30
Drag each number to the correct location on the equation. Each number can be used more than once, but not all numbers will be used.
Balance the equation with the correct coefficients.
2
3
4
5
SIO2 +
HF → SiF4 +
H2O
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Answer:

A substance that increases H3O+ concentration when it is dissolved in water.

Explanation:

Note that H3O+ and H+ are used quite interchangeably in chemistry.

An acid makes the H+ content higher, thereby decreasing the pH.

Answer:

a

A substance that increases H3O+ concentration when it is dissolved in water.

Explanation:

Answer:

Al2(SO4)3 is the limiting reactant

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3Ba + Al2(SO4)3 → 2Al + 3BaSO4

Next, we shall determine the mass of Ba and the mass of Al2(SO4)3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Al2(SO4)3 = 2x27 + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96]

= 54 + 288 = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

411g of Ba reacted with 342g of Al2(SO4)3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above,

411g of Ba reacted with 342g of Al2(SO4)3.

Therefore, 8g of Ba will react with

= (8 x 342/411 = 6.66g of Al2(SO4)3.

From the calculations made above, we can see that it will take a higher mass of Al2(SO4)3 i.e 6.66g than what was given i.e 2.8g to react completely with 8g of Ba.

Therefore, Al2(SO4)3 is the limiting reactant and Ba is the excess reactant.

Answer:

  C

Explanation:

The molecule has 8 carbon atoms joined by 7 C-C bonds.

The first two diagrams show 6 carbon atoms, not 8.

The last two diagrams show line segments representing C-C bonds. Only choice C shows 7 such segments.

The appropriate choice is C.

Answer:

C.

Explanation:

Answer:

polyethylenes

Explanation:

the plastic bottles used to hold potable water and other drinks are made from polyethylene because, the material is both strong and light.

hope this helped!

Answer: Polyethylenes

Explanation: I got 100% on the test :)

Explanation:

A Bronsted-Lowry base is a substance that accepts a proton in the form of a hydrogen (H) atom.

On the other hand;

Bronsted-Lowry acid is the substance that donates the proton.

HF (aq) + SO32- ⇌ F- + HSO3-

In the forward reaction;

Bronsted-Lowry acid : HF

Bronsted-Lowry base: SO32-

In the backward reaction;

Bronsted-Lowry acid : HSO3-

Bronsted-Lowry base: F-

The conjugate base of HF is F-

The conjugate acid of SO32- is HSO3-

Answer:

well the MF is 224.78 g/mol

Explanation:

just times them all by the molor mass and divide it by 3

Answer:

On the particulate level: 6.02 * 10²³ particles of CO(g) reacts with 6.02 * 10²³ particles of Cl₂(g) to form 6.02 * 10²³ particles of COCl2(g).

On the molar level: 1 mole of CO(g) reacts with 1 mole of Cl2(g) to form 1 mole of COCl₂(g).

Explanation:

The particulate level refers to the microscopic or atomic level of substances. It also involves the ions, protons, neutrons and molecules present in substances.

The molar level refers to the quantitative measure of substances in terms of the mole, where a mole represents the amount of substances containing the Avogadro number of particles which is equal to 6.02 * 10³ particles.

Equation of the reaction: CO(g) + Cl₂(g) ----> COCl₂(g)

From the equation above, I mole of CO gas reacts with 1 mole of Cl₂ gas to produce 1 mole of COCl₂ gas.

Since 1 mole of a substance contains 6.02 * 10²³ particles, on a particulate level, 6.02 * 10²³ particles of CO gas reacts with 6.02 * 10²³ particles of Cl₂ gas to produce 6.02 * 10²³ particles of COCl₂ gas.

Answer:

5.73 g Cu

Explanation:

M(CuSO4) = 159.6 g/mol

M(Al) = 27.0 g/mol

M(Cu) = 63.5 g/mol

14.4 g Al * 1 mol/27.0 g = 0.5333 mol Al

14.4 g CuSO4 * 1 mol/159.6 g = 0.0902 mol CuSO4

                                2 Al         +     3CuSO4 ------->  3Cu + Al2(SO4)3

from reaction           2 mol               3 mol

given                     (0.5333 mol )x       0.0902 mol

needed                   0.0601 mol

x= 2*0.0902/3 = 0.0601 mol   Al

Al is excess, CuSO4 is a limiting reactant.

