A balloon is filled to a volume of 1.50L with 3.00 moles of gas at 25.0 c. With pressure and temperature held constant, what will be the volume (in L) of the balloon if .20 moles of gas are added?

Answers

Answer 1

Answer:

1.37L

Explanation:

V2=v1×T2

______

T1


Related Questions

A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen. The decomposition of a sample of diphosphorus pentoxide forms 0.775 g phosphorus to every 1.00 g oxygen.

Required:
How many grams of P205 are formed when 5.89 g of P react with excess oxgen?

Answers

Answer:

There is 13.48 grams of P2O5 formed

Explanation:

Step 1: Data given

A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen.

Mass of P = 5.89 grams

Molar mass of O2 = 32.0 g/mol

atomic mass of P = 30.97 g/mol

molar mass of P2O5 = 141.94 g/mol

Step 2: The balanced equation

4P(s)+5O2(g)⇔ 2P2O5(s)

Step 3: Calculate moles of P

Moles P = Mass P / atomic mass P

Moles P = 5.89 grams / 30.97 g/mol

Moles P = 0.190 moles

Step 4: Calculate moles of P2O5

For 4 moles P we need 5 moles O2 to produce 2 moles P2O5

For 0.190 moles of P we'll have 0.190/2 = 0.095 moles P2O5

Step 5: Calculate mass of P2O5

Mass P2O5 = moles P2O5 * molar mass P2O5

Mass P2O5 = 0.095 moles * 141.94 g/mol

Mass P2O5 = 13.48 grams

There is 13.48 grams of P2O5 formed

Consider a galvanic (voltaic) cell that has the generic metals X and Y as electrodes. If X is more reactive than Y (that is, X more readily reacts to form a cation than Y does), classify the following descriptions by whether they apply to the X or Y electrode.
i. anode
ii. cathode
iii. electrons in the wire flow toward
iv. electrons in the wire flow away
v. cations from salt bridge flow toward
vi. anions from salt bridge flow toward
vii. gains mass
viii. loses mass

Answers

Answer:

X

anode

electrons in the wire flow away

anions from salt bridge flow toward

loses mass

Y

cathode

electrons in the wire flow toward

cations from salt bridge flow toward

gains mass

Explanation:

In a galvanic cell, oxidation occurs at the anode while reduction occurs at the cathode. The metal that is more reactive functions as the anode while the less reactive metal functions as the cathode.

Electrons leave the anode and travel via a wire to the cathode. At the anode cations give up electrons and enter into the solution.

At the cathode, cations pick up electrons and are deposited on the cathode leading to a gain in mass at the cathode.

Positive ions from the salt bridge flow towards the cathode while negative ions from the salt bridge flow towards the anode.

In the reaction HCI + NH4OH --> NH4CI+H2O, which compound has an element ratio of 1:4:1?

H2O

NH4Cl

HCI

ΝΗ4ΟΗ

Answers

NH4Cl has element ratio of 1:4:1

The compound in this reaction which is having the elemental ratio of 1:4:1 is NH₄Cl where nitrogen and chlorine are of one mole each with 4 hydrogens.

What is elemental ratio?

Elemental ratio of a compound is the ratio of number of atoms  of each  elements in that compound. The elemental ratio can be determined from the molecular formula of compounds.

The given reaction is a double displacement reaction. Here, the Cl group is replaced to the ammonia and OH group is replaced to the water. Thus, two species is replaced in the reaction.

In NH₄Cl, there are one nitrogen, 4 hydrogens and one chlorine atom. Therefore, the elemental ratio of the compound is 1:4:1. The elemental ratio of water is 2:1 and HCl is 1:1 and that in NH₄OH is 1:5:1. Hence, option b is correct.

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Manganese-55 has _____neutrons.
55 Mn
25

A. 55
B. 30
C. 25​

Answers

QUESTION:- Manganese-55 has _____neutrons.

