Use average bond enthalpies (linked above) to calculate the enthalpy change for the following gas-phase reaction. CO(g) + Cl2(g) COCl2(g) To analyze the reaction, first draw Lew To analyze the reaction, first draw Lewis structures for all reactant and product molecules. ΔHreaction = ------ kJ Draw the reaction using separate sketchers for each species. Separate multiple reactants and/or products using the + sign from the drop-down arrow. Separate reactants from products using the → symbol from the drop-down menu. Remember to include nonbonding valence electrons in your Lewis structures. To analyze the reaction, first draw Lewis structures for all reactant and product molecules. Draw the reaction using separate sketchers for each species. Separate multiple reactants and/or products using the + sign from the drop-down arrow. Separate reactants from products using the → symbol from the drop-down menu. Remember to include nonbonding valence electrons in your Lewis structures. To analyze the reaction, first draw Lewis structures for all reactant and product molecules. Draw the reaction using separate sketchers for each species. Separate multiple reactants and/or products using the + sign from the drop-down arrow. Separate reactants from products using the → symbol from the drop-down menu. Remember to include nonbonding valence electrons in your Lewis structures.

Answer:

C) Organic Weathering

Explanation:

Heat and Cold

And if some people start reporting and making fun of me in this answer don't listen to them.

They are hunting me down.

Answer:

71.6 g of Ru(CO)₅ is the maximum mass that can be formed.

The limiting reactant is Ag

Explanation:

The reaction is:

RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)

Firstly we determine the moles of each reactant:

169 g . 1mol /481.77g = 0.351 moles of RuI₃

58g . 1mol /28g = 2.07 moles of CO

96.2g . 1mol/ 107.87g = 0.892 moles

Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.

3 moles of Ag react to 1 mol of RuI₃

Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles

We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.

1 mol of RuI₃ react to 3 moles of Ag

Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles

It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.

5 moles of CO react to 3 moles of Ag

Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.

This calculate confirms the theory.

Now, we determine the maximum mass of Ru(CO)₅

3 moles of of Ag can produce 1 mol of Ru(CO)₅

Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles

We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g

Answer: last one is right

Explanation: atoms of same element have always same number of protons but may have several isotopes which have different number of neutrons.

Answer:

yee I'm excellent in maths and science

The name of the given alkene is 2-methyl-1-propene. The correct option is A.

Alkenes are inorganic compounds. They are the series of hydrocarbons, that are formed by carbon-carbon double bonds. To find out alkenes, you have to see that the number of hydrogen atoms is always double the number of carbons.

Propene is an unsaturated compound, which is the second-simplest compound, and it contains a single double bond, and it is a colorless gas and has a texture of petroleum jelly.

CH₃C = CH₂

I

CH₃

Here, the three CH₃ are shown in the diagram, so it will be the methyl because it contains 3 CH₂, and now see the propene means one double bond and four carbon atoms. So it will be the 2-methyl-1-propene.

Thus, the correct option is A. 2-methyl-1-propene.

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Answer:

Explanation: Sodium Nitroprusside, whose molecular formula is  • 2H2O, and whose molecular weight is 297.95. Dry sodium nitroprusside is a reddish-brown powder, soluble in water.

Answer:

72.4%

Explanation:

Step 1: Write the balanced equation

2 Cu + S ⇒ Cu₂S

Step 2: Calculate the theoretical yield, in moles, from 0.0970 moles of Cu

The molar ratio of Cu to Cu₂S is 2:1.

0.0970 mol Cu × 1 mol Cu₂S/2 mol Cu = 0.0485 mol Cu₂S

Step 3, Convert the theoretical yield to mass

The molar mass of Cu₂S is 159.16 g/mol.

0.0485 mol × 159.16 g/mol = 7.72 g

Step 4: Calculate the percent yield

We will use the following expression.

%yield = experimental yield/theoretical yield × 100%

%yield = 5.59 g/7.72 g × 100% = 72.4%

Answer:

[tex]B_3H_6N_3[/tex]

Explanation:

Hello there!

