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3 years ago

Franco was told that he had mycarditis. He went home and researched the disease, and he determined that he had

Chemistry

1 answer:

Reil [10]3 years ago

6 0

Answer:

Myocarditis = C) Inflammation of the muscular wall of the heart.

Explanation:

Myocarditis is a condition usually caused by a virus.

It causes inflammation within the heart, generally along the muscular wall and within the middle layers.

It's definitely not a great condition. It's possible it could weaken his heart's ability to pump properly, which in turn could effect organs and cause organ failure. It'd lead to intense bouts of pain constantly, swelling, problems breathing, or even heart failure.

Feel bad for Franco.

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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]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

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:

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

Calculation for freezing point of solution:

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

Calculation for boiling point of solution:

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

Which subatomic particle do you think is most important to an atom’s identity?

Sonja [21]

The subatomic particle that identifies the atom is the number of protons. This is what distinguishes an element that is is flammmable, hydrogen to one that is essential component in water, oxygen.

5 0

3 years ago

1. A cold front forms when-

Aleks [24]

A cold air mass moves into an area of warm air

4 0

3 years ago

What happens when a chemical is determined to be dangerous?

Vlada [557]

Answer:

Explanation:

8 0

3 years ago

Fritz-Haber process

maks197457 [2]

Answer:

5×10⁵ L of ammonia (NH3)

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above, we can say that:

3 L of H2 reacted to produce 2 L of NH3.

Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:

From the balanced equation above,

3 L of H2 reacted to produce 2 L of NH3.

Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.

Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.

6 0

3 years ago