Can steel turn into gas? ( A random question that came to mind. I have no idea if it is or isn't possible.)

Answers

Answer 1

Answer:

YES STEEL CAN TURN INTO GAS .

Answer 2

Answer: yes it can!

Explanation:


Related Questions

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.

11. An isotope Q has 18 neutrons a mass number of 34. (a) (i) What is an isotope? An isotope is one of two or C (b) Write its electron arrangement. Mass number=34 Number of neutrons=18 Number of Protons = 34-15-16 (c) To which period and group does Q belong? Protors - Electons - Atomic number Period - Group (d) How does Q form its ion?

Answers

An isotope is an element with the same atomic number but different mass number due to differences in number of neutrons.

electron configuration is 2,8,6.

Belongs to group 6 and period group 3.

It forms an ion by accepting 2 electrons

If a hydrogen of an alkane is replaced by NH, the compound becomes_________

a. alcohol

b. carboxylic acid

c. phenol

d. amine​

Answers

Answer:

d. amine.

It becomes an amine.

Explanation:

With general formular

[tex]{ \bf{primary \: amine :R - NH _{2}}} \\ { \bf{secondary \: amine : R {}^{i} - NH - R}} \\ { \bf{tertiary \: amine :R {}^{ii} - N(R {}^{i} ) - R }}[/tex]

R is the aryl group such as alkane

i think it becomes an amine

Which species is the conjugate base of H2SO3

Answers

Explanation:

As you know, the conjugate base of an acid is determined by looking at the compound that's left behind after the acid donates one of its acidic hydrogen atoms.

The compound to which the acid donates a proton acts as a base. The conjugate base of the acid will be the compound that reforms the acid by accepting a proton.

In this case, sulfurous acid has two protons to donate. However, the conjugate base of sulfurous acid will be the compound left behind after the first hydrogen ion is donated.

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.

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

There are _______ alkanes with molecular formula C10H22

a. 74

b. 75

c. 76

d. 77​

Answers

I guess b cause there are 75 alkanes with molecular formula C10H22

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

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).

How many mL of 0.200M KI would contain 0.0500 moles of KI?

Please explain and show work.

Answers

Answer:

250ml

Explanation:

call it V

V*0.2=0.05 (moles)

so V=0.05/0.2 = 0.25l = 250ml

Molarity=0.2MNo of moles=0.05mol

We know

[tex]\boxed{\Large{\sf Molarity=\dfrac{No\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\;\ell}}}[/tex]

[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=\dfrac{0.05}{0.2}[/tex]

[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=0.25L[/tex]

[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=250mL[/tex]

Give the change in condition to go from a gas to a solid. Group of answer choices cool or increase pressure cool or reduce pressure increase heat or reduce pressure increase heat or increase pressure none of the above

Answers

Answer:

cool or increase pressure

Explanation:

For a gas to form solid. There must be reduced heat and pressure. The deposition of gas into solid occurs through the removal of thermal energy. The air looses thermal energy and changes into solid.

Once you have collected 40 mL of distillate, you should ________. turn off your hot plate lower your lab jack carelessly use your hand to remove the heating block turn off the hot plate and carefully lower the lab jack, making sure that no cords or hoses get caught in it

Answers

Answer:

Once you have collected 40 mL of distillate, you should ________.

turn off the hot plate and carefully lower the lab jack, making sure that no cords or hoses get caught in it.

Explanation:

Distillate is the product obtained from the process of distillation.  Distillation is the separation of components of a liquid mixture based on different boiling points. Distillation can be used to purify alcohol, for desalination, refining of crude oil, and for obtaining liquefied gases.  A lab jack is an essential tool that supports and lifts hotplates, glassware, baths, and other small lab equipment requiring stable surfaces at a specific height.

A solution has a OH- concentration of 7.7x10-3. What is the pH of this solution?

Answers

Answer:

11.9 pH

Explanation:

First, we need to find pOH

To find that, we use the formula -log[OH]

-log[7.7x10^-3] = 2.11351

To find the pH, we'll use this formula: 14 = pH + pOH

14 = pH + 2.11351

Subtract boths sides by 2.11351

14 = pH + 2.11351

-2.11351  -2.11351

pH = 11.88649

Na Sa Bant HCL -> 50g Hao pt Soy​

Answers

North America and south africa

The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum.

a. True
b. False

Answers

Answer:

a. True

Explanation:

The main information that gives an infrared absorption spectrum is the type of functional groups that are present in an organic compound. The infrared (IR) spectroscopy is based on the fact that functional groups absorb light in the IR region of the electromagnetic spectrum (approximately at 2,500-16,000 nm) and induces a vibrational excitation of the covalently bonded atoms in the group. The vibration of the atoms can be of different types, such as stretching, bending, etc. Each functional group (such as the carbonyl group) in an organic compound absorbs at a specific IR frequency so they can be distinguished from an IR spectrum.

Preparation the buffer solution: initial pH of buffer solution: ____ Titration of a weak acid with a strong base: initial pH of weak acid: ____ final pH of weak acid: ____ Amount of NaOH added: ____ Titration Curve for Weak Acid with a Strong Base (Paste curve here.)

Answers

Answer:

pH of buffer solution is 7.0

Initial pH of Weak acid is 3.27

Final pH of weak acid is 3.07

Amount of NaOH added is 1ml

Explanation:

Titration is a process in which acid and base are introduced together until a neutral solution is achieved whose pH value is near to buffer solution which is 7.0, the pH value for acid is below 7 while pH value for base is above 7.

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.

https://brainly.com/question/9970247

Aspirin that has been stored for a long time may give a vinegar like odour and give a purple colour with FeCl3. What reaction would cause this to happen
?.

Answers

Answer:

See explanation

Explanation:

The IUPAC name of aspirin is 2-Acetoxybenzoic acid. It is composed of an acetoxy moiety and a benzoic acid moiety.

The compound can be hydrolysed under prolonged storage conditions to yield acetic acid which causes the vinegar like odour.

Also, one of the products of this hydrolysis bears a phenol group which reacts with FeCl3 to give a purple color.

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.

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.

To find more on elemental ratio, refer here:

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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

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

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]

Regards!

Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. Classify each of the following metals by whether they would or would not act as a sacrificial anode to iron under standard conditions.

a. Ag
b. Mg
c. Cu
d. Pb
e. Sn
f. Zn
g. Au

Answers

Answer:

a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

Explanation:

Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. The reactivity series of metals arranges metals based on decreasing order of reactivity. The more reactive metals are found higher up in the series while the least reactive metals are found at the lower ends of the series. Thus, metals above iron in the reactivity series can serve as sacrificial anodes by protecting against corrosion, while those lower than iron cannot.

Based on the reactivity series, the following metals can be classified as either a sacrificial anode for iron or not:

a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

When selling on the street, dealers may not know the purity of the ketamine they have, and thus users do not know exactly how much ketamine they are receiving. It is unlikely that the ketamine is pure, or even that different batches of ketamine have the same purity. Assume the drug the user typically buys is only 25% ketamine, and therefore, the user actually dissolved 0.250 g ketamine in 1/4 cup of water to make the solution instead of 1 g in the previous question. 1 cup = 236.5 mL What volume of this ketamine solution would the 65.0 kg user have to inject to experience a high at 0.400 mg/kg? volume: mL What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg? of use contact us help What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg?

Answers

Answer:

a. 6.15 mL b. 30.73 mL

Explanation:

a. What volume of this ketamine solution would the 65.0 kg user have to inject to experience a high at 0.400 mg/kg?

Since we have 0.250 g of ketamine in 1/4 cup of water and 1 cup of water equals 236.5 mL, we need to find the concentration of ketamine we have.

So concentration of ketamine C = mass of ketamine, m/volume of water, V

m = 0.250 g and V = 1/4 cup = 1/4 × 236.5 mL = 59.125 mL

So, C = m/V = 0.250 g/59.125 mL = 0.00423 g/mL = 4.23 mg/mL

Since the user has a mass of 65 kg and requires a high at 0.400 mg/kg, the mass of ketamine for this high is M = 65 kg × 0.400 mg/kg = 26 mg

Since mass, M = concentration ,C × volume, V

M = CV

V = M/C

The volume of ketamine required for the 0.400 mg/kg high is

V = 26 mg/4.23 mg/mL

V = 6.15 mL

b. What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg?

Since the concentration of ketamine is C = 4.23 mg/mL, and Since the user has a mass of 65 kg and requires an injection of 2.00 mg/kg to be unconscious, the mass of ketamine required to be unconscious is M' = 65 kg × 2.00 mg/kg = 130 mg

Since mass, M' = concentration ,C × volume, V

M' = CV

V = M/C

The volume of ketamine required for the 2.00 mg/kg unconscious injection is

V = 130 mg/4.23 mg/mL

V = 30.73 mL

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.

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

An analytical chemist is titrating of a solution of hydrazoic acid with a solution of . The of hydrazoic acid is . Calculate the pH of the acid solution after the chemist has added of the solution to it.

Answers

Answer:

pH = 12.43

Explanation:

...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it.

To solve this question we need to know that hidrazoic acid reacts with KOH as follows:

HN3 + KOH → KN3 + H2O

Moles KOH:

0.5716L * (0.2900mol /L) =0.1658 moles of KOH

Moles HN3:

0.2127L * (0.6800mol/L) = 0.1446 moles HN3

As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:

0.1658 moles - 0.1446 moles =

0.0212 mol KOH

In 212.7mL + 571.6mL = 784.3mL = 0.7843L

The molarity of KOH = [OH-] is:

0.0212 mol KOH / 0.7843L = 0.027M = [OH-]

The pOH is defined as -log [OH-]

pOH = -log 0.027M

pOH = 1.57

pH = 14 - pOH

pH = 12.43

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

Answers

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

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

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