A student dropped a pea size amount of K2CO3 into a solution of HCl(aq). He observed the formation of gas bubbles and collected the gas into another test tube. The student performed a splint test and observed that the splint was extingished when he placed the splint into the test tube of the gas. What can be said about the results of this students experiment?

a. The student completed the experiment correctly and there were no errors in the experiment.
b. The experiment was performed incorrectly. K2CO3 doesn't react with HCl. Therefore, the student picked up the wrong compound when conducting the experiment.
c. The student performed the splint test incorrectly. He should of observed the splint flare up when the splint was placed in the test tube.
d. The student performed the splint test incorrectly. He should of observed a popping sound when the splint was placed in the test tube.

Answer: the molecular formula of  trioxide is ClOClO3 or Cl2O4

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Answer:

density

Explanation:

The procedure that the student could use to examine an intensive property of a rectangular block of wood is to determine its density. Density is intensive because it is the ration between the mass and the volume.

Answer: find the mass option A

Explanation:

Answer:

Explanation:

Liquid A is poured into Liquid B and a single, clear layer results. More Liquid A is added and two distinct layers form. The solution is now unsaturated and there is a static equilibria occurring

Answer:

When SO

2

is passed through an acidified solution of H

2

S, sulphur is precipitated out according to the reaction.

2H

2

S+SO

2

→2H

2

O+3S

Answer:

0.0900 mol/L

Explanation:

A chemist makes 330. mL of nickel(II) chloride working solution by adding distilled water to 220. mL of a 0.135 mol/L stock solution of nickel(II) chloride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.

Step 1: Given data

Step 2: Calculate the concentration of the final solution

We prepare a dilute solution from a concentrated one. We can calculate the concentration of the working solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁/V₂

C₂ = 0.135 mol/L × 220. mL/330. mL = 0.0900 mol/L

Answer: There are 0.0556 moles present in 44.1 mL of a 1.26 M HCl solution.

Explanation:

Given: Volume = 44.1 mL (1 mL = 0.001 L) = 0.0441 L

Molarity = 1.26 M

Molarity is the number of moles of a substance present in liter of a solution.

Therefore, moles of HCl are calculated as follows.

[tex]Moles = \frac{moles}{Volume (in L)}\\1.26 M = \frac{moles}{0.0441 L}\\moles = 0.0556 mol[/tex]

Thus, we can conclude that there are 0.0556 moles present in 44.1 mL of a 1.26 M HCl solution.

Answer:

When this equation is solved, the two values of the unknown are 0.0643 and -0.082

Explanation:

Given

[tex]1.77 * 10^{-2} = \frac{x^2}{0.298 - x}[/tex] --- the actual equation

Required

The values of x

We have:

[tex]1.77 * 10^{-2} = \frac{x^2}{0.298 - x}[/tex]

Cross Multiply

[tex]1.77 * 10^{-2} * (0.298 - x)= x^2[/tex]

Multiply both sides by 100

[tex]1.77 * (0.298 - x)= 100x^2[/tex]

Open bracket

[tex]0.52746 - 1.77x= 100x^2[/tex]

Rewrite as:

[tex]100x^2 + 1.77x - 0.52746 =0[/tex]

Using quadratic formula:

[tex]x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}[/tex]

Where:

[tex]a = 100; b = 1.77; c = -0.52746[/tex]

So, we have:

[tex]x = \frac{-1.77 \± \sqrt{1.77^2 - 4*100*- 0.52746 }}{2*100}[/tex]

[tex]x = \frac{-1.77 \± \sqrt{214.1169}}{2*100}[/tex]

[tex]x = \frac{-1.77 \± 14.63}{200}[/tex]

Split

[tex]x = \frac{-1.77 + 14.63}{200}\ or\ x = \frac{-1.77 - 14.63}{200}[/tex]

[tex]x = \frac{12.86}{200}\ or\ x = \frac{-16.40}{200}[/tex]

[tex]x = 0.0643\ or\ x = -0.082[/tex]

Answer:

n = 3, l = 2, ml =-2

Explanation:

Quantum numbers are a set of values which can be used to describe the energy and position of an electron in space.

There are four sets of quantum numbers;

1) principal quantum number

2) orbital quantum number

3) spin quantum number

4) magnetic quantum number.

The values of orbital quantum number include; -l to +l;

The set of quantum numbers without error is ; n = 3, l = 2, ml =-2

Answer:

The important characteristics of the particles of matter are – They are very, very small. They have spaces between them. They are constantly moving.

