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2 years ago

. Each antibody is designed to fight off how many pathogens

1 answer:

4 0

Antibodies can destroy pathogens by (i) binding to and blocking the pathogen's receptors, thus causing neutralization of the pathogen, (ii) binding to the pathogen and activating complement, and (iii) binding to the pathogen and facilitating its opsonization and uptake by macrophages, which utilize their Fc receptors ...

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C3H8, + 5O2

Anestetic [448]

Answer:

$from \: the \: equation \\ 1 \: moles \: of \: propane \: produce \: 4 \: moles \: of \: water \\ 2.50 \:moles \: of \: propane \: will \: produce \: ( \frac{(2.50 \times 4)}{1} ) \: moles \\ = 10 \: moles \: of \: water$

4 0

3 years ago

Krypton has a higher melting point than argon because of its:

dezoksy [38]

<h3>Answer:</h3>

Krypton has a higher melting point than argon because of its stronger dispersion forces.

<h3>Explanation:</h3>

Dispersion forces also known as London Dispersion forces are found in non polar compounds. These interactions take place in two steps.

Step 1: Instantaneous Dipole:

In non polar compounds the electron density on atom is symmetrical. When these symmetrical atoms approaches second symmetrical atom, a disturbance in electron densities occur due to repulsion between electrons. Due to repulsion the electrons density change there position and for a small period of time and instantaneous dipole is produced on an atom.

Step 2: Induced Dipole:

The dipole produced in step one when approaches another symmetrical atom, the partial positive part (low electron density site) attracts the electron from symmetrical atom and induces polarity in it. In this way the non polar atoms become polar and interacts with each other.

<h3>Factors Effecting Strength of Dispersion Forces:</h3>

i) Size of Atom / Molecule:

Greater the size of an atom greater will be the dispersion forces and vice versa.

ii) Shape of Molecules:

Greater the branching on molecule, weaker will be the dispersion forces and vice versa.

<h3>Conclusion:</h3>

As the size of Krypton is greater than Argon therefore, it will have stronger dispersion forces and will melt at higher temperature than Argon.

5 0

3 years ago

Why do scientists measure and record the mass of objects rather than their weight?

Zepler [3.9K]

Because the mass is how much of something is in an object, gold and silver can have the same mass but have very different weights.

Hope I helped!

8 0

3 years ago

Oxidation unit test

3241004551 [841]

In this case, according to the given information about the oxidation numbers anf the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.

Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.

Next, we can write the following $x$ , since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:

$Mn^xI_2^-\\\\Mn ^xO_2^{-2}$

Next, we multiply each anion's oxidation number by the subscript, to obtain the following:

$Mn^xI_2^-\rightarrow x-2=0;x=+2\\\\Mn ^xO_2^{-2}\rightarrow x-4=0;x=+4$

Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.

Learn more:

8 0

3 years ago

A 1.00 kg sample of Sb2S3 (s) and a 10.0 g sample of H2 (g) are allowed to react in a 25.0 L container at 713 K. At equilibrium,

Scorpion4ik [409]

<u>Answer:</u> The value of $K_c$ is coming out to be 0.412

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For $Sb_2S_3$ </u>

Given mass of $Sb_2S_3$ = 1.00 kg = 1000 g (Conversion factor: 1 kg = 1000 g)

Molar mass of $Sb_2S_3$ = 339.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of }Sb_2S_3=\frac{1000g}{339.7g/mol}=2.944mol$

<u>For hydrogen gas:</u>

Given mass of hydrogen gas = 10.0 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5mol$

<u>For hydrogen sulfide:</u>

Given mass of hydrogen sulfide = 72.6 g

Molar mass of hydrogen sulfide = 34 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen sulfide}=\frac{72.6g}{34g/mol}=2.135mol$

The chemical equation for the reaction of antimony sulfide and hydrogen gas follows:

$Sb_2S_3(s)+3H_2(g)\rightarrow 2Sb(s)+3H_2S(g)$

Initial: 2.944 5

At eqllm: 2.944-x 5-3x 2x 3x

We are given:

Equilibrium moles of hydrogen sulfide = 2.135 moles

Calculating for 'x', we get:

$\Rightarrow 3x=2.135\\\\\Rightarrow x=\frac{2.135}{3}=0.712$

Equilibrium moles of hydrogen gas = (5 - 3x) = (5 - 3(0.712)) = 2.868 moles

Volume of the container = 25.0 L

Molarity of a solution is calculated by using the formula:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}$

The expression of $K_c$ for above equation, we get:

$K_c=\frac{[H_2S]^3}{[H_2]^3}$

The concentration of solids and liquids are not taken in the expression of equilibrium constant.

$K_c=\frac{(\frac{2.135}{25})^3}{(\frac{2.868}{25})^3}\\\\K_c=0.412$

Hence, the value of $K_c$ is coming out to be 0.412

3 0

3 years ago