OPTIONS :-

A. 55

B. 30

C. 25

ANSWER:- NUMBER OF NEUTRONS IS EQUAL TO THE DIFFERENCE BETWEEN THE MASS IF THE ATOM AND ATOMIC NUMBER

SO DIFFERENCE IS EQUAL TO :- 55-25 = 30 NEUTRONS.

SO THERE IS 30 NEUTRONS IN SINGLE ATOM OF THE MANGANESE-55 ATOM.

Answer:

the mass of an atom is the sum of proton and neutron which are both concentrated in nocleus of an atom. from the question the mass is given as 55 and the proton is 25.

Determine the boiling point of a solution that contains 150.0 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 950 mL of benzene (d = 0.877 g/mL). Pure benzene has a boiling point of 80.1°C and a boiling point elevation constant of 2.53°C/m.

Answers

Answer:

Boiling T° of solution  → 83.6°C

Explanation:

To solve this, we apply Elevation of boiling point, property

ΔT = Kb . m . i

As we talk about organic solute, i = 1. No ions are formed.

m = molality (moles of solute in 1kg of solvent)

We determine mass of solvent by density

D = m /V so D . V = m

950 mL . 0.877 g/mL = 833.15 g

We convert to kg → 833.15 g . 1 kg/ 1000g = 0.833 kg

Moles of solute (naphtalene): 150 g . 1 mol/ 128.16g = 1.17 mol

m = 1.17mol / 0.833 kg = 1.41 mol/kg

We replace data:

Boiling T° of solution - 80.1°C = 2.53°C/m . 1.41 m . 1

Boiling T° of solution = 2.53°C/m . 1.41 m . 1  +  80.1°C → 83.6°C

Answer:

The answer is c or 17.1 g

A monatomic ion with a charge of 2 has an electronic configuration of 1s22s22p6. This ion is a(n) _______ . What is the chemical symbol of the noble gas this ion is isoelectronic with

Answers

Answer:

A. Cation

B. Ne

Explanation:

The ion is positively charged by 2, making it a cation.

The electron configuration of the nearest noble gas Neon is 1s22s22p6

1. A monatomic ion with a charge of 2 has an electronic configuration of 1s22s22p6 is Neon.

2. chemical symbol of the noble gas is Kr (krypton).

Isoelectronic atom or ion has the same number of valence electrons. Krypton has 36 electrons and 36 protons (atomic number 36).

What is Neon?

Neon is a chemical element with the symbol Ne and atomic number 10. It is a noble gas. Neon is a colorless, odorless, inert monatomic gas under standard conditions, with about two-thirds the density of air. It was discovered (along with krypton and xenon) in 1898 as one of the three residual rare inert elements remaining in dry air after nitrogen, oxygen, argon, and carbon dioxide were removed. Neon was the second of these three rare gases to be discovered and was immediately recognized as a new element of its bright red emission spectrum. The name neon is derived from the Greek word, the neuter singular form of  (neos), meaning 'new'. Neon is chemically inert, and no uncharged neon compounds are known. The compounds of neon currently known include ionic molecules, molecules held together by van der Waals forces, and clathrates.

Learn more about Neon

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Given the equation: 2C6H10(l) 17 O2(g) ---> 12 CO2(g) 10 H2O(g) MM( g/mol): 82 32 44 18 If 115 g of C6H10 reacts with 199 g of O2 and 49 g of H2O are formed, what is the percent yield of the reaction

Answers

Answer:

74%

Explanation:

Step 1: Write the balanced equation

2 C₆H₁₀(l) + 17 O₂(g) ⇒ 12 CO₂(g) + 10 H₂O(g)

Step 2: Determine the limiting reactant

The theoretical mass ratio (TMR) of C₆H₁₀ to O₂ is 164:544 = 0.301:1.

The experimental mass ratio (EMR) of C₆H₁₀ to O₂ is 115:199 = 0.578:1.

Since EMR > TMR, the limiting reactant is O₂.

Step 3: Calculate the theoretical yield of H₂O

The theoretical mass ratio of O₂ to H₂O 544:180.

