PLSSSSSS ANSWERRRRR???

when 5.20 × 10⁻³ moles of N₂O₅ gas completely reacts, 2.60 × 10⁻³ mol of NO₂ gas is formed.

What is Moles?

A mole is a unit of measurement of the quantity of a substance in the International System of Units. It is defined as exactly 6.02214076×10²³ particles, which can be atoms, molecules, ions or electrons.

To calculate the quantity of moles of NO₂ gas formed, we need to use the stoichiometric coefficients from the balanced reaction.

The balanced equation is:

2 N₂O₅(g) → 4 NO₂(g) + O₂(g)

According to the balanced equation, for every 2 moles of N₂O₅ that react, 4 moles of NO₂ are produced.

So, we can set up a ratio:

2 moles of N₂O₅ : 4 moles of NO₂

Now, we can use this ratio to calculate the moles of NO₂ gas formed when 5.20 × 10⁻³ mol of N₂O₅ gas reacts.

Given: Moles of N₂O₅ = 5.20 × 10⁻³ mol

Using the ratio:

4 moles of NO₂

------------------- = x moles of NO₂

2 moles of N₂O₅

x = (4/2) × (5.20 × 10⁻³) = 2.60 × 10⁻³ mol

Therefore, when 5.20 × 10⁻³ mol of N₂O₅ gas completely reacts, 2.60 × 10⁻³ mol of NO₂ gas is formed.

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Mendeleev arranged the elements in the periodic table on the basis of atomic masses and stated the behavior of the elements in a periodic law. The periodic law is when elements arranged by the atomic masses they exhibit the presence of similar properties. Mendeleev focused on the chemical properties of the elements and Meyer focused on the physical properties of the elements. The change in the periodic table is that the elements are now arranged in the order of their atomic numbers instead of their atomic masses.

Answer:

6........................

Answer:

B

Explanation:

It is important to stretch your muslces as a warm up prior to any exercise.

Answer:

B

Explanation:

Answer:

heat flows from the object that has more thermal more energy to the object with less thermal energy

Answer: 120 g

Explanation:

The sample that has the largest number of atoms is 1g of Chlorine(Cl). That is option C.

To calculate the sample with the largest number of atoms, the various number of atoms of each sample should be determined.

That is;

For sample A;

Na = 23 g

But 1g of Na= 6.022×10²³

23g of Na= 6.022×23 = 1.39×10²¹

For sample B;

Li = 7g

But 1 g of Na = 6.022×10²³

7 g of Na = 7× 6.022×10²³

= 4.22×10²⁴

For sample C;

Cl2= 71

But 1 g of Cl = 6.022×10²³

71g of Cl = 71×6.022×10²³

= 4.28 × 10²⁵

For sample D;

Au = 197

But 1g of Au = 6.022×10²³

197g = 197×6.022 × 10²³

= 1.186 × 10²⁶

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Complete question:

Which one of the following will have greatest number of atoms?

A.) 1g Na

B.) 1g Li

C.) 1g Cl2

D.) 1g Au

Explanation:

1. If you measure the value of Hydronium concentration is 3.0x10-3 M/L, the solution can be predicted to be acidic. This is because the concentration of hydronium ions, H3O+, is greater than the concentration of hydroxide ions, OH-. The pH value of an acidic solution is less than 7. Since hydronium ions are present in a high concentration, the pH value of the solution will be less than 7.

2. If you measure the number of Hydronium ions is 3.0x1012, the solution can be predicted to be acidic. This is because the concentration of hydronium ions is greater than 1.0 x 10^-7 M/L. The pH value of an acidic solution is less than 7.

3. If you measure the number of Hydroxide ions is 3.0x10-2 moles, the solution can be predicted to be basic. This is because the concentration of hydroxide ions is greater than 1.0 x 10^-7 M/L. The pH value of a basic solution is greater than 7. Hence, if the number of hydroxide ions is greater than 1.0 x 10^-7 M/L, the solution will be basic.

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Answer:  0.548 mol

Explanation:

If the oxidation of glucose 6-phosphate via the pentose phosphate pathway were being used primarily to generate NADPH for biosynthesis, the other product, ribose 5-phosphate would accumulate. The accumulation of ribose 5-phosphate would lead to a problem, i.e. it would quickly deplete NADPH in the cell.

