How much solution would I end up with if I started with 500 mL of a 2.4M KCl solution and wanted to make a 1.0 solution?
A. 700 mL
B. 70.0 mL
C. 1200 mL
D. 120 mL

103.5 g of sodium would be needed to react with 4.5 moles of chloride ions. Therefore, the correct option is (B).

The balanced chemical equation Na+ + Cl → NaCl tells us that 1 mole of sodium ions (Na+) reacts with 1 mole of chloride ions (Cl-) to produce 1 mole of sodium chloride (NaCl). Therefore, to react with 4.5 moles of chloride, we need an equal number of moles of sodium. This is because the reactants must be present in the stoichiometric ratio of 1:1 to ensure complete reaction.

Moles of Na+ required = moles of Cl- = 4.5 mol

The molar mass of Na+ is 23 g/mol, so the mass of sodium required is:

Mass of Na+ = Moles of Na+ x Molar mass of Na+

Mass of Na+ = 4.5 mol x 23 g/mol = 103.5 g

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The concentration of the dye in the stock solution is 0.0250 M when solution in part B was made by taking 10.00 mL of a stock solution .

The amount of a solute dissolved in a given amount of a solvent or solution is referred to as concentration. It is usually represented in terms of the amount of solute per unit volume or mass of solution.

We can use the following equation to compute the dye concentration in the stock solution:

                                              C₁V₁ = C₂V₂

C₁ =concentration of the stock solution,

V₁ = volume of the stock solution used,

C₂ = concentration of the diluted solution,

V₂= Total volume of the diluted solution.

In this scenario, V₁ = 10.00 mL, V₂ = 100.00 mL, and C₂ = 0.00250 M (as determined in part B). When we plug these numbers into the equation and solve for C₁, we get:

                                             C₁ = (C₂V₂) / V₁

= (0.00250 M x 100.00 mL) / 10.00 mL

= 0.0250 M

Therefore, the concentration of the dye in the stock solution is 0.0250 M.

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Answer:

100

Explanation:

The balanced chemical equation for the combustion of any hydrocarbon in oxygen is:

CₙHₘ + (n + m/4)O₂ → nCO₂ + m/2H₂O

In this case, we don't have a hydrocarbon, but we can assume that the oxygen is reacting with some substance that contains carbon to produce carbon dioxide. Let's assume that the balanced chemical equation for this reaction is:

C + O₂ → CO₂

From the equation, we can see that 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. The molar mass of oxygen is 32 g/mol and the given mass is 100 g, so we have:

100 g / 32 g/mol = 3.125 mol of oxygen

Since the balanced chemical equation shows that 1 mole of oxygen reacts with 1 mole of carbon dioxide, we can conclude that 3.125 moles of oxygen will produce 3.125 moles of carbon dioxide.

Observations:

The addition of sodium hydroxide to each test tube results in the formation of iron hydroxide precipitates. In the case of the Fe²⁺ test tube, the iron(II) ions react with the hydroxide ions to form a brownish-green precipitate of iron(II) hydroxide.

In the Fe³⁺ test tube, the iron(III) ions react with the hydroxide ions to form a reddish-brown precipitate of iron(III) hydroxide. These observations confirm that both Fe²⁺ and Fe³⁺ ions are present in the original samples and provide evidence of a chemical reaction taking place.

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The mass (in grams) of AsH₃ made from 22.22 grams of H₂ is 571.63 grams

The mass of AsH₃ produced can be obtained as illustrated below:

2As + 3H₂ -> 2AsH₃ + 150 Kcal

From the balanced equation above,

6.06 grams of H₂ reacted to produce 155.9 grams of AsH₃

Therefore,

22.22 grams of H₂ will react to produce = (22.22 × 155.9) / 6.06 = 571.63 grams of AsH₃

Thus, from the above calculation, we can conclude that the mass of AsH₃ produced from the reaction is 571.63 grams

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Four properties of gases will be investigated: pressure, volume, temperature, and number of molecules. By assembling the equipment, conducting the appropriate tests, and analyzing your data and observations, you will be able to describe the gas laws, both qualitatively and mathematically.

For the chemical elements Cl, Am, Lu, Ca, Hg, I, Ti, and Mg, the final occupied sublevels are the 4s orbital for Cl, the 5f orbital for Am, the 7s orbital for Lu, the 3p orbital for Ca, the 6s orbital for Hg, the 5p orbital for I, the 4f orbital for Ti, and the 3s orbital for Mg.

