Ultraviolet light from the Sun can A. damage nerve cells but not skin cells. B. damage molecules in skin cells, such as DNA. C. cause skin cells to release stinging chemicals. D. help skin cells to repair DNA faster.

Answer:

D

Explanation:

Volcanic eruptions spew large amounts of dust particles from their surroundings into the atmosphere. Accompanying eruptions are also materials like volcanic ash and gases. These gases contribute immensely to the reserve of greenhouse gases in the atmosphere.

Gases in volcanoes are responsible to a large extent for their explosivity when an eruption occurs. The more the gases, the more explosive a volcano can be. These gases are rich in carbon dioxide, methane, etc. A vast number of them are greenhouse gases.

Dust particles also accompany an eruption. These dust can be suspended for an extended period in the atmosphere by the wind.

the best choice is the letter D. Volcanic eruptions add green house gases and  fine dust particles to the atmosphere.

i toke the quiz on gradpoint and the answer was correct

Answer:

a) The charge of one electron is the same charge of another electron.

b) The charge of one electron  has the same magnitude and opposite sign as the charge of a proton. The charge of an electron is negative by convention and the charge of a proton is positive by convention.

c) The mass of a proton is about 1,836 times the mass of an electron.

Explanation:

1) Electron discovery is due to J.J Thompson who is responsible for the famous experiment of the cathode-rays tube.  J.J Thompson, in 1897, found that cathode rays could be deflected by an electric field. He could establish that electrons had negative charge, and also determined the approximate the ratio of its charge to the mass. Yet, he could not determine such magnitudes.

2) Later on, Ernest Rutherford,  with the famous gold foil experiment, found that all of the positive charge (proton) and essentially all of the mass of the atom is concentrated in a tiny region which he named the nucleus.

3) Since, the atom has the same number of electrons and protons, and it is neutral, the charge magnitude of the charges electrons and protons are equal.

4) Robert Millikan, 1908 - 1917, was able to determine the charge of the electron, with which the mass was also determined.

The accepted relative values for the subatomic particles are:

Particle   Relative mass Relative charge

Proton           1                              +1

Neutron        1                                0

Electron        1/ 1,836                    -1

Answer:

all electrons have same charge. Electron is equal and opp to proton

Protean mass of electron

Explanation:

1) moles = mass/mR

CaCO3 Mr = 40 + 12 + (16×3)

= 52 + 48

= 100

mass = 15

so the moles would be 15 ÷ 100

which is 0.15 moles of CaCO3

2) moles = mass ÷ Mr

Mr of Al2O3 = 27 + (16×3)

= 27 + 48

= 75

mass = 204

so the moles would be 204/75 which is 2.72 moles of Al2O3

There are 0.15 moles in 15 grams of CaCO₃ and 2 moles in 204 grams of Al₂O₃.

To determine the number of moles in a given mass of a substance, you can use the formula:

 moles = mass (grams) / molar mass (g/mol)

We know the molar mass of CaCO₃ is 100 g/mol

(Ca: 40.0, C: 12.0, O: 16.0 × 3).

Given mass of CaCO₃ is 15 grams.

Apply the formula: moles = 15 g / 100 g/mol

                                           = 0.15 moles

We calculate the molar mass of Al₂O₃:

 Al: 27.0 × 2 = 54.0

O: 16.0 × 3 = 48.0

Total molar mass of Al₂O₃ = 54.0 + 48.0 = 102 g/mol

Given mass of Al₂O₃ is 204 grams.

Apply the formula: moles = 204 g / 102 g/mol = 2 moles

Complete question.

How many moles are there in :

The equation is:

3AgS2 +  4Al  ---> 2Al2S3 + 3Ag

Answer: The balanced chemical equation is written below.

Explanation:

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

[tex]A+BC\rightarrow AC+B[/tex]

The chemical equation for the reaction of silver sulfide with aluminium follows:

[tex]3Ag_2S+2Al\rightarrow 6Ag+Al_2S_3[/tex]

By Stoichiometry of the reaction:

3 moles of silver sulfide reacts with 2 moles of aluminium metal to produce 6 moles of silver metal and 1 mole of aluminium sulfide

Hence, the balanced chemical equation is written above.

The rate of disappearance of the ClO₃⁻ ion is equal to 9.01 ×10⁻³ M/s.

