How must the hydrogen in an acid be attached (ionically or covalently) to be an acid?

Taking into account the definition of molarity and the stoichiometry of the reaction, the correct option is option C) 0.44 L of 6.9 M NaOH is needed to completely titrate 0.42 L of 2.39 M phosphoric acid.

The balanced reaction is:

H₃PO₄ (aq) + 3 NaOH (aq) → Na₃PO₄ (aq) + 3 H₂O(aq)

Then, by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Molarity is the number of moles of solute that are dissolved in a given volume.

Molarity is determined by:

[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]

 Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].

In this case, 0.42 L of 2.39 M phosphoric acid reacts. So, by definition of molarity, the number of moles that participate in the reaction is calculated as:

[tex]2.39 \frac{moles}{liter}=\frac{number of moles of phosphiric acid}{0.42 liters}[/tex]

Solving:

number of moles of phosphiric acid= 2.39 [tex]\frac{moles}{liter}[/tex]* 0.42 liters

number of moles of phosphiric acid= 1.0038 moles ≅ 1 mole

Approaching 1 mole of the amount of phosphoric acid required, then by stoichiometry of the reaction, 3 moles of NaOH are necessary to react with 1 mole of the acid.

Then by definition of molarity and knowing that 6.9 M NaOH is needed, you can calculate the necessary volume amount of NaOH by:

[tex]6.9 \frac{moles}{liter} =\frac{3 moles}{volume}[/tex]

Solving:

6.9 [tex]\frac{moles}{liter}[/tex]* volume= 3 moles

[tex]volume=\frac{3 moles}{6.9\frac{moles}{liter} }[/tex]

volume= 0.44 L

The correct option is option C) 0.44 L of 6.9 M NaOH is needed to completely titrate 0.42 L of 2.39 M phosphoric acid.

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Answer:

the mas is .291 g

Explanation:

the mass of a object does not change. so when added the substance the beaker. you had the mass of both objects together. you know the mass of the beaker and you know the mass of both. since mass does not change. the beakers mass is still 74.605g. the mass of both objects is 74.896. all you have to do is subtract the mass of the beaker from the total mass. 74.896 - 74.605 equals .291g. so the mass of the unknown substance Is .291g

C₁₀H₁₄N₂ is the empirical formula of the compound whose mass is 40.57 present in 0.2500 moles.

Alkaloid compounds are those naturally present organic compounds in which nitrogen is present.

First we calculate the molecular mass of wanted compound by using the below formula:

n = W/M, where

n = no. of moles = 0.2500 moles (given)

W = given mass = 40.57 grams (given)

M = molar mass = 40.57 grams/= 0.2500 moles = 162.28 g/mole

In the question, given that:

% composition of carbon = 74.02%

% composition of hydrogen = 8.710%

% composition of nitrogen = 17.27%

Now we calculate the mass of these composition by using the below formula:

Composition mass = compound mass * % composition / 100

Mass of carbon = 162.28 * 74.02 / 100 = 120.11g

Mass of hydrogen = 162.28 * 8.710 / 100 = 14.13g

Mass of nitrogen = 162.28 * 17.27 / 100 = 28.02g

Now we calculate the moles of these composition to made empirical formula as:

Moles of carbon = 120.11 / 12 = 10

Moles of hydrogen = 14.13 / 1 = 14

Moles of nitrogen = 28.02 / 14 = 2

Thus, the empirical formula of the compound is C₁₀H₁₄N₂.

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Answer:

K = 3.3

Explanation:

Nitric acid, HNO3, reacts with nitrogen monoxide, NO, to produce nitrogen dioxide, NO2 and water H2O as follows:

2HNO3(g) + NO(g) → 3NO2(g) + H2O(g)

Where equilibrium constant, K, is:

K = [NO2]³[H2O] / [HNO3]²[NO]

[] is the molar concentration of each species at equilibrium.

