An x-ray has a wavelength
of 2.2 x 10-11 m. What is
the frequency of the x-ray? Help asap please

S denotes the ''number of shared electron pairs by an atom'', Hence, Silicon shares its 4 electrons with 4 Cl-atoms. Thus the Value of S is 4 in SiCl₄ .It is also known as covalency

The number of covalent bonds that a particular atom can make with other atoms in forming a molecule.

Covalent bond id formed by the sharing of electron between two atoms.

Hence, Silicon shares its 4 electrons with 4 Cl-atoms. Thus the Value of S is 4 in SiCl₄ . It is also known as covalency

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Answer:

25.91 L

Explanation:

P1V1= P2V2

748 (26.5) =  765 ( V2)      V2 = 25.91 LITERS  

The molar mass of the gas, given that 3.25 g of the gas occupied 2.56 L is 30.66g/mol

To obtain the molar mass of the gas, we shall first obtain the number of mole of the gas. This can be obtained as follow:

PV = nRT

0.95 × 2.56 = n × 0.0821 × 280

Divide both sides by (0.0821 × 280)

n = (0.95 × 2.56) / (0.0821 × 280)

n = 0.106 mole

Haven obtain the mole of the gas, we shall determine the molar mass of the gas as follow:

Molar mass = mass / mole

Molar mass of gas = 3.25 / 0.106

Molar mass of gas = 30.66g/mol

Thus, the molar mass of the gas is 30.66g/mol

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Now that both the air and the carbon have had an eight of electrons, this Lewis electron configuration is acceptable. The C comprises four bonding pairs, while the O has two binding pair and two lone pairs.

A molecule is an object comprised of two or more elements that are linked together chemically; the precise amount of atomic nuclei that make up a molecule is known. One hydrogen atom gets joined with one chlorine atom to form the molecule HCl(g), for instance. It is referred to as a diatomic molecule and has only two atoms.

Real molecules don't overlap and have a defined size. Thus, there is a potent, close-ranged repulsion between them.

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Answer:

The activation energy for the forward reaction is the amount of free energy that must be added to go from the energy level of the reactants to the energy level of the transition state. ... Once a reactant molecule absorbs enough energy to reach the transition state, it can proceed through the remainder of the reaction.

Explanation:

Answer: Activation energy is how much energy it takes for a reaction to occur.

Explanation:

The pH of the solution is 9.95.

The reaction that explains this is given as NH4+ + OH- → NH3 + H2O

The pH of a solution with equal amounts of NH4Cl and NH3 is 12.21.

To solve this problem, we need to first write out the balanced chemical equation for the reaction between NH4Cl and NH3:

NH4Cl + NH3 ⇌ NH4+ + NH2Cl

Next, we need to write out the equilibrium expression:

Kb = [NH4+][OH-]/[NH3]

Since we are given the concentration of NH4Cl and NH3, we can use the initial concentrations to calculate the equilibrium concentrations of NH4+, NH2Cl, and NH3:

[NH4+] = 0.1 M × (0.1 L / 0.18 L) = 0.056 M

[NH2Cl] = 0.056 M

[NH3] = 0.2 M × (0.08 L / 0.18 L) = 0.089 M

Using the equilibrium expression and the value of Kb, we can solve for the concentration of hydroxide ions:

Kb = [NH4+][OH-]/[NH3]

1.79 × 10^-5 = (0.056 M)(x) / (0.089 M)

x = 1.13 × 10^-5 M

Finally, we can use the concentration of hydroxide ions to calculate the pH of the solution:

pH = 14 - pOH = 14 - (-log10(1.13 × 10^-5)) = 9.95

Therefore, the pH of the solution is 9.95.

If some NaOH is added to the solution but the pH barely changes, it means that the added NaOH is being neutralized by the NH4+ ions in the solution, forming more NH3 and water:

NH4+ + OH- → NH3 + H2O

This reaction helps to buffer the pH of the solution.

To calculate the pH of a solution with equal amounts of NH4Cl and NH3, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where A- is the conjugate base of the acid, NH4+, and HA is the acid, NH3.

The pKa of NH4+ is given by:

pKa = pKw - pKb = 14 - 1.79 = 12.21

At the halfway point, the concentration of NH4+ and NH3 are equal:

[NH4+] = [NH3]

Substituting these values into the Henderson-Hasselbalch equation, we get:

pH = 12.21 + log(1) = 12.21

Therefore, the pH of a solution with equal amounts of NH4Cl and NH3 is 12.21.

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The substance combined with germanium that will cause the germanium to emit free electrons is the aluminum. Correct answer: letter D.

