0.402 L = 402 mL<br> 0.0202 L = 20.2 mL<br> 3.6 L = 360 mL<br> 11.1 L = 11,100 mL

Which of the following is an example of maintaining homeostasis?

dsp73

Answer: Drinking water

Explanation:

Your body has a drive to maintain homeostasis in the body that keep it stable. Water, Food, Warmth are all exaamples of things your body needs an ample supply of the maintain homeostasis.

6 0

3 years ago

Write this in a word and skeleton equation:

DiKsa [7]

Answer:

Write this in a word and skeleton equation:

Solid silver chloride and an aqueous solution of nitric acid are produced when a solution of silver nitrate is reacted with a solution of hydrochloric acid.

Explanation:

6 0

2 years ago

A chemist combined 0.440 L of an unknown calcium solution with an excess of ammonium chromate. This resulted in the precipitatio

Rina8888 [55]

The concentration of the original calcium ions is 0.005 M

<h3>What is concentration?</h3>

The term concentration has to do with the amount of substance in solution. We know that the concentration can be measured in a lot of units such as mole/litre, grams per litre, percentage and so on.

As such we have the equation;

Ca^2+(aq) + (NH4)2CrO4(aq) --------> CaCrO4(s) + 2NH4^+(aq)

Number of moles of the precipitate = 346.7 * 10^-3 g/156 g/mol

= 0.0022 moles

Now;

1 mole of Ca^2+ produces 1 mole of CaCrO4 hence 0.0022 moles of CaCrO4 was produced by 0.0022 moles of CaCrO4.

Given that the volume of the solution is 0.440 L, the concentration of the solution is; 0.0022 moles/0.440 L

= 0.005 M

Learn more about molarity:brainly.com/question/8732513

#SPJ1

6 0

1 year ago

The mass percentage of hydrochloric acid within a solution is 28.00%28.00% . Given that the density of this solution is 1.1411.1

sashaice [31]

Answer:

8.76M

Explanation:

Given that

Mass from the density = 1141g

According to the given situation the computation of molarity of the solution is shown below:-

we will took HCL solution which is 1000mL

HCl = 28% by mass

So,

Mass of HCl in 1-litre solution is

$= \frac{28}{100} \times 1141$

Which gives the result of molar mass HCI is

= 319.48g /mol

Now,

Molarity is

$= \frac{319.48}{36.45}$

Which gives results of molarity is

= 8.76M

8 0

3 years ago

A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)

rodikova [14]

Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

$\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }$

<u>Given:</u>

A mysterious white powder could be,

powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

The solute = the powder
The solvent = ethanol
The freezing point of the solvent = −114.6°C
The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

Mass of solute = 82 mg = 0.082 g
Mass of solvent = density x volume, i.e., $\boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg \ }$

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

$\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

Now we combine it with the formula of freezing point depression.

$\boxed{ \ \Delta T_f = K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

$\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }$

$\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }$

$\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }$

We get $\boxed{ \ Mr = 153.65 \ }$

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>

The molality and mole fraction of water brainly.com/question/10861444
About the mass and density of ethylene glycol as an antifreeze brainly.com/question/4053884
About the solution as a homogeneous mixture brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

7 0

3 years ago

Read 2 more answers

0.402 L = 402 mL 0.0202 L = 20.2 mL 3.6 L = 360 mL 11.1 L = 11,100 mL