In this case, since the mass percentages in a compound which is wanted to know the molecular formula, can be assumed to be the masses, we first need to compute the moles they have in the formula unit:

[tex]n_B=40.29gB*\frac{1molB}{10.811gB} =3.73molB\\\\n_H=7.51gH*\frac{1molH}{1.01gH} =7.44molH\\\\n_N=52.20gN*\frac{1molN}{14.01gN} =3.73molN[/tex]

Next, we divide each moles by the fewest ones (3.73 mol) in order to find the subscript in the empirical formula first:

[tex]B:\frac{3.73}{3.73}=1 \\\\H:\frac{7.44}{3.73}=2\\\\N:\frac{3.73}{3.73}=1[/tex]

Then, the empirical formula is BH2N whose molar mass is 26.83 g/mol, so the ratio of molecular to empirical is 80.50/26.83=3; therefore, the molecular formula is three times the empirical one:

[tex]B_3H_6N_3[/tex]

Best regards!

Answer:

b

Explanation:

i know

I haven't done these in some time, so I'm not sure if they are 100% right.

1) 6.3 moles of H2( 2 mol of NH3 / 3  mol of H2)= 4.2 mol of NH3

6.3 moles of H2( 2 mol of NH3/ 3 mol of H2)(17.04 g of NH3/1 mol NH3)= 71.57 g of NH3

2) 2.5 moles of N2(2 mol of NH3/1 mol of N2)= 5 moles of NH3

2.25 moles of N2(2 mol of NH3/ 1 mol of N2)(6.02x10^23 particles/ 1 mol of NH3)= 3.01x10^24 particles of NH3

3) 425 g of NH3(1 mol of NH3/17.04 g NH3)= 24.9 moles of NH3

425 g of NH3(1 mol of NH3/17.04 g of NH3)(1 mol of N2/2 mol of NH3)(28.02 g of N2/1 mol N2)= 349 g of N2

425 g of NH3(1 mol of NH3/17.04 g of NH3)(1 mol of N2/2 mol of NH3)= 12.5 mol of N2

4) 10 moles NH3(3 moles of H2/2 moles of NH3)= 15 moles H2

10 moles NH3(3 mol of H2/2 mol of NH3)(2.02 g of H2/1 mol of H2)= 30.3 g of H2

30.3 g = .0303 liters of H2

Answer:

Performance Tasks 188; Instructional Videos 75

Explanation:

Answer:

N2 + H20

Explanation:

chemical equation balancer

Answer:

F

Explanation:

Answer: Balanced molecular equation :

[tex]2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)[/tex]

Total ionic equation:

[tex]6Na^+(aq)+3PO_4^{2-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)[/tex]  

The net ionic equation:

[tex]2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)[/tex]

Explanation:

Complete ionic equation : In complete ionic equation, all the substances that are strong electrolyte are present in an aqueous state as ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When sodium phosphate and zinc acetate then it gives zinc phosphate and sodium acetate as  product.

The balanced molecular equation will be,

[tex]2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)[/tex]

The total ionic equation in separated aqueous solution will be,

[tex]6Na^+(aq)+2PO_4^{3-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)[/tex]

In this equation, and  are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)[/tex]

Answer:

1

Explanation:

Let's write the equation first:

[tex]C_{4}H_{10} + \frac{13}{2} O_{2} -> 4 CO_2 +5 H_{2}O[/tex]

So 1 mole of butane reacts to give 4moles of Carbon diOxide.

thus , 0.25 mole of butane will react to give = 4 * 0.25= 1 mole Carbon diOxide

Ozone gas can be smelled in the air after a thunderstorm as a result of  reactions between oxides of nitrogen and volatile organic compounds.

Ozone gas can be produced due to chemical reactions between oxides of nitrogen and volatile organic compounds. This reaction occurs when pollututed gas is emitted from vehicles react and other activities with each other in the presence of sunlight.