This is because ribose 5-phosphate is produced in the first step of the pentose phosphate pathway, and if it is not utilized in the next step, it will accumulate. This accumulation of ribose 5-phosphate will have a detrimental effect on the cell's metabolic pathways. In this case, the ribose 5-phosphate will compete with NADP+ as an electron acceptor. Since NADPH is required for the synthesis of fatty acids, nucleotides, and other essential biomolecules, its depletion will lead to an insufficient supply of these biomolecules. Therefore, the accumulation of ribose 5-phosphate will impair the biosynthetic processes that require NADPH.

Thus, it is essential to balance the pentose phosphate pathway to maintain the cellular redox balance, which is necessary for the normal functioning of the cell.

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The nucleus of an atom is small compared to the size of the atom because of Rutherford's nuclear theory. Rutherford's nuclear theory states that most of the mass of an atom and all of its positive charge are contained in a small core called the nucleus.

This statement is consistent with the fact that the nucleus of an atom is small compared to the size of the atom. Rutherford's nuclear theory states that the nucleus of an atom is small compared to the size of the atom, and, therefore, the nucleus has a relatively low mass compared to the mass of an atom.

Rutherford's nuclear theory states that the nucleus is small but contains about half of the mass of an atom. The nuclear theory was discovered by Ernest Rutherford in 1911.

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Answer:

0.013MB

Explanation:

After considering the given data and statements we conclude that the correct option is A , Only one of the statements is true.

The statement that is true for a solution at 25 °C with pOH = 10.53 is:
iii) For this solution, [tex][OH^-] = 10^{(-10.53)} M.[/tex]
Statement i) is false due to the fact that a solution with a pOH greater than 7 is basic, not acidic.
Statement ii) is false due to  the pH of the solution can be evaluated from pOH using the equation [tex]pH + pOH = 14[/tex], not by subtracting pOH from 14. Therefore, the pH of the solution is 14 - 10.53 = 3.47, which is acidic.
Statement iii) is true due to the concentration of hydroxide ions [tex]([OH^-])[/tex] can be calculated from pOH using the equation[tex][OH^-] = 10^{(-pOH)}[/tex]. Therefore, [tex][OH^-] = 10^{(-10.53)} M.[/tex]
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Answer: When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase.

Explanation:

The decomposition reaction is 2[tex]H_{2}O_{2}[/tex] → 2[tex]H_{2}[/tex]O + [tex]O_{2}[/tex]

Hence, the correct option is B.

In this reaction, hydrogen peroxide ([tex]H_{2}O_{2}[/tex] ) decomposes into water ([tex]H_{2}[/tex]O

) and oxygen gas ( [tex]O_{2}[/tex]). Hydrogen peroxide is a compound composed of hydrogen and oxygen. Through a decomposition reaction, the hydrogen peroxide molecules split apart, resulting in the formation of water and oxygen.

The balanced equation indicates that for every two molecules of hydrogen peroxide, two molecules of water and one molecule of oxygen gas are produced. This reaction is an example of a decomposition reaction because a single compound (hydrogen peroxide) breaks down into multiple simpler substances (water and oxygen gas).

The decomposition reaction is 2[tex]H_{2}O_{2}[/tex]  → 2[tex]H_{2}[/tex]O + [tex]O_{2}[/tex]

Hence, the correct option is B.

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Answer:

50?

I think but not 100% sure if not lmk

The value of the double integral ∬(3x - y) dA, where the region R is in the first quadrant enclosed by the circle x² + y² = 9, the line x = 0, and the line y = x, using the transformation to polar coordinates, is 27/4√2.

To evaluate the double integral, we can switch to polar coordinates. In polar coordinates, the variables x and y are expressed in terms of r (radius) and θ (angle).

The region R, which is enclosed by the circle x² + y² = 9, can be expressed in polar coordinates as 0 ≤ r ≤ 3 and 0 ≤ θ ≤ π/4. This represents the first quadrant portion of the circle.

The Jacobian of the transformation from Cartesian to polar coordinates is r (the absolute value of the determinant of the Jacobian matrix). So, dA in polar coordinates is equal to r dr dθ.

