How to calculate the actual volume (ml) of water removed from the burette with water
density 1 g/ml:
i. 5mL of water removed

Answer:

dispersion force

Explanation:

Answer:

108.6 g

Explanation:

First we use the PV=nRT formula to calculate the number of nitrogen moles:

Inputting the data:

Then we convert 2.5 moles of N₂ into moles of NaN₃, using the stoichiometric coefficients of the balanced reaction:

Finally we convert 1.67 moles of NaN₃ into grams, using its molar mass:

Answer:

C₂H₆(g) + 2.5 O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)

Explanation:

Let's consider the unbalanced equation in which gaseous ethane reacts with gaseous oxygen to form carbon monoxide gas and water vapor. This is an incomplete combustion reaction.

C₂H₆(g) + O₂(g) ⇒ CO(g) + H₂O(g)

We will balance it using the trial and error method.

First, we will balance C atoms by multiplying CO by 2 and H atoms by multiplying H₂O by 3.

C₂H₆(g) + O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)

Then, we get the balanced equation by multiplying O₂ by 2.5.

C₂H₆(g) + 2.5 O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)

Answer:

[tex]M_{a}[/tex] = 0.104 m

Explanation:

This expression is used to determine either the mass or volume of an acid or a base used during titration process.

So that;

[tex]M_{a}[/tex][tex]V_{a}[/tex] = [tex]M_{b}[/tex][tex]V_{b}[/tex]

[tex]M_{a}[/tex] is the mass of the acid

[tex]V_{a}[/tex] is the volume of the acid

[tex]M_{b}[/tex] is the mass of the base

[tex]V_{b}[/tex] is the volume of the base

Given that:

[tex]M_{a}[/tex] x 5.0 m l = 5.2 ml x 0.10 m

[tex]M_{a}[/tex] x 5.0 = 0.52 m

[tex]M_{a}[/tex] = [tex]\frac{0.52}{5.0}[/tex]

      = 0.104

[tex]M_{a}[/tex] = 0.104 m

The mass of the acid used is 0.104 m.

Answer:

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

Answer:

C₃H₆O₃

Explanation:

To solve this question we need to find, as first, the moles of each atom in order to find empirical formula (Simplest whole-number ratio of atoms present in a molecule).

With the molar mass of the substance and the empirical formula we can find the molecular formula as follows:

Moles C -Molar mass:12.0g/mol-

0.24g * (1mol/12.0g) = 0.020 moles C

Moles H = Mass H because molar mass = 1g/mol:

0.040 moles H

Moles O -Molar mass: 16g/mol-

Mass O: 0.60g - 0.24g - 0.040g = 0.32g O

0.32g O * (1mol/16g) = 0.020 moles O

Ratio of atoms (Dividing in moles of C: Lower number of moles):

C = 0.020 moles C / 0.020 moles C = 1

H = 0.040 moles H / 0.020 moles C = 2

O = 0.020 moles O / 0.020 moles C = 1

Empirical formula:

CH₂O.

Molar mass CH2O:

12g/mol + 2*1g/mol + 16g/mol = 30g/mol

As molecular formula has a molar mass 3 times higher than empirical formula, the molecular formula is 3 times empirical formula:

The molecular formula of the organic acid would be C3H6O3

Molecular formula = [empirical formula]n

Where n = molar mass/mass of empirical formula

Empirical formula

C = 0.24/12 = 0.02

H = 0.040/1 = 0.04

O = 0.6 - (0.24+0.04) = 0.32/16 = 0.02

Divide by the smallest

C = 1

H = 2

O = 1

Empirical formula = CH2O

Empirical formula mass = 12 + 2 + 16 = 30

n = 90/30 = 3

Molecular formula = [CH2O]3

                               = C3H6O3

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The electronegativity increases across the period and decreases down the group. Thus, option B is correct.

Electronegativity can be defined as the tendency of an atom to gain or attract an electron. The electronegativity has been dependent on the size of the atom, as well as the atomic number and valence electrons.

The atom with the requirement of a less number of atoms to complete its octet can easily gain the electron and thereby have high electronegativity. The atomic size also plays a role in the electronegativity of the atom.

The atom with a bigger size has the lesser force of attraction from the nucleus and thus has difficulty attracting the electron, however, the smaller size atom can easily attract the electron with the attraction force from the nucleus.

Thus, the elements with smaller sizes and a high number of valence electrons are more electronegative. In the periodic table, on moving from left to right the valence electrons increase, thus the electronegativity increases.

On moving down the group, the element size increase, thus the electronegativity decreases down the group.

The electronegativity increases across the period and decreases down the group. Thus, option B is correct.

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The formula that we can use to calculate energy released when a mass of liquid freezes is Grams liquid x mol/g x delta Hfusion.

Stoichiometry deals with the relationship between mass and moles or volume which can be used to make calculations involving chemical reactions. Most chemical calculations involving stoichiometry use the mole concept.

