why does copper Cu (I) have a short lifespan​

Answers

Answer 1

Answer:

Copper (Cu), chemical element, a reddish, extremely ductile metal of Group 11 (Ib) of the periodic table that is an unusually good conductor of electricity and heat. Copper is found in the free metallic state in nature.

Explanation:

hope this helps:)


Related Questions

Do you think the video game "Geometry dash" is dead?

Answers

Answer:

HELL NO!

THAT'S MY CHILDHOOD GAME!

I LOVE THAT GAME!

IT SHOULD NOT BE DEAD!

IT SHOULD LIVE ON FOREVER!

#GemoteryDashShouldLiveOnForever!

[tex]AnimeVines[/tex]

honestly yeah game is only good sometimes but there are so many other games now

Whst type of molecule is
Dehydogenase

Answers

Answer:

A dehydrogenase is an enzyme .

Explanation:

Hope it helps

Mark me as Brainliest plz.

:DD

This is science not chemistry

Answers

That is actually physics because it talks about motion.

HELP! Which atom would be neutral?
A - an oxygen atom with 9 electrons, 8 protons, and 8 neutrons

B - an oxygen atom with 4 electrons, 6 protons, and 4 neutrons

C - an oxygen atom with 8 electrons, 8 protons, and 9 neutrons

D - an oxygen atom with 16 electrons, 18 protons, and 16 neutrons

Answers

Answer:

an oxygen atom with 8 electrons, 8 protons and 9 neutrons

An atom is called as neutral when the number of electron is equal to number of proton . Thus option C is correct.  

What is neutral atom ?

Atom is the smallest unit of an element which can not be seen in with your bare eyes, due to its wavelength.

A visible light wavelength  is about 0.000001 meter, where  the size of an atom is 0.000000001 which is much smaller than 0.000001.

An atom has a nucleus, means it is the mass of the atom.  

An atom is composed of protons, neutrons, and electrons where the protons and neutrons are combinedly form  nucleus and electrons are present surrounding the nucleus.

When the number of proton and electron of an atom will be same  then it is called as neutral only in normal state, but in other condition it either gain or loss electron by developing charge on it.

Neutral atoms are Oxygen, Hydrogen, Lithium, Helium etc.

Hence, option C is correct.  

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All living organisms are made of
They are the smallest units of a living.
thing that can perform all life processes.
A. stones
B. cells
C. water drops
D. bricks

Answers

The answer to the question is cells (B)

A semiconducting material is composed of 105 g of Ga, 80 g of Al, and 92.0 g of As. Which element has the largest number of atoms in this material?

Answers

Answer:

What is Acellus? Acellus is a learning accelerator. It uses video-based lessons with cutting-edge technology to accelerate learning, elevate standardized test scores, reduce dropout rates and transition more students into careers and college.

Bonus 50 points if u got this answer correctly!!


When does a real gas behave like an ideal gas?

Answers

yes i did you do that i’m you did it for you too and i didn’t even know what to say about anything i’m but

Answer:

[tex]at \: low \: pressure \: high \: temparature \\ thank \: you[/tex]

what was daltons motivation for proposing his model of the atom

Answers

Answer:

John Dalton based his partial pressures theory on the idea that only like atoms repel one another, Whereas unlike atoms appear to react indifferently. This notion was erroneous, but it helped to explain why each gas in a mixture behaved independently, serving the purpose of showing that atoms of all kinds are not alike.

The density of gold is 19.3 g/cm3, and the density of iron pyrite is 5.02 g/cm3. What substance do you think you found?

Answers

Answer:

Step 1

1 of 2

V = m / D; V (nugget of gold) = 50 g / 5.0 g / cm 3 = 10 cm3; V (nugget of iron pyrite) = 50 g / 19 g / cm3 = 2.63 cm 3

Result

2 of 2

Due to these results you can conclude that nugget of gold which volume is 10 cm3 is bigger than volume of nugget of iron pyrite which is 2.63 cm3.

Explanation:

Stowing items and come across an aerosol battle of hairspray what should you do?

