What type of equation is it and balance it

The normal range of the pH of blood is between 7.35 and 7.45; variations beyond this range have significant health implications.

patriot [66]

Answer:

The hydrogen ion concentrations associated with these pH value 7.35 is $4.5\times 10^{-8} M$

The hydrogen ion concentrations associated with these pH value 7.45 is $3.6\times 10^{-8} M$ .

Explanation:

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

1) pH = 7.35

Putting values in above equation, we get:

$7.35=-\log[H^+]$

$[H^+]=4.467\times 10^{-8} M\approx 4.5\times 10^{-8} M$

The hydrogen ion concentrations associated with these pH value 7.35 is $4.5\times 10^{-8} M$

2) pH = 7.45

Putting values in above equation, we get:

$7.45=-\log[H^+]$

$[H^+]=3.548\times 10^{-8}M \approx 3.6\times 10^{-8} M$

The hydrogen ion concentrations associated with these pH value 7.45 is $3.6\times 10^{-8} M$ .

8 0

3 years ago

You have a 150mL sample of an aqueous solution at 25C. It contains 15.2mg of an unknown nonelectrolyte compound. If the solution

Serga [27]

The molar mass of the compound:

If the solution has an osmotic pressure of 8.44 torr, then the molar mass of the unknown non-electrolyte is 223.14 g.

What is osmosis?

Osmosis is defined as the flow of solvent molecules through semi-permeable membrane.
Osmotic pressure is the pressure applied to stop the flow of solvent molecules.
It is a colligative property that means osmotic pressure depends on the number of solute particles .

Therefore,

π $V=inRT$ ( for electrolytes)

Where, π= Osmotic pressure

i = Van 't Hoff factor

n= moles

R= Gaseous constant = 62.363577 L torr $mol^{-1}K^{-1}$

T= Temperature

V= Volume of solution

Given:

T= 298K

V= 150 mL= 0.150 L

Given mass of unknown electrolyte= 15.2 mg = 15.2 x $10^{-3}$ g

Osmotic pressure= 8.44 torr

Molar mass= ?

For non-electrolytes:

πV = n RT

πV= $\frac{m}{M}$ RT

Calculations:

Putting the given values in the formula:

8.44 x 0.150 =15.2 x $10^{-3}$ / M x 62.36 x 298

1.266 = 282.5/M

M = 282.5/1.266

M = 223.14 g

Therefore,

The molar mass of the unknown non-electrolyte is 223.14g.

Learn more about Osmotic pressure here,

brainly.com/question/13680877

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7 0

2 years ago

Cl2 + H2 + 2HCI

Cloud [144]

Answer:

1 and 2

Explanation:

The given equation is:

Cl₂ + H₂ → 2HCl

A coefficient is the variable or number before a chemical specie.

In this reaction Cl₂ and H₂ are the reactants;

The coefficient of Cl₂ is 1,

H₂ is 1,

HCl is 2

The subscript is the number to the lower power after a chemical specie is denoted.

For Cl₂, it is 2

6 0

3 years ago

What is the different between a mixture and a compound?

krek1111 [17]

A compound is 2 or more substances chemically combined, mixtures arent

4 0

3 years ago

The equilibrium constant, K., for the following reaction is 83.3 at 500 K. PC13(g) + Cl2(g) = PC13(E) Calculate the equilibrium

lord [1]

Answer:

The equilibrium concentration of $PCl_5=0.228 M$ .

The equilibrium concentration of $PCl_3=0.280 M -0.228 M=0.052M$ .

The equilibrium concentration of $Cl_2=0.280 M -0.228 M=0.052M$ .

Explanation:

Answer:

The equilibrium concentration of HCl is 0.01707 M.

Explanation:

Equilibrium constant of the reaction = $K_c=83.3$

Moles of $PCl_3 = 0.280 mol$

Concentration of $[PCl_3]=\frac{0.280 mol}{1.00 L}=0.280 M$

Moles of $Cl_ = 0.280 mol$

Concentration of $[Cl_2]=\frac{0.280 mol}{1.00 L}=0.280M$

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initial: 0.280 0.280 0

At eq'm: (0.280-x) (0.280-x) x

We are given:

$[PCl_3]_{eq}=(0.280-x)$

$[Cl_2]_{eq}=(0.280-x)$

$[PCl_5]_{eq}=x$

Calculating for 'x'. we get:

The expression of $K_{c}$ for above reaction follows:

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

Putting values in above equation, we get:

$83.3=\frac{x}{(0.280-x)\times (0.280-x)}$

On solving this quadratic equation we get:

x = 0.228, 0.344

0.228 M < 0.280 M< 0.344 M

x = 0.228 M

The equilibrium concentration of $PCl_5=0.228 M$ .

The equilibrium concentration of $PCl_3=0.280 M -0.228 M=0.052M$ .

The equilibrium concentration of $Cl_2=0.280 M -0.228 M=0.052M$ .

4 0

3 years ago