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3 years ago

33 protons 36 electrons 45 neutrons

Chemistry

1 answer:

Artist 52 [7]3 years ago

6 0

As has an atomic number of 33, so it has 33 protons.

It has a charge of 3-, so there are three more electrons than protons. Thus, there are 36 electrons.

It has a mass of 75, which is the sum of neutrons and protons.

33+n=75 ---> n = 75 - 33 = 42 neutrons

The answer is e) 33 protons, 42 neutrons and 36 electrons.

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A purified protein has a molecular mass of 360 kDa when measured by size exclusion chromatography. When analyzed by gel electrop

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Answer:

A protein has four subunits whose molecular masses are 140, 80, and 60 kDa.

A disulfide bond links the two 80 kDa subunits (possibly identical).

Explanation:

Given that:

A protein has four subunits whose molecular masses are 140, 80, and 60 kDa.

A disulfide bond links the two 80 kDa subunits (possibly identical).

As a result of SDS and dithiothreitol analysis treatment, the molecular masses can not be 360 in total. They are 280, which implies that they are in short of 80 kDa. This means that there are possibilities that two groups with a molecular mass of 80 kDa which are joined by a disulfide bond.

The presence of SDS and dithiothreitol acts as a reducing agent, and they can break disulfide bonds whose pH is greater than 7, i.e. those in basic condition.

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In an electroplating apparatus, where does the reduction reaction occur?

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2 years ago

46.6 grams of mercury II sulfate (HgSO4) reacts with an excess of sodium Chloride (NaCl). How many grams of mercury II chloride

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Answer:

$m_{HgCl_2}=42.7gHgCl_2$

Explanation:

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In this case, the undergoing chemical reaction is:

$HgSO_4+2NaCl\rightarrow HgCl_2+Na_2SO_4$

In such a way, the mercury II sulfate (molar mass 296.65g/mol) is in a 1:1 molar ratio with the mercury II chloride (molar mass 271.52g/mol), for that reason the stoichiometry to find mass in grams of mercury II chloride turns out:

$m_{HgCl_2}=46.6gHgSO_4*\frac{1molHgSO_4}{296.65 gHgSO_4}*\frac{1molHgCl_2}{1molHgSO_4} *\frac{271.52gHgCl_2}{1molHgCl_2} \\\\m_{HgCl_2}=42.7gHgCl_2$

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3 years ago

What did ancient astronomers think areas of the moon called mares might be?

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3 years ago

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

5 0

3 years ago