NaCl is ___ formula of common salt
A. Empirical formula
B. Molecular formula
C. Both A and B
D. None of them​

Answers

Answer 1
A because sodium chloride is not a molecule

Related Questions

Analysis of an unknown substance showed that it has a high boiling point and is brittle. It is an insulator as a solid but conducts electricity when melted. Which of the following substances would have those characteristics?
a. HCl
b. Al
c. SiF4
d. KBr
e. I2

Answers

Answer:

The correct option is D (KBr)

Explanation:

Potassium bromide (KBr) is a typical example of an IONIC CRYSTAL. Positive and negative ions are arranged in a regular pattern to give a giant crystal lattice in an ionic solid.

Ionic crystals are hard and have high melting points because the electrostatic forces holding the ions are strong. In the solid state, ionic compounds are poor conductors of electricity because the ions are held rigidly in place and so cannot moves about.

When melted or when dissolved in water, the ions are free to move about and the ionic substances become good conductors of electricity.

Ionic crystals are also BRITTLE. When struck, they tend to shatter because as planes of ions slip by one another, they pass from a condition of mutual attraction to one of mutual repulsion.

Determine whether the stopcock should be completely open, partially open, or completely closed for each activity involved with titration.
Close to the calculated endpoint of a titration ________
At the beginning of a titration _______
Filling the buret with titrant ________
Conditioning the buret with titrant _______

Answers

Answer:

Close to the calculated endpoint of a titration - Partially open

At the beginning of a titration - Completely open

Filling the buret with titrant - Completely closed

Conditioning the buret with the titrant - Completely closed

Explanation:

'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.

As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.

Which compound contains both sigma and pi bonds... HCCl3, H2CO, H2S, or HBr?

Answers

Answer:

H2CO

Explanation:

Becuase it has 2 sigma bonds plus one pi bond and one sigma bond that consitute the double bond between C and O.

Answer:

B. H2CO

Explanation:

13. What would you expect the pH of an aqueous solution of tertiary bromide in water to be (acidic, neutral, or basic)

Answers

Answer:

oshfjidgshsjdh

Explanation:

918474828

The length of a covalent bond depends upon the size of the atoms and the bond order.

a. True
b. False

Answers

Answer:

True

Explanation:

The length of covalent bond depends upon the size of atoms and the bond order.

Determine the kinds of intermolecular forces that are present in each element or compound. Part A KrKr Check all that apply. Check all that apply. dispersion forces dipole-dipole forces hydrogen bonding

Answers

Answer:

The kinds of intermolecular forces that are present in each element Kr-Kr.

Explanation:

Since Kr is an inert gas and in atomic form only it is highly stable.

So, Kr gas does not form molecules.

Between the atoms of inert gas, there exist London dispersion forces.

Hence, the intermolecular forces that are present between Kr-Kr atoms is London dispersion forces.

Determine whether each phrase describes carboxylic acids or esters.

a. Usually have a sour odor _____________
b. Notable for their pleasant fragrances ___________
c. Do not form hydrogen bonds amongst themselves and have higher vapor pressure _____________
d. Their reactions with base are known as neutralizations ____________
e. Their reactions with base are known as saponifications _____________

Answers

the right answer is b

A 3.0-liter sample of an ideal gas is at a pressure of 2.5 atm at 15oC. (i) How many moles of gas are in the sample? (ii) If the volume does not change, what is the pressure of the gas when the temperature is 50oC?

Answers

Explanation:

here are the answers. Note that because the pressure is constant, you can use Gay Lussac's formula

The energy levels of hydrogenlike one-electron ions of atomic number Z differ from those of hydrogen by a factor of Z^2. Predict the wavelength of the 2s--->1s transition in He+.

Answers

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]

When 2.50 moles of oxygen reacts with excess hydrogen gas how many grams of water is
produced?
2 H2 + O2--> 2 H20

Answers

Answer:

90grams

Explanation:

The reaction given in this question is as follows:

2H2 + O2 → 2H20

Based on this equation, 1 mole of oxygen gas (O2) produces 2 moles of water (H2O)

Hence, 2.50moles of oxygen gas will react with excess hydrogen gas to produce (2.5 × 2) = 5.0moles of water.

