Calculate the empirical formula for each of the following stimulants based on their elemental mass percent composition.
adenine (a nucleobase (a purine derivative) with a variety of roles in biochemistry): C 44.44%, H 3.73%, N 51.83%

Answer:

346.g of solution

Explanation:

In this case, if we have 5.2 % by mass it means that in 100 g of the solution we will have 5.2 g of glucose. Therefore we can do the calculation:

5.2 g of glucose = 100 g of solution

[tex]18~g~of~glucose\frac{100~g~of~solution}{5.2~g~of~glucose}=346.15~g~of~solution[/tex]

So, if we need 8 g of glucose we had to have 346.15 g of solution

This logic can work for all types of solutions. By mass (as in this case), by volume or mass/volume.

I hope it helps!

Amount of solution need to get 18 g sucrose is 346 gram (Approx.)

Given that;

Amount of sucrose in solution = 5.2%

Find:

Amount of solution need to get 18 g sucrose

Computation:

Amount of solution need to get 18 g sucrose = [100 / 5.2][18]

Amount of solution need to get 18 g sucrose = [19.23][18]

Amount of solution need to get 18 g sucrose = 346.14

Amount of solution need to get 18 g sucrose = 346 gram (Approx.)

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Answer:

It is the portion of internal energy that can be transferred from one substance to another.

Explanation:

Thermal energy is the energy obtained by an object due to the motion of its particles.

It is also known as internal energy as it is the energy within the particles due to their motion.

Therefore, we can conclude that it is the portion of internal energy that can be transferred from one substance to another.

Hope this helps!

Answer:

C is correct

Explanation:

Answer:

rate=k×[C5H8]^n

k is the rate const.

when the rate is calculated using a few of the above data, (rate=change in concentration/time period) it can be observed that the rate is constant

therefore order of the reaction is 0

Answer:

0.252 milimoles

Explanation:

To convert mass of a substance to moles it is necessary to use the molar mass of the substance.

The formula of morphine is C₁₇H₁₉NO₃, thus, its molar mass is:

C: 17*12.01g/mol = 204.17g/mol

H: 19*1.01g/mol = 19.19g/mol

N: 1*14g/mol = 14g/mol

O: 3*16g/mol = 48g/mol.

204.17 + 19.19 + 14 + 16 = 285.36g/mol

Thus, moles of 71.891 mg = 0.071891g:

0.071891g × (1mol / 285.36g) = 2.5193x10⁻⁴ moles

As 1 mole = 1000 milimoles:

2.5193x10⁻⁴ moles = 0.252 milimoles

Answer:

The volume of the ballon is 325L.

Explanation:

Boyle's law express that the pressure of a gas is inversely proportional to its volume. That means if the pressure increases, the volume decreases. The formula is:

P₁V₁ = P₂V₂

Where P represents pressure and V volume of 1, initial state and 2, final state of the gas.

In the problem, the volume of the tank is 13.0L and the final pressure of the ballon is 1atm -The atmospheric pressure-. As 1atm of gas is in the ballon, the pressure of the tank is 26.0atm - 1.0atm = 25.0atm.

Replacing in Boyle's law expression:

25.0atm*13.0L = 1atmV₂

325L = V₂

The volume of the ballon is 325L.

Answer:

The ratio of pressure in bottle B to that of bottle A is 1 : 4

Explanation:

We'll be by calculating the pressure in both bottles. This is illustrated below below:

For A:

Temperature (T) = T

Volume (V) = V

Number of mole (n) = n

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =...?

PV = nRT

PV = n x 0.0821 x T

Divide both side by V

P = nT0.0821/V

Therefore, the pressure, in bottle A is

PA = nT0.0821/V

For B:

Temperature (T) = the same as that of A = T

Volume (V) = twice that of A = 2V

Number of mole (n) = half that of A = ½n

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =...?

PV = nRT

P x 2V = ½n x 0.0821 x T

Divide both side by 2V

P = ½n x 0.0821 x T/2V

P = nT0.0821/4V

Therefore, the pressure in bottle B is:

PB = nT0.0821/4V

Now, we can obtain the ratio of pressure in bottle B to that of bottle A as follow:

Pressure in bottle A (PA) = nT0.0821/V

Pressure in bottle B (PB) = nT0.0821/4V

PB/PA = nT0.0821/4V ÷ nT0.0821/V

PB/PA = nT0.0821/4V x V/nT0.0821

PB/PA = 1/4

Therefore, the ratio of pressure in bottle B to that of bottle A is 1 : 4.