Adding aluminum to germanium will cause the germanium to emit free electrons. This is because aluminum has one more electron in its outermost shell than germanium does. When the two atoms are combined, the extra electron from the aluminum will fill the outermost shell of the germanium atom, and the germanium atom will then become unstable and emit an electron.

Aluminum and germanium are elements of the periodic table. Some similarities between the two are:

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(b) The chemical equation for the reaction in which silver carbonate is formed is:

AgNO3 + Na2CO3 -> Ag2CO3 + 2NaNO3

(c) The ionic equation for the reaction in which lead(II) iodide is formed is:

Pb(NO3)2 + 2KI -> PbI2 + 2KNO3

Answer: 2.576 G

Explanation:

To determine the remaining amount of strontium-90 after 84 years, we need to consider its half-life. The half-life of strontium-90 is approximately 29 years.

Using the radioactive decay formula:

Amount remaining = Initial amount * (1/2)^(time elapsed / half-life)

We can calculate the remaining amount as follows:

Amount remaining = 10 g * (1/2)^(84 years / 29 years)

Amount remaining = 10 g * (1/2)^(2.896)

Amount remaining ≈ 10 g * 0.2576

Amount remaining ≈ 2.576 g

Therefore, approximately 2.576 grams of the original 10-gram sample of strontium-90 would be left after 84 years.

Answer:

sorry I just need points

Answer:

670,385k

Explanation:

Metals :-

Group 1A - Alkali metals ( highly reactive metals)

Non-metals :-

Group 17 - Halogens ( highly reactive non-metals )

4K+O2-----------2K2O

a. The pH at the halfway point is 4.20, and the pH at the equivalence point is 8.10.
b.  The pH at the halfway point is 11.05 and at the equivalence point is 9.75 for the given titration.
c.  At the halfway point, the pH is 12.18, and at the equivalence point, the pH is 12.77.

For the titration of 100.0 mL of 0.10 M HC₇H₅O₂ (Ka 6.4 x 10^-5) with 0.10 M NaOH,

At the halfway point, half of the HC₇H₅O₂ has been neutralized to form its conjugate base, C₇H₅O₂⁻. The concentration of HC₇H₅O₂ is now 0.05 M and the concentration of C₇H₅O₂⁻ is also 0.05 M. To calculate the pH at the halfway point, use the Henderson-Hasselbalch equation,

pH = pKa + log([C₇H₅O₂⁻]/[HC₇H₅O₂])

pH = -log(6.4 x 10^-5) + log(0.05/0.05)

pH = 4.20

At the equivalence point, all of the HC₇H₅O₂ has been neutralized to form C₇H₅O₂⁻. This means that we have a solution of 0.10 M C₇H₅O₂⁻. To calculate the pH at the equivalence point, we can use the equation for the dissociation of C₇H₅O₂⁻,

C₇H₅O₂⁻(aq) + H₂O(l) ⇌ HC₇H₅O₂(aq) + OH⁻(aq)

Kb = [HC₇H₅O₂][OH⁻]/[C₇H₅O₂⁻]

Kb = 1.6 x 10^-10 (since Kb = Kw/Ka)

[OH⁻] = sqrt(Kb[C₇H₅O₂⁻]) = sqrt(1.6 x 10^-10 x 0.10) = 1.26 x 10^-6 M

pOH = -log[OH⁻] = -log(1.26 x 10^-6) = 5.90

pH = 14.00 - pOH = 8.10

For the titration of 100.0 ml of 0.10 M C₂H₅NH₂ (Kb = 5.6 × 10^-4) with 0.20 M HNO₃, the halfway point occurs when half of the C₂H₅NH₂ has been neutralized by the HNO₃. At this point, the moles of C₂H₅NH₂ are equal to the moles of C₂H₅NH³⁺ formed.

At the halfway point,

Moles of C₂H₅NH₂ = Moles of C₂H₅NH³⁺ formed

0.05 mol = 0.05 mol

Concentration of C₂H₅NH³⁺ = moles / volume

= 0.05 mol / 0.1 L

= 0.50 M

Kb = [C₂H₅NH₂][OH⁻] / [C₂H₅NH³⁺]

5.6 × 10^-4 = (0.10 - 0.05)(x) / (0.05)

x = 1.12 × 10^-3 M

pOH = -log[OH⁻] = -log(1.12 × 10^-3) = 2.95

pH = 14.00 - 2.95 = 11.05

At the equivalence point, all of the C₂H₅NH₂ has been neutralized by the HNO₃, and the solution contains only the conjugate acid, C₂H₅NH³⁺.