Using mole = mass/molar mass

Molar mass of water (H2O) = 1(2) + 16

= 18g/mol

5 = mass/18

mass = 18 × 5

mass of H2O = 90grams

if a=1/2(a+b)h,express a in terms of A,b and h. pls solve with step by step ​

Answers

Answer:

[tex] a = \frac {2A - bh}{h} [/tex]

Explanation:

Given the following mathematical expression;

A = ½(a + b)h

To make a the subject of formula (express a in terms of A, b and h);

First of all, we would cross-multiply;

2A = (a + b)h

Opening the bracket, we have;

2A = ah + bh

Rearranging the mathematical expression, we have;

ah = 2A - bh

[tex] a = \frac {2A - bh}{h} [/tex]

the question is in the attachment

Answers

Answer:

About redox reaction which of the given statements are true?

Explanation:

Redox reaction is the one in which both oxidation and reduction reactions take place simultaneously.

For example:

[tex]C(s)+O_2(g)->CO_2(g)[/tex]

In this reaction, carbon undergoes oxidation and oxygen undergoes reduction simultaneously.

During this reaction, mutual exchange of electrosn take place between the oxidant and the reductant.

Among the given options,

Option B. electrons are transferred

and

option C.They include both oxidation and reduction takes place are the correct answers.

Ice and water constitute a system:
A Chemically heterogeneous and physically heterogeneous
B Chemically homogeneous and physically heterogeneous
C Chemically homogeneous and physically homogeneous
D Chemically heterogeneous and physically homogeneous
E None of the above

Answers

Answer:

B

Explanation:

It's the same substance but in different states.

HETEROGENEOUS mixtures contain substances that are

not uniform in composition. The parts in the mixture can be separated by physical means.

The metal tantalum becomes superconducting at temperatures below 4.483 K. Calculate the temperature at which tantalum becomes superconducting in degrees Celsius.

Answers

Answer:

The correct answer is "-268.667°C".

Explanation:

Given:

Temperature,

= 4.483 K (below)

Now,

The formula of temperature conversion will be:

⇒ [tex]T(^{\circ} C)=T(K)-273.15[/tex]

By putting the values, we get

⇒            [tex]=4.483-273.15[/tex]

⇒            [tex]=-268.667^{\circ} C[/tex]

Thus the above is the correct answer.

Underneath the ocean floor, methane can exist as a solid known as methane hydrate. Methane hydrate forms when large amounts of methane become trapped in the crystal structure of water, which develops into a solid similar to ice. A scientist working for a deep water drilling company is studying the properties of the three different phases of methane: methane hydrate, liquified natural gas, and natural gas. Which of the properties is true

Answers

Answer:

Methane is present in solid, liquid and gaseous form.

Explanation:

Methane hydrate is present in solid state when the hydrogen-bonded water and methane gas come into contact at high pressures and low temperatures in the deep oceans while on the other hand, methane which is present inside the earth surface in gaseous form due to non-availability of water that can combine with methane. Methane is also found in liquid form when it is cool with high pressure and low temperature.

Draw the major product that is obtained when (2S,3S)-2-Bromo-3-phenylbutane is treated with sodium ethoxide.

Answers

Answer:

Explanation:

The mechanism of the reaction is shown in the diagram below. From the reaction, when (2S,3S)-2-Bromo-3-phenylbutane undergoes a reaction with sodium ethoxide (ETONa), the E2 elimination reaction is put into place. Here, the H and the leaving group are antiperiplanar to one another and the reaction mechanism proceeds to form an isomeric (E)-2-phenyl-2butane as the major product.

The products obtained from hydroboration-oxidation of cis-2-butene are identical to the products obtained from hydroboration-oxidation of trans-2-butene. Draw the products and explain why the configuration of the starting alkene is not relevant in this case.