Answer:

Explanation:

use this equation and solve for wavelength (λ)

λ = h/mv.

where h is plancks constant 6.63 × 10−34 J·s

m = mass of lectron

v = velcoeity of electron

Answer: The change in enthalpy associated with the combustion of 322 g of ethanol is [tex]-9567.6kJ[/tex]

Explanation:

To calculate the number of moles we use the equation:

[tex]\text{Moles}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]

[tex]\text{Moles of ethanol}=\frac{322g}{46g/mol}=7moles[/tex]

The balanced chemical reaction is:

[tex]C_2H_5OH(l)+3O_2\rightarrow 2CO_2(g)+3H_2O(l)[/tex] [tex]\Delta H=-1366.8kJ/mol[/tex]

Given :

Energy released when 1 mole of ethanol is combusted = 1366.8 J

Thus Energy released when 7 moles of ethanol is combusted =[tex]\frac{1366.8}{1}\times 7=9567.6kJ[/tex]

Thus the change in enthalpy associated with the combustion of 322 g of ethanol is [tex]-9567.6kJ[/tex]

The  change in enthalpy associated with the combustion is -9567.6KJ

Since there is 322g of ethanol

Also, there is the chemical equation i.e.

C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)ΔH∘c=−1366.8kJ/mol

So, the change should be

= -1366.8kJ *7/1

= -9567.6KJ

Since Energy released at the time when 1 mole of ethanol is combusted = 1366.8 J

So, here Energy released when 7 moles of ethanol is combusted

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Answer:

Red dye

Explanation:

In the given question, the complete question has not been provided but the propanone is used as a solvent in paper chromatography. The paper chromatography was performed for the black ink in which the black ink got separated in the red, blue and yellow colour.

From these three colours that are red, blue and yellow, the dye which is most soluble in propanone was red as red colour moved the most in the given chromatogram and the dye which travelled the most is most soluble in propanone.

Thus, red dye is the correct answer.

Answer:

K₂SO₄ + Pb(C₂H₃O₂)₂ →PbSO₄ + 2KC₂H₃O₂

A chemical equation is a symbolic representation of a chemical reaction. The chemical equation for the reaction between potassium sulfate ([tex]K_2SO_4[/tex]) and lead(II) acetate ([tex]Pb(CH_3COO)_2[/tex]) can be written as follows:

[tex]K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK[/tex]

A basic chemical equation consists of two main parts: the reactant side (left side) and the product side (right side), separated by an arrow indicating the direction of the reaction. Reactants are substances that undergo a chemical change, while products are substances formed as a result of the reaction.

In this reaction, potassium sulfate reacts with lead(II) acetate to form lead(II) sulfate and potassium acetate. It is important to note that the equation is balanced with stoichiometric coefficients, ensuring that the number of atoms of each element is the same on both sides of the equation.

Therefore, the chemical equation for the reaction between potassium sulfate ([tex]K_2SO_4[/tex]) and lead(II) acetate ([tex]Pb(CH_3COO)_2[/tex]) can be written as follows:

[tex]K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK[/tex]

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Answer:

[tex]HI_(_a_q_)~+~NaHCO_3_(_a_q_)~->~NaI_(_a_q_)~+~H_2O_(_l_)~+~CO_2_(_g_)[/tex]

Explanation:

In this case, we will have a neutralization reaction. We have a base ([tex]HI[/tex]) and a base ([tex]NaHCO_3[/tex]). Additionally, we have a strong acid and a strong base, therefore both will be soluble on water, so we will have an aqueous state for these compounds. If we will have a neutralization reaction, we will have as a salt as a product. With this in mind the reaction would be:

[tex]HI_(_a_q_)~+~NaHCO_3_(_a_q_)~->~NaI_(_a_q_)~+~H_2O_(_l_)~+~CO_2_(_g_)[/tex]

All the sodium salts are soluble in water, therefore we will have an aqueous state. Water is a liquid and carbon dioxide is a gas.

I hope it helps!

Answer:

Explanation:

Hello,

We'll be doing some classification of some chemical substances based on molecules, elemental state or ionic or electrovalent properties.

A) Ag = atomic element : silver (Ag) in its elemental state is an atomic element.

