The solid oxide of generic metal M is added to 67 mL of water and reacts to produce a basic solution that is 0.40 M in the resulting compound. Neutralization of the solution required titration with 31 mL of 0.86 M HCl. How many valence electrons must M have

Answer:

They are same.

Explanation:

These are the same measurement; there is no difference in volume. The primary difference is that milliliters are used for fluid amounts while cubic centimeters are used for solids. No matter what is being measured, 1 cc always equals 1 mL.

No it’s not balanced. The Na are not balanced (option A)

the given chemical equation is not balanced because It is missing some coefficients to balance the number of atoms on both sides of the equation.

The balanced chemical equation for the reaction is:

Na2SO4 + Mg(OH)2 → MgSO4 + 2NaOH

This equation shows that two molecules of sodium hydroxide (NaOH) are produced for every one molecule of magnesium hydroxide (Mg(OH)2) consumed, and that the number of sodium (Na) atoms is balanced on both sides of the equation.

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Answer: an atom

Explanation: he smallest particle of an element that still has the element's properties is an atom. All the atoms of an element are alike, and they are different from the atoms of all other elements. For example, atoms of gold are the same whether they are found in a gold nugget or a gold ring

Answer:

enantiomeric excess = 68%

Explanation:

Enantiomeric excess is a value used to determine the purity of chiral molecules. It is possible to determine enantiomeric excess (ee) using:

ee = R - S / R + S * 100

Where R is the mass (In this case percentage) of the R enantiomer and S of the S enantiomer.

Replacing with values of the problem:

ee = 84% - 16% / 84% + 16% * 100

Answer:

Explanation:

In a gaseous reaction mixture partial pressure is proportion to mole of the gas concerned .

Pressure of the reactant gas from gas equation

PV = nRT

P = nRT / V

= .166 x .082 x ( 273+179) / 10

= .615 atm

C₃H₇OH   =     (CH₃)₂CO     +    H₂

before reaction moles in terms of pressure

.615                       0                      0

After reaction

.615 - x                     x                     x

.444 =  x² / ( .615 - x )

.273 - .444 x = x²

x² + .444 x - .273 = 0

x = .361 atm

So partial pressure of acetone is .361 atm at equilibrium.

Answer:

If there are more products than reactants, that means the reaction has shifted towards the left, which is the backward direction. If there are more reactants than products, that means the reaction has shifted towards the right, which is the forward direction.

The balanced equation of the redox reactions by the half-reaction method is as follows:

(a) Zn(s) + 4 H+(aq) + NO₃⁻(aq) ⟶ Zn²⁺ (aq) + 2 H₂O(l) + N₂(g)

(b) Zn(s) + 2 OH⁻(aq) + NO₃⁻(aq) ⟶ Zn(OH)₂(aq) + NH₃(aq)

(c) CuS(s) + 6 H⁺(aq) + 2 NO₃⁻(aq) ⟶ Cu²⁺(aq) + S(s) + 2 NO(g) + 3 H₂O(l)

(d) 4 NH₃(aq) + 5 O₂(g) ⟶ 4 NO₂(g) + 6 H₂O(l)

(e) 2 H₂O₂(aq) + 2 MnO₄⁻(aq) ⟶ 2 Mn²⁺(aq) + 5 O₂(g) + 4 H₂O(l)

(f) 3 NO₂ (g) + 2 OH⁻ (aq) ⟶ 3 NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O(l)

(g) 2 Fe³⁺ (aq) + 2 I⁻ (aq) ⟶ 2 Fe²⁺ (aq) + I₂ (aq)

Half reactions refer to the separate reactions that represent the oxidation and reduction processes occurring in the overall redox reaction.

Considering the given redox reactions:

(a) Zn(s) + 4 H+(aq) + NO₃⁻(aq) ⟶ Zn²⁺ (aq) + 2 H₂O(l) + N₂(g)

Half-reaction (oxidation): Zn(s) ⟶ Zn²⁺(aq) + 2 e⁻

Half-reaction (reduction): 4 H+(aq) + NO₃⁻(aq) + 3 e⁻ ⟶ 2 H₂O(l) + N₂(g)

(b) Zn(s) + 2 OH⁻(aq) + NO₃⁻(aq) ⟶ Zn(OH)₂(aq) + NH₃(aq)

Half-reaction (oxidation): Zn(s) + 4 OH⁻(aq) ⟶ Zn(OH)₂(aq) + 2 e⁻

Half-reaction (reduction): NO₃⁻ aq) + 8 H₂O(l) + 6 e⁻ ⟶ NH₃(aq) + 9 OH⁻ (aq)

