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3 years ago

What is the relative atomic mass of ,A of X?

Chemistry

1 answer:

Fed [463]3 years ago

7 0

11 A n 8 g sample of oxygen atoms contains the same number of atoms as 16 g of element x.
The question does not make sense. does it mean 11 g of A with 8 g Oxygen and 16 g X with 8 g Oxygen.

Does not make sense

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1. A solution is a mixture. The part of<br> the solution in the greater amount is<br> called the

olga_2 [115]

Answer:

Solvent

Explanation:

Many people use the solute to describe the solid being dissolved and the solvent to describe the thing doing the dissolving, but really solvent means the part of the solution with a greater amount/concentration.

For example, if you have 1 gram of salt in 10 liters of water, the water is the solvent.

5 0

3 years ago

How do ions impact your life

Ostrovityanka [42]

Answer:

Increasing your concentration of negative ions is a very important way to improve your health for a variety of reasons. They work to increase serotonin levels in your body, which improves mood and balances your mental health, helping combat depression and anxiety.

Explanation:

7 0

3 years ago

Be sure to answer all parts. What is the effect of each of the following on the volume of 1 mol of an ideal gas? (a) The pressur

aksik [14]

Answer

A)The volume decreases by a factor of 4

B), the volume has increased by factor of 2

Explanation:

A)Given:

P1= 760Kpa

P2 =202Kpa

The temperature changes from37C to155C

There is increase In pressure from P1 to P2

P1= 760torr.

We need to convert to Kpa

But, 1atm= 760torr

Then 760torr 101000pa

Then 101000pa = 101Kpa

We need to convert the temperature from Celsius to Kelvin

T1= 37+273= 310K

But from ideal gas, we know that PV = nRT where nR is constant

Where P= pressure

V= volume

T= temperature

n = number of moles

(P1V1/T1)=(P2V2/T2)

V1/V2 = P2/P1 * T1/T2

V1/V2 = (202/101)*(310/155)

V1/V2=4

$V2= V1/4$

Therefore, the volume has decreased by factor of 4

Given:

P1= 2atm

P2 =101Kpa

The temperature changes from 305K to 32C

There is increase In pressure from P1 to P2

P1= 2atm

We need to convert to Kpa

But, 1atm= 760torr

Then 760torr 101000pa

Then 101000pa = 101Kpa

P1= 202.65kpa

We need to convert the temperature from Celsius to Kelvin

T2= 32+273= 305K

But from ideal gas, we know that PV = nRT

Where P= pressure

V= volume

T= temperature

n = number of moles

(P1V1/T1)=(P2V2/T2)

V1/V2 = P2/P1 * T1/T2

V1/V2 = (202/101)

V1/V2 = (101/202.65)*(305/305)

V1/V2 = 1/2

$V2=2V1$

Therefore, the volume has increased by factor of 2

4 0

3 years ago

Which of the following properties of a protein is least likely to be affected by changes in pH? Tertiary structure Primary struc

chubhunter [2.5K]

Answer:

Primary structure is the correct answer.

Explanation:

The primary structure is the simple level of protein structure.

Primary structure is a basic amino acids sequences in a protein.

In the primary structure, amino acids are attached together by a covalent bond.
Primary structure is when the amino acids are joined together with peptide bonds to produce polypeptide chains

Changes in pH are least likely to change the amino acid sequence or disrupt peptide bonds.

Explanation:

7 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