                               2 Al         +     3CuSO4 ------->  3Cu + Al2(SO4)3

from reaction                               3 mol                  3 mol

given                                            0.0902 mol         x mol

x = 0.0902 mol Cu

0.0902 mol Cu * 1 mol/63.5 g Cu = 5.73 g Cu

Answer:

The answer is "Option b"

Explanation:

In this question first we calculates the moles in F-, HF, and in HCL, which can be defined as follows:

Formula:

[tex]\ Number \ of \ moles\ = \ Molarity \times \ Volume \ in \ litter[/tex]

[tex]\ moles \ in\ F- = 0.100 \ M \times 0.0250 L\\\\[/tex]

                     [tex]=\ 0.0025 \ moles[/tex]

[tex]\ moles \ in \ HF \ = 0.126M \times 0.0250 L[/tex]

                       [tex]= 0.00315 \ moles[/tex]

[tex]\ moles \ in \ HCl = 0.0100M \times 0.00500 L[/tex]

                       [tex]= 0.00005 \ moles[/tex]

[tex]\ Reaction: \\\\F - + H+ \rightarrow HF[/tex]

[tex]\Rightarrow \ moles \ in \ F- = 0.0025 \\\\\Rightarrow \ moles \ in \ H+ = 0.00005 \\\\ \Rightarrow \ moles \ in \ HF = 0.00315\\\\ \ total \ moles = 0.00250 -0.0000500 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00315 + 0.00005\\\\\ total \ moles =0.00245 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00245[/tex]

[tex]\ total \ volume \ in \ the \ solution = \ V = \ 0.0300 L\\\\ after \ addition \ of \ HCl \ the \ concentration \ of \ F- \ = 0.00245\ moles \div V[/tex]

                                                                                [tex]=\frac{ 0.00245 \ moles }{0.0300L}\\\\= \frac{245 \times 10^4}{300 \times 10^5} \\\\= \frac{245}{3000} \\\\ = 0.0817 M[/tex]

Answer:

You would be able to see the ants clearly with the unique body parts.

Explanation:

Convex lens is also known as Converging lens. It helps in converging rays of light to become a principal focus which is usually very clear and visible to the eyes. The converging lens when used to view the ant makes the ant appear very visible and the individual is able to see all the unique body parts of the ant. This type of lens is used in individuals who are longsighted.

Answer:

Explanation:

a. Determine the number of valence electrons for each atom in the molecule

In this case we both atoms are halogens. Therefore we will have 7 electrons for each atom.

b. Draw the Lewis Dot structure

In this case, the formula is [tex]ICl_3[/tex], so the central atom would be "I" and the "Cl" atoms would be placed around "I". See figure 1

c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?)

In this specific case, the "I" atom don't follow the octet rule. We will have an expanded octet for iodine (more than 8 electrons).

d. Show the polarity of each bond and for the molecule by drawing in the dipole +d

The negative dipole would be placed in the atom with higher electronegativity, in this case "Cl". The positive dipole would be placed in the atom with low electronegativity, in this case "I".

I hope it helps!

Explanation:

Benzene is colorless, with a sweet odour.

Color of Bromine is reddish brown .

Hope this helps.

Answer:

It is called 'solar noon'

Explanation:

It's when the sun is at its highest

And also the moment the sun crosses the meridian.

Answer:

The correct answer will be "5.26 × 10⁻⁶".

Explanation:

The given values is:

[tex][H^{+}]=1.9\times 10^{-9} M[/tex]

As we know,

⇒  [tex]pH+pOH=14[/tex]

On taking log, we get

⇒  [tex]-log[H^{+}] + -log[OH^{-}] = 14[/tex]

Now,

Taking "log" as common, we get

⇒  [tex]log[H^{+}][OH^{-}]= -14[/tex]

⇒  [tex][H^{+}][OH^{-}]= 10^{-14}[/tex]

⇒  [tex][OH^{-}]=\frac{10^{-14}}{[H^{+}]}[/tex]

On putting the estimated value of "[tex][H^{+}][/tex]", we get

⇒             [tex]=\frac{10^{-14}}{1.9\times 10^{-9}}[/tex]

⇒             [tex]=5.26\times 10^{-6}[/tex]

Answer:

Na₂SO₄

Explanation:

Barium nitrate, Ba(NO₃)₂ produce precipitate with SO₄²⁻, CO₃²⁻. That means the precipitate could be obtained from Na₂SO₄ and Na₂CO₃.

Also, magnesium nitrate, Mg(NO₃)₂, produce precipitate just with CO₃²⁻. As the unknown solution produce no precipitate, the unknown compound is:

Na₂SO₄

Answer:

0.263M of CH₃COOH is the concentration of the solution.

