How would you combine reactions A-C, shown below, to obtain the overall reaction: NH,(g) +BH,(g) +0,(g)_ 21,0(g) + HBNH(s) Please select all that apply. a. 2H2,(g) +o2,(g) - 2H2O(g) b. H3BNH3(s) - NH3(g) + BH3(g) c. H3BNH3(s) —2H2(g) + HBNH(s)

Answer:

9th:08473 kJ/mol

Explanation:

HAHAHAHAHAHAHAHAHAHA

Answer:

rain

Explanation:

The volcanic eruptions may add salt to ocean waters. option A is correct.

Volcanic eruption produces or releases direct minerals into the water or in the ocean and salt domes are also responsible to increase the salt concentration in the water.

Two of the most commonly found ions in the water of oceans are chloride ions and sodium ions. 85% of the concentration found remaining are the magnesium and sulfate salts.

This process also happens with the mountains but in oceans it invertases the concentration of salt.  

Therefore, volcanic eruptions may add salt to ocean waters. option A is correct.

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#SPJ2

Answer:

D

Explanation:

the charges need to balence out

so finding the LCM which is 12 we find we need 3x's

and 4 zs

so that makes the formula X3Z4 which is D

Answer:

Compound Formula. 1, Carbon dioxide, CO2. 2, Carbon monoxide, CO. 3, Diphosphorus pentoxide, P2O5. 4, Dinitrogen monoxide, N2O.

Explanation:

dont even know by thoAnswer:

Explanation:

Answer:

Explanation:

D or b

Sorry if wrong

Answer:

temporal

Explanation:

Answer:

Well if you want to kiss them go for it, but if you don't then say sorry but I don't feel comfortable with that.

Hope everything goes well <3

Answer:

Explanation:

Solid ice floats over water because ice is less dense than liquid water, or weighs less, by about 10%, than liquid water

Answer:

% recovery

MP range of product

mass of product

Explanation:

Liquid–liquid extraction (LLE) is a process of transferring one (or more) solute(s) which are present in a feed solution to another immiscible liquid (solvent). The other solvent that becomes enriched in the target solute(s) is called extract. The original feed solution that is depleted in solute(s) is subsequently referred to as the raffinate.

This method is used to purify compounds and  separate mixtures of compounds. This is very important when we want to isolate a product from a reaction mixture.

The percent recovery is the amount of solute that is transferred to the extract. This is the most important data to be recorded in an LLE experiment.

The melting point range necessarily helps us to identify the product and the mass of solid tells us the quantity of the solid obtained after extraction.

Answer:

C - Extrusive igneous

Explanation:

Rhyolite is an extrusive igneous due to the high silica content, the lava is very dangerous.

Answer:

Its C

Explanation:

A P E X

Answer:

se que la imaje es un poco confusa con la materia pero

desiaria ayudarte pero CREO que con mi TÉCNICA  es la A

Answer:

Q = 32,007.6 J

Explanation:

Hello there!

In this case, since the energy involved during a heating/cooling process is:

[tex]Q=mC(T_f-T_i)[/tex]

Thus, given the mass, specific heat of water, initial temperature and final one, we plug in obtain:

[tex]Q=450g*4.184\frac{J}{g\°C} (37.0\°C-20.0\°C)\\\\Q=32,007.6J[/tex]

Best regards!

Answer:

A. Beginning of the spring

Explanation:

Pa like, Pa follow, Pa rate

Answer:

c

Explanation:

hope this helps! ^-^

Answer:

C

Explanation:

It's the answer

Answer:

The ATP is broken down into glucose which the cells use for energy.

Answer:

Q = ΔH fusion * mass (g)

when we have:

ΔH fusion (the heat (or enthalpy) of fusion = 0.334 kJ/g

and mass of ice = 22.4 g  

so by substitution, we can get the energy (Q) required to melt this mass of ice:

∴ Q = 0.334KJ/g * 22.4 g  

      = 7.48 KJ

∴ the energy required to melt 22.4 g of ice is = 7.48 KJ

Explanation:

Answer:

true

Explanation:

you are getting closer to the core hope this helps!

The question is missing some parts. Here is the complete question.

Carbon monoxide is a colorless, odorless gas that binds irreversibly to hemoglobin in our blood, causing suffocation and death. CO is formed during incomplete combustion of carbon. One way to represent this equilibrium is:

[tex]2CO_{(g)}[/tex] ⇄ [tex]2C_{(s)}+O_{2}_{(g)}[/tex]

we could also write this reaction three other ways listed below. The equilibrium constant for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.

