b) What is the change in entropy of the reaction if ΔH° = -3.2 kJ mol-1?

Answer:

[tex]n_1=4[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lambda=489.2 nm =>4.86*10^{-7}[/tex]

nf level= Balmer series

nf level= 2

Generally the equation for Wavelength is mathematically given by

[tex]\frac{1}{\lambda}=R[\frac{1}{nf^2}-\frac{1}{n_1^2}][/tex]

Where

[tex]R=Rydberg Constant[/tex]

[tex]R=1.097*10^7[/tex]

Therefore

[tex]\frac{1}{4.86*10^{-7}}=1.097*10^7[\frac{1}{2^2}-\frac{1}{n_1^2}][/tex]

[tex]n_1=4.0021[/tex]

[tex]n_1=4[/tex]

Explanation:

try 96 if that's not right let me know and I'll try to fix it

Answer:24

Explanation:

Answer:

H₂O

Explanation:

Two molecules of Hydrogen and one molecules of Oxygen, when mixed, create H₂O, or water. There is no scientific name for H₂O due to it's common name. It is just refereed to as "water" or H₂O.

Answer:

heat

Explanation:

because it's the cause of change

Answer:

heat

Explanation:

because it is a natural factor that causes the change in Earth's system

Answer:

The correct answer is "654.54 nm".

Explanation:

According to the question,

⇒ [tex]\frac{1}{\lambda}= Rh(\frac{1}{n1^2} -\frac{1}{n2^2} )[/tex]

By substituting the values, we get

       [tex]=1.1\times 10^7(\frac{1}{4} -\frac{1}{9} )[/tex]

       [tex]=1.1\times 10^7(\frac{9-4}{36} )[/tex]

       [tex]=1.1\times 10^7(\frac{5}{36} )[/tex]

       [tex]=654.54\ nm[/tex]

Thus the above is the right solution.

Answer:A) V = 16.892 ml

Explanation:

M1 * V1 = M2 * V2

14.8 M * V1 =0.250 M * 1000 ml

V1 = 16.892 ml

a. The volume of 16.89 milliliters of the stock solution of 14.8 M  should be diluted to make 1000.0 mL of 0.250 M.

b. The concentration of the final solution is 0.296 M.

The concentration or the volume of the concentrated or dilute solution can be calculated by using the equation:

M₁V₁ = M₂V₂

where M₁ and V₁ are the concentration and volume of the concentrated solution respectively and M₂ and V₂ are the concentration and volume of the dilute solution.

A stock solution is a solution that has a high concentration and that will be diluted to a low concentration by the addition of water in it.

Given, a stock solution of concentration, M₁ = 14.8 M

The concentration of the diluted solution, M₂ = 0.250 M

The volume of diluted solution, V₂  = 1000ml

Substitute the value of the molarity and volume in equation (1):

(14.8)× (V₁) = (1000) × (0.250)

V₁ = 16.89 ml

Similarly, for part (b): M₁ = 14.8 M, V₁ = 10 ml and V₂  = 0.5L = 500 ml

(14.8)× (10) = (500) × (M₂)

M₂ = 0.296 M

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Answer:

As a giant star.

Explanation:

A

Answer:

Explanation:

Start with a balanced equation.

2H2 + O2 → 2H2O

Calculate mole H2 using the formula: n = m/M, where:

n = mole

m = mass (g)

M = molar mass (g/mol)

Calculate molar mass of H2.

M H2 = 2 × 1.008 g/mol = 2.016 g/mol

Calculate moles H2.

n H2 = 4.16 g H2/2.016 g/mol = 2.063 mol H2

Calculate moles H2O by multiplying moles H2 by the mole ratio between H2O and H2 from the balanced equation, so that moles H2 cancel.

2.063 mol H2 × (2 mol H2O/2 mol H2) = 2.063 mol H2O

The mass of water will be calculated by rearranging the n = m/M formula to isolate m;

m = n × M

Calculate the molar mass H2O.