Answers

You are stowing items and come across an aerosol bottle of hairspray. W should you do? Please choose all that apply. Stow the hairspray Raise an Andon Remove it and secure it with bubble wrap Place a Flammable sticker on the bottle Submit

Which property altered during a chemical change is not altered during a physical change?

A. composition of the matter

B. temperature of the matter

C. volume of the matter

D. phase of the matter

Answers

A. composition of the matter

The property altered during a chemical change that is not altered during a physical change is composition of the matter.

A physical change is one in which no new substance is formed and it is easily reversible.

A chemical change is one that is not easily reversible and no new substance is formed. It may be accompanied by absorption or evolution of heat.

The composition of a substance changes during a chemical change  because bonds between atoms break and new bonds are formed. This does not occur during a physical change.

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What volume of 0.32M HCl is required to completely neutralize 50.0m mL of a 0.12M Mg(OH)2 solution?​

Answers

Answer:

37.5 mL of HCl will be required for complete neutralization with magnesium hydroxide

Explanation:

[tex]{ \bf{2HCl _{(aq)} + Mg(OH)_{2(s)}→ MgCl_{2(s)} + 2H _{2} O_{(l)}}}[/tex]

we've to first get moles of magnesium hydroxide in 50.0 ml :

[tex]{ \sf{1 \: l \: of \: hydroxide \: contains \: 0.12 \: moles}} \\ { \sf{0.05 \: l \: of \: hydroxide \: contain \: (0.05 \times 0.12) \: moles}} \\ = { \underline{0.006 \: moles}}[/tex]

for complete neutralization:

[tex]{ \sf{1 \: moles \: of \: hydroxide\: react \: with \: 2 \: moles \: of \: acid}} \\ { \sf{0.006 \: moles \: react \: with \: \{0.006 \times 2 \}} \: moles} \\ = { \underline{0.012 \: moles \: of \: acid}}[/tex]

compare with acid molarity:

[tex]{ \sf{0.32 \: moles \: of \: acid \: occupy \: 1 \: litre}} \\ { \sf{0.012 \: moles \: of \: acid \: occupy \: ( \frac{0.012}{0.32}) \: litres }} \\ { \underline{ = 0.0375 \: litre \: \: = \: \: 37.5 \: ml}}[/tex]

Which tool can be used to measure the volume of a liquid to one decimal place?
a thermometer
a graduated cylinder
an Erlenmeyer flask
a test tube

Answers

Answer:B

Explanation:

Answer:

B. a graduated cylinder

How much heat is absorbed by a 112.5 g sample of water when it is heated from 12.5 °C to 92.1 °C? (Specific heat capacity of water is 4.184 J/g °C)

Answers

Answer:

[tex]\boxed {\boxed {\sf37,467.72 \ Joules }}[/tex]

Explanation:

We are asked to find how much heat a sample of water absorbed. Since we are given the mass, temperature, and specific heat, we will use the following formula.

[tex]q=mc \Delta T[/tex]

The mass (m) of the sample is 112.5 grams. The specific heat capacity of water (c) is 4.184 Joules per gram degree Celsius. The difference in temperature (ΔT) is found by subtracting the initial temperature from the final temperature.

ΔT= final temperature - initial temperature

The water was heated from 12.5 degrees Celsius to 92.1 degrees Celsius.

ΔT= 92.1 °C - 12.5 °C= 79.6°C

Now we know three variables and can substitute them into the formula,

m= 112.5 g c= 4.184 J/g °C ΔT= 79.6 °C

[tex]q= (112.5 \ g )(4.184 \ J/g \textdegree C)(79.6 \textdegree C)[/tex]

Multiply the first 2 numbers. Note the units of grams cancel.

[tex]q= (112.5 \ g *4.184 \ J/g \textdegree C)(79.6 \textdegree C)[/tex]

[tex]q= (112.5 *4.184 \ J/ \textdegree C)(79.6 \textdegree C)[/tex]

[tex]q= (470.7 \ J/ \textdegree C)(79.6 \textdegree C)[/tex]

Multiply again. This time the units of degrees Celsius cancel.

