All categories

3 years ago

What system maintain the pH of human blood at very near 7.4?

2 answers:

8 0

The correct answer to this question is "The Buffer System." The system that maintain the pH of human blood at very near 7.4 is called the buffer system. The buffer effect don't allow the blood to change its pH. A buffer solution is typically composed of a weak acid and its conjugate base.

koban [17]3 years ago

5 0

Buffer systems in our body maintain the pH of blood to be at an optimum level which is at a range of 7.3-7.4. A buffer is a solution that resists changes in pH during the addition of an acid or a base. Our blood has a number of buffer systems. Examples are the bicarbonate buffer system and the hemoglobin buffer system.

You might be interested in

Why must the total positive and total negative charge in an ionic compound be equal ?

uranmaximum [27]

Because it must be stable. Everything wants to reach ability when it comes down to atoms and compound/molecules etc.

7 0

4 years ago

Answer the question in the photo.

Schach [20]

Answer:

Explanation:

4 0

4 years ago

250.0 mL of 0.250 M calcium chloride is mixed with 440.0 mL of 0.155 M sodium hydroxide and a precipitation reaction occurs. Wha

Vladimir [108]

Answer:

Solid: 2.52 g

Concentrations: [CaCl₂] = 0.041 M, [NaCl] = 0.100 M

Explanation:

When calcium chloride (CaCl₂) reacts with sodium hydroxide (NaOH), a double replacement reaction occurs, forming NaCl and Ca(OH)₂. NaCl is a soluble salt, and Ca(OH)₂ is a little soluble base, thus, Ca(OH)₂ will be the precipiate.

The balanced reaction equation is:

CaCl₂(aq) + 2NaOH(aq) → 2NaCl(aq) + Ca(OH)₂(s)

The number of moles of the reactants mixed are their volume multiplied by their concentration:

nCaCl₂ = 0.250 L * 0.25 mol/L = 0.0625 mol

nNaOH = 0.440 L*0.155 mol/L = 0.0682 mol

One of the reactants is limiting, and the other is in excess. Let's suppose that CaCl₂ is limiting, then, by the stoichiometry:

1 mol of CaCl₂ ------------- 2 moles of NaOH

0.0625 mol ------------ x

By a simple direct three rule:

x = 0.125 mol of NaOH

Because there's less NaOH than the value found, NaOH must be the limiting reactant and CaCl₂ is in excess. Thus, by the stoichiometry:

1 mol of CaCl₂ ------------- 2 moles of NaOH

x ------------- 0.0682 mol

By a simple direct three rule:

2x = 0.0682

x = 0.0341 mol of CaCl₂ reacts

The number of moles of CaCl₂ that remains is: 0.0625 - 0.0341 = 0.0284 mol. The final volume is 250.0 mL + 440.0 mL = 690. mL = 0.69 L

[CaCl₂] = 0.0284/0.69 = 0.041 M

For the solube product:

2 moles of NaOH ------------ 2 moles of NaCl

0.0682 mol ------------ x

x = 0.0682 mol of NaCl formed

[NaCl] = 0.0682/0.69 = 0.100 M

For the precipitate:

2 moles of NaOH ----------- 1 mol of Ca(OH)₂

0.0682 mol ---------- x

x = 0.0341 mol of Ca(OH)₂ formed

The molar of Ca(OH)₂ is 74.0 g/mol, and the mass is the number of moles multiplied by the molar mass:

mCa(OH)₂ = 0.0341*74 = 2.52 g

8 0

3 years ago

Arrange the following particles from smallest to largest. F, F^-1, Br, Br^-1

Delvig [45]

Explanation:

F< F^-1 < Br < Br^-1

I'm new here

anyone be my friend

3 0

3 years ago

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 2.001 g sample of B

Alborosie

Answer: The empirical formula for the given compound is $C_{15}H_{24}O_1$

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, iron and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=5.995g$

Mass of $H_2O=1.963g$

Mass of sample = 2.001 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 5.995 g of carbon dioxide, $\frac{12}{44}\times 5.995=1.635g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 1.963 g of water, $\frac{2}{18}\times 1.963=0.218g$ of hydrogen will be contained.

Mass of oxygen in the compound = (2.001) - (1.635 + 0.218) = 0.148 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.635g}{12g/mole}=0.136moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.218g}{1g/mole}=0.218moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.148g}{16g/mole}=0.0092moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0092 moles.

For Carbon = $\frac{0.136}{0.0092}=14.78\approx 15$

For Hydrogen = $\frac{0.218}{0.0092}=23.69\approx 24$

For Oxygen = $\frac{0.0092}{0.0092}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 15 : 24 : 1

Hence, the empirical formula for the given compound is $C_{15}H_{24}O_1$

3 0

4 years ago