So we can canclude that Ozone gas can be smelled in the air after a thunderstorm as a result of  reactions between oxides of nitrogen and volatile organic co

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Answer:

Mx will be less than My.

Explanation:

The correct option is - Mx will be less than My.

Because, liquid X have been higher viscosity and vapor pressure than liquid B.

Vapor pressure of liquid A is greater, so the amount of liquid X will vaporized greater and H2 gas has been displaced greater from the beaker as compared to liquid Y.

The amount of H2 in liquid Y remains greater in beaker as compared to liquid X.

So, Mx will be less than My.

Answer:

c substance

Explanation:

Matter is never destroyed or created in chemical reactions. The particles of one substance are rearranged to form a new substance. The same number of particles that exist before the reaction exist after the reaction.

Answer:

c is correct option

according to the question.

Scientists classify organisms into three domains and six kingdoms. In Fungi kingdom are yeasts classified because they are heterotrophic. Therefore, option C is correct.

An organism is referred to be a heterotroph if it is unable to manufacture food on its own and must obtain it from other sources of organic carbon, primarily plant or animal materials.

All fungi are heterotrophic, which means they draw their energy from other living things.

Since they share some traits with fungi, such as the presence of chitin in their cell walls, extracellular digestion, saprophytic feeding, asexual reproduction by budding, and the ability to store food as glycogen, yeast cells are categorised as fungi.

Thus, option C is correct.

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Answer:

Answer: pH = 2.72

Explanation:

Calculate the pH of 0.010 M HNO2 solution. The K, for HNO2 is 4.6 x 104 

Answer:

1. 2Al + Cl2 = Al2Cl2

2 TiCl4 + 2Na = Ti + 2NaCl2

3. H2O2 = H2O + O2

4. Na2S + 2HCl = H2S + 2NaCl

5. Mg(OH)2 + 2HCl = MgCl2 + 2H2O

1. 3O2 = 2O3

Answer:

c. 10–14[H3O+]

Explanation:

On a calculator, calculate 10-8.34, or "inverse" log ( - 8.34). Example: What is the pOH of a solution that has a hydroxide ion concentration of 4.82 x 10-5 M? The hydroxide ion concentration can be found from the pOH by the reverse mathematical operation employed to find the pOH.

Answer:

I honestly don't know but I hope you get more points! ;D

Explanation:

Answer:

A) 1.72 atm

Explanation:

Dalton's Law of Partial Pressures

[tex]P_{t} = P_{1} + P_{2} + P_{3} ....[/tex]

4.43 atm = 1.23 atm + 1.48 atm + x

x = 4.43 atm - (1.23 atm + 1.48 atm)

x = 1.72 atm

Answer:

[tex]1s^2\, 2s^2[/tex].

Explanation:

Electron orbitals in an atom (e.g., [tex]1s[/tex]) are denoted with:

There are two aspects to consider when finding the electron configuration of an atom:

The [tex]s[/tex] orbital in each shell could hold up to [tex]2 \times 1 = 2[/tex] electrons (one [tex]s\![/tex] orbital per shell, with up to two electrons.)

The [tex]p[/tex] orbitals in each shell could hold up to [tex]2 \times 3 = 6[/tex] electrons (three [tex]p\![/tex] orbitals per shell, with up to two electrons in each orbital.)

The [tex]d[/tex] orbitals in each main shell could hold up to [tex]2 \times 5 = 10[/tex] electrons (five [tex]d\![/tex] orbitals per shell, with up to two electrons in each orbital.)

Refer to the order in which the orbitals are filled (Aufbau principle.)

All four electrons of Beryllium are thus assigned to the [tex]1s[/tex] and [tex]2s[/tex] orbitals. In a ground-state Beryllium atom, orbitals [tex]2p[/tex] and beyond would contain no electrons.

Notation:

Write the non-empty orbitals in the order by which they are filled:

[tex]1s^2\, 2s^2[/tex].