B) Cd = atomic element : Cadmium (Cd) is an element of the periodic table and belongs to transition metal.

C) MgCl = ionic compounds: this is a compound formed between magnesium (Mg) and chlorine (Cl) to give MgCl. This compound has ionic or electrovalent properties since electron transfer occurred between the cation (Mg) and anion (Cl).

D) F₂ = moleculer element : Fluorine F₂ is moleculer element since two elements of fluorine combine together to form a molecule.

E) HI = molecular compound : this is a compound formed from the reaction between hydrogen and iodine. It's a molecular compound because they are two different elements combining together to form a compound.

F) NO₂ = molecular compound

G) NaCl = ionic compound

H) Cl₂ = molecular element

Answer:

See figure 1

Explanation:

In this question, we have to remember that a poor electron carbon is a carbon in which we have a positive charge, a carbocation. Therefore we have to start with the production of the carbocation. First, a double bond from the benzene is moved to the carbon in the top to produce a new double bond generating a positive charge in a carbon with ortho position (electron-poor). Then we can move another double bond inside the ring to produce a positive charge in the para carbon. Finally, we can move the last double bond to produce again another positive charge in the second ortho carbon.

See figure 1.

I hope it helps!

[tex]\text{GaBr}_3[/tex]

Answer:

a. NO₂⁻ + H⁺ → HNO₂

b. HNO₂ + OH⁻ → NO₂⁻ + H₂O

Explanation:

A buffer is defined as an aqueous mixture of a weak acid and its conjugate base or vice versa.

The buffer of the problem is HNO₂/NO₂⁻ where nitrous acid is the weak acid and NO₂⁻ is its conjugate base.

a. When a acid is added to a buffer as the buffer of the problem, the conjugate base will react with the acid, to produce the weak acid, thus:

NO₂⁻ + HCl → HNO₂ + Cl⁻

Ionic equation is:

NO₂⁻ + H⁺ + Cl⁻ → HNO₂ + Cl⁻

In the net ionic equation, you avoid the ions that don't react, that is:

b. In the same way, the weak acid will react with the strong acid producing water and the conjugate base, thus:

HNO₂ + NaOH → NO₂⁻ + H₂O + Na⁺

The ionic equation is:

HNO₂ + Na⁺ + OH⁻ → NO₂⁻ + H₂O + Na⁺

And the net ionic equation is:

Answer and Explanation:

Sodium hydroxide (NaOH) is a strong base and nitric acid (HNO₃) is a strong acid. That means that they dissociates in water by giving the ions:

NaOH ⇒ Na⁺(ac) + OH⁻(ac)

HNO₃ ⇒ H⁺(ac) + NO₃⁻(ac)

The reaction between an acid and a base is called neutralization. In this case, HNO₃ loses its proton and it is converted in NO₃⁻ (nitrate anion). NaOH loses its hydroxyl anion (OH⁻) by giving Na⁺ cations.

Na⁺ cations with NO₃⁻ anions form the salt NaNO₃ (sodium nitrate); whereas H⁺ and OH⁻ form water molecules. The complete equation is the following:

HNO₃(ac) + NaOH(ac) ⇒ NaNO₃(ac) + H₂O(l)

The ionic equation is:

H⁺(ac) + NO₃⁻(ac) + Na⁺(ac) + OH⁻(ac) ⇄ Na⁺(ac) + NO₃⁻(ac) + H₂O(ac)

If we cancel the repeated ions at both sides of the equation, it gives the following ionic reaction:

H⁺(ac) + OH⁻(ac) ⇄ H₂O(ac)

Answer:

0.9180 M

Explanation:

Step 1: Write the balanced equation

H₂SO₄(aq) + 2 KOH(aq) ⇒ K₂SO₄(aq) + 2 H₂O(l)

Step 2: Calculate the reacting moles of KOH

27.00 mL of 1.700 M KOH react. The reacting moles of KOH are:

[tex]0.02700L \times \frac{1.700mol}{L} = 0.04590mol[/tex]

Step 3: Calculate the reacting moles of H₂SO₄

The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 0.04590 mol = 0.02295 mol.