At the equivalence point,

Moles of HNO₃ = Moles of C₂H₅NH₂

0.02 mol/L × 0.1 L = 0.01 mol

Concentration of C₂H₅NH³⁺ = moles / volume

= 0.01 mol / 0.1 L

= 0.10 M

Kb = [C₂H₅NH₂][OH⁻] / [C₂H₅NH³⁺]

5.6 × 10^-4 = (0)(x) / (0.10)

x = 5.6 × 10^-5 M

pOH = -log[OH⁻] = -log(5.6 × 10^-5) = 4.25

pH = 14.00 - 4.25 = 9.75

At the halfway point, the moles of acid (HCl) will be equal to the moles of base (NaOH) added. Therefore,

moles of HCl = 0.100 L x 0.50 mol/L = 0.050 mol

moles of NaOH added = 0.050 mol (since NaOH has a 1:1 stoichiometric ratio with HCl)

moles of NaOH remaining = 0.100 mol - 0.050 mol = 0.050 mol

total volume = 100.0 mL + V(NaOH) at halfway point

Using the equation for neutralization,

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

The concentration of OH⁻ can be calculated as follows,

moles of OH⁻ = moles of NaOH remaining + moles of H₂O produced

moles of H₂O produced = moles of HCl reacted = 0.050 mol

moles of OH⁻ = 0.050 mol + 0.050 mol = 0.100 mol

volume at halfway point = 100.0 mL + 0.050 L = 0.150 L

concentration of OH⁻ = 0.100 mol / 0.150 L = 0.667 M

pH = 14 - log[OH⁻] = 14 - log(0.667) = 12.18

At the equivalence point, all of the HCl will have reacted with an equal amount of NaOH, forming NaCl and water.

moles of HCl = 0.100 L x 0.50 mol/L = 0.050 mol

moles of NaOH added = 0.050 mol

moles of NaOH added to reach equivalence point = 0.050 mol / 0.25 mol/L = 0.200 L

total volume at equivalence point = 100.0 mL + 0.200 L = 0.300 L

concentration of NaOH at equivalence point = 0.050 mol / 0.300 L = 0.167 M

since NaOH is a strong base, it will fully dissociate in water to produce OH⁻ ions

moles of OH⁻ = moles of NaOH added = 0.050 mol

concentration of OH⁻ = 0.050 mol / 0.300 L = 0.167 M

pH = 14 - log[OH⁻] = 14 - log(0.167) = 12.77

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Answer: Inertia

Explanation:

The genetic distance between E and G is approximately 50 m.u.

None of the given option is correct.

To determine the genetic distance between the E and G loci, we need to analyze the recombination frequencies between these loci based on the progeny classes provided.

From the table, we can observe the following recombinant progeny classes: Efg (302), eFg (91), EFg (92), and efG (88).

To calculate the genetic distance, we sum up the recombinant progeny classes and divide by the total number of progeny:

Recombinant progeny = Efg + eFg + EFg + efG = 302 + 91 + 92 + 88 = 573

Total progeny = Sum of all progeny classes = 298 + 302 + 99 + 91 + 92 + 88 + 14 + 16 = 1000

Recombination frequency = (Recombinant progeny / Total progeny) x 100

= (500/ 1000) x 100

= 50%

Since 1% recombination is equivalent to 1 map unit (m.u.), the genetic distance between E and G is approximately 50 m.u.

None of the given options (a. 42 m.u., b. 43 m.u., c. 41 m.u., d. 44 m.u., e. 40 m.u.) matches the calculated genetic distance, indicating that none of the provided options is correct.

None of the given option is correct.

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The volume of ethane gas, C₂H₆ in cm³ that will react to produce 40 cm³ of CO₂ is 20 cm³

2C₂H₆ + 7O₂ —> 4CO₂ + 6H₂O

From the balanced equation above,

4 cm³ of CO₂ were obtained from 2 cm³ of C₂H₆

From the balanced equation above,

4 cm³ of CO₂ were obtained from 2 cm³ of C₂H₆

Therefore,

40 cm³ of CO₂ will be obtain from = (40 × 2) / 4 = 20 cm³ of C₂H₆

Thus, 20 cm³ of C₂H₆ are required for the reaction.

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Explanation:

Answer:

The answer is 1

Explanation:

In the core of the Sun hydrogen is being converted into helium. This is called nuclear fusion. It takes four hydrogen atoms to fuse into each helium atom. During the process some of the mass is converted into energy.

Explanation:

Metal is a better conductor than wood, so a metal spoon could burn

Paul's hand.  So, Sue is incorrect. The metal being good conductor of heat, heats up faster than wood spoon and thus, could burn Paul's hand.