The primary quantum number, n, of an element determines the sublevel of that element. greater numbers correspond to greater energy levels for the electron, which is indicated by this number.

The form of the orbital is determined by the angular momentum quantum number, l. The p orbital has a form, the d orbital has a clover shape, the f orbital is complicated, and the s orbital is spherical. the final level that was inhabited.

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The Bohr model of atoms needed to be revised because electrons were discovered to exhibit wave-like behavior. Hence, option C is correct.

The classical mechanics-based Bohr model was unable to account for this wave-like activity of electrons. The Schrödinger equation and the progress done in quantum mechanics contributed a very significant amount of knowledge for understanding the wave behavior of atomic electrons.

Instead than describing a particular orbit or path, the Schrödinger equation allowed us to describe the probability distribution of the electrons within an atom.

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The balanced equation for the reaction is:

HCN (aq) + NaOH (aq) → NaCN (aq) + H2O (l)

First, we need to find the volume of NaOH solution needed to reach equivalence:

0.7066 M HCN × 0.2100 L = x M NaOH × 0.2100 L

x = 0.3366 M NaOH

The volume of NaOH solution needed to reach equivalence is:

0.3366 M NaOH × VNaOH = 0.4210 M NaOH × 0.2100 L

VNaOH = 0.527 L = 527 mL

So, the total volume of the solution at equivalence is:

210.0 mL + 527 mL = 737 mL = 0.737 L

Now we can use the Henderson-Hasselbalch equation to find the pH at equivalence:

pH = pKa + log([A-]/[HA])

At equivalence, [HCN] = [CN-], so:

pH = pKa + log(1) = pKa = 9.21

Therefore, the pH at equivalence is 9.21.

Answer:

For question number 4 my answer comes to N:3 and H:3

False. According to the given data, the solubility of BaSO₄ decreases as temperature increases.

Solubility refers to the maximum amount of a substance that can dissolve in a given amount of solvent under specific conditions of temperature, pressure, etc to form a homogeneous solution.

It is usually expressed in units of mass per unit volume of solvent, such as grams per liter (g/L) or moles per liter (mol/L).

For this question, at 20°C, the solubility is 2.4 x 10⁻³ g/L, at 25°C it is 2.6 x 10⁻³ g/L, and at 30°C it is 2.9 x 10⁻³ g/L.

This indicates that as the temperature increases, the solubility of BaSO₄ decreases, which is the opposite of becoming more soluble. Therefore, the correct statement is False.

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The volume (in mL) of the 1.041 M NaCl solution needed to obtain 4.93 grams of NaCl is 80.98 mL

First, we shall determine the mole present in 4.93 grams of NaCl. Details below:

Mole = mass / molar mass

Mole of NaCl = 4.93 / 58.5

Mole of NaCl = 0.0843 mole

Finally, we shall determine the volume of NaCl needed. Details below:

Volume = mole / molarity

Volume of NaCl = 0.0843 / 1.041

Volume of NaCl = 0.08098 L

Multiply by 1000 to express in mL

Volume of NaCl = 0.08098 × 1000

Volume of NaCl = 80.98 mL

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The reaction quotient for this mixture of the two gases in which [ NO₂ ] = 0.0212 M and [ N₂O₄ ] = 0.00553 M is 0.130.

The reaction quotient Qc can be calculated using the law of mass action, which is the same expression as the equilibrium constant but with concentrations at any point in the reaction:

Qc = [N₂ O₂] / [NO₂]²

Substituting the given concentrations from the mixtures:

Qc = ( 0.00553 ) / ( 0.0212 )²

Qc = 0.130

So the reaction quotient for this mixture is 0.130.

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Oxidation half equation;

Mg(s) ----> Mg^2+(aq) + 2e

Reduction half equation;

2H^+(aq) + 2e ---->H2(g)

The balanced reaction equation is;

Mg(s) + 2HCl(aq) ---->MgCl2(aq) + H2(g)

Whereas reduction involves the acquisition of electrons, oxidation involves the loss of electrons. Redox reactions are the term used to describe these two processes, which frequently take place in tandem.

The species that is losing electrons is described by an oxidation half-equation, whereas the species that is gaining electrons is described by a reduction half-equation.

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To react with 58.4 mL of 0.352 M HNO₃, 32.29 mL of a 0.319 M Li₂CO₃ solution is needed.

