WILL GIVE BRAINLIEST

Why the statement all atoms of one element have the same mass is false

White raven [17]

Some elements have isotopes which have a different number of neutrons, and this means they have different masses.

4 0

3 years ago

Calculate the heat needed to be removed to freeze 45g of water at 0C

alexira [117]

Answer: The heat needed to be removed to freeze 45.0 g of water at 0.0 °C is 15.01 KJ.

Explanation:

Firstly, we need to define the term "latent heat" which is the amount of energy required "absorbed or removed" to change the phase "physical state; solid, liquid and vapor" without changing the temperature.
Types of latent heat: depends on the phases that the change occur between them;

Liquid → vapor, latent heat of vaporization and energy is absorbed.
Vapor → liquid, latent heat of liquification and the energy is removed.
Liquid → solid, latent heat of solidification and the energy is removed.
Solid → liquid, latent heat of fusion and the energy is absorbed.

In our problem, we deals with latent heat of freezing "solidification" of water.
The latent heat of freezing of water, ΔHf, = 333.55 J/g; which means that the energy required to be removed to convert 1.0 g of water from liquid to solid "freezing" is 333.55 g at 0.0 °C.
Then the amount of energy needed to be removed to freeze 45.0 g of water at 0.0 °C is (ΔHf x no. of grams of water) = (333.55 J/g)(45.0 g) = 15009.75 J = 15.01 KJ.

4 0

4 years ago

Phosphoric acid is a triprotic acid with the following pKa values:pKa1=2.148, pKa2=7.198, pKa3=12.375You wish to prepare 1.000 L

laila [671]

Answer:

NaH₂PO₄ = 1.876 g

Na₂HPO₄ = 4.879 g

Combinations: H₃PO₄ and Na₂HPO₄; H₃PO₄ and Na₃HPO₄

Explanation:

To have a buffer at 7.540, the acid must be in it second ionization, because the buffer capacity is pKa ± 1. So, we must use pKa2 = 7.198

The relation bewteen the acid and its conjugated base (ion), is given by the Henderson–Hasselbalch equation:

pH = pKa + log[A⁻]/[HA], where [A⁻] is the concentration of the conjugated base, and [HA] the concentration of the acid. Then:

7.540 = 7.198 + log[A⁻]/[HA]

log[A⁻]/[HA] = 0.342

[A⁻]/[HA] = $10^{0.342}$

[A⁻]/[HA] = 2.198

[A⁻] = 2.198*[HA]

The concentration of the acid and it's conjugated base must be equal to the concentration of the buffer 0.0500 M, so:

[A⁻] + [HA] = 0.0500

2.198*[HA] + [HA] = 0.0500

3.198*[HA] = 0.0500

[HA] = 0.01563 M

[A⁻] = 0.0500 - 0.01563

[A⁻] = 0.03436 M

The mix reaction is

NaH₂PO₄ + Na₂HPO₄ → HPO₄⁻² + 3Na + H₂PO₄⁻

The second ionization is:

H₂PO₄⁻ ⇄ HPO₄⁻² + H⁺

So, H₂PO₄⁻ is the acid form, and its concentration is the same as NaH₂PO₄, and HPO₄⁻² is the conjugated base, and its concentration is the same as Na₂HPO₄ (stoichiometry is 1:1 for both).

So, the number of moles of these salts are:

NaH₂PO₄ = 0.01563 M * 1.000 L = 0.01563 mol

Na₂HPO₄ = 0.03436 M* 1.000 L = 0.03436 mol

The molar masses are, Na: 23 g/mol, H: 1 g/mol, P: 31 g/mol, and O = 16 g/mol, so:

NaH₂PO₄ = 23 + 2*1 + 31 + 4*16 = 120 g/mol

Na₂HPO₄ = 2*23 + 1 + 31 + 4*16 = 142 g/mol

The mass is the number of moles multiplied by the molar mass, so:

NaH₂PO₄ = 0.01563 mol * 120 g/mol = 1.876 g

Na₂HPO₄ = 0.03436 mol * 142 g/mol = 4.879 g

To prepare this buffer, it's necessary to have in solution the species H₂PO₄⁻ and HPO₄⁻², so it can be prepared for mixing the combination of:

H₃PO₄ and Na₂HPO₄ (the acid is triprotic so, it will form the H₂PO₄⁻ , and the salt Na₂HPO₄ will dissociate in Na⁺ and HPO₄²⁻);

H₃PO₄ and Na₃HPO₄ (same reason).

The other combinations will not form the species required.

8 0

3 years ago

Would using a strainer be filtration?

katrin2010 [14]

Answer:

To some extent, yes

Explanation:

8 0

3 years ago

When NEM is added to a purified solution of creatine kinase, Cys 278 is alkylated, but no other Cys residues in the protein are

USPshnik [31]

Answer:

The answer is given below

Explanation:

Cys 278 residue is the only available cysteine which is alkylated by the addition of N-Ethylmaleimide or NEM (alkylating agent). It works by only alkylating the sulfhydryls. In this case, Cys 278 residue is the only one which has exposed cysteine residue.

While the other residues have their sulfhydryls group either involved in the synthesis of disulfide bonds of proteins or their Cys residues are intrinsically placed in the proteins and cannot be alkylated with NEM.

NEM cannot alkylate if its protein is not available in the free form or it is in bounded form. For NEM to alkylate Cys 278, it should be free and should have sulfhydryls available for alkylation.

Alkylation: it is the transfer of alkyl groups. Alkyl groups contain Hydrogen and Carbon in their structure.

5 0

4 years ago