199 g O₂ × 180 g H₂O/544 g O₂ = 65.8 g H₂O

Step 4: Calculate the percent yield of H₂O

%yield = (experimental yield/theoretical yield) × 100%

%yield = (49 g/65.8 g) × 100% = 74%

Answer:

Percentage yield of H₂O = 74.24%

Explanation:

The balanced equation for the reaction is given below:

2C₆H₁₀ + 17O₂ —> 12CO₂ + 10H₂O

Next, we shall determine the masses of C₆H₁₀ and O₂ that reacted and the mass of H₂O produced from the balanced equation. This is can be obtained as follow:

Molar mass of C₆H₁₀ = 82 g/mol

Mass of C₆H₁₀ from the balanced equation = 2 × 82 = 164 g

Molar mass of O₂ = 32 g/mol

Mass of O₂ from the balanced equation = 17 × 32 = 544 g

Molar mass of H₂O = 18 g/mol

Mass of H₂O from the balanced equation = 10 × 18 = 180 g

SUMMARY:

From the balanced equation above,

164 g of C₆H₁₀ reacted with 544 g of O₂ to produce 180 g of H₂O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

164 g of C₆H₁₀ reacted with 544 g of O₂.

Therefore, 115 g of C₆H₁₀ will react to produce = (115 × 544)/164 = 381 g of O₂.

From the calculations made above, we can see that a higher mass (i.e 381 g) of O₂ than what was given (i.e 199 g) is needed to react with 115 g of C₆H₁₀.

Therefore, O₂ is the limiting reactant and C₆H₁₀ is the excess reactant.

Next, we shall determine the theoretical yield of H₂O. This can be obtained by using the limiting reactant as shown below:

From the balanced equation above,

544 g of O₂ reacted to produce 180 g of H₂O.

Therefore, 199 g of O₂ will react to produce = (199 × 180)/544 = 66 g of H₂O.

Thus, the theoretical yield of H₂O is 66 g.

Finally, we shall determine the percentage yield. This can be obtained as follow:

Actual yield of H₂O = 49 g

Theoretical yield of H₂O = 66 g

Percentage yield of H₂O =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of H₂O = 49/66 × 100

Percentage yield of H₂O = 74.24%

GIVING BRAINLIEST
Which equations are used to calculate the velocity of a wave?
O velocity = distance ~ time
velocity = wavelength x frequency
velocity = distance/time
velocity = wavelength/frequency
velocity = distance/time
velocity = wavelength x frequency
velocity = distance ~ time
velocity = wavelength/frequency

Answers

Answer:

velocity = distance/time

velocity = wavelength × frequency

Both of these are commonly known equations to calculate velocity with different variables.


A reactant. Q. decomposes at a second order. The slope of the graph 1/[Q] (1/M) vs time (s) is -0.04556. If the initial
concentration of Q for the reaction is 0.50 M, what is the concentration in M. of Q after 10.0 minutes?

Answers

Answer:

0.034 M

Explanation:

1/[A] = kt + 1/[A]o

[A] = ?

k= 0.04556

t= 10.0 minutes or 600 seconds

[A]o = 0.50 M

1/[A] = (0.04556 × 600) + 1/0.50

[A] = 0.034 M

One crystalline form of silica (SiO2) has a cubic unit cell, and from X-ray diffraction data it is known that the cell edge length is 0.700 nm. If the measured density is 2.32 g/cm3, how many (a) Si4 and (b) O2- ions are there per unit cell

Answers

Answer:

There are 8Si atoms and 16 O atoms per unit cell

Explanation:

From the question we are told that:

Edge length [tex]l=0.700nm=>0.7*10^9nm[/tex]

Density [tex]\rho=2.32g/cm^3[/tex]

Generally the equation for Volume is mathematically given by

[tex]V=l^3[/tex]

[tex]V=(0.7*10^9)m^3[/tex]

[tex]V=3,43*10^-{22}cm[/tex]

Where

Molar mass of  (SiO2) for one formula unit

[tex]M=28+32[/tex]

