1. How is the law of conservation of mass shown by a balanced chemical equation?

A(n) ____________________ is a compound that turns litmus blue and is often found in soaps and detergents.

ehidna [41]

B. base
_______________________________________________________________

8 0

3 years ago

A plot of binding energy per nucleon Eb A versus the mass number (A) shows that nuclei with a small mass number have a small bin

Illusion [34]

Answer:

6He = 4.90 MeV/Nucleon

8Li = 5.18 MeV / Nucleon

62Ni = 8.82 MeV / Nucleon

115 In = 8.54 MeV / Nucleon

Explanation:

Our strategy here is to remember that when the mass of a given nuclei is calculated from the sum of the mass of its protons and neutrons, this mass is greater than the actual value . This is the mass defect.

Now this mass defect we can convert to energy by utilizing Einstein´s equation, E = mc².This is the binding energy.

For 6He with actual mass 6.0189 u ( He has Z = 2, that is 2 protons )

mass protons = 2 x 1.0078 u = 2.0516 u

mass neutrons = 4 x 1.0087 u = 4.0348 u

predicted mass = (2.0516 + 4.0348) u = 6.0504 u

mass defect = (6.0504 - 6.0189) u = 0.0315 u

Now we need to convert this mass expressed in atomic mass units to kilograms ( 1 u = 1.66054 x 10⁻²⁷ Kg )

0.0315 u x 1.66054 x 10⁻²⁷ Kg =5.231 x 10⁻²⁹ Kg

E =5.231x 10⁻²⁸ Kg x (3 x 10⁸ m/s )² = 4.707 x 10⁻¹²J

Finally we will convert this energy in Joules to eV

E = 4.707 x 10⁻¹² J x 6.242 x 10¹⁸ eV/J = 2.94 x 10⁷ eV = 29.4 MeV

E per nucleon for 6He = 29.4 MeV / 6 =4.90 MeV / Nucleon

Now the calculations for the rest of the nuclei are performed in similar manner with the following results:

8Li = 5.18 MeV / Nucleon

62Ni = 8.82 MeV / Nucleon

115 In = 8.54 MeV / Nucleon

8 0

3 years ago

Which statement about the density of warm water is most likely correct?

Damm [24]

Density is a property of a material which describes the mass of a material per unit volume. Density is said to be slightly dependent on temperature. We look at the density of water at different temperatures:
<span>
100 </span>°C: 958.4 kg/m^360 °C: 983.2 kg/m^320 <span>°C</span>: 998.2 kg/m^3

Therefore, warm water has a lower density than water in colder temperature.

3 0

3 years ago

Suppose you need to prepare 141.9 mL of a 0.223 M aqueous solution of NaCl. What mass of NaCl do you need to use to make the sol

Afina-wow [57]

Answer:

1.811 g

Explanation:

The computation of the mass need to use to make the solution is shown below:

We know that molarity is

$Molarity = \frac{Number\ of\ moles}{Volume\ in\ L}$

So,

$Number\ of\ moles = Molarity\ \times Volume\ in\ L$

$= 0.223\times 0.141$

= 0.031 moles

Now

$Mass = moles \times Molecualr\ weight$

where,

The Molecular weight of NaCl is 58.44 g/mole

And, the moles are 0.031 moles

So, the mass of NaCL is

$= 0.031 \times 58.44$

= 1.811 g

We simply applied the above formulas

3 0

2 years ago

Which pair shares the same empirical formula?

ZanzabumX [31]

Answer:- 3. $CH_3$ and $C_2H_6$

Explanations:- An empirical formula is the simplest whole number ratio of atoms of each element present in the molecule/compound.

For example, the molecular formula of benzene is $C_6H_6$ . The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.

In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and the empirical formula is $CH_2$ . In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.

In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is $CH_2$ . The C to H ratio for second molecule is 1:4, so the empirical formula is $CH_4$ . Here also, the empirical formulas are not same and hence it is also not the right choice.

In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is $CH_3$ . In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also $CH_3$ . Hence. this is the correct choice.

In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is $CH_2$ . As the empirical formulas are different, it is not the right choice.

So, the only and only correct pair is the third one. 3. $CH_3$ and $C_2H_6$

4 0

3 years ago

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