Hence, the formula used to obtain the energy released when a mass of liquid freezes is Grams liquid x mol/g x delta Hfusion.

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This question is incomplete, the complete question is;

A certain liquid X has a normal boiling point of 150.4 °C and a molar boiling point elevation constant kb is 0.60 °Ckgmol⁻¹.

A solution is prepared by dissolving some urea (NH22CO) in 750 g of X. This solution boils at 150.9 °C . Calculate the mass of urea that was dissolved. Round your answer to 3 significant digits.

Answer:

the mass of urea that was dissolved is 37.5 g

Explanation:

Given the data in the question;

normal boiling point of X; Tb⁰ = 150.4 °C

boiling point of solution Tb = 150.9 °C

Change in boiling point Δt = Tb - Tb⁰  = 150.9 °C - 150.4 °C = 0.5 °C

Kb = 0.6 °C.kg.mol⁻¹

V = 750 g

Now, we know that

Δt = Kb × molality

so

0.5 = 0.6 × molality

molality = 0.5 / 0.6

molality = 0.833

we know that molar mass of urea is 60 g/mol

so

molality = mass × 1000 / molar mass × volume( g )

we substitute

0.833 = ( mass × 1000 ) / ( 60 × 750 )

0.833 = ( mass × 1000 ) / 45000

0.833 × 45000 = mass × 1000

mass = ( 0.833 × 45000 ) / 1000

mass = 37485 / 1000

mass = 37.485 ≈ 37.5 g   { 3 significance figure }

Therefore, the mass of urea that was dissolved is 37.5 g

Answer:

CHCI3

Explanation:

there are no free CI ions hence it doesnt precipitate with an aqeous solution of AQUO33

Answer:

hope it helps much as you can

Answer:

2

Explanation:

2 alkenes are formed by E2 elimination of HBr from 3-bromo -3, 4 - dimethylhexane using a strong base such as sodium methoxide

Answer:

c

Explanation:

Answer:

1.88 A

Explanation:

Let's consider the reduction of copper in an electrolytic cell.

Cu²⁺ + 2 e⁻ ⇒ Cu

We can calculate the charge used to deposit 12.3 g of Cu using the following relations.

The charge used is:

[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]

We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.

5.50 h × 3600 s/1 h = 1.98 × 10⁴ s

The current used is:

I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A

Answer:

Which piece of equipment would be BEST to measure 5 mL of a liquid?

A 50 mL beaker filled halfway between 0 and 10 mL markings

A 50 mL Erlenmeyer flask filled halfway between 0 and 10 ml marking

A 10 mL graduated cylinder

A disposable dropper

A 100 mL graduated cylinder

Explanation:

To measure 5 mL of a liquid, the best equipment is

a 10 mL graduated cylinder.

The remaining apparatus do not give an accurate measurement because readings are not marked on them.

Answer:

There are 1.195 calories in 5 J.

Explanation:

The relation between calorie and Joules is as follow :

4.184 J = 1 calorie

or

1 J = (1/4.184) calorie

Let the taken number is 5 J

So,

5 J = (5/4.184) calorie

5 J = 1.195 calorie

So, there are 1.195 calories in 5 J.

Answer:

False

Explanation:

An object with a density greater than 1.00g/mL (greater than the density of water) will sink. An object with a density less than the density of water, will float.

If the water has a density of 1.00 g/mL. If you put an object that has a density of 0.79 g/mL into water, it will sink to the bottom, this statement is false.

The density of an actual content is its mass per unit volume. The most common symbol for density is d, but the Latin letter D can also be used.

Three of an object's most fundamental properties are mass, volume, and density. Mass describes how heavy something is, volume describes its size, and density is defined as mass divided by volume.

The density of something is a measure of how heavy it is in relation to its size. When an artifact is more dense than water, it plunges; when an object is less dense than water, it floats.

Density is a property of a substance that is independent of the amount of substance.

As in the given scenario, water is having density 1 g/mL and object in having density less then it so it will float on water.

Thus, the given statement is false as the material will not sink, rather it will float on water.

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#SPJ2

Answer:

10.10

Explanation:

Step 1: Write the basic dissociation reaction for pyridine

C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq)      Kb = 1.9 × 10⁻⁹

Step 2: Calculate [OH⁻]

For a weak base, we will use the following expression.

[OH⁻] = √(Cb × Kb) = √(9.2 × 1.9 × 10⁻⁹) = 1.3 × 10⁻⁴ M

Step 3: Calculate pOH

We will use the definition of pOH.

pOH = -log [OH⁻] = -log 1.3 × 10⁻⁴ = 3.9

Step 4: Calculate pH

We will use the following expression.

pH = 14 - pOH = 14 - 3.9 = 10.10

Answer: The given compounds are arranged according to decreasing boiling point as [tex]CH_{3}OH > CH_{3}Cl > CH_{4}[/tex].