(c) CuS(s) + 6 H⁺(aq) + 2 NO₃⁻(aq) ⟶ Cu²⁺(aq) + S(s) + 2 NO(g) + 3 H₂O(l)

Half-reaction (oxidation): CuS(s) ⟶ Cu²(aq) + S(s) + 2 e⁻

Half-reaction (reduction): 6 H⁺(aq) + 2 NO₃⁻(aq) + 6 e⁻ ⟶ 2 NO(g) + 3 H₂O(l)

(d) 4 NH₃(aq) + 5 O₂(g) ⟶ 4 NO₂(g) + 6 H₂O(l)

Half-reaction (oxidation): 4 NH₃(aq) ⟶ 4 NO₂(g) + 8 H⁺(aq) + 8 e⁻

Half-reaction (reduction): 5 O₂(g) + 10 H₂O(l) + 10 e⁻ ⟶ 20 OH⁻(aq)

(e) 2 H₂O₂(aq) + 2 MnO₄⁻(aq) ⟶ 2 Mn²⁺(aq) + 5 O₂(g) + 4 H₂O(l)

Half-reaction (oxidation): 2 H₂O₂(aq) ⟶ 4 H⁺(aq) + 4 e⁻ + O₂(g)

Half-reaction (reduction): 2 MnO₄⁻(aq) + 16 H⁺ (aq) + 10 e⁻ ⟶ 2 Mn²⁺ (aq) + 8 H₂O (l)

(f) 3 NO₂ (g) + 2 OH⁻ (aq) ⟶ 3 NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O(l)

Half-reaction (oxidation): 3 NO₂(g) + 6 OH⁻(aq) ⟶ 3 NO₃⁻ (aq) + 3 e⁻ + 3 H₂O(l)

Half-reaction (reduction): 3 NO₂ (g) + 2 e⁻ ⟶ 3 NO₂⁻ (aq)

(g) 2 Fe³⁺ (aq) + 2 I⁻ (aq) ⟶ 2 Fe²⁺ (aq) + I₂ (aq)

Half-reaction (oxidation): 2 Fe³⁺ (aq) ⟶ 2 Fe²⁺ (aq) + 2 e⁻

Half-reaction (reduction): 2 I⁻ (aq) ⟶ I₂ (aq) + 2 e⁻

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Faults and problems in three routes of compound Br(bromide) synthesis, and the most feasible route, addressing competition, compatibility, and timing.

The main course includes a nucleophilic replacement of A with MeOH within the sight of HCl, trailed by parchedness with CS₂/Mel/heat. The fundamental issue with this course is the opposition between the replacement and end responses, which could prompt low yields of the ideal item.

The subsequent course includes a comparative nucleophilic replacement of A with MeOH within the sight of TsOH, trailed by a decrease with LiAlH₄. The fundamental issue with this course is the similarity between the acidic TsOH and the diminishing specialist, which could bring about the development of undesirable side items.

The third course includes an immediate buildup of A with ethyl oxalate within the sight of K₂CO₃, trailed by a decarboxylation with H₂SO₄. The fundamental issue with this course is the planning of the decarboxylation step, which could prompt the arrangement of undesirable side items because of overcompensation.

Generally speaking, the most possible course is by all accounts the first, with cautious advancement of the response conditions to limit the opposition among replacement and disposal responses.

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Answer:

New environments for puppies may make them excited, fearful, aggressive, or confused.

Explanation:

"When your dog first experiences a new location or environment, there's no way of knowing how they'll react. New sights, sounds, and smells could make them fearful, aggressive, or over-excited, but with the proper training and introduction, most dogs will quickly adapt and start taking every new location in stride." - Excerpt from *Puppy training textbook*

New environments are also a good thing in puppy growth, to teach them how to learn and adapt to new surroundings.

Hope this helps :)

The mole of the substance, given the data from the question is 0.0014 mole

This is simply defined as the mole of solute per kilogram of water. Mathematically, it is expressed as

Molality = mole / mass (Kg) of water

Mole = molality × mass of water

Mole of substance = 0.056 × 0.0254

Mole of substance = 0.0014 mole

Thus, the mole of the substance is 0.0014 mole

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Heat is added to ice at 0 °C. Explain why the temperature of the ice does not change. What does change?When heat is added to ice at 0°C, the temperature of the ice does not change. This happens because all the heat energy is used up in overcoming the intermolecular forces of attraction (hydrogen bonds) that exist between the water molecules in ice.

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C, A, D, and B is the increasing order of significant digits among the following values.

The number of digits in the value of a number that is meaningful are called significant digits or significant figures. These numbers have an accuracy that is matching to our measurements, simply are all the numbers required.