M H2O = (2 × 1.008 g/mol) + (1 × 15.999 g/mol) = 18.015 g/mol

Calculate the mass H2O.

m = n × M = 2.063 mol H2O × 18.015 g/mol = 37.2 g H2O

4.16 g H2 with excess O2 will produce 37.2 g H2O.

a) HNO₂ + C₂H₇NO → N₂ + C₂H₆O + H₂O

b) H₃O⁺ + F⁻ → HF + H₂O

[tex]\large\color{lime}\boxed{\colorbox{black}{Answer : - }}[/tex]

a) HNO₂ + C₂H₇NO → N₂ + C₂H₆O + H₂O

b) H₃O⁺ + F⁻ → HF + H₂O

Answer:

the number is 7

Explanation:

"Three times" means multiply by 3

"Difference" means subtract

"Sum" means add

3(x - 7) = 23 - (3x + 2)

3x - 21 = 23 - 3x - 2

3x - 21 = 21 - 3x

6x = 42

x = 7

Answer:

79 g/mol

Explanation:

Mass of unknown metal deposited = 3.137 g - 1.4 g = 1.737 g

Number of moles of metal deposited = 0.022 moles

Since;

Number of moles = reacting mass/molar mass

Molar mass = reacting mass/number of moles

Molar mass = 1.737 g/0.022 moles

Molar mass= 79 g/mol

Answer:

C. yes, because it is universally applies to all objects

the effects are: temperature rises, water shortages, and increased fire threats

Answer:

35

Explanation:

is the answer for your question

Answer:

[tex]\boxed {\boxed {\sf 5.56 \times 10^{23} \ molecules \ CH_4}}[/tex]

Explanation:

We are asked to find how many molecules of methane are in 14.8 grams of the substance.

First, we convert grams to moles. We use the molar mass, or the mass of 1 mole of a substance. These values are equivalent to the atomic masses found on the Periodic Table, however the units are grams per mole instead of atomic mass units.

We are given the compound methane, or CH₄. Look up the molar mass of the individual elements (carbon and hydrogen).

Check the formula for subscripts. Hydrogen (H) has a subscript of 4, so there are 4 moles of hydrogen in 1 mole of methane. We must multiply hydrogen's molar mass by 4, then add carbon's molar mass.

Now we use dimensional analysis to convert. To do this, we set up a ratio using the molar mass.

[tex]\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}[/tex]

Since we are converting 14.8 grams of methane to moles, we multiply by this value.

[tex]14.8 \ g \ CH_4 *\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}[/tex]

Flip the ratio so the units of grams of methane cancel.

[tex]14.8 \ g \ CH_4 *\frac{ 1 \ mol \ CH_4} {16.043 \ g \ CH_4 }[/tex]

[tex]14.8 *\frac{ 1 \ mol \ CH_4} {16.043}[/tex]

[tex]\frac {14.8}{16.043} \ mol \ CH_4= 0.9225207256 \ mol \ CH_4[/tex]

Next, we convert moles to molecules. We use Avogadro's Number or 6.022  × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are moles of methane. Set up another ratio using Avogadro's Number.

[tex]\frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]

Multiply by the number of moles we calculated.

[tex]0.9225207256\ mol \ CH_4 * \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]

The units of moles of methane cancel.

[tex]0.9225207256* \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 }[/tex]

[tex]5.55541981 \times 10^{23} \ molecules \ CH_4[/tex]

The original measurement of grams (14.8) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 5 in the hundredth place up to a 6.

[tex]5.56 \times 10^{23} \ molecules \ CH_4[/tex]

14.8 grams of methane is equal to approximately 5.56 × 10²³ molecules of methane.

Answer:

Answer is B) 20.0 moles

Explanation:

From the equation,

1 mole of Fe2O3 = 2 moles of Al

therefore 10.0 moles of Fe2O3 = 10×2

= 20.0 moles.

Answer:

a. True.

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of ethanol:

[tex]{ \bf{CH _{3} CH_{2}OH \: \: \frac{Ag/O_{2} }{500 \degree C} > \: \:CH _{3} CHO}}[/tex]

[tex]{ \sf{CH _{3} CHO \: \: is \: ethanal}} [/tex]

By ozonolysis:

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

Answer:

A. True

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of ethanol:

By ozonolysis:

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

Answer:

It is prepared small amounts of hydrogen cloride for uses in the lab.

It can be  "generated in an HCl generator by dehydrating hydrochloric acid with either sulfuric acid or anhydrous calcium chloride."

Answer:

The fluoride ion is capable of reacting, to a small extent, with water, accepting a proton. The fluoride ion is acting as a weak Brønsted-Lowry base. The hydroxide ion that is produced as a result of the above reaction makes the solution slightly basic.

