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2 years ago

The size of camphor kept in store house gradually decrease in size after few а days?

Chemistry

1 answer:

zzz [600]2 years ago

4 0

Answer:

Because it goes through the process of sublimation.

Explanation:

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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:

Part A: nicotine

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 $C_5H_7N_1=C_5H_7N$

Part B: caffeine

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

When two aqueous solutions are combined the reaction forms a solid that settles out of solution. What is this solid called?

Rama09 [41]

During the chemical reaction, the solid substance that is insoluble formed during the combination of two aqueous solutions is called a precipitate. The answer is letter C. The precipitation reaction occurs when a salt is formed in the reaction of solutions with salts.

6 0

3 years ago

¿Cuál es ion monoatomico más estable formado a partir de aluminio?

Naily [24]

Answer:

Explanation:

3 0

3 years ago

Read 2 more answers

Reynolds number E. What is the mean velocity u. (ft/s) and the Reynolds number Re = pu., D/ for 35 gpm (gallons per minute) of w

Novay_Z [31]

Answer:

The mean velocity is 13 ft/s.

The Reynolds number is 88,583 and it is dimensionless.

Explanation:

We have water flowing in a pipe of 1.05 in diameter.

The density is ρ=62.3 lb/ft and the viscosity is 1.2 cP.

The mean velocity can be calculated as

$u=\frac{Q}{A}=\frac{Q}{\pi*D^2/4}=\frac{35gpm }{3.14*(1.05in)^2/4}\\\\ u=\frac{35}{0.865}*\frac{gal}{min}\frac{1}{in^2}*\frac{231in^3}{1gal}*\frac{1}{60s} \\\\ u=156\,in/s=13\,ft/s$

The Reynolds number now can be calculated for this flow as

$Re=\frac{\rho*u*D}{\mu}$

being ρ: density, u: mean velocity of the fluid, D: internal diameter of the pipe and μ the dynamic viscosity.

To simplify the calculation, we can first make all the variables have coherent units.

Viscosity

$\mu=1.2cP=\frac{1.2}{100}\frac{g}{cm*s}*\frac{1lb}{453.6g}*\frac{30.48cm}{1ft}= 0.0008\frac{lb}{ft*s}$

Diameter

$D=1.05in*(\frac{1ft}{12in} )=0.0875ft$

Then the Reynolds number is

$Re=\frac{\rho*u*D}{\mu}\\\\Re=62.3\frac{lb}{ft^3}*13\frac{ft}{s} *0.0875ft*\frac{1}{0.0008}*\frac{ft*s}{lb}\\\\Re=88,583$

8 0

3 years ago

1. What is the density of a 20 gram of 40 ml liquid?

____ [38]

Answer:

1. $0.5 g/mL$

2. $0.88 g/cm^3$

3. It has the greatest mass to volume ratio

4. Incomplete

5. $H^+ (aq)+OH^-(aq)\rightleftharpoons H_2O (l)$

6. 1

7. Gloves, goggles, coat, mask

8. Flush with tap water for at least 15 minutes

9. The facts are listed below

10. Mass and volume

12. All matter consists of moving particles, the degree of their movement is directly proportional to their kinetic energy

Explanation:

1. In order to solve for density, we should know that density is the ratio between mass and volume of a liquid. In this case, we're given both measures: given mass of m = 20 g and volume of V = 40 mL, we may simply apply the equation of density described here:

$d=\frac{m}{V}$

Substituting the variables, we obtain:

$d=\frac{m}{V}=\frac{20 g}{40 mL}=0.5 g/mL$

2. Given a mixture of several liquids, it's important to understand that liquids with a greater density will tend to form a bottom layer of a solution, while liquids with a lower density will tend to form a top layer of a solution. Here we have a liquid with a density of $d_1 = 1.0 g/cm^3$ and another liquid with a density of $d_2 = 0.88 g/cm^3$ . Notice that $d_1$ > $d_2$ .

This implies that the liquid with a density of $0.88 g/cm^3$ would be on top, as its density is lower than the density of the other liquid with a density of $1.0 g/cm^3$ .

3. The solid phase is not always, but typically denser than liquids or gases. There are some exceptions to this rule, for example, ice, a solid phase of water, is less dense than liquid water.

However, for the majority of cases this statement is true. Remember that solid phases are the most ordered phases with atoms being packed closely to each other. In liquids, atoms are more dispersed with distances between them being greater than those in solids. Similarly, gases have the greatest distances between gas atoms among all three phases.

Since density is directly proportional to mass, let's say we take the same volume of a solid, a liquid and a gas. For the same volume, since we'll have a greater number of solid atoms than for a liquid or a gas (because solid atoms are more closely packed with lower average distances between the atoms), the mass to volume ratio will be the greatest for solids.

4. This seems to be an incomplete question.

5. In order to balance the following ionic equation, we need to follow the mass and charge balancing rules. Firstly, expand a water molecule showing the individual parts of it:

$H-OH$

Secondly, notice that we need to add a hydroxide anion to the proton, so that we obtain the same number of protons and hydroxide anions on the left side, as well as the number of hydrogen and oxygen atoms on the right. This way, the net charge on the left hand side (0) and the net charge on the right hand side (0) are equal, so the charge is balanced as well. We obtain:

$H^+ (aq)+OH^-(aq)\rightleftharpoons H_2O (l)$

6. We should be familiar with the ionization constant of water in the context of this problem. It is defined as the product between the hydronium ions and hydroxide ions and is a constant number at some given temperature. For pure water, the concentration of hydronium ions is balanced by the concentration of hydroxide anions to yield a neutral pH value, meaning the ratio of one with respect to the other would be 1.

For example, at room temperature, the ionization constant of water is defined as:

$K_w=[H^+][OH^-]=10^{-14}$