Step 4: Calculate the molarity of H₂SO₄

0.02295 moles of H₂SO₄ are in 25.00 mL of solution. The molarity of the acid solution is:

[tex]M = \frac{0.02295 mol}{0.02500} = 0.9180 M[/tex]

Answer:

both are the types of mixture and both are impure substances that donot have fixed composition and the composition of constituents  is not uniform

Answer:

Their components van be separated by physical processes

Explanation:

Out of the answers im given, it makes the most sense. I would double check before submitting though

Answer:

Sr

Explanation:

Sr has an ionization of 550 whereas Ba has an ionization of 503

Answer:

Volume of the gass will decrease by three times of the original volume

Explanation:

Volume is inversly propotional to the pressure applied on it.

Answer:

it is decreased to one third of its original volume

Explanation:

Answer:

See the explanation

Explanation:

We have to start by analyzing the molecule. In the molecule, we have two types of isomerism. In the double bond, we can have E or Z isomerism, and additionally, we have a chiral carbon (a carbon with 4 different bonds) (See figure 1). With this in mind, we can have 4 possible isomers.(See figure 2).

a) (3S)-2-(E)-ichthyothereol

b) (3R)-2-(E)-ichthyothereol

c) (3R)-2-(E)-ichthyothereol

d) (3R)-2-(Z)-ichthyothereol

I hope it helps!

Answer:

1.) Option C is correct.

The rate of reverse reaction is greater than zero, but equal to the rate of the forward reaction.

2) Option B is correct.

The rate of reverse reaction is Greater than zero, but less than the rate of the forward reaction.

3) Option C is correct.

The rate of reverse reaction is Greater than zero, and equal to the rate of the forward reaction.

4) Option A is correct.

How much less C2H5CO2CH3 is in the flask when the system has again reached equilibrium? Zero.

Explanation:

HCH,CO2(aq) + C2H5OH(aq) ⇌ C2H,CO2CH3(aq) + H2O

1) Before the main product is removed from the reaction setup, the chemical reaction is at equilibrium.

Chemical equilibrium is a state of dynamic equilibrium such that the concentration of the reactants and the products do not always remain the same but the rate of forward reaction always matches the rate of backward reaction.

2) When 246. mmol of C2HCO2CH3 are removed from the reaction mixture....

And when one of the factors involved in chemical equilibrium changes, Le Chatellier's principle explains that the system then adjusts to remedy this change and takes time to go back to equilibrium again.

When one of the species involved in the chemical reaction at equilibrium, is removed from the reaction mixture, the rate of reaction begins to favour that side of the reaction until equilibrium is re-established.

So, when 246 mmol of one of the products is removed, the response is to cause the rate of forward reaction to be favoured to produce more of products as there are fewer, and the rate of reverse reaction at this moment becomes less than the rate of forward reaction.

3) The rate of the reverse reaction when the system has again reached equilibrium

Like I said in (2) above, the reaction remedies this change in concentration of one of the products until equilibrium is re-established and when chemical equilibrium is re-established the rate of forward reaction once again matches the rate of backward reaction.

4) How much less C2H5CO2CH3 is in the flask when the system has again reached equilibrium?

By the time equilibrium is re-established, the system goes back to how it all was and the concentration of C2H5CO2CH3 goes back to the same as it was at the start of the reaction.

Hope this Helps!!!

Answer:

35.578g or 36g if you round

Explanation:

Q=mc ∆∅ where ∅ is temperature difference

1160= m x 1.716 x (42-23)

m = 1160/ 1.716 x19

m=35.578g

m = 36g to nearest whole number

Answer: C. 36 g

Explanation: I got this right on Edmentum.

Answer: A

Explanation:

q=heat absorbed=+250J

W=−PΔV=−0.5atm×(10−1)

=−4.5L−atm

=−4.5×101J=−454.5J

Now q=ΔE−W

ΔE=q+W=250−454.5

=−204.5J

The change in internal energy (ΔU) is -211 J.

According to the first law of thermodynamics;

ΔU = q + w

ΔU = change in internal energy

q = heat absorbed/ emitted

w = work done on or by the system

We can see from the question that  250 J of heat was absorbed from the surroundings hence q =  250 J

w= PΔV = 0.50 atm(10.1 L -  1.0 L) = 4.55 atm L

But;

1 L atm = 101.325 J

4.55 atm L =  4.55 atm L ×  101.325 J/ 1 L atm

= 461 J

Now;

ΔU = 250 J - 461 J

ΔU = -211 J

Note that work done is negative because the gas expands and does work on the surrounding.

