You combine 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8x10-4 what is the pH of the solution

Answer:

equal to zero is the right answer

Answer:

3:372-88U:771-772828

Answer:

32.0 kJ

General Formulas and Concepts:

Thermochemistry

Specific Heat Formula: q = mcΔT

Explanation:

Step 1: Define

Identify variables

[Given] m = 1.00 g

[Given] ΔT = 1.48 °C

[Given] c = 21.6 kJ/g °C

[Solve] q

Step 2: Find Heat

Step 3: Check

Follow sig fig rules and round. We are given 3 sig figs.

31.968 kJ ≈ 32.0 kJ

The correct answer is mixture

Answer:

Substance 1 will diffuse at a faster rate.

Explanation:

We can solve this problem by keeping in mind Fick's law, which states:

Where:

As (dc/dx) is equal for both substances, as stated by the problem, the substance with the higher diffusion constant will diffuse at a faster rate.

Thus the answer is substance 1.

Answer:

1.104 J/g°C

Explanation:

Using Q = m × c × ∆T

Where;

m = mass of substance (g)

c = specific hear capacity (J/g°C)

∆T = change in temperature (°C)

For a colorimeter,

Q(water) = - Q(metal)

m. c. ∆T (water) = - m. c. ∆T (metal)

According to the information provided;

For water:

m = 28.0g

c = 4.184 J/g°C

∆T = (21.23 - 19.73°C)

For the metal:

m = 2.05g

c = ?

∆T = (21.23 - 98.88°C)

m. c. ∆T (water) = - m. c. ∆T (metal)

[28 × 4.184 × (21.23 - 19.73°C)] = -[2.05 × c × (21.23 - 98.88°C)]

[117.152 × 1.5] = -[2.05 × c × (-77.65)]

175.728 = -[-159.1825c]

175.728 = 159.1825c

c = 175.728 ÷ 159.1825

c = 1.104

c = 1.104 J/g°C

Answer:

1.06 °C

Explanation:

From the question given above, the following data were obtained:

Mass of gold (M₉) = 10 g

Specific heat capacity of gold (C₉) = 0.129 J/gºC

Initial temperature of gold (T₉) = 24 °C

Mass of water (Mᵥᵥ) = 118 g

Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Initial temperature of water (Tᵥᵥ) = 1 °C

Equilibrium temperature (Tₑ) =?

The equilibrium temperature of the system can be obtained as follow:

Heat loss by the gold = heat gained by the water

M₉C₉(T₉ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Cᵥᵥ)

10 × 0.129 (24 – Tₑ) = 118 × 4.184 (Tₑ – 1)

1.29(24 – Tₑ) = 493.712 (Tₑ – 1)

Clear bracket

30.96 – 1.29Tₑ = 493.712Tₑ – 493.712

Collect like terms

30.96 + 493.712 = 493.712Tₑ + 1.29Tₑ

524.672 = 495.002Tₑ

Divide both side by 495.002

Tₑ = 524.672 / 495.002

Tₑ = 1.06 °C

Therefore, the temperature of the system is 1.06 °C

The amount of heat of the system is measured by a device called a calorimeter. The final temperature of the system will be 1.06 degrees celsius.

The equilibrium temperature is the temperature that follows the law of thermodynamics and is said to be the system that has alike temperatures.  

Given,

Mass of Ag [tex]\rm (M_{g})[/tex] = 10g

Specific heat capacity of Ag [tex](\rm C_{g})[/tex] = [tex]\rm 0.129 J/g^{\circ}C[/tex]

The initial temperature of Ag [tex](\rm T_{g})[/tex] = [tex]24 ^{\circ}\;\rm C[/tex]

Mass of water [tex](\rm M_{w})[/tex] = 118 g

Specific heat capacity of water [tex](\rm C_{w})[/tex] = [tex]4.184 \rm \;J/g^{\circ}\;\rm C[/tex]

The initial temperature of water [tex](\rm T_{w})[/tex] = [tex]1 ^{\circ}\;\rm C[/tex]

Equilibrium temperature = [tex](\rm T_{e})[/tex]

The equilibrium temperature can be shown as, heat loss by the gold = heat gained by the water:

[tex]\rm \rm M_{g}C_{g}(T_{g} - T_{e}) = M_{w}C_{w}(T_{e}-C_{w})[/tex]

Substituting values in the equation:

[tex]\begin{aligned} 10 \times 0.129 (24 - \rm T_{e}) &= 118 \times 4.184 (\rm T_{e} - 1)\\\\\rm 1.29(24 - T_{e}) &= 493.712 (\rm T_{e} - 1)\\\\524.672 &= 495.002 \;\rm T_{e}\end{aligned}[/tex]

Now divide both the sides by 495.002:

[tex]\begin{aligned} \rm T_{e} &= \dfrac{524.672 }{495.002}\\\\\rm T_{e} &= 1.06 \;^{\circ}\rm C\end{aligned}[/tex]

Therefore, the final temperature of the system is 1.06 degrees celsius.