[tex]q= (470.7 \ J/ \textdegree C *79.6 \textdegree C)[/tex]

[tex]q= (470.7 \ J *79.6 )[/tex]

[tex]q= 37467.72 \ J[/tex]

37, 467.72 Joules of heat are absorbed by the sample fo water.

calculate the number of each atom in 2.5 gram of caco3 ​

Answers

Answer:

[tex]2.5\; \rm g[/tex] of [tex]\rm CaCO_{3}[/tex] would contain:

Approximately [tex]1.5 \times 10^{22}[/tex] calcium atoms (approximately [tex]0.025\; \rm mol[/tex],)Approximately [tex]1.5 \times 10^{22}[/tex] carbon atoms (approximately [tex]0.025\; \rm mol[/tex],) andApproximately [tex]4.5 \times 10^{22}[/tex] oxygen atoms (approximately [tex]0.075\; \rm mol[/tex].)

Explanation:

Look up the Avogadro constant: [tex]N_{\rm A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}[/tex].

For example, "[tex]1\; \rm mol[/tex] of carbon atoms" would contain [tex]N_{\rm A}[/tex] carbon atoms (approximately [tex]6.022\times 10^{23}[/tex]) by definition.

Look up the relative atomic mass of carbon, calcium, and oxygen on a modern periodic table:

Calcium: [tex]40.078[/tex].Carbon: [tex]12.011[/tex].Oxygen: [tex]15.999[/tex].

In other words, the mass of [tex]1\; \rm mol[/tex] of calcium atoms would be [tex]40.078\; \rm g[/tex]. The mass of [tex]1\; \rm mol\![/tex] of carbon atoms would be [tex]12.011\; \rm g[/tex], and the mass of [tex]1\; \rm mol \!\![/tex] of oxygen atoms would be [tex]15.999\; \rm g[/tex].

As the formula [tex]\rm CaCO_{3}[/tex] suggests, every formula unit of this ionic compound includes one calcium atom, one carbon atom, and three oxygen atoms. The formula mass of [tex]\rm CaCO_{3}\![/tex] would give the mass of every mole of [tex]\rm CaCO_{3}\!\![/tex] formula units.

Calculate the formula mass of [tex]\rm CaCO_{3}[/tex] from the relative atomic mass data:

[tex]\begin{aligned} & M({\rm CaCO_{3}}) \\ =\; & 40.078\; \rm g \cdot mol^{-1} \\ & + 12.011\; \rm g \cdot mol^{-1} \\ & + 3 \times (15.999\; \rm g \cdot mol^{-1}) \\ =\; & 100.086\; \rm g \cdot mol^{-1}\end{aligned}[/tex].

Calculate the number of [tex]\rm CaCO_{3}[/tex] formula units in that [tex]2.5\; \rm g[/tex] of this compound:

[tex]\begin{aligned}& n({\rm CaCO_{3}}) \\ =\; & \frac{m({\rm CaCO_{3}})}{M({\rm CaCO_{3}})} \\ =\; & \frac{2.5\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \\ \approx\; & 0.025\; \rm mol\end{aligned}[/tex].

In other words, [tex]2.5\; \rm g[/tex] of [tex]\rm CaCO_{3}[/tex] would contain approximately [tex]0.025\; \rm mol[/tex] [tex]\rm CaCO_{3}\![/tex] formula units.

Again, there are one calcium atom, one carbon atom, and one oxygen atom in every [tex]\rm CaCO_{3}[/tex] formula unit. That approximately [tex]0.025\; \rm mol[/tex] [tex]\rm CaCO_{3}\![/tex] formula units would thus contain:

Approximately [tex]1 \times 0.025\; \rm mol = 0.025\; \rm mol[/tex] calcium atoms, Approximately [tex]1 \times 0.025\; \rm mol = 0.025\; \rm mol[/tex] carbon atoms, andApproximately [tex]3 \times 0.025\; \rm mol = 0.075\; \rm mol[/tex] oxygen atoms.

Make use of the Avogadro constant to convert the numbers.