Explanation:

The reaction of acetic acid (CH₃COOH) with NaOH is:

CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O

1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.

In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:

0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.

As the sample of acetic acid had a volume of 25.0mL = 0.025L:

6.56x10⁻³ moles of CH₃COOH / 0.0250L =

0.263M of CH₃COOH is the concentration of the solution

Answer:

The empirical formula of dimethylmercury is C2H6Hg

Explanation:

Dimethylmercury, as it says in the name, presents not only the mercury metal in its structure (Hg) but also two radical groups called methyl, which is why its name begins with the prefix DI, referring to the fact that there are two methyl.

Answer:

CaSO4

Explanation:

Calcium sulfate (or calcium sulphate) is the inorganic compound with the formula CaSO4 and related hydrates.

Answer:

Explanation:boom

Answer: 1. [tex]H^++OH^-\rightarrow H_2O[/tex]  Lewis acid : [tex]H^+[/tex], Lewis base : [tex]OH^-[/tex]

2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex] Lewis acid : [tex]BCl_3[/tex], Lewis base : [tex]Cl^-[/tex]

3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex] Lewis acid : [tex]K^+[/tex], Lewis base : [tex]H_2O[/tex]

Explanation:

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

1. [tex]H^++OH^-\rightarrow H_2O[/tex]

As [tex]H^+[/tex] gained electrons to complete its octet. Thus it acts as lewis acid.[tex]OH^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]H^+[/tex].

2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex]

As [tex]BCl_3[/tex] is short of two electrons to complete its octet. Thus it acts as lewis acid. [tex]Cl^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]BCl_3[/tex].

3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex]

As [tex]K^+[/tex] is short of electrons to complete its octet. Thus it acts as lewis acid. [tex]H_2O[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]K^+[/tex].

Answer:

A- 25.9 kJ

Explanation:

ΔH of formation is defined as the amount of energy that is involved in the formation of 1 mole of substance.

ΔH of rust is -826kJ/mol, that means when 1 mole of rust is formed, there are released -826kJ.

Moles of 5.00g of Fe₂O₃ (Molar mass: 159.69g/mol) are:

5.00g ₓ (1 mole / 159.69g) = 0.0313 moles of Fe₂O₃.

If 1 mole release -826kJ, 0.0313 moles release:

0.0313 moles ₓ (-826kJ / 1 mole) = -25.9kJ

Thus, heat involved is:

Answer:

1 mole of the gas occupies 22. 4L at O0C at 1atm pressure. Hence, the correct option is C.

Explanation:

Answer:

[tex]Q=-3.11x10^5kJ[/tex]

Explanation:

Hello,

In this case, for the given reaction, we are given the standard enthalpy of reaction per mole of ammonia that is -92.6 kJ, it means, that forming one mole of ammonia will release 92.6 kJ of energy. In such a way, for the formation of 5.71x10⁴ g of ammonia, the following amount of heat will be released:

[tex]Q=5.71x10^4gNH_3*\frac{1molNH_3}{17gNH_3}*-92.6\frac{kJ}{molNH_3}\\ \\Q=-3.11x10^5kJ[/tex]

Best regards.

The amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ

We'll begin by calculating the number of mole in 5.71×10⁴ g of NH₃

Mass of NH₃ = 5.71×10⁴ g

Molar mass of NH₃ = 14 + (3×1) = 17 g/mol

Mole = mass / molar mass

Mole of NH₃ = 5.71×10⁴ / 17

N₂(g) + 3H₂(g) —> 2NH₃(g) ΔH°rxn = −92.6 kJ

Since reaction took place at standard conditions, it means:

1 moles of NH₃ required −92.6 kJ

Therefore,

3358.82 moles of NH₃ will require = 3358.82 × –92.6 = –311026.732 KJ

Thus, the amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ

Learn more: https://brainly.com/question/17332795

You need to know the amount of heat generated by the combustion reaction.

Assuming propane as fuel, you can use thiis data:

C3H8(g)+5O2(g)---3CO2(g)+4H2O(g) ΔH= -2217 KJ

So when 3 moles of CO2 is emmitted 2217 kJ of heat is produced.

The molar wegiht of CO2 is 12 g/mol + 2 * 16 g/mol = 44 g/mol.

Then 3 mol * 44 g / mol = 132 g of CO2 are produced with 2217 kJ of heat.