Learn more about equilibrium temperature here:

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Answer:

1 Mole = 84.007 g/mol

Explanation:

Sodium bicarbonate (IUPAC name: sodium hydrogen carbonate), commonly known as baking soda or bicarbonate of soda, is a chemical compound with the formula NaHCO3

Explanation:

Grignard reagent reacts with ketones. Upon chemical reaction of acetone and Grignard reagent there will be formation of tertiary alcohol.

1). So the statement, acetone cannot be used as a solvent because the Grignard reagent will react with its carbonyl, instead of reacting with the planned synthesis carbonyl is true.

2). Tetrahydrofuran is not a suitable solvent for the Grignard reaction because his oxygen may form complexes with the Mg, deactivating the Grignard is a false statement.

3). Phenol can be used as a solvent in Grignard reaction as long as is anhydrous and moisture is kept out of the system with a drying tube with drierite is false statement.

4). Syringes are used in the Grignard experiment to avoid mixing all the reagents is false statement.

5). We use syringes to inject the reagents through a septum preventing moisture to enter the system is true statement.

The fluorine atom is partially negatively charged while chlorine is partially positively charged.

Polar molecules are molecules whose molecules are partially charged due to the electronegative differences between the atoms in the molecule of the compound.

Chlorine monofluoride is a polar molecule.

Fluorine is more electronegative than the chlorine atom.

Therefore, the fluorine atom is partially negatively charged while chlorine is partially positively charged.

Learn more about polar molecules at: https://brainly.com/question/1433127

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Answer:

i didnt understand

Explanation:

Answer:

x = 2 (C₂H₆)

Explanation:

The general formula for alkanes is CₓH₂ₓ₊₂

2x + 2 = 6

Simply solve for n:

2x = 4

x = 2

Answer: [tex]Fe_2O_3[/tex] and [tex]C_2H_4O_2[/tex] are the compounds.

Explanation:

A chemical compound is defined as a chemical substance that is formed by the combination of two or more atoms of different elements which cannot be separated by any physical means but when chemically treated, they decompose into their parent elements.

For example, water is made up of hydrogen and oxygen. This compound is a liquid and its individual components are gases. When water is decomposed, it forms hydrogen and oxygen gas.

For the given options:

He(Helium) is an element formed by the combination of only type of atoms.

[tex]O_2[/tex] and [tex]P_4[/tex] are molecules of same element.

[tex]Fe_2O_3[/tex] is a compound fomed by the combination of iron and oxygen atoms.

[tex]C_2H_4O_2[/tex] is a compound fomed by the combination of carbon, hydrogen, and oxygen atoms.

Hence, [tex]Fe_2O_3[/tex] and [tex]C_2H_4O_2[/tex] are the compounds.

Answer:

Group 1A: alkali metals, or lithium family.

Group 2A: alkaline earth metals, or beryllium family.

Group 7A: the manganese family.

Group 8A: the iron family.

Explanation:

Answer:

1A: Alkali Metals

2A: Alkaline Earth Metals

7A: Halogens

8A: Noble Gases

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

[tex]rate = k[P]^{2} [Q][/tex]

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

[tex]rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }[/tex]

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

[tex]k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4[/tex]

Second row:

2. Rate value:

[tex]rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1[/tex]

3.Third row:

[tex][Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}[/tex]

4. Fourth row:

[tex][P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}[/tex]

Answer:

[tex]v_{solute}=160mL[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the volume of methyl alcohol solute by using the definition of by-volume percentage:

[tex]\%v=\frac{v_{solute}}{v_{solution}} *100\%[/tex]

Whereas we solve for the volume of the solute as shown below:

[tex]v_{solute}=\frac{\%v*v_{solution}}{100\%} \\\\v_{solute}=\frac{80\%*200mL}{100\%}\\\\ v_{solute}=160mL[/tex]

Regards!

Answer:

a) 0.15 mol.

b) 8.95 g.

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to infer this problem is solved by using the ideal gas equation:

[tex]PV=nRT[/tex]

And proceed as follows:

a) Here, we solve for the moles, n,  as follows:

[tex]n=\frac{PV}{RT} \\\\n=\frac{0.50atm*4.5L}{0.08206\frac{atm*L}{mol*K}*178K} \\\\n=0.15mol[/tex]

b) for the calculation of the mass, we recall the molar mass of butane, 58.12 g/mol, to obtain:

[tex]0.15mol*\frac{58.12g}{1mol} =8.95g[/tex]

Regards!

Answer:

Al2o3 is 101

(nh4)2O IS 52

S8 is 256.56

Ba(oh)2 is 171.35

Cacl2 is 110.98

H2O is 18.01

Explanation:

Answer:

Explanation:

578kj/mol

Answer:

I has 2 double carbon carbon bonds

Answer:

it is...

Explanation: Pure copper(II) sulfate is white. It is also known as anhydrous copper(II) sulfate because it has no water in it. When water is present in a sample of copper(II) sulfate it turns blue.