Answer:

They are very very small

Answer:

the product is Mg(Clo3)2 + AgI

Answer:

C₆H₆

Explanation:

Each border of the figure represents 1 atom of carbon. We have 6 borders = 6 atoms of carbon.

Each atom of carbon form 4 bonds. All the carbons are doing a double bond and a single bond with other carbons. That means are bonded 3 times. The other bond (That is not represented in the figure. See the image) comes from hydrogens. As we have 6 carbons that are bonded each 1 with one hydrogen. There are six hydrogens and the molecular formula is:

This structure is: Benzene

Answer:

I expect to observe a change in colour from reddish brown to a colourless solution

Explanation:

Bromine (Br2) attacks the electron rich carbon-carbon triple bond in but-2-yne, an alkyne to form an initial product 2, 3 dibromobut-2-ene; which reacts with excess bromine to form a final product 2,2,3,3 -tetrabromobutane.

The reaction occurs in two steps. On approaching but-2-yne, bromine molecule becomes polarised forming an induced dipole containing a bromonium ion.

Br - Br → Br+ - Br-

The bromonium ion (Br+) formed then attacks the carbon - carbon triple bond to form the initial product

2,3- dibromobut-2-ene

CH2-C≡C-CH2 + Br+ →

CH2 - CBr =CBr-CH2

(2,3- dibromobut-2-ene)

Which in the presence of excess bromine gives the final product

2,2,3,3 - tetrabromobutane.

CH2 - CBr =CBr-CH2 + Br2 →

CH3 -CBr2-CBr2 - CH3

2,2,3,3 - tetrabromobutane.

A visible change in colour from the reddish-brown colour of Bromine to a colourless solution is observed during the reaction.

Answer:

= 35.7 amu

Explanation:

            Isotopic                fractional                                Wt Avg mass

          mass (amu)      abundance (= % / 100)       = Isotopic mass X (= % / 100)

X-35          35                         0.30                                           10.5 amu

X-36          36                         0.70                                           25.2 amu

__________________________________________________________

Average Atomic Mass of Element X = ∑ Wt. Avg. Masses = 35.7 amu

Answer:

The correct options are a, b and d

Explanation:

A catalyst is a substance that increases the rate of a chemical reaction by reducing the activation energy. Le Catelier's  principle explains how a substance or an "action" can affect a reaction in equilibrium.

The principle states that when a change is made to the conditions of a reacting system at equilibrium, the position of the equilibrium moves to counteract the change made. These changes are change in temperature, pressure, volume and/or concentration. These changes will either cause the equilibrium to shift forward or backward.

However, the presence of a catalyst DOES NOT affect a chemical equilibrium/equilibrium constant nor does it affect the reaction quotient because the same amount of reactants and products are available just as in uncatalyzed reaction except that the reaction proceeds faster (which does not affect equilibrium).

The rate of reaction is given as the time required by the reactant to convert into the product. The addition of catalyst increases the rate of reaction, while the reaction quotient and the equilibrium remain unaffected.

A catalyst is a chemical or compound that adds to the reaction and lowers the activation energy by providing an alternative path to the reaction.

The catalyst takes part in the reaction but did not consume in the chemical reaction.

The equilibrium and the reaction quotient are dependent on the conversion of the reactant to the product. The catalyst is not used in the reaction and thus did not affect the reaction quotient or the equilibrium.

Hence, options A, B, and D are correct for the use of catalysts in the chemical reaction.

Learn more about catalysts, here:

https://brainly.com/question/17052831

Answer:

At higher temperatures, particles move faster and collide more, increasing solubility rates.

Agitation increases solubility rates as well, by bringing fresh solvent into contact with the undissolved solute

The smaller the particle size, the higher (faster) solubility rate. Vice versa, the bigger the particle size, the lower (slower) solubility rate.

Explanation:

Answer:

Te

K

Explanation:

Main group elements are elements designated and located on the s-block and p-block on the periodic table. The core group or prominent elements are the elements that make up those groups. These groupings comprise the most naturally rich and abundant elements. The main group elements are those whose characteristics are more predicted as a function of their periodic table location. From the given options, only Te and K are elements of the main-group.

Answer:

double decomposition reaction

Answer:

pKa = 4.89.

Explanation:

We can solve this problem by using the Henderson-Hasselbach equation, which states:

pH = pKa + log [tex]\frac{[A^-]}{[HA]}[/tex]

In this case [A⁻] is the concentration of sodium benzoate and [HA] is the concentration of benzoic acid.

We input the given data:

4.63 = pKa + log [tex]\frac{0.16}{0.29}[/tex]

And solve for pKa:

pKa = 4.89

Answer:

2s2 2p5

Rb < Sr< Sn< Te<I

Explanation:

Electron affinity is the ability of an atom to accept electrons to form negative ions.