Now you have to calculate how much energy you need to produce if only 12% is abosrbed by the pork

Energy absorbed by the pork = 12% *  total energy =>

total energy = energy absorbed by the pork / 0.12 = 1700 kJ / 0.12 = 14,166.67 kJ.

Now, state the proportion:

132 g CO2 / 2217 kJ = x / 14,166.7 kJ =>

x = 14,166.67 * 132 / 2217 = 843.48 g CO2.

Answer: 843 g of CO2

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

Answer:

an atom with a half-filled subshell - hydrogen

an atom with a completely filled outer shell - argon

an atom with its outer electrons occupying a half-filled subshell and a filled subshell- copper

Explanation:

The outermost shell or the valence shell of the atom is the last shell in the atom. Chemical reactions occur at this outer most shell. The number of electrons on the outermost shell of an atom determines the group to which it belongs in the periodic table as well as its chemical properties.

Hydrogen has a half filled 1s sublevel. Only one electron is present in this sublevel.

Let us consider argon

1s2 2s2 2p6 3s2 3p6

The outermost ns and np levels are completely filled. Thus the outermost shell is completely filled.

In the last case; let us look at the electronic configuration of nitrogen;

1s2 2s2 2p3

The outermost 2p subshell is exactly half filled while the 2s sublevel is fully filled. The outermost shell of nitrogen is made up of 2s2 and 2p3 sublevels.

Answer:

Less than 0.033 M:

[tex][Z]_{eq}=2.4x10^{-3}M[/tex]

Explanation:

Hello,

In this case, the described equilibrium is:

[tex]2A+B\rightarrow 2Z[/tex]

Thus, the law of mass action is:

[tex]K=\frac{[Z]^2}{[A]^2[B]}=0.43[/tex]

Nevertheless, given the initial concentration of Z that is 0.033 M, we should invert the equilibrium since the reaction will move leftwards:

[tex]\frac{1}{K}=\frac{[A]^2[B]}{[Z]^2}=\frac{1}{0.43}=2.33[/tex]

Know, by introducing the change [tex]x[/tex] due to the reaction extent, we can write:

[tex]2.33=\frac{(2x)^2*x}{(0.033-2x)^2}[/tex]

Which has the following solution:

[tex]x_1=2.29M\\x_2=0.0181M\\x_3=0.0153M[/tex]

But the correct solution is [tex]x_3=0.0152M[/tex] since the other solutions make the equilibrium concentration of Z negative which is not possible. In such a way, its concentration at equilibrium is:

[tex][Z]_{eq}=0.033M-2(0.0153M)[/tex]

[tex][Z]_{eq}=2.4x10^{-3}M[/tex]

Which is clearly less than 0.033 M since the addition of a product shift the reaction leftwards in order to reestablish equilibrium (Le Chatelier's principle).

Regards.

Answer:

30.0 g.

Explanation:

Hello,

In this case, for us to round the given number off to three significant figures, we firstly realize it has initially five significant figures. Thus, cutting at the third digit, which is the second nine, we will have 29.9 g, nonetheless, as a five is after such nine, we should round the nine to ten, so the result is 30.0 g.

Best regards.

Answer:

[tex]1.66~V[/tex]

Explanation:

We have to start with the half-reactions for both ions:

[tex]Zn^+^2~+2e^-~->Zn[/tex] V= -0.76

[tex]Ag^+~e^-~->~Ag[/tex] V= +0.80

If we want a spontaneous reaction (galvanic cell) we have to flip the first reaction, so:

[tex]Zn~->~Zn^+^2~+2e^-~[/tex] V= +0.76

[tex]Ag^+~+~e^-~->~Ag[/tex] V= +0.80

If we want to calculate ºE we have to add the two values, so:

ºE=0.76+0.80 = 1.56 V

Now, we have different concentrations. So, if we want to calculate E we have to use the nerts equation:

[tex]E=ºE~+~\frac{0.059}{n}LogQ[/tex]

On this case, Q is equal to:

[tex]Q=\frac{[Zn^+^2]}{[Ag^+]^2}[/tex]

Because the total reaction is:

[tex]Zn~+~2Ag^+~->~Zn^+^2~+~2Ag[/tex]

So, the value of "Q" is:

[tex]Q=\frac{[0.052 M]}{[0.0042]^2}=2947.84[/tex]

Now, we can plug all the values in the equation (n=2, because the amount of electrons transferred is 2). So:

[tex]E=1.56~V~+~\frac{0.059}{2}Log(2947.84)=1.66~V[/tex]

I hope it helps!