Determine whether each chemical substance would remain the same color or turn pink in the presence of phenolphthalein.

Answer:

yes it is

Explanation:

because there are some acid which really harm skin.

Answer:

The total photons required for this radiation = 5.1938 × 10²⁸ photons

Explanation:

Given that:

A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat.

If the radiation wavelength is 12.5 cm,

density of water = 1g/cm³

volume of the container = 0.250 L = 250 cm³

density = mass/volume

mass of the water = density × volume

mass of the water =  1g/cm³  × 250 cm³

mass of the water = 250 g

specific heat capacity of water = 4.182 J/g°C

The change in temperature was from 20.0° C to 99° C

ΔT =( 99 -20.0)° C

ΔT = 79.0° C

The heat absorbed in the process is calculated by using the formula,

q = mcΔT

q = 250 g × 4.182 J/g°C ×  79.0° C

q = 82594.5 Joules

Recall that the radiation wavelength λ = 12.5 cm = 0.125 m

The amount of energy of one photon of the radiation wavelength is determined by using the formula:

E = hv  

since v = c/λ

E = hc/λ

where;

h = Planck's constant = 6.626 × 10⁻³⁴ J.s

c = velocity of light = 3.0 × 10⁸ m/s

∴

E = (6.626 × 10⁻³⁴ J.s × 3.0 × 10⁸ m/s)/ 0.125 m

E = 1.59024⁻²⁴ Joules

The total photons required for this radiation = total heat energy/energy of radiation

The total photons required for this radiation = 82594.5 Joules/1.59024⁻²⁴ Joules

The total photons required for this radiation = 5.1938 × 10²⁸ photons

Answer:

.217, .311, and .472, respectively.

Explanation:

The total number of moles of gas is 3.50 + 5.00 + 7.60 = 16.10 (to preserve significant digits).

X of helium=3.50/16.10 = .217

X of krypton=5.00/16.10 = .311

X of neon=7.60/16.10 = .472

Answer:

А It is hard because its ions are held together by strong electrostatic attractions.

B It is malleable because its atoms can easily move past one another without disrupting the bonding.  

Explanation:

These are correct explanations of the properties of magnesium.

C is wrong. Mg is a good conductor of electricity and it has metallic bonds.

D is wrong. Mg has no molecules. It has no intermolecular forces.

Answer:

Explanation:

Answer in attached file .

Answer:

water molecules

Explanation:

Redox reactions are carried out under acidic or basic conditions as the case may be.

If the reaction is carried out in an acid medium, then we must balance the hydrogen ions on the lefthand side of the reaction equation with water molecules on the righthand side of the reaction equation.

For instance, the equation for reduction of MnO4^- under acidic condition is shown below;

MnO4^-(aq) + 5e + 8H^+(aq) --------> Mn^2+(aq) + 4H2O(l)

Answer:

Explanation:

For pH of a buffer solution , the formula is

pH = pKa + log [ Base ] / [ conjugate acid ]

=   pKa + log [ NH₃ ] / [ NH₄⁺ ]

Ka = Kw / Kb

Kb for NH₄OH = 1.8 x 10⁻⁵

Ka = 10⁻¹⁴ / 1.8 x 10⁻⁵

= 5.6 x 10⁻¹⁰

pH = - log ( 5.6 x 10⁻¹⁰ ) + log 0.2 / 0.2

= 10 - log 5.6

= 9.25

Effect of addition of HCl

H⁺ of HCl will react with NH₃ to produce NH₄⁺

25 mL of .1 HCl = 2.5 mM of HCl

25 mL of .1 NH₄⁺ = 2.5 mM of NH₄⁺

65 mL of .2 M NH₃ = 13 mM of NH₃

65 mL of .2 M NH₄⁺ = 13 mM of NH₄⁺

NH₃ + H⁺ = NH₄⁺

NH₄⁺ formed = 2.5 + 13  mM

15.5 mM of NH₄⁺

NH₃ = 13 mM

Concentration of NH₃ = 13 / 90

Concentration of NH₄⁺ = 15.5 / 90

pH of final buffer mixture

= 9.6 + log 13 / 15.5

= 9.25 - .076

= 9.174

The pH value  is mathematically given as

pH= -6.332.

Question Parameters:

the pH of a 0.20 M NH3/0.20 M NH4Cl buffer

the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.

Generally, the equation for the Chemical Reaction  is mathematically given as

HCl + NH3 --> NH4^+ + Cl^-

Therefore

pH= pka + log(13/14).

pH= -6.3 + log 0.93.

pH= -6.3+ (-0.032).

pH= -6.332.

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Answer:

false, it is not an example of facilitated diffusion

Answer:

Explanation:

When a drop of ink added into the water gradually moves in the whole quantity of water due to this entire water turns into blue color. This is nothing but the diffusion of ink particles into the water molecules. This is because water, as well as ink molecules, are in random motion due to the motion of ink substance.

Answer:

ΔHrxn = -635.14kJ/mol

Explanation:

We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:

(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol

(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole

(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole

Reaction (1) - (3) produce:

Ca(s) + H2O(l) → H2(g) + CaO(s)

ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol

Now this reaction + 1/2(2):

Ca(s) + ½ O2(g) → CaO(s)

ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)

There are various variants of Cov id virus. The Brazilian variant P also known as Gamma variant is the third variant of the original SARS-CoV2.

The correct answer is more negative

This variant has raised concerns since it has ability to spread more quickly then previous variants and this is more negative variant.

It is assumed that Cov id variant Gamma and Delta has ability to absorb move UV light but this is not proved yet and research is underway.

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Answer:

[tex]V_2=1.17L[/tex]

Explanation:

Hello,

In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:

[tex]V_1C_1=V_2C_2[/tex]

Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:

[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]

Best regards.

Answer:

Each chlorine atom shares a pair of electrons with the sulfur atom. ... Each chlorine atom shares all its valence electrons with the sulfur atom

Answer:

AgI I the limiting reactant.

Explanation:

The balanced equation for the reaction is given below:

2AgI + HgI2 → Ag2HgI4

Next, we shall determine masses AgI and HgI2 that reacted from the balanced equation.

This is illustrated below:

Molar mass of Agl = 108 + 127 = 235 g/mol

Mass of AgI from the balanced equation = 2 x 235 = 470 g

Molar mass of HgI2 = 201 + (2x127) = 455 g/mol

Mass of HgI2 from the balanced equation = 1 x 455 = 455 g

From the balanced equation above,

470 g of AgI reacted with 455 g of HgI2.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above,

470 g of AgI reacted with 455 g of HgI2.

Therefore, 2 g of AgI will react with

= (2 x 455)/470 = 1.94 g of HgI2.

From the calculations made above, we can see that only 1.94 g out of 3.50 g of HgI2 given is needed to react completely with 2 g of AgI.

Therefore, AgI I the limiting reactant.

Answer:

THE FINAL TEMPERATURE OF WATER IS -4.117 °C

Explanation:

Mass of the aluminium = 50 g

c = 0.88 J/g C

Initial temperature of aluminium = 225 °C

Volume of water = 100 ml

Density of water = 1 g/ml

Mass of water = density * volume of water

Mass of water = 1 * 100 = 100 g of water

Initial temperature of water = 20 C

It is worthy to note that the heat of a system is constant and conserved as no heat is lost or gained by a closed system,

So therefore,

heat lost by aluminium = heat gained by water

H = mass * specific heat capacity * temeprature change

So:

m c ( T2- T1) = m c (T2-T1)

50 * 0.88 * ( T2 - 225) = 100 * 4.18 *( T2 - 20)

44 ( T2 - 225 ) = 418 ( T2 - 20)

44 T2 - 9900 = 418 T2 - 8360

-9900 + 8360 = 418 T2 - 44 T2

-1540 = 374 T2

T2 = - 4.117

So therefore the final temperature of water is -4.117 °C

Answer:

CO3^2-

Explanation:

In qualitative analysis, we try to use chemical reactions to determine the composition of an unknown substance. The addition of certain reagents to the unknown solution gives certain results that show the presence or absence of certain species from the unknown sample.

When dilute HCl is added to an unknown sample and effervescence is observed, then the unknown sample must contain CO3^2- or HCO3^-. The presence of these species is confirmed if the gas evolved is passed through limewater and the gas turns limewater milky.

Answer:

Half-life at 629K = 252.4min

Explanation:

Using Arrhenius equation:

[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]

And as Half-life in a first order reaction is:

[tex]t_{1/2}=\frac{ln2}{K}[/tex]

We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:

[tex]58.0min=\frac{ln2}{K}[/tex]

K = 0.01195min⁻¹ = K₁

[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]

[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]

[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]

K₂ = 2.75x10⁻³ min⁻¹

And, replacing again in Half-life expression:

[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]

The half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.   

The activation energy of a reaction is related to its rate constant as follows:  

[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex]   (1)

Where:

We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:

[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex]   (2)

Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min

Hence, the rate constant at 652 K is:                            

[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]

Now, from equation (1) we can find the pre-exponential factor (A):

[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]  

With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):

[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]

Finally, the half-life at 629 K is (eq 2):

[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]

Therefore, the half-life at 629 K is 251.1 min.

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Kc require (aqueous/gaseous) products to be on the numerator and (aqueous/gaseous) reactants to be in the denominator, whereas Ksp will require (aqueous) products to be on the numerator and (aqueous) reactants to be in the denominator. Both require products on top and reactants in the bottom.

K = [products] / [reactants]

Kc is used when a reaction reaches dynamic equilibrium, whereas Ksp is used when an insoluble ionic solid dissolved by a tiny amount in a solution, as well as in determining whether or not a precipitate will form.

Kc can be used to measure equilibrium concentration for all reactions, whereas Ksp is limited to only ionic compounds' solubility.

The decomposition of HI (g) will required the use of Kc since the species are all gaseous, and gases cannot be ionic.

Answer:

1) 6.524779402×10^(-17)  

2)521.1g

3)113

Explanation:

Answer: 1) 6.524779402×10^(-17)

2)521.1g

Explanation:

Answer:

[ClO] = 3.48×10¯⁹ M.

Explanation:

The following data were obtained from the question:

Equilibrium constant (Kc) = 4.96×10¹¹

Concentration of Cl2O2, [Cl2O2] = 6x10¯⁶ M.

Concentration of ClO, [ClO] =.?

The equation for the reaction is given below:

2ClO(g) ⇌ Cl2O2(g)

The equilibrium constant for a reaction is simply defined as the ratio of the concentration of product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant, Kc for the reaction is given by:

Kc = [Cl2O2] / [ClO]²

Thus, we can calculate the concentration of ClO, [ClO] as follow:

Kc = [Cl2O2] / [ClO]²

4.96×10¹¹ = 6x10¯⁶ / [ClO]²

Cross multiply

4.96×10¹¹ × [ClO]² = 6x10¯⁶

Divide both side by 4.96×10¹¹

[ClO]² = 6x10¯⁶ / 4.96×10¹¹

[ClO]² = 1.21×10¯¹⁷

Take the square root of both side

[ClO] = √ (1.21×10¯¹⁷)

[ClO] = 3.48×10¯⁹ M

Therefore, the concentration of ClO, [ClO] is 3.48×10¯⁹ M.

Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.

hydro = water

Answer:

Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!

Explanation:

The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)

Answer:

Explanation:

We shall apply Arrhenius equation which is given below .

[tex]ln\frac{k_2}{k_1} = \frac{E_a}{R} [\frac{1}{T_1} -\frac{1}{T_2} ][/tex]

K₂ and K₁ are rate constant at temperature T₂ and T₁ , Ea is activation energy .

Putting the given values

[tex]ln\frac{5.9}{1} = \frac{39500}{8.3} [\frac{1}{298} -\frac{1}{T_2} ][/tex]

[tex].000373= [\frac{1}{298} -\frac{1}{T_2} ][/tex]

T₂ = 335.27 K

= 62.27 °C

The higher temperature is 62.27°C.

Given that the activation energy of the reaction is:

Eₐ =  39.5 kJ/mol

initial temperature T₁ = 25°C = 298K

Let the final temperature be T₂

The rate constant at temperature T₁ be K₁, and at a temperature T₂ be K₂.

According to the question: K₂/K₁ = 5.9

Now, applying the Arrhenius equation we get:

[tex]\ln\frac{K_2}{K1}=\frac{E_a}{R}[\frac{1}{T_1} -\frac{1}{T_2}]\\\\\ln(5.9)= \frac{39.5}{8.3}[\frac{1}{298} -\frac{1}{T_2}]\\\\0.000373=\frac{1}{298} -\frac{1}{T_2}[/tex]

T₂ = 335.27K

T₂ =  335.27 -273

T₂ = 62.27°C

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Answer:

option (A) is right answer

Answer:

c

Explanation:

trust me I took the exam

Answer:

Atomic mass = 127.198 amu

Explanation:

The average atomic mass is obtained by summing the masses of the isotopes each multiplied by its abundance.

Atomic mass = (97.62 * 0.0825) + (109.3 * 0.2671) + (138.3 * 0.6504)

Atomic mass = 8.05365 +  29.19403 + 89.95032

Atomic mass = 127.198 amu

Answer:

The magnetic field shows (D)

Explanation:

Because like poles repel eachother

The magnetic field is represented by two poles of a magnetic with magnetic filed lines from the poles. The image C represents the magnetic filed .

A magnetic filed is a generated by strong filed lines from a magnet with two poles. A magnet have two poles namely magnetic south and north poles.

The magnetic south poles of of a magnet is always attracted towards the magnetic north poles. Magnetic north poles of a magnet is generally shown in the left side and the field lines will be outward towards the geographic north pole.

The magnetic field lines from the south pole is inward and attracted towards the north. Therefore, right figure representing the magnetic filed is option C.

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Answer:

when equal moles of an acid and base are mixed,after reaction the two are compounds are said to be at the Equivalent point.

The mass of magnesium chloride produced from 2.30 moles of chlorine gas is 218.99 grams.

Stoichiometry refers to the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions.

According to this question, magnesium reacts with chlorine gas to form magnesium chloride as follows:

Mg + Cl₂ → MgCl₂

Based on the above chemical equation, 1 mole of chlorine gas forms 1 mole of magnesium chloride.

This means that 2.30 moles of chlorine gas will 2.30 moles of magnesium chloride.

Next, we convert moles of magnesium chloride to mass as follows:

molar mass of magnesium chloride = 95.211g/mol

mass of magnesium chloride = 95.211 × 2.30 = 218.99 grams.

Therefore, 218.99 grams of magnesium chloride will be formed.

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Explanation:

1. Nucleus is a memberane bound organelle that contains cell,s chromosomes.

Answer:

D. Rare

Explanation:

Qualitative research has to do with non-numerical data and is used to understand the beliefs of a group of people which can be gotten from surveys, questionnaires, interviews, etc.

A qualitative researcher may select a RARE case that is unusually rich in information pertaining to the research question.

This is because he wants to get an insight into the why, how, where and when of that particular ase as it's not a usual occurrence.

Answer:

Explanation:

a) 4-nitrobenzoic acid         pKa= 3.41

    benzoic acid                   pKa= 4.19

    4-chlorobenzoic acid     pKa= 3.98

b) benzoic acid                    pKa= 4.19    

   cyclohexanol                   pKa= 18.0

   phenol                              pKa= 9.95

c) 4-Nitrobenzoic acid             pKa= 3.41

   4-nitrophenol                       pKa= 7.15

   4-nitrophenylacetic acid     pKa= 3.85