For Example, measure the area of a park to within 1 meter. The calculations can give an area of approximately 58.37215 m^2.

But here, there are more digits than the accuracy of measurement.

Thus, it can be decided to utilize only 2 of the significant digits. The final result is 58 m^2, while 5 and 8 are the two significant digits.

Some more examples can be

A plus sign or a minus sign can be added to the value to show the level of accuracy. For example, 300 ± 10 shows that it lies somewhere between 290 and 310.

Therefore, in the above-mentioned values, C, A, D, and B is the increasing order of significant digits among the following values.

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Answer:

Explanation:

2.2 g/mL is the answer.

The pH of the solution can be obtained as 8.8.

We know that the pH tells us the amount of the hydrogen ions that we can be able to find in the solution. We now have the following that can be obtained from the reaction.

Number of moles of the NaOH = 30/1000 * 0.4

= 0.012 moles

Number of moles of HF = 25/1000 * 0.48

= 0.012 moles

Total volume present = 25 mL + 30 mL = 55 mL or 0.055 L

Molarity of NaF = 0.012 moles/0.055 L

= 0.22 M

Kb = 1 * 10^-14/Ka

Kb =  1 * 10^-14/ 6.76×10^−4

Kb = 1.5 * 10^-10

Then we have;

Kb = [HF][OH-] / [F-]

= (x)(x) / 0.22

1.5 * 10^-10 = x^2/0.22

x =  1.5 * 10^-10 * 0.22

x = √ 1.5 * 10^-10 * 0.22

x = 5.7 * 10^-6 M

pOH = - log( 5.7 * 10^-6 )

= 5.2

pH = 14 - 5.2

= 8.8

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The cations that are more likely to undergo rearrangement are  II and III Option C

Cation rearrangement is a phenomena that causes cations, which are positively charged ions, to reorganize themselves under specific conditions. The term "cation rearrangement " describes the movement of ions inside a substance or solution that is caused by a variety of elements, including concentration gradients, electric fields, or chemical reactions.

Rearrangements influence the mobility and redistribution of cations inside substances or solutions, which can have a significant impact on the system's behavior and characteristics.

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Answer:

C.) protons and neutrons

Explanation:

Most atoms contain proton(s), neutron(s), and electron(s). Within the nucleus of an atom, there are protons and neutrons. Electrons are located outside of the nucleus.

Writing and balancing complex half-reactions in acidic solutions involves a systematic approach. Here are the general steps to follow:

Identify the oxidation and reduction half-reactions: Determine which species is being oxidized (losing electrons) and which is being reduced (gaining electrons).

Balance the atoms: Begin by balancing all atoms except hydrogen and oxygen in each half-reaction.

Balance the charges: Add electrons (e-) to one side of each half-reaction to balance the charges.

Balance the oxygen atoms: Add water (H2O) to the side of the equation that lacks oxygen atoms.

Balance the hydrogen atoms: Add hydrogen ions (H+) to the side of the equation that lacks hydrogen atoms. Keep in mind that the solution is acidic.

Balance the charges: Adjust the number of electrons (e-) on each side of the equation to ensure that the charges are balanced.

Multiply the half-reactions: Multiply each half-reaction by an appropriate factor so that the number of electrons gained and lost are equal.

Combine the half-reactions: Add the two balanced half-reactions together and cancel out common terms on both sides of the equation.

By following these steps, you can effectively write and balance complex half-reactions in acidic solutions, ensuring that the overall reaction is balanced in terms of both atoms and charges.

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Answer:

x/2

Explanation:

X = 2 molecules of H2

For 2 molecules of H2, there's only 1 molecule of O2. Meaning, there's twice the amount of H2, so O2 = x/2 molecules.

I hope I'm understanding this question right.

Answer:

60ug

Explanation:

If it has experienced two half lives, that means it has been halved twice. in that case, to undo it, just multiply it by two twice. 0.15ug * 2 = 0.30ug. 0.30ug * 2 = 0.60ug. Hope this helps.

Answer:

rhrhrhrhrhrhrrhrhrhrh

Explanation:

jrjrut5jt vrm VIP fo fo di xiu xiu cdi Deㅗ도돋ㅎㄷㄷㅎㄷㅎㄷㅍㄷuehehrgrvrrvhi UV co cu tu sew ccue

Answer:

9.51 × 10⁴ kL

Explanation:

Step 1: Given data

Volume of the sample (V): 9.51 × 10⁹ cL

Step 2: Convert "V" to liters

We will use the conversion factor 1 L = 100 cL.