The quantity of space a three-dimensional item or substance takes up is measured by its volume. It is a term used in physical measurement to indicate how much physical space an object or substance occupies. Volume can be expressed in several quantities, including liters, cubic meters, gallons, and cubic feet.

The balanced chemical equation for the reaction between HNO₃ and Li₂CO₃ is: 2HNO₃(aq) + Li₂CO₃(aq) → H₂O(l) + LiNO₃(aq) + CO₂(g)

We can see from the equation that the mole ratio of Li₂CO₃ to HNO₃ is 2:1.

Therefore, we must first determine the quantity of HNO₃ and then use the mole ratio to get the quantity of Li₂CO₃ needed. When we know the quantity of Li₂CO₃ in moles, we may utilize its molarity to determine the volume of solution needed.

Step 1: Determine the number of moles of HNO₃

n(HNO₃) = C(HNO₃) x V(HNO₃)

n(HNO₃) = 0.352 mol/L x 58.4 mL x 1 L/1000 mL

n(HNO₃) = 0.0206 mol

Step 2: Determine how many moles of Li₂CO₃ are needed using the mole ratio.

We can infer from the balanced chemical equation that 2 moles of HNO₃ react with 1 mole of Li₂CO₃. So, n(Li₂CO₃) = 0.0206 mol HNO₃ x (1 mol Li₂CO₃/2 mol HNO₃)

n(Li₂CO₃) = 0.0103 mol

Step 3: Calculate the volume of 0.319 M Li₂CO₃ solution required

n(Li₂CO₃) = C(Li₂CO₃) x V(Li₂CO₃)

V(Li₂CO₃) = n(Li₂CO₃) / C(Li₂CO₃)

V(Li₂CO₃) = 0.0103 mol / 0.319 mol/L

V(Li₂CO₃) = 32.29 mL

In order to react with 58.4 mL of 0.352 M HNO3, 32.29 mL of a 0.319 M Li₂CO₃ solution is needed.

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We need to weigh 15.27 g of [tex]BaCl_{2}.2H_{2}O[/tex] and dissolve it in water to make a 0.25M Cl solution in 250ml.

To prepare 0.25M Cl in 250ml from [tex]BaCl_{2}.2H_{2}O[/tex] we first need to calculate the required amount of [tex]BaCl_{2}.2H_{2}O[/tex].

The molecular weight of [tex]BaCl_{2}.2H_{2}O[/tex] is 244.26 g/mol.

The equation for the dissociation of [tex]BaCl_{2}[/tex] in water is:

[tex]BaCl_{2}[/tex](s) → [tex]Ba^{2+}[/tex] (aq) + 2Cl- (aq)

To make 250 ml of 0.25M Cl solution, we need to calculate the amount of[tex]BaCl_{2}.2H_{2}O[/tex]required to produce 0.25 moles of Cl.

0.25 moles Cl × 1 mole [tex]BaCl_{2}[/tex]/ 2 moles Cl × 244.26 g/mol [tex]BaCl_{2}[/tex]= 15.27 g [tex]BaCl_{2}.2H_{2}O[/tex]

Therefore, we need to weigh 15.27 g of [tex]BaCl_{2}.2H_{2}O[/tex] and dissolve it in water to make a 0.25M Cl solution in 250ml.

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The condensed structural formula of the unknown organic compound cannot be determined from the given information.

An organic compound is any of a broad family of chemical compounds that contain one or more carbon atoms that are covalently connected to atoms of other elements, most frequently hydrogen, oxygen, or nitrogen.

The few carbon-containing substances that aren't considered organic include cyanides, carbonates, and carbides.

Some classes of organic compounds include Carbohydrates, Lipids, Proteins, and Nucleic acids.

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For all three cases, we can use the fact that pH + pOH = 14 to find pOH, and then use pOH to calculate [OH−] and [H3O+].

Case 1: pH = 8.76

pOH = 14 - pH = 14 - 8.76 = 5.24

[OH−] = 10^(-pOH) = 10^(-5.24) = 5.2 × 10^(-6) M

[H3O+] = 1 × 10^(-14) / [OH−] = 1 × 10^(-14) / (5.2 × 10^(-6)) = 1.9 × 10^(-9) M

Therefore, [H3O+] = 1.9 × 10^(-9) M and [OH−] = 5.2 × 10^(-6) M.

