When 91.96g of Na reacts with 32.o g of O2 how many grams of NaO2 are produced

The reaction of 2-bromo-2-methylpropane and naphthalene is stoichiometrically  in a mole ratio of 1:1.

Notice that in the question, the moles of 2-bromo-2-methylpropane placed in the flask wasn't mentioned. if I assume that it was 1.24 moles then;

1 mole of naphthalene reacts with 1 mole of 2-bromo-2-methylpropane

x moles of naphthalene reacts with 1.24 mole of 2-bromo-2-methylpropane

x = 1 * 1.24/1

= 1.24 moles of naphthalene

Molar mass of naphthalene  = 128.2 g/mol

Mass of naphthalene  = 128.2 g/mol * 1.24 moles of naphthalene

Mass of naphthalene  =159 g

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Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.

hydro = water

Answer:

Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!

Explanation:

The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)

The compound does not undergo an aldol addition reaction in presence of aqueous sodium hydroxide is 2, 2-dimethylbutanal.

The Aldol Reaction occurs when the enolate of an aldehyde or ketone combines with the carbonyl of another molecule at the aplha-carbon under basic or acidic circumstances to produce beta-hydroxy aldehyde or ketone.

For the formation of enolate ion, compound should contain alpha hydrogen in it and among the given compound only 2, 2-dimethylbutanal will not have alpha hydrogens.

Hence 2, 2-dimethylbutanal does not undergo an aldol addition reaction.

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Answer:

yes it is

Explanation:

because there are some acid which really harm skin.

Answer:

[ClO] = 3.48×10¯⁹ M.

Explanation:

The following data were obtained from the question:

Equilibrium constant (Kc) = 4.96×10¹¹

Concentration of Cl2O2, [Cl2O2] = 6x10¯⁶ M.

Concentration of ClO, [ClO] =.?

The equation for the reaction is given below:

2ClO(g) ⇌ Cl2O2(g)

The equilibrium constant for a reaction is simply defined as the ratio of the concentration of product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant, Kc for the reaction is given by:

Kc = [Cl2O2] / [ClO]²

Thus, we can calculate the concentration of ClO, [ClO] as follow:

Kc = [Cl2O2] / [ClO]²

4.96×10¹¹ = 6x10¯⁶ / [ClO]²

Cross multiply

4.96×10¹¹ × [ClO]² = 6x10¯⁶

Divide both side by 4.96×10¹¹

[ClO]² = 6x10¯⁶ / 4.96×10¹¹

[ClO]² = 1.21×10¯¹⁷

Take the square root of both side

[ClO] = √ (1.21×10¯¹⁷)

[ClO] = 3.48×10¯⁹ M

Therefore, the concentration of ClO, [ClO] is 3.48×10¯⁹ M.

Answer:

The total photons required for this radiation = 5.1938 × 10²⁸ photons

Explanation:

Given that:

A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat.

If the radiation wavelength is 12.5 cm,

density of water = 1g/cm³

volume of the container = 0.250 L = 250 cm³

density = mass/volume

mass of the water = density × volume

mass of the water =  1g/cm³  × 250 cm³

mass of the water = 250 g

specific heat capacity of water = 4.182 J/g°C

The change in temperature was from 20.0° C to 99° C

ΔT =( 99 -20.0)° C

ΔT = 79.0° C

The heat absorbed in the process is calculated by using the formula,

q = mcΔT

q = 250 g × 4.182 J/g°C ×  79.0° C

q = 82594.5 Joules

Recall that the radiation wavelength λ = 12.5 cm = 0.125 m

The amount of energy of one photon of the radiation wavelength is determined by using the formula:

E = hv  

since v = c/λ

E = hc/λ

where;

h = Planck's constant = 6.626 × 10⁻³⁴ J.s

c = velocity of light = 3.0 × 10⁸ m/s

∴

E = (6.626 × 10⁻³⁴ J.s × 3.0 × 10⁸ m/s)/ 0.125 m

E = 1.59024⁻²⁴ Joules

The total photons required for this radiation = total heat energy/energy of radiation

The total photons required for this radiation = 82594.5 Joules/1.59024⁻²⁴ Joules

The total photons required for this radiation = 5.1938 × 10²⁸ photons

Answer:

1) 6.524779402×10^(-17)  

2)521.1g

3)113

Explanation:

Answer: 1) 6.524779402×10^(-17)

2)521.1g

Explanation:

Answer:

I would say c) a closed circuit.