To solve this question we need to find molarity of each gas and replace these in the equation as follows:

[NO2] -Molar mass NO2-46.0g/mol-

18.6g * (1mol/46.0g) = 0.404mol / 7.7L = 0.0525M

[H2O] -Molar mass:18.01g/mol-

236.7g * (1mol/18.01g) = 13.14 moles / 7.7L = 1.707M

[HNO3] -Molar mass:53.01g/mol-

16.2g * (1mol/53.01g) = 0.3056 moles / 7.7L = 0.0397M

[NO] -Molar mass: 30.0g/mol-

11.0g * (1mol/30.0g) = 0.367 moles / 7.7L = 0.0476M

Replacing:

K = [NO2]³[H2O] / [HNO3]²[NO]

K = [0.0525M]³[1.707M] / [0.0397M]²[0.0476M]

An indicator usually signals the endpoint of a neutralization reaction by undergoing a color change. They aid in discovering the point of equivalence of a titration.

The kind of indicator used depends on the nature of acid/base reacted.

In the case of  CH3NH2 with HBr which  strong acid and weak base titration, suitable indicators include; bromophenol blue, methyl  orange, methyl red, and chlorophenol blue.

In the case of HNO3 with NaOH, this is a strong acid, strong base titration hence phenolphthalein, methyl red, chlorophenol, and bromothymol blue cresol red blue are suitable indicators.

In the case of  HNO2 with KOH, this a weak acid, strong base titration and the suitable indicators are cresol red and  phenolphthalein.  

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Answer:

natural mineral water.

N₂ + 3H₂ ⇄ 2NH₃ + heat

In the given equilibrium, we notice that the heat is on the right. which means that if the heat requirements don't meet, the reactants on the right will no longer react due to the lack of heat

but because the reactants on the left don't have such weaknesses, they will keep reacting hence producing more and more ammonia until a new equilibrium is reached

where there will be more ammonia and less nitrogen and hydrogen as compared to the equilibrium we had initially

Answer:

Explanation:

heat is given out as 1 of the products, along w/ NH3 in the forward reaction. so its an exothermic reaction

decreasing temperature favors exothermic reaction as more heat can be absorbed by the environment

so equilibrium will shift towards the products

Answer:

a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. i. NO₂⁻ is the oxidizing agent

ii. NO₃⁻ is the reducing agent.

Explanation:

a. Balance the following skeleton reaction

The reaction is

NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)

The half reactions are

NO₂ (g) → NO₃⁻ (aq)  (1) and

NO₂ (g) → NO₂⁻  (aq) (2)

We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)

We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms

NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)

The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.

So, NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

We now add equation (4) and (5)

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

+  NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + e⁻  (4)

2NO₂ (g) + H₂O (l)  → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq)  

We now add two hydroxide ions to both sides of the equation.

So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + 2OH⁻ (aq)

The hydrogen ion and the hydroxide ion become a water molecule

2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H₂O (l)

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

So, the required reaction is

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. Identify the oxidizing agent and reducing agent

Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = 0

x - 4 = 0

x = 4

Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = -1

x - 4 = -1

x = 4 - 1

x = 3

Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = -1

x + 3(-2) = -1

x - 6 = -1

x = 6 - 1

x = 5

i. The oxidizing agent

The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus  NO₂⁻ is the oxidizing agent

ii. The reducing agent

The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and  NO₃⁻ is the reducing agent.

the rate would decrease because the reactants are being depleted.

Answer:

to the left

Explanation:

If the concentration of products is increased for a reaction that is in equilibrium, the equilibrium would shift to the left side of the reaction (the reactant's side).

For a reaction that is in equilibrium, the reaction is balanced between the reactants and the products. According to Le Cha telier's  principle, if one of the constraints capable of influencing the rate of reactions is applied to such a reaction that is in equilibrium, the equilibrium would shift so as to neutralize the effects created by the constraint.

Hence, in this case, if the concentration of the products of a reaction in equilibrium is increased, the equilibrium would shift in such a way that more reactants are formed so as to annul the effects created by the increase in the concentration of the products. Since reactants are always on the left side of chemical equations, it thus means that the equilibrium would shift to the left.