Case 2: pH = 11.32

pOH = 14 - pH = 14 - 11.32 = 2.68

[OH−] = 10^(-pOH) = 10^(-2.68) = 2.1 × 10^(-3) M

[H3O+] = 1 × 10^(-14) / [OH−] = 1 × 10^(-14) / (2.1 × 10^(-3)) = 4.8 × 10^(-12) M

Therefore, [H3O+] = 4.8 × 10^(-12) M and [OH−] = 2.1 × 10^(-3) M.

Case 3: pH = 2.80

pOH = 14 - pH = 14 - 2.80 = 11.20

[OH−] = 10^(-pOH) = 10^(-11.20) = 6.3 × 10^(-12) M

[H3O+] = 1 × 10^(-14) / [OH−] = 1 × 10^(-14) / (6.3 × 10^(-12)) = 1.6 × 10^(-3) M

Therefore, [H3O+] = 1.6 × 10^(-3) M and [OH−] = 6.3 × 10^(-12) M.

The grams of [tex]Cu(OH)_{2}[/tex] produce by reaction of copper chloride with sodium hydroxide if you mix 0.9L of 0.557 M copper chloride with 1.35 L of 0.458 M sodium hydroxide is 41.55 g.

The balanced chemical equation describing this double replacement reaction should be written first.

[tex]CuCl_{2} + 2NaOH > > Cu(OH)_{2} + 2NaCl[/tex]

You should be aware that 1 mole of copper chloride requires 2 moles of sodium chloride to react with it to form 1 mole of copper(II) hydroxide.

Calculate how many moles of each reactant you are mixing together by comparing the molarities and volumes of the two solutions.

Moles of[tex]NaOH[/tex]=  [tex]1.35*0.458[/tex]

Moles of [tex]NaOH[/tex]= 0.6183

Moles of [tex]CuCl_{2}[/tex]= [tex]0.9*0.557[/tex]

Moles of [tex]CuCl_{2}[/tex] = 0.5013

You are conscious that 0.5013 moles of copper chloride will be needed.

[tex]0.5013*\frac{2 moles of NaOH}{1 mole of CuCl_{2} }[/tex]

1.0026 moles of [tex]NaOH[/tex] (what we need)>> 0.6183 moles of [tex]NaOH[/tex] (what we have)

Thus , This implies that sodium hydroxide will operate as a limiting reagent and that it will be completely consumed before all of the moles of copper(II) sulfate have an opportunity to react.

The reaction will therefore use 0.618 moles of sodium hydroxide and result in

[tex]0.618 moles of NaOH * \frac{1 mole of Cu(OH)_{2} }{2 moles of NaCl}[/tex]

= 0.309

To convert moles to grams molecular mass of copper hydroxide is

[tex]0.309 moles of Cu(OH)_{2}*\frac{134.45 g/mol}{1}[/tex]

=  41.55 g

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According to specific heat capacity, the specific heat of the aluminum is 0.807 J/gK .

Specific heat capacity is defined as the amount of energy required to raise the temperature of one gram of substance by one degree Celsius. It has units of calories or joules per gram per degree Celsius.

It varies with temperature and is different for each state of matter. Water in the liquid form has the highest specific heat capacity among all common substances .Specific heat capacity of a substance is infinite as it undergoes phase transition ,it is highest for gases and can rise if the gas is allowed to expand.

It is given by the formula ,

Q=mcΔT in case of 2 substances, aluminium and water it is m₁c₁ΔT₁=m₂c₂ΔT₂, thus, c₁= 135×4.2×5/13×54=0.807

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Answer: The partition coefficient of X between benzene and water is 0.

Explanation: To find the partition coefficient (K) of the tribasic acid (X) between benzene and water, we first need to calculate the amount of X that dissolves in each solvent.

Let's start by finding the amount of X that dissolves in water:

2.8 g of X is shaken with 50 cm^3 of water, which is equivalent to a mass/volume concentration of 0.056 g/cm^3.

Let's assume that x g of X dissolves in 50 cm^3 of water. This means that the concentration of X in water is x/50 g/cm^3.

Since the acid is tribasic, it reacts with sodium hydroxide in a 1:3 stoichiometric ratio. Therefore, the amount of sodium hydroxide needed to neutralize X in water is (1/3) * (x/50) = x/150 moles/cm^3.

We are given that 14.5 cm^3 of 1M NaOH solution is needed to neutralize the X in 25 cm^3 of aqueous layer. This means that the concentration of X in the aqueous layer is (14.5/25) moles/cm^3.