1) [tex]2C_{(s)}+O_{2}_{(g)}[/tex] ⇄ [tex]2CO_{(g)}[/tex]       K₁ =

2) [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex] ⇄ [tex]CO_{(g)}[/tex]    K₂ =

3) [tex]CO_{(g)}[/tex] ⇄ [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex]    K₃ =

Answer: 1) [tex]K_{1}=\frac{1}{K}[/tex]

              2) [tex]K_{2}=\frac{1}{K^{1/2}}[/tex]

             3) [tex]K_{3}=K^{1/2}[/tex]

Explanation: A chemical reaction can be reversible, i.e., can proceed in both directions: to the right of the arrow (forward) or towards the left of the arrow (backward).

When the rates of forward and backward reactions are the same, the reaction is in equilibrium. In that state, we can determine the equilibrium constant, [tex]K_{c}[/tex].

For the first way to represent equilibrium of CO formed, the [tex]K_{c}[/tex] is calculated

[tex]2CO_{(g)}[/tex] ⇄ [tex]2C_{(s)}+O_{2}_{(g)}[/tex]

[tex]K=\frac{[O_{2}]}{[CO]^{2}}[/tex]

in which the symbol [ ] is concentration of the compound.

In equilibrium constant, solids are not included.

Equilibrium constants for the other reactions:

1) [tex]2C_{(s)}+O_{2}_{(g)}[/tex] ⇄ [tex]2CO_{(g)}[/tex]  

[tex]K_{1}=\frac{[CO]^{2}}{[O_{2}]}[/tex]

Comparing K₁ and K, the first one is the inverse of K, so writing in terms of K

[tex]K_{1}=\frac{1}{K}[/tex]

2) [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex] ⇄ [tex]CO_{(g)}[/tex]

[tex]K_{2}=\frac{[CO]}{[O_{2}]^{1/2}}[/tex]

In terms of K, K₂ is

[tex]K_{2}=\frac{1}{K^{1/2}}[/tex]

3) [tex]CO_{(g)}[/tex] ⇄ [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex]

[tex]K_{3}=\frac{[O_{2}]^{1/2}}{[CO]}[/tex]

This constant in terms of K will be

[tex]K_{3}=K^{1/2}[/tex]

In conclusion, K₁, K₂ and K₃ in terms of K is [tex]\frac{1}{K}[/tex],[tex]\frac{1}{K^{1/2}}[/tex] and [tex]K^{1/2}[/tex], respectively.

Answer:

Explanation:

One farad of charge is capable of depositing one gram equivalent of a metal

One gram equivalent of Ag = 108 grams

One gram equivalent of Cd = 112 / 2  grams

= 56 grams . [ for cadmium equivalent mass = atomic mass / s ]

electric charge flowing = current x time = 5 x 600 = 3000 coulomb

one farad = 96500 coulomb

96500 coulomb  deposits 108 gram of Ag

3000 coulomb deposits 108 x 3000 / 96500 gram

= 3.35 grams of Ag

Similarly ,

96500 coulomb  deposits 56 gram of Cd

3000 coulomb deposits 56 x 3000 / 96500 gram

= 1.74  grams of Cd

So there will be much larger increase in the spoon of Ag due to larger deposit of Ag by charge .

A per the statement the true and false have been stated.

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Answer:

yes solution is always a mixture but not all mixtures are solution

Explanation:

A solution.is a homogeneous mixture of substance that have uniform composition throughout

And a mixture hVe twoo or more substances that are not chemically.combine

Answer:

a. Endothermic

b. 26.37kJ/mol

Explanation:

a. As we can see, the temperature of the water is decreasing when the reaction is occurring, that means the reaction is absorbing heat and is endothermic

b. To find the enthalpy we must find the change in heat when 12.1g of KCl are dissolved. Using the equation:

Q = -m*ΔT*C

Where Q is change in heat

m the mass of solution (250g + 12.1g = 262.1g)

ΔT is change in heat (17.1°C - 21.0°C = -3.9°C)

And C is specific heat of the solution (4.184J/g°C assuming is the same than the specific heat of water).

Replacing:

Q = -262.1g*-3.9°C*4.184J/g°C

Q = 4277J = 4.28kJ

As reaction enthalpy is the change in heat per mole of reaction, we must find the moles of 12.1g of KCl:

Moles KCl -Molar mass: 74.55g/mol-:

12.1g KCl * (1 mol / 74.55g) = 0.1623 moles KCl

The reaction enthalpy us:

4.28kJ / 0.1623mol = }

Answer:

B

Explanation:

Answer:

it gives base, it is process of neutralization,when acid reacts with base it gives salt and water

Answer:

a. pH raises

b. pH neutral

c. pH raises

Explanation:

First of all we need to dissociate these compounds.

Ca(OH)₂ →  Ca²⁺  +  2OH⁻

This is the calcium hydroxide, a strong base.

In this case, we are giving OH⁻ to medium, so the pH will be increased.

OH⁻  +  H⁺  ⇄  H₂O

Hydroxides will neutralize the protons in order to make water, but if we have many OH, we talk about a basic solution where pH is lower than 7.

b. CuBr → Cu⁺  +  Br⁻

This is similar to NaCl, a ionic salt which is neutral.

The ions from this salt, do not make hydrolisis, that's why pH does not change.

c. Ca₃(PO₄)₂ → 3Ca²⁺  +  2PO₄⁻³

Calcium cathion comes from a strong base. It means, that the ion can not make hydrolisis because it is the conjugate weak acid.

Phosphate anion comes from a weak acid, it can gives OH⁻ to medium so pH will increase; the PO₄⁻³ anion is the conjugate strong base of the phosphoric acid, that's why it can react:

PO₄⁻³ + H₂O  ⇄  H₂PO⁻²  +  OH⁻

Answer:

mesure it

Explanation:

Answer:

A- pH = 13.12

B- pH = 12.91

C- pH = 12.71

D- pH = 12.43

E- pH = 11.55

F- pH = 7

G- pH = 2.46

H- pH = 1.88

Explanation:

This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) →  H₂O(l) + KCl(aq)

Our pH at the equivalence point is 7, because we have made a neutral salt.

To determine the volume at that point we state the formula for titration:

mmoles of base = mmoles of acid

Volume of base  . M of base = Volume of acid . M of acid

50mL . 0.129M = 0.258 M . Volume of acid

Volume of acid = (50mL . 0.129M) / 0.258 M →  25 mL (Point F)

When we add 25 mL of HCl, our pH will be 7.

A- At 0 mL of acid, we only have base.

KOH → K⁺ + OH⁻

[OH⁻] = 0.129 M

To make more easy the operations we will use, mmol.

mol . 1000 = mmoles → mmoles / mL = M

- log 0.129 = 0.889

14 - 0.889 = 13.12

B-  In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺

Initially we have  0.129 M . 50 mL = 6.45 mmoles of OH⁻

1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:

6.45 mmol - 1.81 = 4.64 mmoles of OH⁻

This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.

[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M

- log 0.0815 M = 1.09 → pOH

pH = 14 - pOH → 14 - 1.09 = 12.91

C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺

Our initial mmoles of OH⁻ would not change through all the titration.

Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.

6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻

Total volume is: 50 mL of base + 12.5 mL = 62.5 mL

[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M

- log  0.0517 = 1.29 → pOH

14 - 1.11 = 12.71

D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.

6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻

Total volume is: 50 mL of base + 18 mL = 68 mL

[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M

- log  0.0265 = 1.57 → pOH

14 - 1.57 = 12.43

E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.

6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻

Total volume is: 50 mL of base + 24 mL = 74 mL

[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M

- log  3.51×10⁻³  = 2.45 → pOH

14 - 2.45 = 11.55

F- This the equivalence point.

mmoles of OH⁻ = mmoles of H⁺

We add (25 mL . 0.258M) = 6.45 mmoles of H⁺

All the OH⁻ are neutralized.

OH⁻  +  H⁺  ⇄   H₂O              Kw

[OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷  →  pOH = 7

pH → 14 - 7 = 7

G- In this case we have an excess of H⁻

We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺

[H⁺] = 0.26 mmol / Total volume

Total volume is: 50 mL + 26 mL → 76 mL

[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M

- log 3.42×10⁻³ = 2.46 → pH

H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons

Total volume is 50 mL + 29 mL = 79 mL

[H⁺] = 1.03 mmol / 79 mL → 0.0130 M

- log 0.0130 = 1.88 → pH

After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.