Answers

Answer:

a) Attached below

b)  The presence of racemic mixture found as product in both  cases shows that   products are identical ( i.e. they have same configuration

Explanation:

Diagrams of the products obtained from hydroboration-oxidation of cis-2-butene , hydroboration-oxidation of trans-2-butene.

attached below

The presence of racemic mixture found as product in both  cases shows that   products are identical ( i.e. they have same configuration )

How do you predict the geometrical shape of NH3 on VSEPR model

Answers

Answer:

NH3 Ammonia

Explanation:

Ammonia has 4 regions of electron density around the central nitrogen atom (3 bonds and one lone pair). These are arranged in a tetrahedral shape. The resulting molecular shape is trigonal pyramidal with H-N-H angles of 106.7°.

The Li2+ ion is very similar to the hydrogen atom, in that it has one electron and energy levels similar
to the hydrogen atom. However, the relation = − (
1

2 −
1

2) cannot be used for this ion but
rather the relation = −2+ (
1

2 −
1

2) where the constant 2+=1.96x10-17J.
Use this relation to determine the third ionization energy, which is energy required to remove the last
electron from a Li2+ ion in kJ/mol, if the ion starts off in the ground state (Li2+ → Li3+
+ e-
).

Answers

Hhhhccffdfgcfghbvvbh


The speed of light_____ meters per second

Answers

299,792,458 meters per second.

A student observes a chemical
reaction where two liquids are mixed together. After the liquids are mixed the beaker feels cold to touch. This reactions is an
example of a
reaction.
Synthesis
Combustion
Exothermic
Endothermic

Answers

Answer:

Endothermic

Explanation:

Endothermic reaction is one in which the enthalpy increases. What this implies is that, it is a closed system which absorbs heat from its surroundings and thus after reaction become cooler than prior to the reaction.

In this case, the mixture of the two liquids feel colder to touch. Thus, it is an endothermic reaction from the definition earlier given.

what is the qualitative analysis of (nh4) 2co3 using NaOH, HCL, BaCL2, and AgNO3​

Answers

Answer:

qualatatiev is fs-hj_jakakak

Why the catalytic and optical properties of nanomaterial are different from bulk material

Answers

Answer:

The material properties of nanostructures are different from the bulk due to the high surface area over volume ratio and possible appearance of quantum effects at the nanoscale. ... Yu; they found that the structural distortions on the quantum dots depend both on the kind of dopant and on the size of the dots.

Explanation:

hope it helps

determine the number of atoms of H in 35.0 grams of C2H4O2

Answers

Answer:

1.40x10^24 atoms of H

Explanation:

define saturated and unsaturated fats​

Answers

Explanation:

Saturated fats are defined as the fat where fatty acid chains contain only single bonds.

For example, stearic acid, palmitic acid etc.

Unsaturated fats are defined as the fat where fatty acids contain one or more number of double bonds on the carbon atoms.

For example, oleic acid, palmitoleic acid etc.

22. Glycerol, C3H8O3, is a substance used extensively in the manufacture of cosmetics, foodstuffs, antifreeze, and plastics. Glycerol is a water-soluble liquid with a density of 1.2656 g/mL at 15 0C. Calculate the molarity of a solution of glycerol made by dissolving 50.000 mL glycerol at 15 0C in enough water to make 250.00 mL of solution. Show work.

Answers

Answer:

2.75M

Explanation:

Density = mass/volume

Density of glycerol (C3H8O3) at 15°C is given as 1.2656 g/mL.

Volume of glycerol at this same temperature before dissolving in water = 50mL

Hence, using Density = mass/volume

1.2656 = mass/50

mass = 1.2656g/mL × 50mL

mass = 63.28g

Molarity of solution = number of moles (n) ÷ volume (V)

Volume of glycerol dissolved in water = 250mL = 0.250L

number of moles = mass/molar mass

Molar mass of C3H8O3 = 12(3) + 1(8) + 16(3)

= 36 + 8 + 48

= 44 + 48

= 92g/mol

n = 63.28g ÷ 92g/mol

n = 0.688mol

* Molarity = n/V

Molarity = 0.688/0.250

Molarity = 2.75M

Classify each of the following as a strong acid or a weak acid and indicate how each should be written in aqueous solution. Classify ... In solution this acid should be written as: weak 1. hydrocyanic acid H3O CN- _______ 2. hydrobromic acid

Answers

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⁻).

