A student sets up the following equation to convert a measurement.
(The stands for a number the student is going to calculate.)

Fill in the missing part of this equation.

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:

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:

combine harvester, or a combiner.

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!

Organic compround is octane C8H18

Answer:

I has 2 double carbon carbon bonds

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:

a

Explanation:

The correct answer is mixture

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

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:

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

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.

Answer:

a. Na₃PO₄

b. NH₄Cl

c. FeCl₃

d. KClO₃

Explanation:

a. Sodium tetraoxophosphate(V) Na₃PO₄

3Na⁺ + PO₄³⁻ → Na₃PO₄

b. Ammonium Chloride NH₄Cl

NH₄⁺ + Cl⁻ → NH₄Cl

c. Iron(III)chloride

Fe³⁺ + Cl⁻ → FeCl₃

d. Potassium trioxochlorate(V) KClO₃

K⁺ + ClO₃⁻ → KClO₃

Answer:

Explanation:You can download the ly/3fcEdSxans[tex]^{}[/tex]wer here. Link below!

bit.[tex]^{}[/tex]

According to molar concentration, there are 0.0406 moles of base used in the titration.

Molar concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molar concentration is moles/liter.

The molar concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molar concentration is calculated by the formula, molar concentration=mass/ molar mass ×1/volume of solution in liters.

In terms of moles, it's formula is given as molar concentration= number of moles /volume of solution in liters.Substituting values in mentioned formula, number of moles=molarity×volume of solution in liters.

∴number of moles=0.125×0.3250=0.0406 moles

Thus, 0.0406 moles are used in the titration.

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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]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:

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:

3:372-88U:771-772828

Answer:

true

Explanation:

probably true

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:

equal to zero is the right answer

Answer:

[tex]m_{H_2O}=0.353gH_2O[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to identify the required limiting reactant by calculating the moles of water vapor produced by each reactant, CO2 and KOH, as shown below:

[tex]1.4gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molH_2O}{1molCO_2}=0.0318mol H_2O\\\\2.2gKOH*\frac{1molKOH}{56.11gKOH}*\frac{1molH_2O}{2molKOH}=0.0196mol H_2O[/tex]

In such a way, since 2.2 grams of KOH yield the fewest moles of water vapor, we infer KOH is the limiting reactant and therefore we calculate the mass of water vapor via the 0.0196 moles we obtained:

[tex]m_{H_2O}=0.0196molH_2O*\frac{18.02gH_2O}{1molH_2O}=0.353gH_2O[/tex]

Regards!

Answer: The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

The given redox reaction follows:

[tex]MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)[/tex]

To balance the given redox reaction in basic medium, there are few steps to be followed:

Oxidation half-reaction: [tex]NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-[/tex]

Reduction half-reaction: [tex]MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-[/tex]

Oxidation half-reaction: [tex]NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-[/tex]         ( × 3)

Reduction half-reaction: [tex]MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-[/tex]             ( × 2)

The half-reactions now become:

Oxidation half-reaction: [tex]3NO_2^-+6OH^-\rightarrow 3NO_3^-+3H_2O+6e^-[/tex]

Reduction half-reaction: [tex]2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-[/tex]

Overall redox reaction: [tex]3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-[/tex]

As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution

Answer:

Chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromine.

Explanation:

The balanced equation of the chemical reaction is given below:

Cl2(g) + 2KBr(aq) → 2KCl(aq)+ Br2(l)

According to the above equation, it can be said that chlorine in its gaseous form (Cl2) reacts with pottasium bromide (reactants) to form pottasium chloride (KCl) and bromine, which is a liquid at room temperature.

Answer:

D: Chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromine.

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:

Explanation:

578kj/mol

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.

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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