9.51 × 10⁹ cL × (1 L / 100 cL) = 9.51 × 10⁷ L

Step 3: Convert "V" to kL

We will use the conversion factor 1 kL = 1000 L.

9.51 × 10⁷ L × (1 kL / 1000 L) = 9.51 × 10⁴ kL

9.51 × 10⁹ cL is equal to 9.51 × 10⁴ kL.

Answer:

-0.0132 atm

Explanation:

Step 1: Calculate the atmospheric pressure change if altitude increases 100. m

Atmospheric pressure decreases by 1.00 mmHg as altitude increases 10.0 m. The decrease in pressure when the altitude increases 100. m is:

100. m × (-1.00 mmHg/10.0 m) = -10.0 mmHg

Step 2: Convert "-10.0 mmHg" to atm

We will use the conversion factor 1 atm = 760 mmHg.

-10.0 mmHg × 1 atm/760 mmHg = -0.0132 atm

Answer:

it is true bc i looked it up

Answer:

joules

Explanation:

i think

Explanation:

The SI unit of electric current is the ampere, or amp, which is the flow of electric charge across a surface at the rate of one coulomb per second. The ampere (symbol: A) is an SI base unit Electric current is measured using a device called an ammeter.

Answer:

The trachea, also called the windpipe, is a cartilaginous tube that connects the larynx to the bronchi of the lungs, allowing the passage of air, and so is present in almost all air-breathing animals with lungs. The trachea extends from the larynx and branches into the two primary bronchi. At the top of the trachea the cricoid cartilage attaches it to the larynx. The trachea is formed by a number of horseshoe-shaped rings, joined together vertically by ligaments over their substance and by the trachealis muscle at their ends. The epiglottis closes the opening to the larynx during swallowing.

Explanation:

You asked I answered.

Hope this helped! :)

P.S. If you still don't get where the trachea is, look at the graph below. :)

Explanation:

is responsible for interrupting the interaction between pectins and solvent molecules

The enthalpy change (ΔH) for the reaction CaC(s) + H2O(I) → Ca(OH)(ag) + CH4(g) is -3617.6 kJ.

To calculate ΔH for the reaction CaC(s) + H2O(l) → Ca(OH)2(ag) + CH4(g), we can use Hess's law, which states that the enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states.

We can manipulate the given reactions to obtain the desired reaction:

(i) Ca(s) + 2C(graphite) → CaC2(s) ΔH = X (unknown value)

(ii) Ca(s) + 1/2O2(g) → CaO(s) ΔH = -635.5 kJ

(iii) CaO(s) + H2O(l) → Ca(OH)2(ag) ΔH = -653.1 kJ

(iv) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = -1300.0 kJ

(v) C(graphite) + 1/2O2(g) → CO(g) ΔH = -393.5 kJ

Now, let's manipulate these equations to cancel out the common reactants and products and obtain the desired reaction:

(i) Ca(s) + 2C(graphite) → CaC2(s) ΔH = X

(ii) Ca(s) + 1/2O2(g) → CaO(s) ΔH = -635.5 kJ

(iii) CaO(s) + H2O(l) → Ca(OH)2(ag) ΔH = -653.1 kJ

(iv) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = -1300.0 kJ

(v) C(graphite) + 1/2O2(g) → CO(g) ΔH = -393.5 kJ

Now, let's sum up the equations to obtain the desired reaction:(i) Ca(s) + 2C(graphite) → CaC2(s) ΔH = X

(ii) 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1271 kJ

(iii) CaO(s) + H2O(l) → Ca(OH)2(ag) ΔH = -653.1 kJ

(iv) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = -1300.0 kJ

(v) C(graphite) + 1/2O2(g) → CO(g) ΔH = -393.5 kJ

By adding equations (ii), (iii), (iv), and (v), we can cancel out CaO(s), H2O(l), and O2(g):

2Ca(s) + 2C(graphite) + CH4(g) → 2Ca(OH)2(ag) + CO(g) ΔH = X -1271 -653.1 -1300.0 -393.5

2Ca(s) + 2C(graphite) + CH4(g) → 2Ca(OH)2(ag) + CO(g) ΔH = X -3617.6 kJ

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Answer: phosphoric acid, H3PO4 (gives pop a tart taste)

% O: 1 mol H3PO4 (4 mol O / 1 mol H3PO4) (16 g / 1 mol O) = (64 g O / 98 g H3PO4) x 100 = 65.3%

% P: 1 mol H3PO4 (1 mol P / 1 mol H3PO4) (31 g / 1 mol P) = (31 g P / 98 g H3PO4) x 100 = 31.6%