Calculate how many moles of NO2 form when each
quantity of reactant completely reacts.
2 N2O5(9) + 4NO2(g) + O2(9)
Part B
6.4 mol N205
Express your answer using two significant figures.
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n =
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Part C
16.2 g N205
Express your answer using three significant figurer

Answer:

A Brønsted-Lowry acid is any species that can donate a proton (H +) to another molecule. A Brønsted-Lowry base is any species that can accept a proton from another molecule. In short, a Brønsted-Lowry acid is a proton donor (PD), while a Brønsted-Lowry base is a proton acceptor (PA).

Answer:

resources that you can use more than 1 time like rechargeable battery or solar energy

Explanation:

Answer:

I think it would be B.

Explanation:

Answer: potential energy is stored

Explanation:

Answer:

(8.3×40)+(8.3×71)

921.3grames

Answer:

[tex]\Large \boxed{\sf 921.3 \ g}[/tex]

Explanation:

Use formula

[tex]\displaystyle moles=\frac{mass}{M_r}[/tex]

[tex]M_r=40+35.5 \times 2 = 111[/tex]

Solve for mass

[tex]mass=8.3 \times 111=921.3[/tex]

Answer:

B

Explanation:

Answer:

B

Explanation:

Edge 2022

Answer:

3224 kJ/mol

Explanation:

The combustion of benzoic acid occurs as follows:

C₇H₆O₂ + 13/2O₂ → 7CO₂ + 3H₂O + dE

The change in temperature in the reaction is the change due the energy released, that is:

3.256K * (10.134kJ / K) = 33.00kJ are released when 1.250g reacts

To find the heat released per mole we have to find the moles of benzoic acid:

Moles benzoic acid -Molar mass: 122.12g/mol-:

1.250g * (1mol / 122.12g) = 0.0102 moles

The dE combustion per mole of benzoic acid is:

33.00kJ / 0.0102moles =

3224 kJ/mol

Explanation:

It is represented using the ideal gas equation , or PV = nRT, where P is the pressure in atmospheres, V is the volume in liters, n represents the quantity of particles in the container, T represents the temperature in Kelvin, and R is the ideal gas constant equal to 0.0821 liters atmospheres per moles Kelvin.

Answer:

[tex]\mathbf{a \ strong \ base \ such \ as\ LiOH \ or \ NaOH.}[/tex]

Explanation:

In the condensation reactions of carbonyl compounds, it is essential to establish the order of events in advance to minimize or suppress the possibilities of self-condensation and the occurrence of cross-condensation, which unfortunately are an obvious threat in these reactions.

Self-condensation:

Any carbonyl compound that has one or more alpha hydrogens, on the carbons adjacent to the carbonyl group, runs the risk of undergoing a self-condensation reaction if the corresponding rigor is not carried out.

Cross-condensation:

Ideally, in the condensation reactions of carbonyl compounds, one of the reacting molecules should quickly enolise, while the other preferably should not have Hα, to ensure that no other by-products are formed.

[tex]\text{To achieve this process;}[/tex] [tex]\text{A self-aldol or crossed aldol requires a strong base such as LiOH or NaOH.}[/tex]

Answer:

It is 98 degrees C.

Explanation:

Answer:

3.45E24

Explanation:

calculator I'm not that great at math but I'm good at science but that isn't really science but it's aix between it tho

Answer:

Promethium

Pm 61

Discovered by Jacob .A. Marinsky

aswer: protactinium-234 is more dang erous than uranium

There can be emissions of radiations like gamma radiation. There can be emission of particles too like alpha particle. Therefore protactinium-234 is more dangerous than uranium-234.

Nuclear decay is process in which the radioactive element releases particles or radiations. Alpha particles is ⁴₂He. Alpha particle is nothing but helium particle.  Protactinium-234 and uranium-234 are radioactive element.

If we were sitting near a gram of protactinium-234 and a gram of uranium-234, both solid materials then protactinium-234 is more dangerous than uranium-234. The half-life of protactinium-234 is smaller than that of uranium-234. So, protactinium-234 is more active than uranium-234 and hence more harmful than uranium-234.

Therefore, protactinium-234 is more dangerous than uranium-234.

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Answer: I believe it's C

Hope this helped<3

Can you please make my answer brainly

Answer:

stress

Explanation:

people cn get stressed and not want to do work

[tex]\huge\underline\mathbb\pink{ANSWER\::}\\\\[/tex]

_BaS + _PtF2 → _BaF2 + _Pts

Barium Sulfide + Platinum(II) Fluoride = Barium Fluoride + Platinum(II) Sulfide.

=> 1 BaS + 1 PtF2 → 1 BaF2 + 1 PtS

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HOPE IT HELPS

PLEASE MARK ME BRAINLIEST☺️

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=T_f^0-T_f=8.2^0C[/tex] = Depression in freezing point

[tex]K_f[/tex] = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality =[tex]\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9[/tex]

[tex]8.2^0C=1\times K_f\times 1.9[/tex]

[tex]K_f=4.32^0C/m[/tex]

Now Depression in freezing point for sodium chloride is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=20.0^0C[/tex] = Depression in freezing point

[tex]K_f[/tex] = freezing point constant  

m= molality = [tex]\frac{136g\times 1000}{950g\times 58.5g/mol}=2.45[/tex]

[tex]20.0^0C=i\times 4.32^0C\times 2.45[/tex]

[tex]i=1.9[/tex]

Thus vant hoff factor for sodium chloride in X is 1.9

Answer:

MgS + KI = K2S + MgI2

Answer:

your answer is C

Explanation:

because when u make P1 the subject formula then P2 V2 will be divided with V1

Answer:

The three components of a chemical formula are the empirical formula, the molecular formula, and the structural formula.Hope this answer helps.