The rate of reaction can be defined as the speed at which the products are produced or the reactants are consumed in a chemical reaction. The rate provides information about the time frame under which a reaction can be completed.

The rate of reaction can be described as the speed of a chemical reaction at which reactants are converted into products. Some reactions are instantaneous, while some take a little longer to reach the final equilibrium.

Given the chemical reaction is:

[tex]ClO_3^-(aq) + 9I^- (aq)+ 6 H^+ (aq) \longrightarrow 3I^{-3} (aq) + Cl^-(aq) + 3H_2O[/tex]

The initial rate of disappearance of I⁻ ion = 81.1 ×10⁻³ M/s.

The rate of disappearance of iodide ion is given by:

[tex]-\frac{D[I^-]}{Dt} =81.1\times 10^{-3}[/tex]

The relation between the rate of disappearance of I⁻ and ClO₃⁻ ion:

[tex]-\frac{D[ClO_3^-]}{Dt} =-\frac{1}{9} \frac{D[I^-]}{Dt}[/tex]

[tex]-\frac{D[ClO_3^-]}{Dt} =-\frac{1}{9} \times 81.1 \times 10^{-3}[/tex]

[tex]\frac{D[ClO_3^-]}{Dt} = 9.01 \times 10^{-3}M/s[/tex]

Therefore, the rate of the disappearance of ClO₃⁻ ion is equal to 9.01 ×10⁻³ M/s when the initial rate of disappearance of I⁻ is  81.1 ×10⁻³ M/s.

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The number of valence electrons influences whether an atom will gain or lose electrons during compound formation, following the octet rule. Metals typically lose electrons to form cations, while nonmetals gain electrons to form anions. This behavior underpins the formation of ionic and covalent bonds.

The number of valence electrons in an atom is fundamentally related to its tendency to gain or lose electrons during compound formation, adhering to the octet rule. Atoms seek to complete their outermost electron shell, typically aiming to have eight electrons in this valence shell for stability, mirroring the electron configuration of the nearest noble gas. Metals often lose electrons to form positively charged ions (cations), while nonmetals tend to gain electrons, forming negatively charged ions (anions). This transfer and sharing of electrons give rise to different types of chemical bonds, primarily ionic and covalent, which are pivotal in the formation of compounds. The electrostatic attraction between oppositely charged ions in ionic bonding, and the sharing of electrons in covalent bonding, exemplify the underlying principles driving atoms to gain or lose electrons for compound formation.

A physical change occurs when something is altered. Such as cutting a piece of paper or chopping a piece of wood into pieces.  Melting ice, breaking glass, evaporating liquids are go through a physical change.

Answer:

(D) Na₂SO₄•10H₂O (M = 286).

Explanation:

ΔTf = i.Kf.m,

Where, ΔTf is the depression in freezing point of water.

i is van't Hoff factor.

Kf is the molal depression constant.

m is the molality of the solute.

(A) CuSO₄•5H₂O:

CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(B) NiSO₄•6H₂O:

NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(C) MgSO₄•7H₂O:

MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(D) Na₂SO₄•10H₂O:

Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.

So, i = dissociated ions/no. of particles = 3/1 = 3.

∴ The salt with the high (i) value is Na₂SO₄•10H₂O.

So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.

Answer:

Explanation:

The percent ionization of a weak acid is the percent of the original acid that ionizes.

You may calculate the percent ionization of an acid of known concentration, from the equilibrium constant.

So, to determine the percent ionization of a 0.215 M solution of benzoic acid, you must look for the equilibrium (dissociation or ionization) constant.

Equilibrium constants depend on temperature. So, you must know the temperature.

For this question, I will assume 25°C, for which you can find that the dissociation constant, Ka, for benzoic acid is 6.25×10⁻⁵.