Electron affinity is a periodic trend that decreases down the group but increases across the period.

This accounts for the trends observed in the answer. The atom having the electronic configuration, 2s2 2p5 must be a halogen and it exhibits the highest value of electron affinity.

Also, since electron affinity increases across the period, the electron affinities of the elements increases. Therefore, the arrangement of atoms as shown in the answer depends on increasing electron affinity.

Answer:

The reaction is the combustion of methanol (CH3OH).

Write the balanced chemical equation.

Draw Lewis structures for each structure.

Explanation:

The balanced chemical equation for the combustion of methane is shown below:

[tex]2CH_3OH(g)+3O_2(g)->2CO_2(g)+ 4 H_2O(g)[/tex]

Lewis structures of the given molecules are shown below:

Answer:

False

Explanation:

Answer:

Determining the Slope on a p-t Graph. It was learned earlier in Lesson 3 that the slope of the line on a position versus time graph is equal to the velocity of the object. ... If the object has a velocity of 0 m/s, then the slope of the line will be 0 m/s. The slope of the line on a position versus time graph tells it all.

Explanation:

#carryonlearning

Answer:

See explanation and image attached

Explanation:

The reactions of the alkynes involved are shown in the image attached to this answer.

First of all, the reaction of two equivalents of bromine with the alkyne converts it to a saturated compound as shown. One equivalent of HBr converts the alkyne to alkene while excess HBr completely reduces the compound to a saturated compound.

Li/NH3 reduces the alkyne to an alkeneby anti addition to the triple bond.

Reaction of the alkyne with H2O, H2SO4, HgSO4 converts it to an aldehyde as shown.

Answer:

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Answer:

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Answer: A volume of 37.5 mL of 0.300 M NaF would be required to make a 0.0450 M solution of NaF when diluted to 250.0 mL with water.

Explanation:

Given: [tex]M_{1}[/tex] = 0.300 M,   [tex]V_{1}[/tex] = ?

[tex]M_{2}[/tex] = 0.0450 M,    [tex]V_{1}[/tex] = 250.0 mL

Formula used is as follows.

[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]

Substitute the values into above formula as follows.

[tex]M_{1}V_{1} = M_{2}V_{2} \\0.300 M \times V_{1} = 0.0450 M \times 250.0 mL\\V_{1} = 37.5 mL[/tex]

Thus, we can conclude that a volume of 37.5 mL of 0.300 M NaF would be required to make a 0.0450 M solution of NaF when diluted to 250.0 mL with water.

The forces in gases are weaker than forces in solids and liquids.

Because they have more kinetic energy that overcomes the force of attraction.

Answer:

41264 g of CO₂

Explanation:

Combustion reaction is:

C₃H₈ + 5O₂ →  3CO₂  +  4H₂O

1 mol of propane react to 5 moles of oxygen in order to proudce 3 moles of carbon dioxide and 4 moles of water.

In a combustion reaction, our reactant reacts to oxygen and the products are always CO₂ and water.

We have the volume of propane but we need moles of it, so we need to apply density.

Density = mass / volume so mass = density . volume.

Density of propane is: 493 g/L

Mass of propane is 493 g/L . 27.9L = 13754.7 g

We convert mass to moles: 13754.7 g . 1 mol/ 44g = 312.6 moles

According to reaction, 1 mol of propane can produce 3 moles of CO₂

Our 312.6 moles will produce 312.6 . 3 = 937.8 moles

We convert moles to mass: 937.8 mol . 44 g/mol = 41264 g

Answer:

1.6 x 10⁻²⁴ moles

Explanation:

We have 0.984 molecules of F₂. We know that 1 mol is equal to 6.022 x 10²³ molecules. Thus, we have a conversion factor: 1 mol/6.022 x 10²³ molecules

So, we multiply the molecules of F₂ by the conversion factor to calculate the moles:

0.984 molecules x  1 mol/6.022 x 10²³ molecules = 1.6 x 10⁻²⁴ moles

Answer:

15 mL of the solute

Explanation:

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

Solution = 50 mL

Solvent = 35 mL

Solute =?

Solution is simply defined as:

Solution = solute + solvent

With the above formula, we can easily obtain the solute in the solution as follow:

Solution = 50 mL

Solvent = 35 mL

Solute =?

Solution = solute + solvent.

50 = solute + 35

Collect like terms

50 – 35 = solute

15 = solute

Solute = 15 mL

Therefore, 15 mL of the solute is required.