A scientist that carelessly pours cultures of living microbes down the drain represents a type of bacterial pollution.

Bacterial pollution refers to the uncontrolled proliferation of bacteria in the surrounding environment and ecosystems.

In conclusion, a scientist that carelessly pours cultures of living microbes down the drain represents a type of bacterial pollution.

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Answer:

20. ²⁴¹₉₅Am —> ²³⁷₉₃Np + ⁴₂He

21. ⁴²₁₉K —> ⁰₋₁β + ⁴²₂₀Ca

Explanation:

20. ²⁴¹₉₅Am —> __ + ⁴₂He

Let ⁿₘX be the unknown.

Thus, the equation becomes:

²⁴¹₉₅Am —> ⁿₘX + ⁴₂He

Next, we shall determine n, m and X. This can be obtained as follow:

241 = n + 4

Collect like terms

n = 241 – 4

n = 237

95 = m + 2

Collect like terms

m = 95 – 2

m = 93

Thus,

ⁿₘX => ²³⁷₉₃X => ²³⁷₉₃Np

Therefore, the complete equation is:

²⁴¹₉₅Am —> ²³⁷₉₃Np + ⁴₂He

21. __ —> ⁰₋₁β + ⁴²₂₀Ca

Let ʸₓZ be the unknown.

Thus, the equation becomes:

ʸₓZ —> ⁰₋₁β + ⁴²₂₀Ca

Next, we shall determine y, x and Z. This can be obtained as follow:

y = 0 + 42

y = 42

x = –1 + 20

x = 19

Thus,

ʸₓZ => ⁴²₁₉Z => ⁴²₁₉K

Therefore, the complete equation is:

⁴²₁₉K —> ⁰₋₁β + ⁴²₂₀Ca

This question is incomplete, the complete question is;

The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m² at 500C.

Assume a diffusion coefficient of 1.0 × 10⁻⁸ m²/s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained

Answer:

the number of kilograms of hydrogen that pass per hour through is 2.592 × 10⁻³ kg

Explanation:

Given the data in the question;

thickness of sheet t = 5 mm

Area A = 0.20 m²

Temperature T = 500°C

diffusion coefficient ∝ = 1.0 × 10⁻⁸ m²/s

concentration high pressure side C[tex]_A[/tex] = 2.4

concentration low pressure side C[tex]_B[/tex] = 0.6 kg

from the question, we calculate the concentration gradient

dc/dx = (C[tex]_A[/tex] -  C[tex]_B[/tex])/dt

so we substitute

dc/dx = (2.4 - 0.6) / ( - 5 × 10⁻³ )

dc/dx = -360

now, mass of hydrogen per hour that diffuse through a pd sheet

M = -∝AT(dc/dx)

where time t is 1 hour ( 3600 sec )

we substitute

M = -(1.0 × 10⁻⁸) × 0.20 × 3600 (-360)

M = 0.002592

M = 2.592 × 10⁻³ kg per one hour

Therefore, the number of kilograms of hydrogen that pass per hour through is 2.592 × 10⁻³ kg

The number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m² at 500°C is 2.592 × 10⁻³ kg.

Purification: It refers to the process of removing impurities or contaminants from a substance in order to make it cleaner, purer, or more refined. It also involves the elimination or reduction of unwanted substances or components that may be present in the original material.

According to the question, given data is:

Thickness of sheet t = 5 mm

Area A = 0.20 m²

Temperature T = 500°C

Diffusion coefficient ∝ = 1.0 × 10⁻⁸ m²/s

Concentration high pressure side [tex]C_A[/tex] = 2.4

Concentration low pressure side[tex]C_B[/tex]= 0.6 kg

from the question, we calculate the concentration gradient,

[tex]dc/dx = (C _A- C_B)/dt[/tex]

so, we substitute the values,

[tex]dc/dx[/tex] = (2.4 - 0.6)/ ( - 5 × 10⁻³)

[tex]dc/dx = -360[/tex]

Now, mass of hydrogen per hour that diffuse through a Palladium sheet,

M = -∝[tex]AT(dc/dx)[/tex]

where time t is 1 hour (3600 sec)

We substitute,

M = -(1.0 × 10⁻⁸) × 0.20 × 3600 (-360)

M = 0.002592

M = 2.592 × 10⁻³ kg per hour.

Therefore, the number of kilograms of hydrogen that pass per hour through a palladium sheet is approximately 2.592 × 10⁻³ kg.

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This question is incomplete, the complete question is;

The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m² at 500C.

Assume a diffusion coefficient of 1.0 × 10⁻⁸ m²/s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.

Answer:

Explanation:

It processes the environment and sends out signals.