With that, you follow these steps:

1. Write the ionization equation:

2. Calculate the concentration of the ion C₆H₅CO₂⁻ at equilibrium

                    C₆H₅CO₂H ⇄ C₆H₅CO₂⁻ + H⁺

       I           0.215                  0              0

      C            - x                   + x           + x

      E          0.215 - x              x              x

        Ka = x² / 0.215 - x =  6.25×10⁻⁵

           x² = 0.215 × 6.25×10⁻⁵ = 0.0000078125

           x = 0.00367 M

           If you do not make the assumption but solve the quadratic equation you will get x = 0.00363 M

3. Calculate the percent ionization:

With x = 0.00363 M (exact calculation)

With x = 0.00367 M

The percentage ionization is equal to 1.74%

Data;

                [tex]HC_7H_5O_2 + H_2O \to C_7H_5O_2^- + H_3O^+\\[/tex]

initial           0.215           -             -                -

change        -x                               +x             +x

equilibrium   0.215 - x                    x                x

The equilibrium concentration of the acid

[tex]K_a = \frac{[C_7H_5O_2^-][H_3O^+}{[HC_7H_5O_2]}[/tex]

Let's substitute the value in this

[tex]K_a = \frac{x.x}{0.215-x}\\ K_a = \frac{x^2}{0.215-x} \\[/tex]

The Ka for benzoic acid is 6.5*10^-5

[tex]6.5*10^-^5 = \frac{x^2}{0.215 - x} \\[/tex]

since the value of Ka is very small.

0.215 - x = 0.215

[tex]x^2 = 6.5*10^-^5 * 0.215\\x = \sqrt{1.3975*10^-^5} \\x = 3.738*10^-^3[/tex]

[tex][H_3O^+] = x = 3.738*10^-^3[/tex]

The percentage ionization would be

[tex]\frac{[H_3O^+}{[HC_7H_5O_7]} * 100= \frac{3.738*10^10^-^3}{0.215} * 100 = 1.74 \%[/tex]

The percentage ionization is equal to 1.74%

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Atoms that are bunched together the closest are solids.

Answer:

Question 1: 1) Increasing the pressure          C) Shift to the right

                   2) Removing hydrogen gas        A) Shift to the left  

                   3) Adding a catalyst                     B) No effect

Question 2: This reaction is exothermic because the system shifted to the right on cooling.

Question 3: Shift it toward the reactants.

Question 4: Adding more of gas C to the system.

Question 5: It will shift toward the reactant side because the reactant side has one more mole of gas than the product side.

Question 6: True.

Question 7: there is no suitable choice is provided.

We can shift the equilibrium toward the right via:

Increasing N2O3 concentration,

decreasing NO and/or NO2 concentration,

decreasing the pressure,

lowering the T (cooling the system).

Explanation:

Question 1: Match the action to the effect on the equilibrium position for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g).

Match Term Definition

Removing ammonia A) No effect

Removing hydrogen gas B) Shift to the left

Adding a catalyst C) Shift to the right

1) Increasing the pressure:

so, the right match is: C) Shift to the right.

2) Removing hydrogen gas:

so, the right match is: A) Shift to the left.

3) Adding a catalyst:

so, the right match is: B) No effect.

Question 2: Nitrogen dioxide gas is dark brown in color and remains in equilibrium with dinitrogen tetroxide gas, which is colorless.

2NO2(g) ⇌ N2O4(g)

When the light brown color equilibrium mixture was moved from room temperature to a lower temperature, the mixture turned lighter brown in color. Which of the following conclusions about this equilibrium mixture is true?

So, the right choice is: This reaction is exothermic because the system shifted to the right on cooling.

Question 3: According to Le Châtelier's principle, how will a decrease in concentration of a reactant affect the equilibrium system?

So, the right choice is: Shift it toward the reactants.

Note: The answer of Q 4, 5, 6 & 7 and all answers are in the attached word file.

These questions are based on Le Châtelier's principle and relate to the effects of changes in concentration, temperature, and pressure on chemical equilibrums. The reaction will always try to offset the disturbance by shifting towards the side that will help restore equilibrium.

The answers are based on Le Châtelier's principle, which states that equilibrium in a system will adjust to offset any changes made.

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About 234 grams .nA 2M NaCl solution contains approximately 58.5g of salt. So to make two liters of a 2M solutions you would need 117g of salt dissolved in 2 liters of water.

= 1.23 moles

V1/n1 = V2/n2

V1 = 55 L, n1 =2.46 moles

1/2 of the gas was used and thus; 1/2 or 27.5 L remained.