So

we get a basic solution when you dissolve NaF in water.

Na is a alkaline earth metal

Metallic compound dissolved in water acts as base

Also there is another reason

Na F is ioniC

Fluorine is known as having highest electron affinity in world

It can accept a line pair of OH-

So it's Bronsted Lowry base .

Hence its basic

QUESTION:- Manganese-55 has _____neutrons.

OPTIONS :-

A. 55

B. 30

C. 25

ANSWER:- NUMBER OF NEUTRONS IS EQUAL TO THE DIFFERENCE BETWEEN THE MASS IF THE ATOM AND ATOMIC NUMBER

SO DIFFERENCE IS EQUAL TO :- 55-25 = 30 NEUTRONS.

SO THERE IS 30 NEUTRONS IN SINGLE ATOM OF THE MANGANESE-55 ATOM.

Answer:

the mass of an atom is the sum of proton and neutron which are both concentrated in nocleus of an atom. from the question the mass is given as 55 and the proton is 25.

Answer:

Part A

HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)

Part B

ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)

Explanation:

The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;

RCOOH + NaOH ----> RCOONa + H2O

We have to note the fact that the net ionic reaction still remains;

H^+(aq) + OH^-(aq) ---> H2O(l)

In both cases, the reaction can occur and they actually do occur as written.

Answer:

dsgsdfd

Explanation:

Answer:

However, there has to be 2 electrons on the first shell, and 8 on the others.

Explanation:

Hope this helps :)

Answer:

D. beta particle

Explanation:

Pu has 94 protons which is only one more than Np (93), that means beta decay. Atomic mass stays the same in beta decay. Beta is very small particle so atomic mass stays at 239.

The given reaction is an example of beta decay where the atomic number of the nuclei increases by one unit and mass number remains unchanged. Therefore, the particle X is a beta particle or an electron.

Heavy unstable isotopes of atoms undergoes nuclear decay by the emission of charged particles such as alpha or beta particle. In alpha decay, the helium nuclei is emitted and in beta decay, electrons are emitted.

In alpha decay, the mass number of the nuclei decreases by 4 units and atomic number by 2 units. In beta decay, the mass number does not change but the atomic number increases by one unit.

Neptunium undergoes beta decay and forms the plutonium nuclei. Thus X is a beta particle. The reaction is written as follows:

[tex]\rm _{93} ^{239}Np \rightarrow _{94}^{239}Pu + _{-1}^{0}e[/tex]

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Answer:

17

Explanation:

Step 1: Calculate the needed concentrations

[A]i = 1.00 mol/5.00 L = 0.200 M

[B]i = 1.80 mol/5.00 L = 0.360 M

[B]e = 1.00 mol/5.00 L = 0.200 M

Step 2: Make an ICE chart

        A(aq) + 2 B(aq) ⇄ C(aq)

I       0.200    0.360        0

C        -x           -2x         +x

E     0.200-x  0.360-2x   x

Then,

[B]e = 0.360-2x = 0.200

x = 0.0800

The concentrations at equilibrium are:

[A]e = 0.200-0.0800 = 0.120 M

[B]e = 0.200 M

[C]e = 0.0800 M

Step 3: Calculate the concentration equilibrium constant (K)

K = [C] / [A] × [B]²

K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17

Answer:

D. An alkene ​

Explanation:

because Ethane is C2H4

Answer:

It's a alkANE. C.

Explanation:

The easiest way to memorize this is to look at the endings. Substances that end in -ANE are alkANEs. Substances that end in -ENE are alkENEs. Substances that end in -YNE are alkYNEs.

Answer:

Element

Element : A pure substance composed of the same type of atom throughout. Compound : A substance made of two or more elements that are chemically combined in fixed amounts.

Explanation:

Answer:

B

Explanation:

In order to create spontaneity, an endothermic process has to occur along with positive entropy and high temperature

Answer:

The lewis structure for NO₃⁻ is shown in the attachment below

For the central nitrogen atom:

The number of lone pairs = 0

The number of single bonds = 2  

The number of double bonds= 1

Explanation:

The lewis structure for NO₃⁻ is shown in the attachment below.

From the Lewis structure

For the central nitrogen atom:

The number of lone pairs = 0

The number of single bonds = 2  

The number of double bonds= 1