Answer:

put it in a beaker then tne beaker in a boiling water bath and when it temperature has reached 100°c condense that gas ... if there is no gas evaporating by 100° then there is no water

Explanation:

water evaporates at 100°c

Answer:

combine harvester, or a combiner.

So, we know the reduction potentials of the electrodes :

Electrode A = -0.21 V

Electrode B = -0.15 V

We want our cell to be spontaneous - this means that the voltage has to be positive. In order to do so, we need to turn the more negative reduction half reaction to be an oxidation half reaction.

-0.21 is more negative than -0.15, so electrode A will be the oxidation half reaction.

Anode is where the oxidation half reaction takes place, so electrode A is the correct answer . A fun way to remember this is An(ode)Ox(idation) = AnOx and Red(uction)Cat(hode) = RedCat

The E° cell will be 0.21 -0.15 V = 0.6 V

Answer:

2.37 (w/w)% of NH3 in the fertilizer

Explanation:

The HCl reacts with NH3 as follows:

HCl + NH3 ⇄ NH4Cl

To solve this question we must find the  moles of HCl used in the titration = Moles NH3. With its molar mass we can find mass of NH3 and using the dilutions we can find weight percent as follows:

Moles HCl = Moles NH3

32.37mL = 0.03237L * (0.1077mol/L) =

Mass NH3 in the dilution -Molar mass: 17.031g/mol-

0.003486moles NH3 * (17.031g/mol) = 0.05937g NH3

Mass NH3 in the sample:

0.05937g NH3 * (79.733g + 24.803g) / 10.560g =

0.588g NH3

Weight percent:

0.588g NH3 / 24.803g * 100 =

Answer:

5.34275*10²³

Explanation:

Molar mass of O2 is 32g/mol

mass of the sample is 28.4g

number of moles = (mass of the sample) / (molar mass of O2)

=28.5g / 32g/mol

=0.8875mol

number of molecules = number of moles * Avogadro's Constant

= 0.8875 * (6.02*10²³)

= 5.34275 molecules

Answer:

[tex]m_{CO_2}=75.6gCO_2[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out mandatory for us to calculate the reacting moles of both C and O2 because we are given grams and pressure, temperature and volume, respectively:

[tex]n_C=36gC*\frac{1molC}{12gC}=3.0molC \\\\n_{O_2}=\frac{3.0atm*14L}{0.08206\frac{atm*L}{mol*K}*298K}=1.72molO_2[/tex]

Thus, since C and O2 react in a 1:1 mole ratio, we infer C is in excess, and the grams of CO2 can be calculated with the moles of O2:

[tex]m_{CO_2}=1.72molO_2*\frac{1molCO_2}{1molO_2}*\frac{44.01gCO_2}{1molCO_2} \\\\ m_{CO_2}=75.6gCO_2[/tex]

Best regards!

Answer:

67 L

Explanation:

Step 1: Write the balanced equation

Na₂S(aq) + 2 HCl(aq) → 2 NaCl(aq) + H₂S(g)

Step 2: Calculate the moles corresponding to 234 g of Na₂S

The molar mass of Na₂S is 78 g/mol.

234 g × 1 mol/78 g = 3.0 mol

Step 3: Calculate the moles of H₂S produced from 3.0 moles of Na₂S

The molar ratio of Na₂S to H₂S is 1:1. The moles of H₂S formed are 1/1 × 3.0 mol = 3.0 mol.

Step 4: Calculate the volume occupied by 3.0 moles of H₂S at STP

At STP, 1 mole of H₂S occupies 22.4 L.

3.0 L × 22.4 L/1 mol = 67 L

Answer:

The reactive nucleophile is Ketone.

Explanation:

In organic chemistry, The process of acid - catalyzed aldol condensation starts from when ketone (or any aldehyde) is converted to an -enol, after which it attacks another ketone/aldehyde that has already been activated by parbonyl oxygen protonation.

The process of this is that first of all the ketone undergoes tautomerization to form -enol. Thereafter, the other carbonyl will undergo protonation which makes the carbon activated towards attack. Now, the nucleophilic enol will be added to the carbonyl in a [1,2]-addition reaction and we will now use deprotonation to obtain the neutral Aldol product.

Now, since only the ketone can produce an -enol, thus it is the nucleophile as aldehydes are better electrophiles

Answer:

d. 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Explanation:

Aluminum metal reacts with oxygen gas in a combination reaction that forms a product that coats the metal preventing it from further oxidation: aluminum oxide. Aluminum is a cation with charge 3+ (Al³⁻) and oxide is an anion with charge 2- (O²⁻). Thus, the neutral compound aluminum oxide has the chemical formula Al₂O₃. The unbalanced chemical equation is:

Al(s) + O₂(g) → Al₂O₃(s)

We can balance using the trial and error method. First, we will balance O atoms by multiplying Al₂O₃ by 2 and O₂ by 3.

Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Finally, we get the balanced equation by multiplying Al by 4.

4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Organic compround is octane C8H18

Answer:

a

Explanation: