Calculate the pH for a 1.0 x 10-5 M solution of OH at 25°C.
pH = -log[H*), pOH = -log[OH-]
14 = pH + POH
0 pH = 11.00
0 pH = 9.00
0 pH = 6.00
0 pH = 3.00

Answers

Answer 1

Answer:

ph=9.00

Explanation:

Given 1.0x10^-15M of OH at 25c

first find Poh from Poh=-log[OH]

then Poh=-log[1.0x10^-5]

from above Ph=5

then find Ph from ph+poh =pw

where pw=14

so

ph+5=14

ph=9.00

Answer 2

The pH for a 1.0 x 10-5 M solution of OH at 25°C is 9.

What is pH?pH is a measure of acidity and basicity of aqueous solution. The range of pH goes from 0 - 14, with 7 being neutral. pH of less than 7 indicate acidity, whereas a pH of greater than 7 indicates a base.pH is the measure of the relative amount of free hydrogen and hydroxyl ions in aqueous or other solutions.

In water pH + pOH=14

pH = 14 - pOH = 14 - 5 = 9

From definition, pOH = - log₁₀ [OH⁻]  

Thus pOH = -log₁₀ (1 x 10⁻⁵)

                  = - (-5) = 5

pH + pOH = 14

pH = 14 - pOH

     = 14 - 5 = 9  

pH = 9

Hence, pH = 9 is the correct answer.

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Explanation:

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔG°f (kJ/mol) -110.9 87.6 51.3 -237.1

Answers

Answer:

[tex]\Delta G_{rxn}=51.0kJ/mol[/tex]

Explanation:

Hello there!

In this case, according to the definition of the Gibbs free energy of reaction in terms of the Gibbs free energy of formation of the involved reactants and products, it is possible for us to compute it as shown below:

[tex]\Delta G_{rxn}=3\Delta _f G_{NO_2}+\Delta _f G_{H_2O}-2\Delta _f G_{HNO_3}-\Delta _f G_{NO}[/tex]

Thus, by plugging in the given data, it is possible to obtain:

[tex]\Delta G_{rxn}=3(51.3)+(-237.1)-2(-110.9)-(87.6)\\\\\Delta G_{rxn}=51.0kJ/mol[/tex]

Best regards!

The standard Gibbs free energy change (ΔG°rxn) for the reaction is 76.7 kJ/mol.

The balanced equation for the reaction is:

2 HNO₃(aq) + NO(g) → 3 NO₂(g) + H₂O(l)

The standard free energy change for a reaction ( ΔG0rxn) can be found out using the equation,

ΔG° = Σ ΔG°fproducts - Σ ΔGf° ​​​​​reactants

ΔG°rxn = ΔH°rxn - TΔS0rxn  

∆H°rxn = change in enthalpy of reaction

∆S° = Change in entropy of reaction

T = Temperature ( 298 K )

ΔG°rxn = (3 × ΔG°f(NO₂(g)) + ΔG°f(H₂O(l))) - (2 × ΔG°f(HNO₃)aq)) + ΔG°f(NO(g)))

Substituting the given ΔG°f values:

ΔG°rxn = (3 × 51.3 kJ/mol) + (-237.1 kJ/mol) - (2 × (-110.9 kJ/mol)) + (87.6 kJ/mol)

ΔG°rxn = 153.9 kJ/mol - 237.1 kJ/mol + 221.8 kJ/mol + 87.6 kJ/mol

ΔG°rxn = 226.2 kJ/mol - 149.5 kJ/mol

ΔG°rxn = 76.7 kJ/mol

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How are the testes and the ovaries similar?

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They are reproductive organs
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Answer:

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Answer:

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Explanation:

Which a balanced chemical equation ? O C 7 H 16 +5O 2 6CO 2 +4H 2 O OB C 7 H 16 +11O 2 7CO 2 +8H 2 O C 7 H 16 +14O 2 7CO 2 +5H 2 O OD . C 7 H 16 +22O 2 14CO 2 +16H 2 O

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Answer: [tex]C_7H_{16}+11O_2\rightarrow 7CO_2+8H_2O[/tex]

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced chemical equation is:

[tex]C_7H_{16}+11O_2\rightarrow 7CO_2+8H_2O[/tex]

i need to write a brief paragraph on why this lab is important, please help!! here is the question!
How can the properties of acids and bases be used to classify solutions?

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Answer:

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A student weighs an empty flask and stopper and finds the mass to be 53.256 g. She then adds about 5 mL of an unknown liquid and heats the flask in a boiling water bath at 98.8°C. After all the liquid is vaporized, she removes the flask from the bath, stoppers it, and lets it cool. After it is cool, she momen- tarily removes the stopper, then replaces it and weighs the flask and condensed vapor, obtaining a mass of 53.780 g. The volume of the flask is known to be 231.1 mL. The absolute atmospheric pressure in the laboratory that day is 728 mm Hg. a. What was the pressure of the vapor in the flask in atm?

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Answer:

a. The pressure in the flask open to the atmosphere during the vaporization of the unknown liquid is equal to the prevailing atmospheric pressure equivalent to 0.957734 atm

Explanation:

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The volume of the unknown liquid she adds = 5 mL

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a. We note that the student stoppers the flask after all the liquid has evaporated. Therefore, given that the flask was open to the atmospheric pressure as the liquid evaporates, the pressure of the vapor in the flask is equal to the prevailing atmospheric pressure, or 728 mmHg

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Strong acids have mostly ions in solution, therefore the bonds holding H and A together must be weak. Strong acids easily break apart into ions. Weak acids exist mostly as molecules with only a few ions in solution, therefore the bonds holding H and A together must be strong.Aug 15, 2020

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dioxide and 1.50 grams of water vapor. It was also determined that 279 mL of 0.1 M KOH was needed to titrate the acid completely.

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Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

[tex]n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC[/tex]

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Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

[tex]m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO[/tex]

Thus, the subscripts in the empirical formula are:

[tex]C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2[/tex]

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

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Which are equal to the moles of the acid:

[tex]n_{acid}=0.0279mol[/tex]

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(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

[tex]C_7H_6O_2[/tex]

Best regards!

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HOPE THIS HELPS ^^

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Fairly certain it's C. The particles In his hands vibrate faster because of friction :)

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Please do not answer from links
Please answer correctly
Urgent

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[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \underline{ \large \orange{\tt{ ꧁ A \: N\: S \: W \: E \: R ꧂}}}}[/tex]

✧ [tex] \large{ \red{\tt{E\:X \: O \: T \: H \: E \: R \: M \: I \: C \: \: R \: E \: A \: C \: T\: I \: O \: N}} : }[/tex]

☃ The chemical reaction in which heat is released to the surroundings is called exothermic reaction. The amount of heat energy released is written along with products. This indicates that heat is given out.

[tex] \pink{ \text{Any \: two \: balanced \: equation \: for \: Exothermic \: changes}} : [/tex]

C + O₂ ⇾ CO₂ + HeatCaO + H₂O ⇾ Ca ( OH )₂ + Heat

ツ Hope I helped! ♡

☂ Have a wonderful day / night ! ☄

[tex] \underbrace{ \overbrace{ \mathfrak{Carry \: On \: Learning}}}[/tex] ✎

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

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I looked it up and that's what I found

hope it helps

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78.96 g mol

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78.96 g mol

in the period table the mass of Se is 78.96 g mol

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does anyone know this please?

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Answer:

1.) b

2.) c

3.) e

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5.) j

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-Todo <3

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