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Answer:

31.2g

Explanation:

The following data were obtained from the question:

Volume of bromine = 10mL

Density of bromine = 3.12 g/mL

Mass of bromine =...?

The Density of the substance is related to it's mass and volume by the following equation:

Density = Mass /volume

With the above equation, we can calculate the mass of bromine as follow:

Density = Mass /volume

Volume of bromine = 10mL

Density of bromine = 3.12 g/mL

Mass of bromine =...?

Density = Mass /volume

3.12 = Mass /10

Cross multiply

Mass of bromine = 3.12 x 10

Mass of bromine = 31.2g

Therefore, the mass of bromine is 31.2g

Answer: Zn (s) |Zn2+ || Au1+| Au(s)

Explanation:

Answer:

0.0025Litters

Explanation:

2.5ml= 2.5x10^-3l

2.5ml= 0.0025l

Answer:

AAAAAAAA

Explanation:

Answer: 2. 2.0 atm

3. 6.2 L

4. 2.6 atm

Explanation:

2. To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.

[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.

We are given:

[tex]P_1=1.0atm\\V_1=20.0mL\\P_2=?\\V_2=10.0mL[/tex]

Putting values in above equation, we get:

[tex]1.0\times 20.0mL=P_2\times 10.0mL\\\\P_2=2.0atm[/tex]

Thus new pressure in the syringe at 20°C if the plunger is depressed to 10.0 mL is 2.0 atm

3. To calculate the final volume of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where,

[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.

[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.

We are given:

[tex]V_1=5.2L\\T_1=30.0^oC=(30.0+273)K=303K\\V_2=?\\T_2=90.0^oC=(90.0+273)K=363K[/tex]

Putting values in above equation, we get:

[tex]\frac{5.2L}{303K}=\frac{V_2}{363}\\\\V_2=6.2L[/tex]

Thus  its new volume is 6.2 L.

4. To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant volume.

Mathematically,

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

where,

[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.

[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.

We are given:

[tex]P_1=2.3atm\\T_1=0^0C=273K\\P_2=?\\T_2=37^0C=(37+273)K=310K[/tex]

Putting values in above equation, we get:

[tex]\frac{2.3atm}{273K}=\frac{P_2}{310K}\\\\P_2=2.6atm[/tex]

Hence, the pressure of the gas if its temperature is warmed to 37°C is 2.6 atm

Answer:

2. Based on the Boyle's law, the pressure of the sample of a gas is inversely proportional to the volume at a constant temperature. Mathematically, the law can be written as,

P₁V₁ = P₂V₂

Here P₁ and V₁ are the pressure and volume initially, while P₂ and V₂ are the eventual pressure and volume.

Based on the given information, P₁ is atm, V₁ is 20 ml, V₂ is 10 ml and there is a need to find P₂. The volumes V₁ and V₂ can also be written as,

V₁ = 20 ml × 1L / 1000 ml = 0.02 L

V₂ = 10 ml × 1L / 1000 ml = 0.01 L

Now putting the values, we get P₂ as,

P₂ = P₁V₁/V₂

P₂ = 1 atm × 0.02 L / 0.01 L

P₂ = 2 atm

3) Based on the given question, the initial volume (V₁) of the sample is 5.2 L, the initial temperature (T₁) is 30 degree C or 273 + 30 = 303 K, the final temperature (T₂) is 90 degree C or 273 + 90 = 363 K.

Based on Charle's law, the volume of the gas is directly proportional to the temperature at constant pressure,

V = KT, here K is the proportionality constant

K or constant = V/T

Now using the formula,

V₁/T₁ = V₂/T₂

5.2/303 = V₂/363

V₂ = (5.2 × 363) / 303

V₂ = 6.22 L

4. Based on the given information, the T₁ or initial temperature of halothane is 0 degree C or 273 + 0 = 273 K, the T₂ or the final temperature of halothane is 37 degree C or 273 + 37= 310 K.

The P₁ or initial pressure of halothane is 2.3 atm, and as volume is constant so V₁ = V₂

Now using the combined gas law equation, we can find P₂,

P₁V₁/T₁ = P₂V₂/T₂

P₂ = P₁V₁T₂/T₁V₂

P₂ = (2.3 atm) (310 K) / (273 K)

P₂ = 2.6 atm

Hence, the final pressure of halothane is 2.6 atm.

Answer:

Two

Explanation:

2

rough endoplasmic reticulum

smooth endoplasmic reticulum