Therefore;

55 L/2.46 moles = 27.5 L/ n2

n2 = (2.46 × 27.5)/55

    = 1.23 moles

The number of moles if 1/2 of the gas is used is mathematically given as

n2= 1.23 moles

Question Parameter(s):

a propane tank containing 55l

has 2.46 moles of the gas C3 is propane

Generally, the equation for the volume  is mathematically given as

V1/n1 = V2/n2

Therefore

55 L/2.46  = 27.5 L/ n2

n2 = (2.46 * 27.5)/55

n2= 1.23 moles

In conclusion, the number of moles

n2= 1.23 moles

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The pressure would be 6000 Newton’s

The pressure of I when the system returns to equilibrium is 0.589 atm.

Equilibrium is defined as a state of a reversible chemical process in which there is no net change in the quantity of reactants and products.

To calculate the pressure we considered the equation

I2 -> 2Ig

Kp = I² / I2

     = (0.23)² / 0.21 = 0.25

I2 at initial = 2 x 0.21 = 0.42

I2 at equilibrium = 0.42 - x

Ig at initial = 2 x 0.23 = 0.46

Ig at equilibrium = 0.46 - 2x

Solving x using equilibrium constant

0.25 = (I)² / I2 = (0.46 + 2x)² / 0.42-x

0.25 (0.42 - x) = (0.46 + 2x)²

0.105 - 0.25x = 0.1764 + 4x²

-0.25 - 4x² = 0.1764 - 0.105

4.25 x² = 0.0714

x² = 0.714 / 4.25

x² = 0.0168

x = 0.129 atm

PI2 = 0.42 + 2x0.129

PI2 = 0.678 atm

PI = 0.46 + 0.129

PI = 0.589 atm

Thus, the pressure of I when the system returns to equilibrium is 0.589 atm.

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Applied chemistry is key in making ice cream, particularly through managing the crystallization process and emulsion stability. Adjusting the mixture with substances like glucose or corn syrup helps maintain a smooth texture by preventing large ice crystals. Hydrocolloid stabilizers are used to improve emulsion stability, ensuring a creamy consistency.

Applied chemistry plays a crucial role in making ice cream, a process filled with fascinating chemical reactions and principles. Making ice cream involves understanding the behavior of mixtures, the crystallization process, and the effects of temperature changes. Particularly, the process of crystallization is central to achieving the smooth texture of ice cream. As the ice cream mixture cools, small ice crystals form. To ensure these crystals remain small, preventing coarse texture, substances like glucose or corn syrup can be added as interferents. These substances disrupt the crystallization of water and fats, maintaining a smooth texture.

In addition to crystallization, emulsion stability is also crucial. The ice cream mix is an emulsion of fat in water, and maintaining this emulsion is key to preventing separation and ensuring a creamy consistency. Hydrocolloid stabilizers, such as locust bean gum, are often added to improve this stability. Therefore, applied chemistry in ice cream making involves manipulating the properties and interactions of ingredients to achieve the desired texture, appearance, and taste of the final product.

Answer:

Explanation:

1) Chemical equation:

2) Oxidation states in the reactants

a) C₆H₁₂O₆

Here, you will see that the oxidation state of carbon in glucose is zero.

There are 6 atoms of C which may have different oxidation number.

You can work with a more developed formula such as:

     O    OH   OH    OH   OH  OH

     ║     |        |         |       |       |

H - C¹ -  C²  -  C³  -  C⁴ - C⁵  - C⁶ - H

             |        |         |       |       |

             H     H       H      H     H

        ∴ C¹ 's oxidation state may be determined from x + (−2) + (0) + (+1) = 0

            ⇒ x = + 1

       ∴ Their oxidation states are given by:

            x + (−1) + 2(0) + (+1) = 0 ⇒ x=0

        ∴ C⁶ 's oxidation state may be determined from:

             x + (−1) + (0) + 2(+1) = +1 ⇒ x = −1

b) O₂ : since it is alone, its oxidation state is zero.

3) Oxidation states in the products

c)  CO₂

d) H₂O

4) Changes in the oxidation state

5) Conclusion:

Answer:

C₆H₁₂O₆ is oxidized and O₂ is reduced.

Explanation:

1-b

2-c

3-a

4-d

5-d

6-b

7-a

a catabolic pathway. Cellular respiration is a catabolic pathway.  