[tex]M=60g/mol[/tex]

Therefore

Density of Si per unit length is

[tex]\rho_{si}=\frac{9.96*10^{23}}{3.43*10^22}[/tex]

[tex]\rho=0.29[/tex]

Molar mass of  (SiO2) for one formula unit

[tex]M=28+32[/tex]

[tex]M=60g/mol[/tex]

Therefore

There are 8Si atoms and 16 O atoms per unit cell

công thức của định lý pytago

Answers

The sum of the squares of two sides of a right angle is equal to the square of the hypotenuse

Determine the mmol of both starting materials (factoring in that formic acid is not pure, but rather 88% weight/volume, or 88g/100 ml), showing your work. Determine the limiting reagent in this synthesis. Lastly, calculate the theoretical yield of benzimidazole that you could expect to form.

Answers

Solution :

Molecular      Molar Mass       Volume      Density       Mass      Moles      nmoles

formula            (g/mol)               (mL)          (g/mL)           (g)

[tex]$C_6H_8N_2$[/tex]            108.14                                                    0.108      0.001          1

HCOOH           46.02                0.064          1.22     0.07808     0.0017       1.7

mmoles of o-phenylenediamine = 1 mmoles

mmoles of formic acid = 1.7 [tex]\approx[/tex] 2 mmoles

From the reaction of o-phenylenediamine and formic acid, we see,

1 mmole of o-phenylenediamine reacts with 1 mmole of formic acid.

But here, 2 mmoles of the formic acid , this means that the formic acid is an excess reagent and the o-phenylenediamine is the limiting reagent here.

The amount of product depends on the limiting reagent that is o-phenylenediamine. So, 1mmole of o-phenylenediamine will give 1mmole of product.

molar mass of Benzimidazole = [tex]118.14[/tex] g/mol

mmoles of Benzimidazole formed = [tex]1[/tex] mmol

Mass of benzimidazole formed = molar mass x [tex]\frac{nmoles}{1000}[/tex]

                                                    [tex]$=\frac{118.14 \times 1}{1000}$[/tex]

                                                     = 0.11814 g

So the theoretical yield of Benzimidazole is = 0.118 g = 118mg

Na Sa Bant HCL -> 50g Hao pt Soy​

Answers

North America and south africa

Ammonia reacts with oxygen to produce nitrogen monoxide and water:
4 NH3(g) + 5 O2(g) ---> 4 NO(g) + 6 H2O(g)
Which of the following are stoichiometric amounts of the two reactants?
a) 1.0 g, 1.25 g
b) 0.75 mol, 0.9375 mol

Answers

Answer:

b) 0.75 mol, 0.9375 mol

Explanation:

According to this question, ammonia reacts with oxygen to produce nitrogen monoxide and water. The chemical equation is as follows:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Based on this balanced equation, 4 moles of ammonia (NH3) reacts with 5 moles of oxygen (O2).

A stoichiometric amount of the two reactants (NH3 and O2) must represent the ratio 4:5.

Given the provided options;

0.75 mol of ammonia (NH3) will react with (0.75 × 5/4) = 0.935 mol of O2 for them to be in stoichiometry.

N.B: 1 mol of NH3 will react with 1.25mol of O2 and not 1g, 1.25g.

Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and products in your balanced equation. You do not need to include the states with the identities of the oxidizing and reducing agents.
NO_2(g) rightarrow NO_3^-(aq) +NO_2^- (aq) [basic]
The oxidizing agent is:______.
The reducing agent is:_______.

Answers

Answer:

a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. i. NO₂⁻ is the oxidizing agent

ii. NO₃⁻ is the reducing agent.

Explanation:

a. Balance the following skeleton reaction

The reaction is

NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)

The half reactions are

NO₂ (g) → NO₃⁻ (aq)  (1) and

NO₂ (g) → NO₂⁻  (aq) (2)

We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)

We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms

NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)

The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.

So, NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

We now add equation (4) and (5)

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

+  NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + e⁻  (4)

2NO₂ (g) + H₂O (l)  → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq)  

We now add two hydroxide ions to both sides of the equation.

So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + 2OH⁻ (aq)

The hydrogen ion and the hydroxide ion become a water molecule

2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H₂O (l)

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

So, the required reaction is

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. Identify the oxidizing agent and reducing agent

Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = 0

x - 4 = 0

x = 4

Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = -1

x - 4 = -1

x = 4 - 1

x = 3

Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = -1

x + 3(-2) = -1

x - 6 = -1

x = 6 - 1

x = 5

i. The oxidizing agent

The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus  NO₂⁻ is the oxidizing agent

ii. The reducing agent

The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and  NO₃⁻ is the reducing agent.

At 445oC, Kc for the following reaction is 0.020. 2 HI(g) <--> H2 (g) + I2 (g) A mixture of H2, I2, and HI in a vessel at 445oC has the following concentrations: [HI] = 1.5 M, [H2] = 2.50 M and [I2] = 0.05 M. Which one of the following statements concerning the reaction quotient, Qc, is TRUE for the above system?
a. Qc = Kc; the system is at equilibrium.
b. Qc is less than Kc; more H2 and I2 will be produced.
c. Qc is less than Kc; more HI will be produced.
d. Qc is greater than Kc; more HI will be produced.

Answers

Explanation:

The given balanced chemical equation is:

[tex]2 HI(g) <--> H_2 (g) + I_2 (g)[/tex]

The value of Kc at 445oC is 0.020.

[HI]=1.5M

[H2]=2.50M

[I2]=0.05M

The value of Qc(reaction quotient ) is calculated as shown below:

Qc has the same expression as the equilibrium constant.

[tex]Qc=\frac{[H_2][I_2]}{[HI]^2} \\Qc=(2.50Mx0.05M)/(1.5M)^2\\Qc=0.055[/tex]

Qc>Kc,

Hence, the backward reaction is favored and the formation of Hi is favored.

Among the given options, the correct answer is option d. Qc is greater than Kc; more HI will be produced.

The specific heat capacity of lead is 0.13 J/g-K. How much heat (in J) is required to raise the temperature of 15 g of lead from 22 °C to 37 °C? a. 5.8 × 10-4 J b. 0.13 J c. 29 J d. 2.0 J e. -0.13 J

Answers

Answer:

c. 29 J

Explanation:

Step 1: Given data

Specific heat capacity of Pb (c): 0.13 J/g.K (= 0.13 J/g.°C)Mass of Pb (m): 15 gInitial temperature: 22 °CFinal temperature: 37 °C

Step 2: Calculate the temperature change

ΔT = 37 °C - 22 °C = 15 °C

Step 3: Calculate the heat (Q) required to raise the temperature of the lead piece

We will use the following expression.

Q = c × m × ΔT

Q = 0.13 J/g.°C × 15 g × 15 °C = 29 J

Which of the following have only a -C-O-C- functional group?

Answers

Answer:

B) ethers

Explanation:

The functional group of an organic compound defines its specificity. The functional group is responsible for the chemical behavior of an organic compound. For example, alkenes are known to have a carbon-carbon double bond (C=C) functional group.

Likewise, organic compounds known as ETHERS are known to possess an ethoxy functional group i.e. oxygen atom bonded to two alkyl groups (R- OR; where R is an alkyl group). Members of ether functional group includes dimethyl ether (CH3-O-CH3), diethyl ether (C2H5-O-C2H5).

A 0.204 g sample of a CO3 2- antacid is dissolved with 25.0ml of 0.0981 M HCL. The hydrochloric acid that is not neutralized by the antacid is titrated to a bromophenol blue endpoint with 5.83 ml of 0.104 M NaOH. Assuming the active ingredient in the antsacid sample is CaCO3, calculate the mass of CaCO3 in the sample.