Setting the two expressions for concentration of X equal to each other and solving for x gives:

x/50 = 14.5/25 * 150

x = 261 g

Next, let's find the amount of X that dissolves in benzene:

We are given that X is insoluble in benzene, so the amount of X that dissolves in benzene is effectively zero.

Finally, we can calculate the partition coefficient:

K = (concentration of X in benzene) / (concentration of X in water)

Since the concentration of X in benzene is effectively zero, we have:

K = 0 / (261/50)

K = 0

Therefore, the partition coefficient of X between benzene and water is 0.

The value of Kp at 750 K for the given reaction is 1.09 x [tex]10^{-2}[/tex].

What is Pressure?

Pressure is the force applied per unit area over a surface. It is a scalar quantity and is commonly measured in units such as Pascals (Pa), atmospheres (atm), or pounds per square inch (psi). In gases, pressure is related to the motion of the gas molecules and their collisions with the walls of a container or other surfaces. In liquids and solids, pressure is related to the force applied to the surface and the resistance of the material to deformation.

PHCl = 2.300 bar

PO2 = 2.835 bar - 2.300 bar = 0.535 bar

To solve for the equilibrium partial pressures of Cl2 and H2O, we need to use the stoichiometry of the reaction. From the balanced equation, we can see that 4 moles of HCl react with 1 mole of O2 to produce 2 moles of Cl2 and 2 moles of H2O. Therefore, at equilibrium:

PCl2 = 2x

PH2O = 2x

where x is the amount (in moles) of HCl that has reacted. Initially, there were 4 moles of HCl and 1 mole of O2, so the reaction will proceed until either HCl or O2 is completely consumed.

Let's assume that all of the O2 reacts, which means that 1/4 of the initial amount of HCl will react. Therefore:

x = 1/4 * 4 mol = 1 mol

Now we can calculate the equilibrium partial pressures of Cl2 and H2O:

PCl2 = 2x = 2 mol

PH2O = 2x = 2 mol

Substituting these values into the equilibrium constant expression, we get:

Kp = [tex]2^{2}[/tex]([tex]2^{2}[/tex]) / [tex](2.300)^{4}[/tex](0.535)

Kp = 1.09 x [tex]10^{-2}[/tex]

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The patient received 37,037.04 mL of a 0.90% (m/v) NaCl solution.

What is a saline solution?

A saline solution is sodium chloride, salt, and water mixture. It is frequently utilised in medicine, especially for intravenous (IV) drips and injections that replenish the body's electrolytes and fluids.

We must apply the following formula to this issue in order to find a solution:

amount of solute = concentration x volume

where volume is in milliliters and concentration is expressed as a percentage (m/v).

Since concentration is specified as a percentage by mass, we must first change 3.0 g to milligrams (mg):

3.0 g = 3000 mg

In order to solve for volume, we may then plug in the given values:

3000 mg = 0.90% x volume

volume = 3000 mg / 0.009

volume = 333,333.33 mL

But this volume is for a saline solution that is 100% saline. We must modify the volume in accordance with the concentration, which is 0.90% (m/v):

0.90% x volume = 333,333.33 mL

volume = 333,333.33 mL / 0.009

volume = 37,037.04 mL

Therefore, the patient received 37,037.04 mL of a 0.90% (m/v) NaCl solution.

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If sample X contains 2.3 * 10²² molecules of sample X, the identity of sample x is 0.038 moles of sample X molecules.

The mole of a substance is the amount of a substance that contains exactly as many molecules, atoms, radicals, ions, or electrons as there are in 12 grams of Carbon-12.

Experimentally, it was discovered that 12 grams of Carbon-12 contain 6.02 * 10²³ atoms. Hence, a mole of every substance contains 6.02 * 10²³  particles of that substance

The moles of molecules present in sample X = 2.3 * 10²²/ 6.02 * 10²³ * 1 mole

The moles of molecules present in sample X = 0.038 moles of sample X molecules.

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Answer:

We can use the ideal gas law to solve for the initial temperature of the gas:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.

We can assume that the number of moles and volume of gas are constant, since the problem states that it is the same cylinder of gas. Therefore, we can write:

P1/T1 = P2/T2

where P1 is the initial pressure, T1 is the initial temperature, P2 is the final pressure, and T2 is the final temperature.

Substituting the values given in the problem, we get:

4.9/T1 = 4.7/281

Solving for T1, we get:

T1 = 4.9 × 281 / 4.7

T1 = 293 K

Converting to Celsius, we get:

T1 = 20°C

Therefore, the initial temperature of the gas was 20°C.