Answer:

A

Explanation:

Autotrophs utilize the energy from  sunlight to reduce carbon dioxide to carbohydrates (glucose). The energy from the sunlight is used to split water into H+ and O2- and the H+ used in the reduction process. The labeled carbon in the carbon dioxide will, therefore, be incorporated by the autotrophs in the carbohydrates made in photosynthesis.  

Im not sure what it is but i will try      

BaCl₂(aq) + Na₂SO₄(aq) = BaSO₄(s) + 2NaCl(aq)

Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) = BaSO₄(s) + 2Na⁺(aq) + 2Cl⁻(aq)

Ba²⁺(aq) + SO₄²⁻(aq)= BaSO₄(s)

The ionic reaction is

Ba2+ + 2 Br- + 2 Na+  + SO42----->   2 Na+ + 2Br- + BaSO4.

Na+ and Br- are on both sides of the equation so they are spectator ions.

The answer is A Br- (aq).

Answer:2:3

Explanation: on edginuity

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Answer:

1038.96 kPa

Explanation:

We’ll use the ideal gas law; P1V1/T1 = P2V2/T2

P1*14.8/75.5 = 101.3*16.5/70.2

P1 = (101.3 * 16.5 * 75.5) / (70.2 *14.8)

P1 = 1038.96

Answer:

This is a form of peer review

Explanation:

This is used to validate the work of the other scientists and add veracity to findings or question the findings. This peer review is used to improve the quality of research in science. Peer-reviewed articles provide a trusted form of scientific communication.

Answer:

Reproducibility

Explanation:

Reproducibility or replication means that the results of scientific studies can be exactly obtained by other scientists working independently but following the same procedure as described in a previous scientific study.

Scientific research is generally expected to be utterly reproducible in all ramifications.

However, due to insufficient information regarding research methodologies,bias in reportage of research findings and poor experimental design, some scientific studies may not be readily reproducible.

That will make a gold-202 nucleus.

Refer to a periodic table. The atomic number of mercury Hg is 80.

Step One: Bombard the [tex]\displaystyle ^{202}_{\phantom{2}80}\text{Hg}[/tex] with a neutron [tex]^{1}_{0}n[/tex]. The neutron will add 1 to the mass number 202 of [tex]^{202}_{\phantom{2}80}\text{Hg}[/tex]. However, the atomic number will stay the same.

[tex]^{202}_{\phantom{2}80}\text{Hg} + ^{1}_{0}n \to ^{203}_{\phantom{2}80}\text{Hg}[/tex].

Double check the equation:

Step Two: The [tex]^{203}_{\phantom{2}80}\text{Hg}[/tex] nucleus loses a proton [tex]^{1}_{1}p[/tex]. Both the mass number 203 and the atomic number will decrease by 1.

Refer to a periodic table. What's the element with atomic number 79? Gold Au.

[tex]^{203}_{\phantom{2}80}\text{Hg} \to ^{202}_{\phantom{2}79}\text{Au} + ^{1}_{1}p[/tex].

Double check the equation:

A gold-202 nucleus is formed.

Answer: 1.30

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

[tex]pH=-\log [H^+][/tex]

[tex]pH+pOH=14[/tex]

Given: 0.050 M [tex]HNO_3[/tex]

[tex]HNO_3\rightarrow H^++NO_3^-[/tex]

Concentration of [tex]H^+[/tex] = 0.050 M

[tex]pH=-log[0.050M][/tex]

[tex]pH=1.30[/tex]

Thus pH of 0.050 M of [tex]HNO_3[/tex] is 1.30.

Answer:

1.30 is the pH balance

Explanation:

The increase in temperature when solid X is added to solution Y indicates an exothermic reaction, and the further increase with additional solid could be due to supersaturation, the disturbance of equilibrium, and additional exothermic reactions.

When solid X is added to solution Y and the temperature rises, this indicates an exothermic reaction, or a reaction that releases heat. The solubility of the solid in the solution may also be temperature-dependent, potentially leading to a supersaturated solution if the temperature is cooled after the solute is added. This supersaturated solution is relatively stable, but adding more solid or initiating other forms of agitation can disturb this equilibrium and trigger further reactions, potentially causing additional temperature changes. Note that this behavior is similar to certain hand warmer mechanisms that take advantage of such exothermic reactions and solubility behaviors to generate heat.

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