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2 years ago

Describe a second degree burn

Chemistry

1 answer:

Fittoniya [83]2 years ago

6 0

Answer:

Second-degree burns, or partial thickness burns, are more severe than first-degree burns. They affect the outer layer of skin, called the epidermis, and part of the second layer of skin, called the dermis. Second-degree burns can be very painful and often take several weeks to heal.

Explanation:

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What is the hardest subject u ever taking, mine is chemistry

artcher [175]

Answer:

Mine is language arts brainliest?

Explanation:

7 0

2 years ago

Which of the following is a list of the minimum amount of data needed for determining the molar enthalpy of solution of KCl(s) i

Gre4nikov [31]

Answer:

Explanation:

molar enthalpy of solution of KCl(s) is heat evolved or absorbed when one mole of KCl is dissolved in water to make pure solution . The heat evolved or absorbed can be calculated by the following relation.

Q = msΔt where m is mass of solution or water , s is specific heat and Δt is change in temperature of water .

So data required is mass of water or solution , initial and final temperature of solution , specific heat of water is known .

Now to know molar heat , we require mass of solute or KCl dissolved to know heat heat absorbed or evolved by dissolution of one mole of solute .

3 0

3 years ago

some inkjet printers produce picoliter-sized drops. how many water molecules are there in one picoliter of water? the density of

GrogVix [38]

3.37 x 10¹⁰ molecules

Explanation:

Given parameters:

Volume of water = 1pL = 1 x 10⁻¹²L

Density of water = 1.00g/mL = 1000g/L

Unknown:

Number of water molecules = ?

Solution:

To solve this problem, we first find the mass of the water molecule in the inkjet.

Mass of water = density of water x volume of water

Then, the number of molecules can be determined using the expression below:

number of moles = $\frac{mass of water}{molar mass of water}$

Number of molecules = number of moles x 6.02 x 10²³

Solving:

Mass of water = 1 x 10⁻¹² x 1000 = 1 x 10⁻⁹g

Number of moles:

Molar mass of H₂O = 2 + 16 = 18g/mol

Number of moles = $\frac{1 x 10^{-12} }{18}$ = 5.6 x 10⁻¹⁴moles

Number of molecules = 5.6 x 10⁻¹⁴ x 6.02 x 10²³ = 33.7 x 10⁹

= 3.37 x 10¹⁰ molecules

Learn more:

Number of molecules brainly.com/question/4597791

#learnwithBrainly

4 0

3 years ago

Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc

Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is $C_5H_7N$

(b) The empirical formula for the given compound is $C_4H_5N_2O$

Explanation:

<u>Part A: nicotine </u>

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles$

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 1.23 moles.

For Carbon = $\frac{6.17}{1.23}=5.01\approx 5$

For Hydrogen = $\frac{8.70}{1.23}=7.07\approx 7$

For Nitrogen = $\frac{1.23}{1.23}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is $C_5H_7N_1=C_5H_7N$

<u>Part B: caffeine</u>

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles$

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 1.03 moles.

For Carbon = $\frac{4.12}{1.03}=4$

For Hydrogen = $\frac{5.19}{1.03}=5.03\approx 5$

For Nitrogen = $\frac{2.06}{1.03}=2$

For Nitrogen = $\frac{1.03}{1.03}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is $C_4H_5N_2O_1=C_4H_5N_2O$

6 0

3 years ago

If a car is traveling 100 km/h and comes to a stop in 3 minutes,what is acceleration of a passenger who is using vehicle restrai

Rama09 [41]

We are given –

Final velocity of car is, v= 0
Initial velocity of car is, u= 100 km/hr
Time taken, t is = 3 minutes or 180 sec

Here–

$\qquad$ $\pink{\bf \longrightarrow Initial\: velocity = 100 \:km/hr}$

$\qquad$ $\sf \longrightarrow Initial\: velocity = \dfrac{ 100 \times 1000}{3600} \:m/s$

$\qquad$ $\pink{ \bf \longrightarrow Initial\: velocity = 27.78\: m/s}$

Now –

$\qquad$ ____________________________

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{Final\: Velocity -Initial \:Velocity }{Time}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{v -u}{t}}$

$\qquad$ $\sf \longrightarrow Acceleration = \dfrac{(0- 27.78)}{1800}$

$\qquad$ $\sf \longrightarrow Acceleration =\cancel{ \dfrac{- 27.78}{1800}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = -0.15 \: m/s^2}$

$\qquad$ _______________________________

6 0

2 years ago

Describe a second degree burn ​

Describe a second degree burn