A 0.50 mol sample of COBr2 is transferred to a 9.50-L flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species.

Answers

Answer:

Equlibrium concentration for each species ae as follows:

[CO] = 0.043 mol/L

[Br₂] = 0.043 mol/L

[COBr₂] = 0.01 mol/L

Explanation:

Let take a look at the chemical equation taking place at equilibrium

COBr2(g) ⇄ CO(g) + Br2(g)

The concentration of COBr2 i.e.

[COBr2] = no of moles/volume

= 0.50 mol/9.50 L

[COBr2] = 0.0530 mol/L

At standard conditions

Kc for COBr2 = 0.190

Now, the ICE table for the above reaction can be computed as follows:

                       COBr2(g)              ⇄       CO(g)         +           Br2(g)

Initial               0.053                                   0                           0

Change             -x                                       +x                          +x

Equilibrium     (0.053 - x)                              x                           x

[tex]\mathsf{K_c = \dfrac{[CO][Br_2]}{[COBr_2]}}[/tex]

[tex]K_c = \dfrac{(x) (x)}{(0,053 -x)}[/tex]

[tex]0.190= \dfrac{x^2}{(0.053 -x)}[/tex]

x² = 0.190(0.053 - x)

x² = 0.01007 - 0.190x

x² + 0.190x - 0.01007 = 0

Using quadratic formula:

x ≅ 0.043 mol/L

SInce: x = [CO][Br₂] = 0.043 mol/L

[COBr₂] = 0.053 - x

[COBr₂] = 0.053 - 0.043 mol/L

[COBr₂] = 0.01 mol/L

Ammonia and oxygen react to form nitrogen monoxide and water. Construct your own balanced equation to determine the amount of NO and H2O that would form when 2.78 mol NH3 and 5.19 mol O2 react.

Answers

Answer:

The amount of NO formed s 2.78 moles or 83.4 grams

The amount of H2O formed is 4.17 moles or 75.1 grams

Explanation:

Step 1: Data given

Ammonia = NH3

Oxygen = O2

nitrogen monoxide = NO

water = H2O

Number of moles NH3 = 2.78 moles

Number of O2 = 5.19 moles

Step 2: The balanced reaction

4NH3 + 5O2 → 4NO + 6H2O

Step 3: Calculate moles of products

For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

NH3 is the limiting reactant

All the NH3 will react. There will be 0 moles of NH3 left

For 4 moles NH3 we need 5 moles O2

For 2.78 moles NH3 we need 5/4 * 2.78 = 3.475 moles

There will be left 5.19 - 3.475 = 1.715 moles O2

For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 2.78 moles NH3 we'll have 2.78 moles NO and 6/4 * 2.78 = 4.17 moles H2O

Step 4: Calculate mass of NO and H2O

Mass = moles * molar mass

Mass NO = 2.78 moles * 30.01 g/mol

Mass NO = 83.43 grams

Mass H2O = 4.17 moles * 18.02 g/mol

Mass H2O = 75.14 grams

The amount of NO formed s 2.78 moles or 83.4 grams

The amount of H2O formed is 4.17 moles or 75.1 grams

What effect would a decrease in volume have on pressure, assuming that temperature (T) and moles of gas (n) are kept constant

Answers

Answer:

Pressure increases

Explanation:

Boyle's law states that; '' the volume of a given mass of ideal gas is inversely proportional to its pressure at constant temperature.

Hence, when the volume of a given mass of ideal gas is decreased, the molecules of the gas come closer together so they collide with each other and the walls of the container more frequently.

This implies that the pressure of the gas increases as volume decreases in accordance with Boyle's law.

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