g aqueous barium hydroxide (ba(oh)2) and nitric acid (hno3) participate in a complete neutralization reaction. in the molecular equation, what are the products

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:

[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.

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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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.

Answer:

C

Explanation:

if we were to followw the IUPAC

Answer:

C.

Explanation:

I believe that it should be A and B.

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

Answer:

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

Explanation:

Explanation:

here's the answer to your question

Answer:

0.8017

Explanation:

Find the molar Mass of KF

K = 39

F = 19

Total = 58

Note: these numbers are approximate. Use your periodic table to get the exact numbers.

mols = given mass / molar mass

given mass = 46.5

molar mass = 58

mols = 46.5 / 58

mols = 0.8017

Answer:

Exothermihic chart

Explanation:

Explanation:

The standard entropy change of a reaction has a positive value. This reaction results in an increase in entropy.

Positive entropy means the system has increased its degree of disorderness.

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

Copper tube is used in EBRT.

Copper tube is used in EBRT.

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Answer:

7.38*10^21

Explanation:

2Na+2H20=2NaOH+H2

nNa=0.0123

number of water moles: 0.012*6*10^23=7.38*10^21

According to Arrhenius definition of acids, [tex]NH4^+[/tex] is an acid.

According  to Arrhenius  definition of acids and bases, acid is any substance that produces hydrogen ion in solution as its only positive ion.

Following this definition, let us now consider what happens when [tex]NH4^+[/tex] is introduced into a water;

[tex]NH4^+[/tex](aq)-------> NH3(aq) + [tex]H^+[/tex](aq)

Hence, according to Arrhenius definition of acids, [tex]NH4^+[/tex] is an acid.

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Answer:B) 1s 2s 2p 3s 3p 4s 3d

Explanation:

The ground state electron configuration shows how the electrons in the atomic orbitals of an atom are in their lowest , most stable energy arrangements and since Electrons must be filled following the Aufbau's principle(electrons fill lowest energy shells first)

Now, Titanium lies in period IV and  group 4 of the periodic table with 22 as its atomic number

Thus, the ground-state electron configuration of a neutral atom of titanium is 1s²2s²2p⁶3s²3p⁶4s²3d².  

Answer:

we don't understand why humans have only 46 chromosomes

Answer:

46 chromosomes is what we don't understand

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:

This question is asking to find the new temperature

The answer for the final temperature is 429.73K

Explanation:

Using Charles law equation as follows:

V1/T1 = V2/T2

Where;

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to this question;

V1 = 3.0L

V2 = 4.4L

T1 = 20°C = 20 +273 = 293K

T2 = ?

Using V1/T1 = V2/T2

3/293 = 4.4/T2

Cross multiply

293 × 4.4 = 3 × T2

1289.2 = 3T2

T2 = 1289.2 ÷ 3

T2 = 429.73K

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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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:

ag and au are sure not to react. but hg and sn might or might not

The question is incomplete, the complete question is;

When determining the amount of an oxidant present by titration, you can use iodine and starch as an indicator.

First, the oxidant, like hypochlorite, oxidizes

Choose...

neutral iodine into iodide ion

iodide ion into neutral iodine

iodate polyatomic ion into iodide ion

When starch and iodine are both present, the solution is

Choose...

blue-black

brownish yellow

clear

During the titration, the titrant, like thiosulfate, reduces the

Choose...

iodide ion into iodate polyatomic ion

neutral iodine into iodide ion

iodide ion into neutral iodine

When the iodine has completely reacted at the endpoint of the titration, the solution should become

Choose...

clear

blue-black

brownish yellow

Answer:

1. iodide ion into neutral iodine

2. blue-black

3. neutral iodine into iodide ion

4. clear

Explanation:

Hypochlorite oxidizes the iodide ion to iodine molecule according to the reaction equation;

ClO-(aq) + 2H+(aq) + 2I-(aq) ---------> 6 I2(l) + Cl- (aq)+ H2O(l)

When iodine is added, the colour of the starch solution immediately changes to blue-black.

A reduction reaction occurs when the titrant, thiosulfate is added as follows;

I2 + 2S2O32- → 2I- + S4O62-

The solution at end point is found to become clear again.

Answer:

c

Explanation:

it takes 10,000 years to just reduce down the decay