactions for the given equation with the correct number of electrons

Oxidation half-reaction: $NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-$

Reduction half-reaction: $MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-$

Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions

Oxidation half-reaction: $NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-$ ( × 3)

Reduction half-reaction: $MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-$ ( × 2)

The half-reactions now become:

Oxidation half-reaction: $3NO_2^-+6OH^-\rightarrow 3NO_3^-+3H_2O+6e^-$

Reduction half-reaction: $2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-$

Add the equations and simplify to get a balanced equation

Overall redox reaction: $3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-$

As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution

3 0

3 years ago