Answer:

[tex]\boxed {\boxed {\sf 0.14 \ mol \ CaCl_2}}[/tex]

Explanation:

We are asked to calculate the number of moles of 15 grams of calcium chloride (CaCl₂).

To convert from grams to moles, we use the molar mass, or the mass of 1 mole of a substance. Molar masses are found on the Periodic Table because they are equivalent to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up the individual elements in the compound: calcium and chloride.

Notice the chemical formula has a subscript of 2 after Cl or chlorine. There are 2 moles of chlorine in every 1 mole of calcium chloride. We must multiply chlorine's molar mass by 2 before adding calcium's molar mass.

We will convert using dimensional analysis, so we must create a ratio using the molar mass.

[tex]\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]

We are converting 15 grams of calcium chloride to moles, so we must multiply the ratio by this value.

[tex]15 \ g \ CaCl_2 *\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]

Flip the ratio so the units of grams of calcium chloride cancel.

[tex]15 \ g \ CaCl_2 *\frac { 1 \ mol \ CaCl_2}{110.98 \ g \ CaCl_2}[/tex]

[tex]15 *\frac { 1 \ mol \ CaCl_2}{110.98}[/tex]

[tex]\frac { 15}{110.98} \ mol \ CaCl_2[/tex]

[tex]0.1351594882\ mol \ CaCl_2[/tex]

The original measurement of grams (15) has 2 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 3 up to a 4.

[tex]0.14 \ mol \ CaCl_2[/tex]

15 grams of calcium chloride is approximately 0.14 moles of calcium chloride.

A leaking tap that drops water at the rate of 3 drops every second, will leak 259.2 L in a day.

We know that a leaking tap drops water at the rate of 3 drops every second and that each drop is approximately 1 ml. The milliliters of water dropped every second are:

[tex]\frac{3drop}{1s} \times \frac{1mL}{1drop} = \frac{3mL}{1s}[/tex]

We want to know the number of seconds in 1 day. We will use the following conversion factors:

[tex]1day \times \frac{24h}{1day} \times \frac{60min}{1h} \times \frac{60s}{1min} = 86400 s[/tex]

3 mL of water are dropped every second. The mL of water dropped in 86400 s are:

[tex]86400 s \times \frac{3mL}{1s} = 259200 mL[/tex]

Finally, we will convert mL to L using the conversion factor 1 L = 1000 mL.

[tex]259200 mL \times \frac{1L}{1000 mL} = 259.2 L[/tex]

Approximately 259.2 L of water will be dropped in 1 day.

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Answer:

C.

Explanation:

I believe that it should be A and B.

Explanation:

in the terms of interval,thats a major sixth up or a minor third using the circle of fifths simply move three position clock wise around the to find the relative minor! so starting from C,count one( G),two( D)and then three:A is related to minor key.

Copper tube is used in EBRT.

Copper tube is used in EBRT.

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Answer:

ethyl - 3- heptanol mouthing

Answer:

A

Explanation:

I'm assuming this question implies that the surface area is in relation to the volume of the pieces. In that case, the SMALLER the size of each piece, the larger the surface area. This is because more particles are able to fit into the container if they are smaller, leading to more surface area. Since more pieces can fit into the container, MORE collisions happen according to collision theory. I cannot add a link, but for a helpful analogy, look up "How To Speed Up Chemical Reactions (and get a date) - Aaron Sams.

Answer: A. smaller, more

Explanation: Founders Educere answer

Answer:

d. north pole on the top edge and south pole on the bottom edge

Explanation:

Answer:

A) XY3

In both of the atom, they want to have 8 valence electrons on the outer shell so they can become stable.

Answer:

1) Photosynthesis

Photosysnthesis is the process in which plantsand some other organisms usethe energy in sunlight to make food.

2) Chloroplast

Chloroplast is a green pigment that traps energy from the sun.

3) Cell Wall

Cell wall is a stiff structure that surrounds and protects a cell; found in plant, fungus, and some bacteria cells.

4) DNA

DNA is found in the nucleus of a cell, a long nucleic acid molecule containingthe

genetic instructions for the development and functioning of all living organisms.

5) Cell Organeles

Cells is living things consisting of many cell organeles

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The freezing of water into ice and the dehydration of copper sulfate are both reversible. The correct options are C and D.

They are reactions in which the reverse can occur.

The freezing of water into ice can be undone. That is, the ice can be thawed back to water.

The dehydration of copper sulfate involves the removal of water molecules. As soon as water becomes available again, the reaction is reversed.

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not 100% sure but I think its C

The correct name for the given alkane molecule is 2,2-dimethylpropane. (Option C)

The given alkane molecule is composed of three carbon atoms. The central carbon atom is bonded to two methyl groups (CH₃) on either side. This results in a branched structure where the methyl groups are attached to the same carbon atom.

According to the rules of IUPAC nomenclature, the prefix "2,2-" indicates that the methyl groups are attached to the second carbon atom in the main chain. The parent chain, which consists of three carbon atoms, is named propane. Since there are two identical methyl groups attached to the second carbon atom, the compound is named 2,2-dimethylpropane.

Hence, the name of the given alkane molecule is 2,2-dimethylpropane.

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The concentration of CO₃²⁻ ions will be higher

To explain, I want you to imagine H₂CO₃ in water.

we know that it will lose 2 of it's protons, and form 2 ions

The ion which is more stable will have a higher concentration because that ion will refuse to react with anything else, so once something turns into that specific ion, it's not going back... unless there's a more stabler ion possible

In this case, the 2 ions formed are: HCO₃⁻ and CO₃⁽²⁻⁾, drawing the structures of both the ions tells us that both of them have resonance, but the CO₃⁽²⁻⁾ ion has more resonance structures and hence is more stable

Answer:

we don't understand why humans have only 46 chromosomes

Answer:

46 chromosomes is what we don't understand

Answer:

6.4×10¯³ g of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Next, we shall determine the masses of CH₄ and O₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of CH₄ = 12 + (4×1)

= 12 + 4

= 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Finally, we shall determine the mass of O₂ needed to react with 1.6×10¯³ g of CH₄. This can be obtained as illustrated below:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Therefore, 1.6×10¯³ g of CH₄ will react with = (1.6×10¯³ × 64) / 16 = 6.4×10¯³ g of O₂

Thus, 6.4×10¯³ g of O₂ is needed for the reaction.

Answer:

molecules of real gases occupy no volume.

Explanation:

Answer:

See detailed explanation.

Explanation:

Hello there!

In this case, for the given scenario, we will proceed as follows:

1. Here, we infer that the products are iron (III) hydroxide (precipitate) and calcium chloride:

[tex]3Ca(OH)_2+2FeCl_3\rightarrow 3CaCl_2+2Fe(OH)_3[/tex]

2. In this step we firstly calculate the moles of both reactants, by using their molar masses 74.093 and 162.2 g/mol respectively:

[tex]14.0gCa(OH)_2*\frac{1molCa(OH)_2}{74.093gCa(OH)_2}=0.189molCa(OH)_2 \\\\17.0gFeCl_3*\frac{1molFeCl_3}{162.2gFeCl_3}=0.105molFeCl_3[/tex]

Now, we calculate the moles of calcium hydroxide consumed by 0.105 moles of iron (III) chloride by using the 3:2 mole ratio between them:

[tex]0.105molFeCl_3*\frac{3molCa(OH)_2}{2molFeCl_3} =0.157molCa(OH)_2[/tex]

Thus, we infer that calcium hydroxide is in excess as 0.189 moles are available for it but just 0.157 moles react and therefore, iron (III) chloride is the limiting reactant.

3. Here, we use the moles of iron (III) chloride we've just computed, the 2:2 mole ratio with iron (III) hydroxide and its molar mass (106.867 g/mol) as shown below:

[tex]0.105molFeCl_3*\frac{2molFe(OH)_3}{2molFeCl_3} *\frac{106.867gFe(OH)_3}{1molFe(OH)_3} \\\\=11.2gFe(OH)_3[/tex]

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