For example, the number of calcium atoms in that approximately [tex]0.025\; \rm mol[/tex] of calcium atoms would be:

[tex]\begin{aligned} & N({\text{calcium}) \\ = \; & n({\text{calcium}) \cdot N_{\rm A} \\ \approx \; & 0.025\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 1.5 \times 10^{22} \end{aligned}[/tex].

Likewise, the number of carbon atoms in that approximately [tex]0.025\; \rm mol[/tex] of carbon atoms would be:

[tex]\begin{aligned} & N({\text{carbon}) \\ = \; & n({\text{carbon}) \cdot N_{\rm A} \\ \approx \; & 0.025\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 1.5 \times 10^{22} \end{aligned}[/tex].

The number of oxygen atoms in that approximately [tex]0.075\; \rm mol[/tex] of oxygen atoms would be:

[tex]\begin{aligned} & N({\text{oxygen}) \\ = \; & n({\text{oxygen}) \cdot N_{\rm A} \\ \approx \; & 0.075\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 4.5 \times 10^{22} \end{aligned}[/tex].

A 25.0 mL sample of 0.723 M HClO4 is titrated with a 0.273 M KOH solution. The H3O+ concentration after the addition of 66.2 mL of KOH is __________ M.
a. 0.723
b. 1.00 × 10-7
c. 2.81 × 10-13
d. 0.439
e. 0.273

Answers

B 2048% q100-10000(10)

In rutherfords gold foil experiment, most alphas particles passed through the gold foil without deflection and were detected on the screen. What caused the particles to pass through without any deflection?

Answers

Answer:

The vast majority of the volume in an atom consists of the electron orbitals with electrons that have too little mass to deflect the alpha particles.  So most pass right through the foil.  Only the ones hitting the nucleus are deflected.

Explanation:

The whole reason Rutherford and his students conducted this experiment was to test the then current hypothesis that an atom consists of a homogeneous blend of atomic particles, known as the "plum pudding" model.  If so, there should be no, or minimal, deflection of particles.  That's not what they found.  Since particles were found to be deflected, the group concluded that atoms must contain a large object, which became known as the nucleus.

Which word equation best describes the following formula equation:
4Al + 302 + 2Al2O3?

Answers

Answer:

Explanation:

The equation is not correctly written, so let's start with that.

4Al + 3O2 ====> 2Al2O3?

Totally correct, including the balance numbers, would be

4moles aluminum + 3 moles Oxygen ===> 2 moles of Aluminum Oxide

Anything else is incomplete.

Answer:

aluminum reacts with oxygen to produce aluminum oxide.

I hope this helps

The volume of a gas sample is 15 L while the temperature is 50 °C. If the temperature is changed to 30 °C and the pressure remains constant, what will the new volume be? Round your answer to the correct number of significant figures. (Don’t forget to convert the temperature to the Kelvin scale by adding 273 to the Celsius temperature.)

Answers

T1=50°CT2=30°CV1=15LV2=?

Convert temperatures to Kelvin

[tex]\\ \sf\longmapsto 50+273=323K[/tex]

[tex]\\ \sf\longmapsto 30+273=303K[/tex]

Using Charles law

[tex]\\ \sf\longmapsto \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}[/tex]

[tex]\\ \sf\longmapsto \dfrac{15}{323}=\dfrac{V_2}{303}[/tex]

[tex]\\ \sf\longmapsto V_2=\dfrac{15(303)}{323}[/tex]

[tex]\\ \sf\longmapsto V_2=\dfrac{4545}{323}[/tex]

[tex]\\ \sf\longmapsto V_2=14.07L[/tex]

A beggar wrapped himself with a few layers of newspaper on a cold winter night. This helped him to keep himself warm because:
(a) friction between the layers of newspaper produces heat.
(b) air trapped between the layers of newspaper is a bad conductor of heat.
(c) newspaper is a conductor of heat.
(d) newspaper is at a higher temperature than the temperature of the surrounding.

the one who gets its correct will be marked as the brainliest and also be followed by me!!

P.S. have a great Sunday :)​

Answers

Answer:

(b)The air trapped between the layers of newspaper is a bad conductor of heat.Thus the layers of newspaper helped the beggar to keep himself warm

Hey there!