When preparing a plate, it is necessary to flame the metal loop between each step because it helps to kill any bacteria present. The metal loop is first heated in the flame until it is glowing red-hot.

This sterilizes it and burns off any organic matter that may be present. After that, the loop is allowed to cool for a few seconds to avoid killing any microbes in the sample. The loop is then used to transfer the inoculum from the source, such as a colony on an agar plate or a liquid culture, to the plate.

Once the inoculum is transferred, the loop is flamed again before being used for the next sample. This process is repeated for each sample to ensure that each one is not contaminated with the previous sample. It is also important to ensure that the loop is not too hot before touching the agar or culture, as this can kill the microbes and prevent them from growing.

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Answer:

\( T_{1}= 94.75 K\)

Explanation:

Given the following data;

Final pressure, P2 = 760 Torr

Initial pressure, P1 = 320 Torr

Final temperature, T2 = 225K

To find the initial temperature, we would use Gay Lussac's law;

Gay Lussac law states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Gay Lussac's law is given by;

\( PT = K\)

\( \frac{P1}{T1} = \frac{P_{2}}{T_{2}}\)

Making T1 as the subject formula, we have;

\( T_{1}= \frac{P_{1}}{P_{2}} * T_{2}\)

Substituting into the equation, we have;

\( T_{1}= \frac{320}{760} * 225 \)

\( T_{1}= 0.4211 * 225 \)

\( T_{1}= 94.75 K\)

Therefore, the initial temperature of the gas is 94.75 Kelvin.

Phosphoric acid has three equivalence points, corresponding to its three dissociable protons. In this lab, you should be able to see all three equivalence points if you perform a complete titration of the acid.

Phosphoric acid, which has the chemical formula H3PO4, is a triprotic acid. This means it has three acidic hydrogen atoms that can be donated to a base in an acid-base reaction.
Therefore, phosphoric acid has three equivalence points. An equivalence point is reached when the number of moles of the base added to the acid is equal to the number of moles of acidic hydrogens in the acid.
In a lab setting, you should be able to observe all three equivalence points if you carefully titrate the phosphoric acid with a strong base, such as sodium hydroxide (NaOH), and use an appropriate indicator or a pH meter to monitor the changes in pH during the titration.

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In a close system, the total energy of the system will decrease as the ice causes the warm water molecules to lose kinetic energy.

It is a system that is enclosed and would not allow the flow of any material form itself to another or form another system to itself.

A close system also does not allow transfer of energy between system.

Therefore, the total energy of the system will decrease as the ice causes the warm water molecules to lose kinetic energy.

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Answer:

O Al + Mg(NO3)2 -->

Explanation:

A. False (The pH of an alkaline solution is greater than 7.)

B. True

C. True

D. False (Alkalis are usually soapy to touch and have a bitter or metallic taste.)

Corrected versions:

A. The pH of an alkaline solution is greater than 7.

D. Alkalis are usually soapy to touch and have a bitter or metallic taste.

Calculation:

To calculate the density of the essential oil, you would use the following equation:

density = mass/volume

Plugging in the given values, you would get:

density = 40 g/50 ml = 0.8 g/ml

The correct option that describes the periodic trend of the first ionization energy for metals is: option 2.

The first ionization energy exhibits a decreasing trend from left to right within a period. This can be attributed to the increasing number of protons and electrons as we move across the period.

The greater number of protons exerts a stronger pull on the electrons, leading to a smaller atomic radius.

Consequently, the electrons are more tightly held by the nucleus, requiring more energy to remove them, thus increasing the first ionization energy.

In contrast, the first ionization energy shows an increasing trend from top to bottom within a group. This is primarily due to the phenomenon of electron shielding.

As we move down a group, additional electron shells are added, increasing the distance between the outermost electrons and the nucleus.

The inner electrons act as a shield, reducing the effective nuclear charge experienced by the outermost electrons.

As a result, the outermost electrons are more loosely held and require less energy to be removed, leading to a decrease in the first ionization energy.

To summarize, the first ionization energy generally decreases from left to right within a period and increases from top to bottom within a group.

Options 1, 3, and 4 contain incorrect statements or do not accurately describe the periodic trend of the first ionization energy for metals.

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Answer:

c. a hypothesis  is the correct answer

Explanation:

Answer:

Explanation:

Nitrate of most of the elements are water soluble . Aluminium nitrate is also one of them . Aluminium nitrate is easily dissolved in water . It forms hydrated salt whose formula is as follows

Al( NO₃ )₃ . 9H₂O

It is a white crystalline salt . It is also soluble in alcohol.