Explanation:

The temperature at which vapor pressure of a substance becomes equal to the atmospheric pressure is called boiling point.

Stronger is the intermolecular forces present the atoms of a molecule more heat will be required by it to break the bond between its atoms. Hence, more will the boiling point of the molecule.

In [tex]CH_{3}OH[/tex] (methanol), there is hydrogen bonding present which is a stronger force. So, it will have highest boiling point as compared to [tex]CH_{3}Cl[/tex] and [tex]CH_{4}[/tex].

In [tex]CH_{3}Cl[/tex] (chloroform), there is more electronegative atom attached (Cl) is attached to less electronegative atom (C and H). So, electrons are more pulled towards the chlorine atom. So, boiling point of  [tex]CH_{3}Cl[/tex] is more than methane [tex](CH_{4})[/tex].

Thus, we can conclude that given compounds are arranged according to decreasing boiling point as [tex]CH_{3}OH > CH_{3}Cl > CH_{4}[/tex].

Answer:

0.0238M SbCl3, 1.07M H+, 1.14M Cl-

Explanation:

The total volume of the solution is:

4mL + 5.00mL + 12.0mL = 21mL

As the volume of the SbCl3 is 5.00mL, the dilution factor is:

21mL / 5.00mL = 4.2 times

The concentration of SbCl3 is:

0.10M SbCl3 / 4.2 times = 0.0238M SbCl3

The concentration of H+ = [HCl]:

4.5M / 4.2 times = 1.07M H+

The initial concentration of Cl- is:

3 times SbCl3 + HCl = 0.10M*3 + 4.5M =

3 times SbCl3 because 1 mole of SbCl3 contains 3 moles of Cl-

4.8M Cl- / 4.2 times = 1.14M Cl-

The direction of heat flow is increased which means blocks temperature is higher and hotter than it was before

Answer:

Hence among the options a and c, option d is that the correct answer because it has rock bottom energy ( as n value increases, energy decreases as energy levels come closer).

Explanation:

The relation between energy and wavelength is:

[tex]\Lambda = hC/E[/tex]  

From this equation, it's clear that wavelength and energy are inversely proportional to every other. The Lower the energy of a specific transition, the longest will the wavelength be of that specific transition.  

Among the given options, options b and d are often ruled out, since those transitions produce to release of a photon because it is coming down from an excited state.

Answer:

Explanation:

Target organ toxins are chemicals that can cause adverse effects or disease states manifested in specific organs of the body. Toxins do not affect all organs in the body to the same extent due to their different cell structures.

Answer:

True

Explanation:

The valence orbitals of boron are 2s2 2p1. We have to recall that all the valence orbitals whether full or empty are involved in the formation of molecular orbitals.

The number of molecular orbitals formed is equal to the number of atomic orbitals that are combined.

Since there are two valence orbitals and there is only one p orbital among the valence orbitals, it is true that only one p orbital is needed to form molecular orbitals in boron.

Answer:

a)calculated molarity of NaOH would be lower

b) calculated molarity of NaOH would be lower

c) calculated molarity of NaOH would be lower

d) calculated molarity of NaOH would be unaffected

Explanation:

Let us recall that the reaction of NaOH and HCl is as follows;

NaOH(aq) + HCl(aq) ----> NaCl(aq) + H2O(l)

Since the reaction is 1:1, when the number of moles of HCl reacting with NaOH is low due to dilution, the calculated molarity of NaOH also becomes less than it's accurate value.

When 40mL of water is added to the titration flask rather than 25ml of water, the acid is more dilute hence less number of moles of acid than necessary reacts with the base thereby yielding a less than accurate value of the molarity of NaOH.

If the burette wet with water is not rinsed with NaOH solution, the concentration of the NaOH in the burette decreases due to dilution with water and a less than accuracy value is calculated for the molarity of NaOH.

If five drops of phenolphthalein is used instead of one or two drops, there is no qualms since enough phenolphthalein may be added to ensure that a sharp end point is obtained.

Answer:

See explanation

Explanation:

Liquids are known to have a definite volumes but not a definite shape. This means that a liquid takes on the volume of the container in which it is found.

Hence, when a liquid is transferred from one container to another, the volume of the liquid remains the same but the shape of the liquid changes.

This happens when the two containers do not possess the same shape.

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.

Answer:

In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 the Tert butyl occupies the axial and the bromine occupies equatorial positions. Compound 1 reacts faster than compound 2.

Explanation:

In cyclic organic compounds, substituents may occupy the axial or equatorial positions. The axial positions are aligned parallel to the symmetry axis of the ring while the equatorial positions are around the plane of the ring.

Bulky substituents have more room in the equatorial than in the axial position. This means that compound 1 is more stable than compound 2.

This is clear on the basis of stability of the molecules because compound 1 will react faster than compound 2 since the bulky tertiary butyl group in compound 1 occupy equatorial and not axial positions.