I believe your answer would be "Air trapped between the layers of newspaper is a bad conductor of heat."

Hope this helps!

Have a great day!

Illustrate the law of law of multiple proportions using the formation of SO2 and SO3 by sulphur and oxygen

Answers

As illustrated from law of multiple proportion, the mass of the oxygen which combined with a fixed mass of the sulphur is in simple whole number ratio of 2 is to 3 (2:3).

The law of Multiple proportion states that when two elements combine to form more than one compound, the mass of the second element, which combines with a fixed mass of the first element, will always be ratios of small whole numbers.

To illustrate the law of multiple proportion from sulphur and oxygen, we use the given compounds.

The reaction of oxygen with sulphur to form SO₂ and SO₃:

[tex]S \ + \ O_2 \ -----> SO_2\\\\2SO_2 \ + \ O_2 \ ---> \ 2SO_3[/tex]

atomic mass of oxygen = 16 g

atomic mass of sulphur = 32 g

In the formation of SO₂:

mass of sulphur = 32 g and mass of oxygen = (2x16) = 32 g

In the formation of SO₃:

mass of sulphur = 32 g and mass of oxygen = (3 x 16) = 48 g

Notice, while mass of sulphur is constant, mass of oxygen changed.

The ratio of mass of oxygen in the given two compounds = 32g : 48g = 2:3

Thus, based on the law of multiple proportion, the mass of the oxygen which combined with a fixed mass of the sulphur is in simple whole number ratio of 2 is to 3 (2:3).

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What is the major organic product obtained from the reaction of ethylbenzene with nitronium ion, followed by benzylic oxidation?
a. 3-ethyl-aniline.
b. 4-nitro-benzoic acid.
c. 3-nitro-toluene.
d. 3-nitrobenzoic acid.

Answers

Answer:

b

Explanation:

for the equation given, how many grams of methane will react with a 125g of oxygen
CH4 (g) + 2O2(g) -> CO2(g) +2H2O​

Answers

Answer:

16

Explanation:

1) in the given equation CH₄+2O₂⇒ 2H₂O+CO₂;

M(CH₄)=12+4=16 (g/mol); M(O₂)=32 (g/mol);

2) if m(O₂)=125 gr., then ν(O₂)=m(O₂)/M(O₂)=125/32≈3.90625 (moles);

3) if in the given equation 'ν' of O₂ is 2, then the 'ν' of CH₄ is 1;

4) m(CH₄)=ν(CH₄)*M(CH₄)=1*16=16 (gr.).

what ways we can prevent rusting? Explain.​

Answers

Answer:

By electroplating

Explanation:

The metal which requires to be protected from rust is connected to the anode while a lower metal in the series is connected to the cathode, the a complete circuit is made and current is passed through the circuit there by resulting to the coating of the metal at the anode.

true or false? scientists who study processes that take millions of years cannot conduct experiments

Answers

Answer:

true

Explanation:

earth haven't been around for 1 million years yet

Answer:

true

Explanation:

they wouldnt be alive to complete the process

Which of the following statements best defines the actual yield of a reaction? (5 points)

Select one:
a.The amount of product measured after a reaction
b.The ratio of measured yield over theoretical yield
c.The maximum amount of product that can be obtained Incorrect
d.The ratio of measured yield over stoichiometric yield

Answers

Answer:

its d

. . . . . . . .. . . . . . . . .. .

Are ALL acids and based
harmful to humans?

Answers

No they aren’t all harmful there are acids like citric acid that are ok to consume

a hurricane that is rated category 5 would have which of the characteristics below?
A winds of 119-153 miles per an hour with little to no damage to structers
B winds of 154-177 miles per an hour with some damage to roofs doors or windows
C winds of 178-209 miles per hour with damage to homes
D winds of 251 miles per hour or higher with complete roof failure and possible complete structure failure

Answers

D is the right answer for this one

Fine the volume in dm³ and in Ml of a 0.505 mol/dm of NaCH required to react with 40Ml of 0.505 molar concentration of H2SO4​

Answers

watch u know bout rolling down in the deep...

Hello please I need help

Answers

It is d hope that helps!!
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