Hope I was right.

Answer:

Half-life at 629K = 252.4min

Explanation:

Using Arrhenius equation:

[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]

And as Half-life in a first order reaction is:

[tex]t_{1/2}=\frac{ln2}{K}[/tex]

We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:

[tex]58.0min=\frac{ln2}{K}[/tex]

K = 0.01195min⁻¹ = K₁

[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]

[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]

[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]

K₂ = 2.75x10⁻³ min⁻¹

And, replacing again in Half-life expression:

[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]

The half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.   

The activation energy of a reaction is related to its rate constant as follows:  

[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex]   (1)

Where:

We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:

[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex]   (2)

Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min

Hence, the rate constant at 652 K is:                            

[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]

Now, from equation (1) we can find the pre-exponential factor (A):

[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]  

With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):

[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]

Finally, the half-life at 629 K is (eq 2):

[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]

Therefore, the half-life at 629 K is 251.1 min.

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Answer:

CO3^2-

Explanation:

In qualitative analysis, we try to use chemical reactions to determine the composition of an unknown substance. The addition of certain reagents to the unknown solution gives certain results that show the presence or absence of certain species from the unknown sample.

When dilute HCl is added to an unknown sample and effervescence is observed, then the unknown sample must contain CO3^2- or HCO3^-. The presence of these species is confirmed if the gas evolved is passed through limewater and the gas turns limewater milky.

Answer:

Atomic mass = 127.198 amu

Explanation:

The average atomic mass is obtained by summing the masses of the isotopes each multiplied by its abundance.

Atomic mass = (97.62 * 0.0825) + (109.3 * 0.2671) + (138.3 * 0.6504)

Atomic mass = 8.05365 +  29.19403 + 89.95032

Atomic mass = 127.198 amu

Answer:

Explanation:

Answer in attached file .

Answer:

false, it is not an example of facilitated diffusion

Answer:

Explanation:

When a drop of ink added into the water gradually moves in the whole quantity of water due to this entire water turns into blue color. This is nothing but the diffusion of ink particles into the water molecules. This is because water, as well as ink molecules, are in random motion due to the motion of ink substance.

Answer:

AgI I the limiting reactant.

Explanation:

The balanced equation for the reaction is given below:

2AgI + HgI2 → Ag2HgI4

Next, we shall determine masses AgI and HgI2 that reacted from the balanced equation.

This is illustrated below:

Molar mass of Agl = 108 + 127 = 235 g/mol

Mass of AgI from the balanced equation = 2 x 235 = 470 g

Molar mass of HgI2 = 201 + (2x127) = 455 g/mol

Mass of HgI2 from the balanced equation = 1 x 455 = 455 g

From the balanced equation above,

470 g of AgI reacted with 455 g of HgI2.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above,

470 g of AgI reacted with 455 g of HgI2.

Therefore, 2 g of AgI will react with

= (2 x 455)/470 = 1.94 g of HgI2.

From the calculations made above, we can see that only 1.94 g out of 3.50 g of HgI2 given is needed to react completely with 2 g of AgI.

Therefore, AgI I the limiting reactant.

Answer:

.217, .311, and .472, respectively.

Explanation:

The total number of moles of gas is 3.50 + 5.00 + 7.60 = 16.10 (to preserve significant digits).

X of helium=3.50/16.10 = .217

X of krypton=5.00/16.10 = .311

X of neon=7.60/16.10 = .472

Acetic acid is basically known as CH_3COOH.

Chemical Equation:-

[tex]\boxed{\sf {CH_3COOH\atop Acetic\:acid}+{KOH\atop Potassium\:Hydroxide}\longrightarrow {CH_3COOK\atop Potassium\: Acetate}+{H_2O\atop Water}}[/tex]

Solution:-2

[tex]\boxed{\sf {HCN\atop Hydrogen\:Cyanate}+{Na_2CO_3\atop Sodium\:Carbonate}\longrightarrow {NaCN\atop Sodium\:Cyanate}+{H_2O\atop water}+{CO_2\atop Carbon\:dioxide}}[/tex]

Answer:

– 844 kJ/mol.

Explanation:

The following data were obtained from the question:

N2(g) + 3 F2(g) –––> 2 NF3(g)

Enthalpy of N≡N (N2) = 945 kJ/mol

Enthalpy of F–F (F2) = 155 kJ/mol

Enthalpy of N–F3 (NF3) = 283 kJ/mol

Enthalpy change (∆H) =?

Next, we shall determine the enthalpy of reactant.

This is illustrated below:

Enthalpy of reactant (Hr) = 945 + 3(155)

Enthalpy of reactant (Hr) = 945 + 465

Enthalpy of reactant (Hr) = 1410 kJ/mol

Next, we shall determine the enthalpy of the product.

This is illustrated below:

Enthalpy of product (Hp) = 2 x 283

Enthalpy of product (Hp) = 566 kJ/mol

Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:

Enthalpy of reactant (Hr) = 1410 kJ/mol

Enthalpy of product (Hp) = 566 kJ/mol

Enthalpy change (∆H) =?

Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)

Enthalpy change (∆H) = 566 – 1410

Enthalpy change (∆H) = – 844 kJ/mol

Answer:

– 844 kJ/mol.

Explanation:

The following data were obtained from the question:

N2(g) + 3 F2(g) –––> 2 NF3(g)

Enthalpy of N≡N (N2) = 945 kJ/mol

Enthalpy of F–F (F2) = 155 kJ/mol

Enthalpy of N–F3 (NF3) = 283 kJ/mol

Enthalpy change (∆H) =?

Next, we shall determine the enthalpy of reactant.

This is illustrated below:

Enthalpy of reactant (Hr) = 945 + 3(155)

Enthalpy of reactant (Hr) = 945 + 465

Enthalpy of reactant (Hr) = 1410 kJ/mol

Next, we shall determine the enthalpy of the product.

This is illustrated below:

Enthalpy of product (Hp) = 2 x 283

Enthalpy of product (Hp) = 566 kJ/mol

Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:

Enthalpy of reactant (Hr) = 1410 kJ/mol

Enthalpy of product (Hp) = 566 kJ/mol

Enthalpy change (∆H) =?

Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)

Enthalpy change (∆H) = 566 – 1410

Enthalpy change (∆H) = – 844 kJ/mol

Explanation:

Answer:

ΔHrxn = -635.14kJ/mol

Explanation:

We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:

(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol

(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole

(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole

Reaction (1) - (3) produce:

Ca(s) + H2O(l) → H2(g) + CaO(s)

ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol

Now this reaction + 1/2(2):

Ca(s) + ½ O2(g) → CaO(s)

ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)

The mass of magnesium chloride produced from 2.30 moles of chlorine gas is 218.99 grams.

Stoichiometry refers to the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions.

According to this question, magnesium reacts with chlorine gas to form magnesium chloride as follows:

Mg + Cl₂ → MgCl₂

Based on the above chemical equation, 1 mole of chlorine gas forms 1 mole of magnesium chloride.

This means that 2.30 moles of chlorine gas will 2.30 moles of magnesium chloride.

Next, we convert moles of magnesium chloride to mass as follows:

molar mass of magnesium chloride = 95.211g/mol

mass of magnesium chloride = 95.211 × 2.30 = 218.99 grams.

Therefore, 218.99 grams of magnesium chloride will be formed.

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Answer:

- photography     (captures images that can be shared and

compared by scientists)

- space telescope    (allows access to images taken from outside

Earth's atmosphere)

- radio telescope     (detects electromagnetic frequencies outside

the visible spectrum that reaches Earth)

- optical telescope    (obtains a magnified and clear view of a part

of the sky to observe celestial objects)

Explanation:

Photography is used to capture still images based on the principle that some compounds react in the presence of optical energy.

Space telescope is a type of observatory telescope positioned in outer space to observe distant planets, galaxies and other astronomical objects. Space telescopes reduces the interference from ultraviolet frequencies, X-rays and gamma rays; as well as light pollution which ground-based observatories encounter.

A radio telescope is a specialized antenna and radio receiver used to receive radio waves from astronomical radio sources in the sky. Radio telescope is used to study radio frequencies  emitted by astronomical objects, that fall outside the visible light spectrum.

An optical telescope is used to gather and focuses light, from a far distant object. Optical telescope is used within the visible light spectrum of the electromagnetic spectrum, to create a magnified image, for direct view, or to make a photograph, or to collect data through electronic image sensors.