Now, we can express the integral in terms of polar coordinates:

∬(3x - y) dA = ∫∫(3r cos θ - r sin θ) r dr dθ

Next, we need to set up the limits of integration. Since the region R is bounded by x = 0, y = x, and the circle x² + y² = 9, we can set up the limits as follows:

0 ≤ θ ≤ π/4 (corresponding to the angle that spans the first quadrant)

0 ≤ r ≤ 3 (corresponding to the radius of the circle)

Now, we can evaluate the integral:

∫[0,π/4] ∫[0,3] (3r cos θ - r sin θ) r dr dθ

By integrating with respect to r first, we get:

∫[0,π/4] [(3/4) r³ cos θ - (1/2) r³ sin θ] |[0,3] dθ

Simplifying further, we obtain:

∫[0,π/4] [(27/4) cos θ - (27/2) sin θ] dθ

Finally, we integrate with respect to θ:

[27/4 sin θ + (27/2) cos θ] |[0,π/4]

Plugging in the limits of integration, we have:

(27/4 sin(π/4) + (27/2) cos(π/4)) - (27/4 sin(0) + (27/2) cos(0))

Evaluating the trigonometric functions at the respective angles, we find:

(27/4 (1/√2) + (27/2) (1/√2)) - (27/4 (0) + (27/2) (1))

Simplifying further, we get the final result:

(27/4√2 + 27/2√2) - (0 + 27/2) = (27/4√2 + 27/2√2) - 27/2 = 27/4√2

Therefore, the value of the double integral ∬(3x - y) dA over the region R, using the transformation to polar coordinates, is 27/4√2.

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The diameter of the capillary tube is approximately 8.12 micrometers.

To find the diameter of the capillary tube, we can use the relationship between the height of the liquid column in a capillary tube and the surface tension.

The formula for the height of the liquid column in a capillary tube is given by:

h = (2 × T) / (ρ × g × r)

where:

h is the height of the liquid column,

T is the surface tension of the liquid,

ρ is the density of the liquid,

g is the acceleration due to gravity, and

r is the radius of the capillary tube.

In this case, we are given:

h = 8.4 cm = 0.084 m (converted to meters),

T = 0.07199 kg/s²,

ρ = 1.0 g/cm³ = 1000 kg/m³ (converted to kg/m³), and

g = 9.8 m/s².

We can rearrange the formula to solve for the radius of the capillary tube:

r = (2 × T) / (ρ × g × h)

Substituting the given values:

r = (2 × 0.07199 kg/s²) / (1000 kg/m³ × 9.8 m/s² × 0.084 m)

Calculating this expression gives us:

r ≈ 4.06 x 10⁻⁶ m

Finally, we can find the diameter of the capillary tube by doubling the radius:

diameter = 2 × r = 2 × 4.06 x 10⁻⁶ m = 8.12 x 10⁻⁶ m

Therefore, the diameter of the capillary tube is approximately 8.12 micrometers.

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Answer:

Explanation:

Remember, matter can neither be created nor destroyed, so every atom of reactant the leftt side of the equation must be accounted for in the products on the right side of the equation. With  magnesium and hydrochloric acid we have:  Mg  +  2HCl   →  MgCl2  +  H2(g).  Magnesium is divalent, and so requires two chloride ions to form magnesium chloride.  Hydrogen gas is a diatomic molecule, so the reaction requires the product to be H2.  Going across the equation from left to right we have 1 atom of magnesium (accounted for on the right side of the equation), and on the left, 2 moles of hydrochloric acid; the two hydrogens represented on the left of the equation are accounted for on the right of the equation.  Likewise, the two chloride anions on the left of the equation are accounted for in the formation of MgCl2 on the right. When all the atoms are accounted for on both sides of the equation, we say the equation is "balanced".

Answer: The red on is attracted to The north pole

Explanation:

Answer:

11.3m/s

Explanation:

Answer: 15 grams of salt are there in 5 L of solution.

Explanation:

Concentration of a solution is defined as the number of grams of solute dissolved per liter of the solution.

Given : Concentration of solution : 3 g/L

1 Liter of solution contains = 3 gram of salt

Thus 5 L of solution contains = [tex]\frac{3g}{1L}\times 5L=15g[/tex]

Thus 15 grams of salt are there in 5 L of solution.

Answer:

H2

Explanation:

I would say H2 cause CH4 have less hydrogen gas .

At standard temperature and pressure conditions, both the balloons have equal grams of gas.

Standard temperature and pressure are defined as a standard set of conditions required for experimental measurements which are established to allow comparison between different sets of data.

Standards which are commonly used are those International Union of pure and applied chemistry and national institute of standards and technology.These are not universally accepted standards but are the ones which are commonly used.

Standard conditions of pressure and temperature are necessary to define standard reference conditions used to express volumes of liquids and gases.

These values are important to physicists, chemists ,engineers ,pilots and navigators.

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Answer:

1.67 × 10^9

Explanation: