Select the best answer from the choice below. When the wavelength decreases, the frequency:

Answer:

ion is missing

Explanation:

Answers:

1. Chemistry

2. Cold

3. Heat

4. Protons and neutrons

5. Positive

6. Neutral

7. True

8. Nucleus

9. Protons

10. Nobel gasses

11. Dmitri Mendeleev

12. the bottom of the square

13. True

14. Helium

Answer:

I am sure that the C one is correct

Answer:

how is this college level..im doing this in middle school..

Explanation:

Answer:

Sphere

Explanation:

The shape of the s-sublevel is spherical in shape.

This sublevel has an azimuthal quantum number of 0 and it is spherical in shape;

Answer:

residue is whatever remains after something else has been removed while filtrate is the liquid or solution that has passed through a filter, and which has been separated from the filtride.

filtration id done by placing a filter paper on the beaker or container then pour the filtride then let it settle and it will pass through and you will have the fitrate and residue

Explanation:

Answer:

The temperature of the Nitrogen after throttling is  [tex]T_2 = 300 \ K[/tex]

Explanation:

From the question we are told that

   The  temperature is  [tex]T_1 = 300 \ K[/tex]

    The pressure is  [tex]P = 200 \ kPa = 200 * 10^{3} \ Pa[/tex]

    The pressure after being  [tex]P_1 = 100 \ kPa = 100 * 10^{3} \ Pa[/tex]

Generally from the first law of thermodynamics we have that

      [tex]Q - W = \Delta U + \Delta K[/tex]

Here [tex]\Delta U[/tex]  is the change internal  energy which is mathematically represented as

            [tex]\Delta U = C_p (T_2 - T_1)[/tex]

Here      [tex]C_p[/tex] is the specific heat of the gas at constant pressure

        [tex]\Delta K[/tex]  is the change kinetic energy which is negligible

         Q  is the thermal  energy  which is Zero for an adiabatic process

         W  is the work done and the value is zero given that the gas was throttled adiabatically

So

      [tex]0= \Delta U +0[/tex]

=>   [tex]\Delta U = 0[/tex]

=>  [tex](T_2 - 300) = 0[/tex]

=>   [tex]T_2 = 300 \ K[/tex]

Answer:

Its B

Explanation:

Answer:

53.5%

Explanation:

Step 1: Given data

Step 2: Calculate the mass corresponding to 0.00875 moles of KClO₃

The molar mass of KClO₃ is 122.55 g/mol.

0.00875 mol × 122.55 g/mol. = 1.07 g

Step 3: Calculate the percent by mass of KClO₃ in the mixture

We will use the following expression.

%m/m = mass of KClO₃/mass of the sample × 100%

%m/m = 1.07 g/2.00 g × 100%

%m/m = 53.5%

The percent by mass of KClO3 in the original mixture is 53.5%.

Since

The mass of the sample should be 2.00 g

Moles of KClO₃: 0.00875 mol

The molar mass of KClO₃ is 122.55 g/mol.

Now the mass is

= 0.00875 mol × 122.55 g/mol.

= 1.07 g

Now the percentage should be

= mass of KClO₃/mass of the sample × 100%

= 1.07 g/2.00 g × 100%

= 53.5%

hence, The percent by mass of KClO3 in the original mixture is 53.5%.

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Answer:

i can only help with 2 :( atimic mass is: 35.453  and number of electrons is:17

Explanation:

Answer: sulfur

Explanation:

Answer:

its sulfur babes

Explanation:

Complete Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M trimethylamine, (CH3)3N(aq), with 0.2000 M HClO4(aq) after 9.48 mL of the acid have been added.Kb of trimethylamine = 6.5 x 10-5.

Answer:

The pH is  [tex]pH = 9.84[/tex]  

Explanation:

From  the question we are told that

   The  volume of trimethylamine, (CH3)3N(aq) is [tex]V_{t} = 20.00mL[/tex]

    The concentration of trimethylamine is [tex]C_t = 0.1000 \ M[/tex]  

    The  volume of HClO4(aq) is [tex]V_{h} = 9.48 mL[/tex]

    The  concentration of  HClO4(aq) is  [tex]C_h = 0.200 M[/tex]

      The Kb  value is  [tex]K_b = 6.5 * 10^{-5}[/tex]

Generally the the pOH of this reaction is mathematically represented as

       [tex]pOH = pK_b + log [\frac{N_h}{N_b} ][/tex]

Here [tex]N_h[/tex] is the number of moles of acid which is evaluated as

      [tex]N_h = C_h * V_h[/tex]

=>    [tex]N_h = 0.200 * 9.48[/tex]

=>    [tex]N_h = 1.896[/tex]

Here [tex]N_t[/tex] is the number of moles of acid which is evaluated as

      [tex]N_t = C_t * V_t[/tex]

=>    [tex]N_t = 0.100 * 20[/tex]

=>    [tex]N_t = 2[/tex]

So

     [tex]pOH = -log(K_b) + log [\frac{N_h}{N_b} ][/tex]

     [tex]pOH = -log(6.5*10^{-5}) + log [\frac{1.896}{2} ][/tex]

=>   [tex]pOH = 4.1639[/tex]

Generally the pH is mathematically represented as

       [tex]pH = 14 - pOH[/tex]

=>    [tex]pH = 14 - 4.1639[/tex]  

=>    [tex]pH = 9.84[/tex]  

Answer:

Pentyl Ethanoate, CH3COOC5H11, which smells like bananas, is produced from the esterification reaction: CH3COOH(aq) +C5H11OH(aq) → CH3COOC5H11(aq) + H … 2O(l) A reaction uses 3.58g of CH3COOH and 4.75g of C5H11OH and has a yield of 45.00%. Determine the mass of ester that foorms

Answer:

A heating curve graphically represents the phase transitions that a substance undergoes as heat is added to it. The plateaus on the curve mark the phase changes. The temperature remains constant during these phase transitions.

Answer:

B

Explanation:

Newton's third law. states that:

Action and reaction are equal and opposite.

Answer:

0.16 L

Explanation:

From the question given above, the following data were obtained:

Mole = 0.0806 mole

Molarity = 0.490 M

Volume =?

Molarity is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is expressed as:

Molarity = mole /Volume

With the above formula, we can obtain the volume of the solution as illustrated below:

Mole = 0.0806 mole

Molarity = 0.490 M

Volume =?

Molarity = mole /Volume

0.490 = 0.0806/volume

Cross multiply

0.490 × volume = 0.0806

Divide both side by 0.490

Volume = 0.0806 / 0.490

Volume of solution = 0.16 L

Thus, the volume of the solution is 0.16 L.

A solution is a mixture of solute in the solvent. The volume of the solution is 0.16 L.

A solution is a mixture of solute in the solvent.

An example of a solution is a mixture of salt in water.

Given,

Moles = 0.0806 mol [tex]\rm CaCl_2[/tex]

Molarity = 0.490 M

To find: the volume of the solution

[tex]Molarity = \dfrac{n}{V}[/tex]

Putting the values

[tex]0.490 M = \dfrac{ 0.0806}{V}\\\\0.490 M \times V = 0.0806\\\\Divide\; both\;side\; with\; 0.490\\\\V= \dfrac{ 0.0806}{0.490}=0.16 L[/tex]

Thus, the volume so the solution is 0.16 l.

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Answer:

[tex]0.192\ \text{M}[/tex]

Explanation:

A = Absorbance of solution

E = Molar absorptivity

l = Length of cuvette = 1 cm

c = Concentration of solution

Beer's law is given by

[tex]A=Elc[/tex]

The equation of a straight line is given by

[tex]y=mx+c[/tex]

Comparing the above equations we get

Value on [tex]y[/tex] axis = A = Absorbance of solution = 0.23

[tex]m[/tex] = Slope of line = El = Molar absorptivity multiplied with length = [tex]1.2\times 1\ \text{M}^{-1}[/tex]

[tex]x[/tex] = Value of x axis = c = Concentration of solution

So we get

[tex]c=\dfrac{A}{El}\\\Rightarrow c=\dfrac{0.23}{1.2\times 1}\\\Rightarrow c=0.192\ \text{M}[/tex]

The molar concentration of the sample is [tex]0.192\ \text{M}[/tex]

Answer:

I believe its 1,2, and 5

Explanation:

C D E

temperature of medium

type of wave

type of medium

Answer:

75603.86473 K

Explanation:

Given that:

The 1st excited electronic energy level of He atom = 3.13 × 10⁻¹⁸  J

The objective of this question is to estimate the temperature at which the ratio of the population will be 5.0 between the first excited state to the ground state.

The formula for estimating the ratio of population in 1st excited state to the ground state can be computed as:

[tex]\dfrac{N_2}{N_1} = e ^{^{-\dfrac{(E_2-E_1)}{KT}}} = e ^{^{-\dfrac{(\Delta E)}{KT}}}[/tex]

From the above equation:

Δ E = energy difference =  3.13 × 10⁻¹⁸  J

k = Boltzmann constant = 1.38 × 10⁻²³  J/K

[tex]\dfrac{N_2}{N_1} = 0.5[/tex]

Thus:

[tex]0.05 =e^{^{ -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}[/tex]

[tex]In (0.05) = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}[/tex]

[tex]-3.00 = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}[/tex]

[tex]-3.00 = -226811.5942 \times \dfrac{1}{T}[/tex]

[tex]T = \dfrac{-226811.5942}{-3.00 }[/tex]

T = 75603.86473 K

Answer:

adding a catalyst will increase the reaction rate.

Explanation:

im like 95% sure thats right.

Answer:

A. C. D

Explanation:

ap ex

Answer:

[tex]23.093\ \text{torr}[/tex]

Explanation:

[tex]M_g[/tex] = Molar mass of glucose = 180.2 g/mol

[tex]M_w[/tex] = Molar mass of water = 18 g/mol

[tex]m_g[/tex] = Mass of glucose = 76.6 g

[tex]m_w[/tex] = Mass of water = [tex]250\times 1\ \text{g/mL}=250\text{g}[/tex]

[tex]P_0[/tex] = Vapor pressure of pure water at 25°C = 23.8 torr

The mole fraction of glucose is

[tex]x_g=\dfrac{\dfrac{m_g}{M_g}}{\dfrac{m_g}{M_g}+\dfrac{m_w}{M_w}}=\dfrac{\dfrac{76.6}{180.2}}{\dfrac{76.6}{180.2}+\dfrac{250}{18}}\\\Rightarrow x_g=0.0297[/tex]

Mole fraction of the solute would be

[tex]\dfrac{P_0-P}{P_0}=x_g\\\Rightarrow 0.0297=\dfrac{23.8-P}{23.8}\\\Rightarrow P=23.8-0.0297\times23.8\\\Rightarrow P=23.093\ \text{torr}[/tex]

The vapor pressure of the solution is [tex]23.093\ \text{torr}[/tex].

Answer:

c. Potential decreases and kinetic increases.

Explanation:

As the boulder is rolling down the hill, potential energy decreases as kinetic energy increases.

Please Brainliest me

The enthalpy : 320.3 kJ/mol

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

Based on the principle of Hess's Law,

the change in enthalpy of a reaction will be the same even though it is through several stages or ways  

Reaction

2Fe (s) + 3Cl₂ (g) → 2FeCl₃ (s) ∆H° = -800.0 kJ/mol  x 2

4Fe (s) + 6Cl₂ (g) → 4FeCl₃ (s) ∆H° = -1600.0 kJ/mol

Si(s) + 2 Cl₂ (g) → SiCl₄ (s) ∆H° = -640.1 kJ/mol

Reverse

SiCl₄ (s) → Si(s) + 2 Cl₂ (g) ∆H° = 640.1 kJ/mol x 3

3SiCl₄ (s) → 3Si(s) + 6Cl₂ (g) ∆H° = 1920.3 kJ/mol

------------------------------------------------------------------------ +

3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s) ∆H° = 320.3 kJ/mol

The enthalpy for the formation of 4 moles of [tex]\rm FeCl_3[/tex] has been 320.3 kJ/mol.

The enthalpy of formation of 1 mole at standard temperature and pressure has been termed the standard enthalpy of formation.

The reaction enthalpy to be calculated has reactants of 3 moles of [tex]\rm SiCl_4[/tex], and 4 moles of Fe.

The enthalpy for the formation of [tex]\rm SiCl_4[/tex] has been -640.1 kJ/mol.

The enthalpy for the dissociation of [tex]\rm SiCl_4[/tex] has been the reverse of the formation reaction = +640.1 kJ/mol.

The enthalpy for the dissociation of 3 moles of [tex]\rm SiCl_4[/tex] has been = 3 [tex]\times[/tex] 640.1 kJ/mol.

The enthalpy for the dissociation of 3 moles of [tex]\rm SiCl_4[/tex] has been = 1920.3 kJ/mol.

The enthalpy for the formation of 2 moles of [tex]\rm FeCl_3[/tex] = -800 kJ/mol

The enthalpy for the formation of 4 moles of [tex]\rm FeCl_3[/tex] = -1600 kJ/mol.

The final reaction has 3 moles of  [tex]\rm SiCl_4[/tex], and 4 moles of [tex]\rm FeCl_3[/tex].

Thus, the final enthalpy of reaction will be: 1920 kJ/mol + (-1600 kJ/mol)

The final enthalpy of reaction will be = 320.3 kJ/mol.

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Answer:

This means that a solution must be cooled to a lower temperature than the pure solvent in order for freezing to occur.

Explanation:

Answer:

Freezing Point Depression. The freezing point of a solution is less than the freezing point of the pure solvent. This means that a solution must be cooled to a lower temperature than the pure solvent in order for freezing to occur

Answer:

Cultivated rice paddies

Drilling of natural deposits

Fossil fuel use

Burning of biomass

Landfills

Explanation:

The bulk of the methane released into the atmosphere are as a result of various human activities.

Answer:

[tex]$Cu_2+(SO_3)^2 \rightleftharpoons CuSO_3 \ (insoluble)$[/tex]

[tex]$Cu_2+2(SO_3)^2 \rightleftharpoons [Cu(SO_3)_2]_2 \ (soluble)$[/tex]

Explanation:

A ligand may be defined as a molecule or an ion or that binds to the central metal atom in order to form a more coordination complex. Sulfite is one such ligand and it behaves similarly as hydroxide as a ligand.

Now, according to the question, when we react copper with sulfite ion, it forms copper sulfite. The equation is

[tex]$Cu_2+(SO_3)^2 \rightleftharpoons CuSO_3 \ (insoluble)$[/tex]

Now when excess of the sulfite ion is used in the reaction, we get a complex formation, which is shown by

[tex]$Cu_2+2(SO_3)^2 \rightleftharpoons [Cu(SO_3)_2]_2 \ (soluble)$[/tex]

The way the sulfite reacts is quite similar to hydroxide ion where they form a complex ion when hydroxide ion in excess is used in the reaction with metal cation.

Answer:

Throughout the overview section following portion, the description and according to particular circumstance is defined.

Explanation:

As per the question,

⇒ [tex]NH_3(aq)+H_2O\leftrightarrow NH_4+(aq)+OH-(aq)[/tex]

⇒ [tex]NH_3+(aq) \leftrightarrow NH_3(aq)+H+(aq)[/tex]

However, it is also available in NH3 form throughout the aqueous solution much of the moment.