Answer:

- photography     (captures images that can be shared and

compared by scientists)

- space telescope    (allows access to images taken from outside

Earth's atmosphere)

- radio telescope     (detects electromagnetic frequencies outside

the visible spectrum that reaches Earth)

- optical telescope    (obtains a magnified and clear view of a part

of the sky to observe celestial objects)

Explanation:

Answer:

Explanation:

For pH of a buffer solution , the formula is

pH = pKa + log [ Base ] / [ conjugate acid ]

=   pKa + log [ NH₃ ] / [ NH₄⁺ ]

Ka = Kw / Kb

Kb for NH₄OH = 1.8 x 10⁻⁵

Ka = 10⁻¹⁴ / 1.8 x 10⁻⁵

= 5.6 x 10⁻¹⁰

pH = - log ( 5.6 x 10⁻¹⁰ ) + log 0.2 / 0.2

= 10 - log 5.6

= 9.25

Effect of addition of HCl

H⁺ of HCl will react with NH₃ to produce NH₄⁺

25 mL of .1 HCl = 2.5 mM of HCl

25 mL of .1 NH₄⁺ = 2.5 mM of NH₄⁺

65 mL of .2 M NH₃ = 13 mM of NH₃

65 mL of .2 M NH₄⁺ = 13 mM of NH₄⁺

NH₃ + H⁺ = NH₄⁺

NH₄⁺ formed = 2.5 + 13  mM

15.5 mM of NH₄⁺

NH₃ = 13 mM

Concentration of NH₃ = 13 / 90

Concentration of NH₄⁺ = 15.5 / 90

pH of final buffer mixture

= 9.6 + log 13 / 15.5

= 9.25 - .076

= 9.174

The pH value  is mathematically given as

pH= -6.332.

Question Parameters:

the pH of a 0.20 M NH3/0.20 M NH4Cl buffer

the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.

Generally, the equation for the Chemical Reaction  is mathematically given as

HCl + NH3 --> NH4^+ + Cl^-

Therefore

pH= pka + log(13/14).

pH= -6.3 + log 0.93.

pH= -6.3+ (-0.032).

pH= -6.332.

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I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.

Answer:

D. Rare

Explanation:

Qualitative research has to do with non-numerical data and is used to understand the beliefs of a group of people which can be gotten from surveys, questionnaires, interviews, etc.

A qualitative researcher may select a RARE case that is unusually rich in information pertaining to the research question.

This is because he wants to get an insight into the why, how, where and when of that particular ase as it's not a usual occurrence.

Answer:

when equal moles of an acid and base are mixed,after reaction the two are compounds are said to be at the Equivalent point.

Answer:

The magnetic field shows (D)

Explanation:

Because like poles repel eachother

The magnetic field is represented by two poles of a magnetic with magnetic filed lines from the poles. The image C represents the magnetic filed .

A magnetic filed is a generated by strong filed lines from a magnet with two poles. A magnet have two poles namely magnetic south and north poles.

The magnetic south poles of of a magnet is always attracted towards the magnetic north poles. Magnetic north poles of a magnet is generally shown in the left side and the field lines will be outward towards the geographic north pole.

The magnetic field lines from the south pole is inward and attracted towards the north. Therefore, right figure representing the magnetic filed is option C.

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There are various variants of Cov id virus. The Brazilian variant P also known as Gamma variant is the third variant of the original SARS-CoV2.

The correct answer is more negative

This variant has raised concerns since it has ability to spread more quickly then previous variants and this is more negative variant.

It is assumed that Cov id variant Gamma and Delta has ability to absorb move UV light but this is not proved yet and research is underway.

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Answer:

Explanation:

a) 4-nitrobenzoic acid         pKa= 3.41

    benzoic acid                   pKa= 4.19

    4-chlorobenzoic acid     pKa= 3.98

b) benzoic acid                    pKa= 4.19    

   cyclohexanol                   pKa= 18.0

   phenol                              pKa= 9.95

c) 4-Nitrobenzoic acid             pKa= 3.41

   4-nitrophenol                       pKa= 7.15

   4-nitrophenylacetic acid     pKa= 3.85

Answer:

option (A) is right answer

Answer:

c

Explanation:

trust me I took the exam

Answer:

water molecules

Explanation:

Redox reactions are carried out under acidic or basic conditions as the case may be.

If the reaction is carried out in an acid medium, then we must balance the hydrogen ions on the lefthand side of the reaction equation with water molecules on the righthand side of the reaction equation.

For instance, the equation for reduction of MnO4^- under acidic condition is shown below;

MnO4^-(aq) + 5e + 8H^+(aq) --------> Mn^2+(aq) + 4H2O(l)