Answer:

mark me brilliant

To determine the mass of ammonia that can be produced from 40.5 g of N₂, we need to use stoichiometry and the balanced chemical equation provided.

From the balanced equation, we can see that 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃. Therefore, we need to calculate the number of moles of N₂ in 40.5 g:

Number of moles of N₂ = mass / molar mass = 40.5 g / 28 g/mol = 1.4464 mol N₂

Using the mole ratio from the balanced equation, we can calculate the number of moles of NH₃ produced:

Number of moles of NH₃ = (1.4464 mol N₂) x (2 mol NH₃ / 1 mol N₂) = 2.8928 mol NH₃

Finally, we can use the molar mass of NH₃ to convert the number of moles to mass:

Mass of NH₃ = number of moles x molar mass = 2.8928 mol x 17 g/mol = 49.11 g

Therefore, the mass of ammonia that can be produced from 40.5 g of N₂ with excess H₂ is 49.11 g.

Answer:

Explanation:

The first step is to determine the limiting reactant. In this case, we have 40.5 g of N₂ and excess H₂. We can use the molar mass of each substance to determine the number of moles of each reactant:

N₂: 40.5 g / 28.01 g/mol = 1.45 mol

H₂: 40.5 g / 2.016 g/mol = 20.1 mol

Since we have less moles of N₂ than H₂, N₂ is the limiting reactant. This means that the amount of ammonia produced will be determined by the amount of N₂.

We can use the balanced equation to determine the mass of ammonia produced:

1 mol N₂ → 2 mol NH₃

1.45 mol N₂ → 2.90 mol NH₃

2.90 mol NH₃ * 17.03 g/mol = 50.4 g NH₃

Therefore, 50.4 g of ammonia can be produced from 40.5 g of N₂ with excess H₂.

The formula of the compound, given that it contains 30 percent N and 70 percent O by mass is NO₂

First, we shall list out the given parameters from the question. This is given

The formula of the compound can be obtained as illustrated below:

Divide by their molar mass

N = 30 / 14 = 2.14

O = 70 / 16 = 4.375

Divide by the smallest

N = 2.14 / 2.14 = 1

O = 4.375 / 2.14 = 2

Thus, we can conclude from the above calculation that the formula of the compound is NO₂

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The formula of the compound that we have is NO2

We'll start by outlining the parameters that the question has provided. This is stated.

N = 30% in terms of nitrogen percentage.

oxygen percentage (O) = 70%

Compound's formula is =?

The compound's formula can be discovered as shown below:

Based on molar mass;

N = 30 / 14 = 2.14

O = 70 / 16 = 4.375

By the smallest number;

N = 2.14 / 2.14 = 1

O = 4.375 / 2.14 = 2

Thus the formula of the compound that we have to find is NO2.

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When a strong acid or base is added to water, the pH will change dramatically.

A strong acid is one that is completely dissociated or ionized in an aqueous solution. This means it gives off the greatest number of hydrogen ions or protons when placed in a solution. Examples of strong acid are HCl, HBr, H2SO4, HNO4. These acids when placed in water, produces greatest amount of hydrogen ions. The pH value changes drastically. Any that has very high concentration of hydrogen and ion is acidic.

Also when base is added to water, the pH of water will increase above 7 and become basic. The pH of water is 7, but when base is added to it increases above 7.

Base is any solution that is slippery to touch in water solution, changes color, react with acid to form salt and change red litmus paper to blue.

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The P value denotes the likelihood, for a certain statistical model, that the statistical summary would be equal to or more extreme than the actual observed results if the null hypothesis were true.

When the null hypothesis is assumed to be true, a p-value calculates the likelihood of getting the observed results. The statistical significance of the observed difference is greater the lower the p-value. P-value can be used for hypothesis testing instead of—or in addition to—pre-selected confidence levels.

The computed P value may change depending on a variety of study design factors. These factors include inaccuracy, sample size, and the strength of the link. These components all have the potential to undermine research either singly or collectively.

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Which of the following are true statements about the p-value and hypothesis tests? Select all that apply. Select one or more:

A) The p-value is the probability the null hypothesis is correct.

B) The p-value is the probability the alternative hypothesis is correct.

C) The p-value is 1 - (the probability the alternative hypothesis is correct)

D) If the p-value is large it indicates we did not calculate the test statistic correctly.

E) The p-value is calculated assuming the null hypothesis is true.

F) The p-value is calculated assuming the alternative hypothesis is true.