The pH value is given by 14 less the logarithm of the hydronium ion

concentration or the hydrogen ion concentration in the solution.

Responses (approximate values):

The pH values are;

A) 13.11

B) 12.91

C) 12.71

D) 12.42

E) 11.54

F) 7

G) 2.469

H) 1.884

A) Concentration of the KOH = 0.129 M

Amount of HCl added = 0.00 ml

The pH = -log[H⁺] = 14 - pOH

pOH = -log[OH⁻]

Which gives;

pH = 14 - (-log[OH⁻] )

B) Volume of acid added = 7.00 mL = 0.007 L

Concentration of the acid = 0.258 M HCl

Number of moles of acid, H⁺ = 0.007 × 0.258 moles = 0.001806 moles

Number of moles of KOH remaining, OH⁻= 0.05 × 0.129 - 0.001806 = 0.004644

Number of moles of OH⁻ = 0.004644 moles

[tex]Concentration, \ [OH^-] = \dfrac{0.004644 \, moles}{0.05 7 \, L} \approx \mathbf{ 0.0815 \, M}[/tex]

C) 12.5 mL HCl contains, 0.0125 × 0.258 moles = 0.003225 moles

OH⁻ remaining = 0.05 × 0.129 - 0.003225 = 0.003225 moles

[tex]Concentration \ of \ [OH^-]= \dfrac{0.003225\, moles}{(0.05 + 0.0125) \, L} = \mathbf{0.0516 \, M}[/tex]

D) 18.0 mL HCl contains, 0.018 × 0.258 moles = 0.004644 moles

OH⁻ remaining = 0.05 × 0.129 - 0.004644 = 0.001806 moles

[tex]Concentration \ of \ [OH^-] = \dfrac{0.001806\, moles}{(0.05 + 0.018) \, L} \approx \mathbf{0.0266\, M}[/tex]

E) 24.0 mL HCl contains, 0.024 × 0.258 moles = 0.006192 moles

OH⁻ ion remaining = 0.05 × 0.129 - 0.006192 = 0.000258 moles

[tex]Concentration \ of \ [OH^-] = \dfrac{0.000258\, moles}{(0.05 + 0.024) \, L} \approx \mathbf{0.0035\, M}[/tex]

F) 25.0 mL HCl contains, 0.025 × 0.258 moles = 0.00645 moles

[OH⁻] remaining = 0.05 × 0.129 - 0.00645 = 0 moles

[tex]Concentration \ of \ [OH^-] = \dfrac{0\, moles}{(0.05 + 0.024) \, L} \approx 0\, M[/tex]

Therefore;

Number of moles of KOH = 0, or the solution is neutralized

[OH⁻] = [H⁺]

Which gives;

G) 26.0 mL HCl contains, 0.026 × 0.258 moles = 0.006708 moles

[OH⁻] remaining = 0.05 × 0.129 - 0.006708 = -0.000258 moles

Therefore

Number of moles of H⁺ = 0.000258

[tex]\mathbf{Concentration} \ of \ \mathbf{[H^+] }= \dfrac{0.000258\, moles}{(0.05 + 0.026) \, L} \approx 0.003395\, M[/tex]

H) 29.0 mL HCl contains, 0.029 × 0.258 moles = 0.007482 moles

H⁺ remaining = 0.007482 - 0.05 × 0.129 = 0.001032 moles

Therefore

Number of moles of H⁺ = 0.001032

[tex]Concentration \ of \ [H^+] = \mathbf{ \dfrac{0.001032\, moles}{(0.05 + 0.029) \, L}} \approx 0.01306\, M[/tex]

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Answer:

performing

Explanation:

Answer:

1 - Weak electrolyte

2- Non electrolyte

3- Weak electrolyte

4- Strong electrolyte

Explanation:

A strong electrolyte refers to an electrolyte that decomposes completely in solution. This means that there are more charge carriers in solution when a strong electrolyte is dissolved in water. A strong electrolyte produces a strong glow. LiOH is a strong electrolyte.

A weak electrolyte is not completely dissociated in water. Only a small amount dissociates in water. HF is a weak electrolyte. A weak electrolyte does not produce a bright light.

A non-electrolyte does not dissociate in solution at all hence it does not power a bulb E.g  C12H22O11.

Answer:

C9H19 C5H12 C6H17 C7H20