All categories

4 years ago

Describe white blood cells in comparison to red blood cells

1 answer:

5 0

The main difference between red blood cells and white blood cells is in their respective functions. While red blood cells are responsible for transport of respiratory gases (oxygen and carbon dioxide), white blood cells provide defense mechanisms for fighting foreign microorganisms entering the human organism.

You might be interested in

Fill in the coefficients that will balance the following reaction:

Kazeer [188]

$\\ \rm\Rrightarrow H_2SO_4+2KOH\longrightarrow K_2SO_4+2H_2O$

On both sides

Hence balanced

6 0

2 years ago

Why does a heating curve have horizontal plateaus at certain points during the heating process? What is happening at those point

iren [92.7K]

The reason is because of the direction the plateaus is, so this affects it in the degree of the end result

6 0

3 years ago

Given the following at 25°C, calculate <img src="https://tex.z-dn.net/?f=%5CDelta%20H_f" id="TexFormula1" title="\Delta H_f" alt

N76 [4]

Answer: The enthalpy of formation of HCN(g) is 135.1 kJ/mol

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN(g))})+(6\times \Delta H_f_{(H_2O(g))})]-[(2\times \Delta H_f_{(NH_3(g))})+(3\times \Delta H_f_{(O_2(g))})+(2\times \Delta H_f_{(CH_4(g))})]$

We are given:

$\Delta H_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-80.3kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.6kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$

Putting values in above equation, we get:

$-870.8=[(2\times \Delta H_f_{(HCN(g))})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN(g))}=135.1kJ/mol$

Hence, the enthalpy of formation of HCN(g) is 135.1 kJ/mol

7 0

3 years ago

What 987kj to calories

wariber [46]

I think the answer is 233.748 calories

8 0

3 years ago

Solve for Va MaVa=MbVb

PolarNik [594]

Answer:

Va = (MbVb)/Ma

Explanation:

Divide both sides by Ma and voila!

8 0

3 years ago