Answers

Answer:

0.0922 g

Explanation:

Number of moles of acid present = 25/1000 × 0.0981

= 0.00245 moles

Number of moles of base = 5.83/1000 × 0.104

= 0.000606 moles

Since the reaction of HCl and NaOH is 1:1

Number of moles of HCl that reacted with antacid = 0.00245 moles - 0.000606 moles

= 0.001844 moles

From the reaction;

CaCO3 + 2HCl ----> CaCl2 + H2O + CO2

1 mole of CaCO3 reacts with 2 moles of HCl

x moles of CaCO3 reacts with 0.001844 moles ofHCl

x = 1 × 0.001844/2

= 0.000922 moles

Mass of CaCO3 = 0.000922 moles × 100 g/mol

= 0.0922 g

To draw a Lewis structure for a polyatomic ion, begin by calculating A, the available electrons, and N, the needed electrons. What is N for CIO3-, the chlorate ion?
A = 26
N = ?​

Answers

Answer:

16

Explanation:

Because the sum of all electron in that compound should be 41 and as it has one electron extra ,total no. of electrons are 42 .

So if we add 26 +16 we get 42

Hence it's correct answer

explain in details how triacylglycerol have an advantage over carbohydrates as stored fuel

Answers

Answer:

As stored fuels, triacylglycerols have two significant advantages over polysaccharides such as glycogen and starch. The carbon atoms of fatty acids are more reduced than those of sugars, and oxidation of triacylglycerols yields more than twice as much energy, gram for gram, as that of carbohydrates.

Explanation:

What is the concentration of Htions at a pH = 11?
mol/L
What is the concentration of Htions at a pH = 6?
mol/L
How many fewer Htions are there in a solution at a
pH = 11 than in a solution at a pH = 6?

Answers

Answer:
a) 1 x 10^-11 mol/L
b) 1 x 10^-6 mol/L
c) 1 x 10^-5 fewer H+ ions

Explanation

pH stands for Power of Hydrogen, the more acidic a substance is, the more H+ ions it has rendering the substance acidic. a pH of 1 means the concentration of H+ ions is 1 x 10^-1. A pH of 7 means the concentration of H+ ions is 1 x 10^-7 and so on.

10^-11 has 10^-5 more H+ ions than 10^-6

Hope this helps :)

Three peptides were obtained from a trypsin digestion of two different polypeptides. Indicate the possible sequences from the given data.

a. Val-Gly-Arg
b. Ala-Val-Lys
c. Ala-Gly-Phe

Answers

Answer:

A) Val-Gly-Arg-Ala-Val-Lys-Ala-Gly-Phe

B) Ala-Val-Lys-Val-Gly-Arg-Ala-Gly-Phe

Explanation:

The possible sequences that could be obtained from the available dat provided are :

A) Val-Gly-Arg-Ala-Val-Lys-Ala-Gly-Phe

B) Ala-Val-Lys-Val-Gly-Arg-Ala-Gly-Phe

Trypsin is generally a catalyst  for the breakdown of proteins into smaller peptides ( during the hydrolysis of peptide bonds )  

who much the velocity of a body when it travels 600m in 5 min​

Answers

Answer:

2 m/s

Explanation:

Applying the formulae of velocity,

V = d/t............. Equation 1

Where V = Velocity of the body, d = distance, t = time

From the question,

Given: d = 600 m, t = 5 minutes = (5×60) = 300 seconds.

Substitute these values into equation 1

V = 600/300

V = 2 m/s.

Hence the velocity of the body when it travels is 2 m/s

write the balanced equation for
[B]⁴[C][D]/[A]²​

Answers

A2-34=56 this is the equation to your expression

A reaction vessel is charged with phosphorus pentachloride, which partially decomposes to phosphorus trichloride and molecular chlorine according to the following reaction:

PCl5(g)â PCl3(g)+Cl2(g)

When the system comes to equilibrium at 250.0°C, the equilibrium partial pressures are: PPCl5 = 0.688 atm and PPCl3 = PCl2 = 0.870 atm.

Required:
What is the value of Kp at this temperature?

Answers

Answer:

Kp = 1.10.

Explanation:

Hello there!

