Ask question

Ask question

All categories

2 years ago

10

Bicarbonate is a solution present in human blood. Bicarbonate helps to maintain our bodies pH and resists pH change. Bicarbonate

is an example of a(n) ______________________.
Indicator

Buffer

Acid

2 answers:

Elena L [17]2 years ago

8 0

Answer:

buffer

Explanation:

yulyashka [42]2 years ago

3 0

Yeah what they said^

You might be interested in

Name a Solid, Liquid, and a Gas. Solid: Liquid: Gas:

crimeas [40]

Answer:

Gas - Steam

Solid - Rock

Liquid - Juice

3 0

3 years ago

Read 2 more answers

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

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

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

<u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

<u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

3 years ago

Carbon, hydrogen and ethane each burn exothermically in an excess of air. AHⓇ =-393.7 kJ mol. C(s) + O2(g) → CO2(g) H2(g) + % O2

Salsk061 [2.6K]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is 51.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the reaction of carbon and water follows:

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_1=-393.7kJ$ ( × 2)

(2) $H_2+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_2=-285.9kJ$ ( × 2)

(3) $2C_2H_4(s)+2O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_3=-1411kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[2\times \Delta H_1]+[2\times \Delta H_2]+[1\times (-\Delta H_3)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(2\times (-393.7))+(2\times (-285.9))+(1\times -(-1411))]=51.8kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is 51.8 kJ.

6 0

3 years ago

Please help? Liquid, Chemical Properties, Compounds or Products?

Rainbow [258]

Answer:

They are compounds.

Explanation:

..........

3 0

2 years ago

Read 2 more answers

How many molecules of O2 will be required to produce 28.8 g of water?

Mrac [35]

I think 1.67 x 10^20

5 0

3 years ago