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vfiekz [6]
3 years ago
13

Energy conversion means reducing energy use True or false

Chemistry
1 answer:
I am Lyosha [343]3 years ago
3 0
False, energy conversion just means the energy is going to be used by another force
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Adding sugar to your tea physical or chemical change ​
ExtremeBDS [4]

Answer:

physical

Explanation:

no chemical reaction occurs

6 0
3 years ago
Read 2 more answers
If you are given a 0. 31 g piece of sodium metal to react with water, how many moles of hcl would it take to neutralize the sodi
Sphinxa [80]

The moles of HCl to neutralize the sodium hydroxide produced is<u> 0.0135 mole.  </u>

Neutralization or neutralization is a chemical response wherein an acid and a base react quantitatively with each other. In a reaction in water, neutralization outcomes in there being no excess of hydrogen or hydroxide ions gift in the answer.

<u>calculation:-</u>

<u />

2Na + 2H₂O  -----> 2NaOH + H₂

2 mol or 46g of Na produces 80 grams of NaOH

∴ 0.31 g of Na will produce = 80/46 × 0.31

                                              =  0.54 gram of NaOH.

mol of NaOH = 0.54/40

                      = 0.0135

Since both Hcl and NaOH have the same valance factor,

1 mole NaOH is needed to neutralize 1 mol HCl

∴ 0.0135 mole of NaOH will need = 0.0135 mole of HCl

mass = 0.0135 × 36.5

         =<u> 0.493 grams of HCL.</u>

Learn more about neutralizing here:-brainly.com/question/23008798

#SPJ4

4 0
1 year ago
Sighting along the C2-C3 bond of 2-methylbutane, the least stable conformation (Newman projection) has a total energy strain of
natima [27]

Answer:

21 KJ/mol

Explanation:

For this question, we have to start with the <u>linear structure</u> of 2-methylbutane. With the linear structure, we can start to propose all the <u>Newman projections</u> keep it in mind that the point of view is between carbons 2 and 3 (see figure 1).

Additionally, we have several <u>energy values for each interaction</u> present in the Newman structures:

-) Methyl-methyl <em>gauche: 3.8 KJ/mol</em>

-) Methyl-H <em>eclipse: 6.0 KJ/mol</em>

-) Methyl-methyl <em>eclipse: 11.0 KJ/mol</em>

-) H-H <em>eclipse:</em> 4.0 KJ/mol

Now, we can calculate the energy for each molecule.

<u>Molecule A</u>

In this molecule, we have 2 Methyl-methyl <em>gauche </em>interactions only, so:

(3.8x2) = 7.6 KJ/mol

<u>Molecule B</u>

In this molecule, we have a Methyl-methyl <em>eclipse </em>interaction a Methyl-H <em>eclipse </em>interaction and an H-H <em>eclipse</em> interaction, so:

(11)+(6)+(4) = 21 KJ/mol

<u>Molecule C</u>

In this molecule, we have 1 Methyl-methyl <em>gauche </em>interaction only, so:

3.8 KJ/mol

<u>Molecule D</u>

In this molecule, we have three Methyl-H <em>eclipse </em>interaction, so:

(6*3) = 18 KJ/mol

<u>Molecule E</u>

In this molecule, we have 1 Methyl-methyl <em>gauche </em>interaction only, so:

3.8 KJ/mol

<u>Molecule F</u>

In this molecule, we have a Methyl-methyl <em>eclipse </em>interaction a Methyl-H <em>eclipse </em>interaction and an H-H <em>eclipse</em> interaction, so:

(11)+(6)+(4) = 21 KJ/mol

The structures with higher energies would be less stable. In this case, structures B and F with an energy value of 21 KJ/mol (see figure 2).

I hope it helps!

3 0
3 years ago
What label is given to an orbital with quantum numbers n=4 and l=1?
shtirl [24]

The principal quantum number, <span>nn</span>, designates the principal electron shell. Because n describes the most probable distance of the electrons from the nucleus, the larger the number n is, the farther the electron is from the nucleus, the larger the size of the orbital, and the larger the atom is. n can be any positive integer starting at 1, as <span><span>n=1</span><span>n=1</span></span> designates the first principal shell (the innermost shell). The first principal shell is also called the ground state, or lowest energy state. This explains why <span>nn</span> can not be 0 or any negative integer, because there exists no atoms with zero or a negative amount of energy levels/principal shells. When an electron is in an excited state or it gains energy, it may jump to the second principle shell, where <span><span>n=2</span><span>n=2</span></span>. This is called absorption because the electron is "absorbing" photons, or energy. Known as emission, electrons can also "emit" energy as they jump to lower principle shells, where n decreases by whole numbers. As the energy of the electron increases, so does the principal quantum number, e.g., n = 3 indicates the third principal shell, n = 4 indicates the fourth principal shell, and so on.

3 0
3 years ago
2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an
Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression d=\frac{m}{V} that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL

d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL

d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL

d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL

The problem says that all the samples have the same mass, so:

m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

V_{lithium}=\frac{m}{d_{lithium}}

V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m

V_{lithium}=1.88\frac{mL}{g}*m

V_{gold}=\frac{m}{d_{gold}}

V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m

V_{gold}=5.18*10^{-2}\frac{mL}{g}*m

V_{aluminum}=\frac{m}{d_{aluminum}}

V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m

V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m

V_{lead}=\frac{m}{d_{lead}}

V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m

V_{lead}=8.85*10^{-2}\frac{mL}{g}*m

If we assume m = 1g, we find that:

V_{lithium}=1.88mL

V_{gold}=5.18*10^{-2}mL

V_{aluminum}=3.70*10^{-1}mL

V_{lead}=8.85*10^{-2}mL

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

4 0
2 years ago
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