In this case, according to the given information about the chemical reaction at equilibrium, it turns out possible for us to find the partial pressures-based equilibrium expression for the decomposition of phosphorous pentachloride by applying the law of mass action whereas the pressure of products is divided by that of the reactants as shown below:

[tex]Kp=\frac{p_{PCl_3}p_{Cl_2}}{p_{PCl_5}}[/tex]

Now, we plug in the given pressures to obtain:

[tex]Kp=\frac{0.870}{0.688} \\\\Kp=1.10[/tex]

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When 3-methyl-1-pentene is treated with in dichloromethane, the major product is 1-bromo-3-methyl-2-pentene.

a. True
b. False

Answers

Answer:

True

Explanation:

When Methyl Pentene is introduced in a chemical reaction with dichloromethane then the major product will be bromomethylpentene. There can be small amount of bromo methyl pentene than the amount of methyl pentene introduced for reaction.

Un sistema formado por una única sustancia, ¿será siempre homogéneo? ¿Porqué? Piensa a partir de las definiciones y trata de corroborar o negar usando ejemplos concretos.

Answers

Una sustancia homogénea es una sustancia que se compone de una sola fase.

Recordemos que definimos una fase en química como "cantidad química y físicamente uniforme u homogénea de materia que se puede separar mecánicamente de una mezcla no homogénea y que puede consistir en una sola sustancia o una mezcla de sustancias" según Ecyclopedia Britiannica.

El hecho de que un sistema esté compuesto por una sola sustancia no lo hace es autóctono. A veces, un sistema puede estar compuesto por partículas sólidas de una sustancia en equilibrio con su líquido. El sistema contiene solo una sustancia pero en diferentes fases, por lo tanto, el sistema contiene una sustancia pero no es homogéneo.

Por tanto, el hecho de que un sistema contenga una sola sustancia no significa necesariamente que sea homogéneo.

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How many grams of sodium chloride are contained in 250.0 g of a 15% NaCl solution?

Please explain and show work.

Answers

Given concentration of NaCl=15%

Means ,

In every 100g of Solution 15g of NaCl is present .

Now

Given mass=250g

So ,

[tex]\\ \Large\sf\longmapsto 250\times 15\%[/tex]

[tex]\\ \Large\sf\longmapsto 250\times \dfrac{15}{100}[/tex]

[tex]\\ \Large\sf\longmapsto 37.5g[/tex]

37.5g of NaCl present in 250g of solution.

Answer:

Given concentration of NaCl=15%

Means ,

In every 100g of Solution 15g of NaCl is present .

Now

Given mass=250g

So ,

➡250 × 15%

➡250×15/100

➡37.5g

37.5g of NaCl present in 250g of solution.

g 32.53 g of a solid is heated to 100.oC and added to 50.0 g of water in a coffee cup calorimeter and the contents are allowed to sit until they finally have the same temperature. The water temperature changes from 25.36 oZ to 34.4 oC. What is the specific heat capacity (in J/goC) of the solid

Answers

Answer:

0.886 J/g.°C

Explanation:

Step 1: Calculate the heat absorbed by the water

We will use the following expression

Q = c × m × ΔT

where,

Q: heatc: specific heat capacitym: massΔT: change in the temperature

Q(water) = c(water) × m(water) × ΔT(water)

Q(water) = 4.184 J/g.°C × 50.0 g × (34.4 °C - 25.36 °C) = 1.89 × 10³ J

According to the law of conservation of energy, the sum of the energy lost by the solid and the energy absorbed by the water is zero.

Q(water) + Q(solid) = 0

Q(solid) = -Q(water) =  -1.89 × 10³ J

Step 2: Calculate the specific heat capacity of the solid

We will use the following expression.

Q(solid) = c(solid) × m(solid) × ΔT(solid)

c(solid) = Q(solid) / m(solid) × ΔT(solid)

c(solid) = (-1.89 × 10³ J) / 32.53 g × (34.4 °C - 100. °C) = 0.886 J/g.°C

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