Draw the Lewis structure for the polyatomic formate anion. Be sure to include all resonance structures that satisfy the octet rule.

Answer:

answer number C is the correct answer for this

She goes away.

Negative

She doesn't go away.

Answer:

Kp=Kc *(RT)+-3

Explanation:

The relation between Kp and Kc is given below:

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant , R = 0.082057 L atm.mol⁻¹K⁻¹

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

Δn = (0)-(2+1) = -3

Thus, Kp is:

Kp=Kc *(RT)+-3

Answer:

Gases act more ideal at lower pressure beacuse the attractive forces between molecules will decrease or become less significant compared to the empty space between them.

Explanation:

Generally, a gas behaves more like an ideal gas at higher temperature and lower pressure as the potential energy due to intermolecular forces becomes less significant compared with the particles "kinetic energy" and the size of the molecules become less significant compared to the empty space between them.

Attractive forces between molecules, decrease the pressure of a reak gas, slowing the molecules and reducing collisions with the walls.The higher the value of a gas, the greater the attraction between molecules and the more easily the gas will compress.

answered by: John Glenly Pillazo Mahusay

The internal energy of system is raised by 3 times

Answer:

6 mins

Explanation:

The time taken for Ar to effuse can be obtained as follow:

Time for Br₂ (t₁) = 12 mins

Molar mass of Br₂ (M₁) = 2 × 80 = 160 g/mol

Molar mass of Ar (M₂) = 40 g/mol

Time for Ar (t₂) =?

t₂/t₁= √(M₂/M₁)

t₂ / 12 = √(40/160)

Cross multiply

t₂ = 12 × √(40/160)

t₂ = 12 × 0.5

t₂ = 6 mins

Therefore, it will take 6 mins for the same amount of Ar to effused out.

Answer:

2.9 × 10²⁴ atoms Cu; 3.3 × 10²¹ atoms C; 1.44 × 10²⁵ atoms Hg; 9.63 × 10²² atoms Na

Explanation:

To convert moles to atoms, we need a conversion factor: Avogadro's number. There are 6.02 × 10²³ atoms in 1 mole of atoms.

4.9 mol Cu × 6.02 × 10²³ atoms Cu/1 mol Cu = 2.9 × 10²⁴ atoms Cu

5.4 × 10⁻³ mol C × 6.02 × 10²³ atoms C/1 mol C = 3.3 × 10²¹ atoms C

24.0 mol Hg × 6.02 × 10²³ atoms Hg/1 mol Hg = 1.44 × 10²⁵ atoms Hg

0.160 mol Na × 6.02 × 10²³ atoms Na/1 mol Na = 9.63 × 10²² atoms Na

Answer:

.

Explanation:

Explanation:

Answer:

hydrogen bonds

Explanation:

similarly CH3OH also has a OH group and H hydrogen thus it will also form h-bonding.

Answer:

isn't it 0.998

Explanation:

cause 0.998 is in the same grams / mole. I don't get

A calorimeter has been filled with 75 mL of water. The number of grams of water is 74.85 grams.

A calorimeter is an instrument that is used to measure calorimetry. It is used to measure the heat of chemical reactions and physical and chemical changes.

Mass is the quantity of matter in a physical body. Volume is the space occupied by a three-dimensional object. Mass is the quantity of matter in a physical body. The product of the compound's molar mass and the substance's moles are defined as mass.

Given, that the volume of the water is 75 mL

The temperature of the water is 25 °C

The density of water is 0.998 g/mL

To calculate the mass, the volume is multiplied by density.

Mass = volume x density

putting the values in the formula

75 x  0.998 = 74.85 grams

Thus, the grams of water that have at 25°C is 74.85 grams.

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Answer:

Draw the skeletal structure for: (E)-hept-5-en-2-one

Explanation:

The root word hept indicates that the given compound has seven carons in its longest chain.

-en- primary suffix indicates that the compound has one double bond in it.

2-one indicates that the compound has -C=O bond in the second carbon.

The prefix (E) indicates that the highest priority groups are on the opposite direction of the double bond.

The structure of the given molecule is:

Answer:

An unknown compound's 2,4-DNP product melting range is between 144-146'C. It does not give a silver mirror on the Tollens test and is slow to react to the chromic acid test. What would this compound be

Explanation:

From the given data it is clear that the unknown compound gives positive test with 2,4-DNP reagent.

That means it has a carbonyl group.Either aldehyde or ketone.

It does not give a silver mirror on the Tollens test and is slow to react to the chromic acid test.

That means aldehyde is absent.

So, the other carbonyl group that is ketone is present in the given unknown compound.

Answer:

i think 1. law of muliple proportion

Explanation:

please like

Answer:

hey friend

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Answer:

Filtration is a technique used as a remedy to separate mixes

Explanation:

If you have a sodium, then you will explore that salt dissolves but the sand is still the same.

If the salt in the resin water solution scanners, the sand remains the residue and passes through the filter paper.

All you have to do now is pleasant the salty water so that the water can evaporate, leaving the salt behind.

Answer:

1920 Joules

Explanation:

* means multiply

^ means exponent

kinetic energy is in Joules or J

kinetic energy = 1/2 * mass * (velocity)^2

kinetic energy = 1/2 * 60 * (8)^2

kinetic energy = 1/2 * 60 * (8*8)

kinetic energy = 1/2 * 60 * (64)

kinetic energy = 1920 Joules or 1920 J

Answer:

KE = 1920 J

General Formulas and Concepts:

Energy

Kinetic Energy Formula: [tex]\displaystyle KE = \frac{1}{2}mv^2[/tex]

Explanation:

Step 1: Define

[Given] m = 60 kg

[Given] v = 8 m/s

Step 2: Find KE

Topic: AP Physics 1 Algebra-Based

Unit: Energy

Answer:

HCN, weak acid

H⁺, Br⁻, strong acid

Explanation:

Hydrocyanic acid is a weak acid, according to the following equation.

HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)

Thus, it should be written in the undissociated form (HCN).

Hydrobromic acid is a strong acid, according to the following equation.

HBr(aq) ⇒ H⁺(aq) + Br⁻(aq)

Thus, it should be written in the ionic form (H⁺, Br⁻).

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:

Higher than the actual value

Explanation:

Titration is a volumetric process in which a known volume of solution is dispensed from a burette to react with a known volume of solution in a conical flask.

When acid-base titration is carried out in such a way that the base is in the burette and the acid is in the conical flask and drops of the base adhere to the inner walls of the flask, but do not actually mix with the solution, the calculated acid concentration would be higher than the actual value.

This is because;

From CA= CBVBnA/VAnB

When VB(volume of base) that reacted is lower than the actual volume recorded, then the calculated volume of CA(concentration of acid) is much higher than the actual value since drops of the base adhere to the inner walls of the flask.

Answer:

d.

Explanation:

H2 + Cl2 = 2HCL

From the equation 2g hydrogen combine with 71g of chlorine.

So 35.5 g Cl2 combines with 1g of H2

There are 100g of Cl2 so this will, by proportion, react with 100/35.5 g hydrogen.

This is 2.8 g hydrogen so the mass of HCl formed = 102.8 g.

The true statement is that d. between 100 and 190 g of HCl is produced.

To find mass of HCL:

H2 + Cl2 = 2HCL

From the equation, 2g of hydrogen combines with 71g of chlorine.

So 35.5 g Cl2 combines with 1g of H2

There are 100g of Cl2 so this will, by proportion, react with 100/35.5 g of hydrogen.

This is 2.8 g hydrogen so the mass of HCl formed = 102.8 g.

Hydrogen chloride may be formed by the direct combination of chlorine (Cl2) gas and hydrogen (H2) gas.

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Answer: The fractional abundance of lighter isotope is 0.518

Explanation:

Average atomic weight is the sum of the masses of the individual isotopes each multiplied by its fractional abundance. The equation used is:[tex]\text{Average atomic weight}=\sum_{i=1}^{n}\text{(Atomic mass of isotope)}_i\times \text{(Fractional abundance)}_i[/tex]           ......(1)

Let the fractional abundance of Ag-107 isotope be 'x'

Atomic mass = 106.90509 amu

Fractional abundance = x

Atomic mass = 108.9047 amu

Fractional abundance = (1 - x)

Average atomic mass of silver = 107.8682 amu

Plugging values in equation 1:

[tex]107.8682=(106.90509 \times x) + (108.9047 \times (1-x))\\\\107.8682=106.90509x+108.9047-108.9047x\\\\1.99961x=1.0365\\\\x=0.518[/tex]

Fractional abundance of Ag-107 isotope (lighter) = x = 0.518

Hence, the fractional abundance of lighter isotope is 0.518

Answer:

Human life has become very simple by using organic chemicals. The importance of organic chemicals in the daily life and industrial area can be explained as follows. (i) Food Vitamins, proteins, sugar, flour, fats etc. ... (ii) Agriculture is an important place for organic chemical for the growth of agricultural production.

Answer:

6.023*10^23........ .......

Answer:

the house found in a very cold are Igloos or the ice house

Answer:

This is an oxidation-reduction (redox) reaction:

2 Ni0 - 4 e- → 2 NiII

(oxidation)

2 O0 + 4 e- → 2 O-II

(reduction)

Ni is a reducing agent, O2 is an oxidizing agent.

Answer:

[tex]\mathbf{\lambda \simeq 3.039 \times 10^{-8} \ m}[/tex]

Explanation:

For a hydrogen-like atom, the spectral line wavelength can be computed by using the formula:

[tex]\bar v = Z^2 R_H \Big(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\Big)[/tex]

where:

emitted radiation of the wavenumber  [tex]\bar v[/tex] = ???

atomic no of helium Z = 2

Rydberg's constant [tex]R_H = 1.097*10^7 \ m^{-1}[/tex]

the initial energy of  the principal quantum [tex]n_1[/tex] = 2

the initial energy of  the principal quantum [tex]n_1[/tex] = 2

Now, the emitted radiation of the wavenumber can be computed as:

[tex]\bar v = (2)^2 (1.097*10^7 \ m^{-1} ) \Big(\dfrac{1}{1^2}-\dfrac{1}{2^2}\Big)[/tex]

[tex]\bar v = 3.291 \times 10^ 7/m[/tex]

Now, the wavelength for the transition can be computed by using the relation between the wavelength λ and the emitted radiation of the wavenumber  [tex]\bar v[/tex], which is:

[tex]\bar v = \dfrac{1}{\lambda}[/tex]

[tex]\lambda = \dfrac{1}{\bar v}[/tex]

[tex]\lambda = \dfrac{1}{3.291 \times 10^{7}}\times \dfrac{m}{1}[/tex]

[tex]\mathbf{\lambda =3.03859 \times 10^{-8} \ m}[/tex]

[tex]\mathbf{\lambda \simeq 3.039 \times 10^{-8} \ m}[/tex]