Mass number symbol =

Answer: The mass

Explanation: ability to rust, flammability, and ability to combust are chemical properties.

The platinum-rhodium catalyst used in most modern catalytic converters in automobiles is primarily employed for the oxidation of harmful pollutants. It facilitates the conversion of carbon monoxide (CO) and unburned hydrocarbons (HC) into carbon dioxide (CO2) and water (H2O).

The platinum-rhodium catalyst in catalytic converters is specifically designed to promote the oxidation reactions of carbon monoxide (CO) and unburned hydrocarbons (HC). These reactions are crucial for reducing the emission of harmful pollutants from automobile exhaust gases.

Carbon monoxide (CO) is a toxic gas produced during incomplete combustion. The platinum-rhodium catalyst assists in the oxidation of CO, converting it into carbon dioxide (CO2). This reaction is represented by the equation:

2 CO + O2 → 2 CO2

Unburned hydrocarbons (HC) are volatile organic compounds (VOCs) that contribute to smog formation. The platinum-rhodium catalyst aids in their oxidation, transforming them into carbon dioxide (CO2) and water (H2O). The general reaction can be expressed as:

HC + O2 → CO2 + H2O

The platinum-rhodium catalyst is essential in facilitating these oxidation reactions, promoting more complete combustion of harmful pollutants and reducing their negative environmental impact.

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Answer:

V = 5000 mL

Explanation:

¡Hola!

En este caso, dado que la molaridad de una solution se calcula al dividir las moles por el volume de solución en litros, es posible calcular el volumen cuando se dividen las moles por la molaridad, tal y como se muestra a continuación:

[tex]M=\frac{n}{V}\\\\V=\frac{n}{M}[/tex]

Así, podemos reemplazar la molaridad y moles dadas para obtener:

[tex]V=\frac{1.5mol}{0.3mol/L}=5L[/tex]

Que en mililitros sería:

[tex]V=5L*\frac{1000mL}{1L}\\\\V=5000mL[/tex]

¡Saludos!

To determine the empirical formula of a compound, the given masses of the elements (carbon and oxygen) are used to calculate the moles of each element. The mole ratio between the elements is then determined to find the simplest whole number ratio, which represents the empirical formula.

First, we calculate the moles of carbon and oxygen using their respective molar masses. The molar mass of carbon (C) is approximately 12.01 g/mol, and the molar mass of oxygen (O) is approximately 16.00 g/mol.

Moles of carbon = 20.4 g / 12.01 g/mol ≈ 1.70 mol

Moles of oxygen = 54.4 g / 16.00 g/mol ≈ 3.40 mol

Next, we determine the simplest whole number ratio by dividing the number of moles of each element by the smallest number of moles. In this case, the smallest number of moles is approximately 1.70 mol (carbon).

Carbon: 1.70 mol / 1.70 mol ≈ 1

Oxygen: 3.40 mol / 1.70 mol ≈ 2

Therefore, the empirical formula for this compound is C1O2, which can be simplified to CO2.

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Answer:

It would be C. Wind blowing against a rock can reduce the rock´s size.

Explanation:

Ask your parents or guardian for a tutor.

It's very dangerous to give someone your snap online that you don't know!

Have a nice day <3

There are 5.50 equivalents of Mg^2+ present in a solution containing 2.75 mol of Mg^2+.

The concept of equivalents is used to quantify the number of reactive entities or charges present in a solution. In the case of Mg^2+, each Mg^2+ ion carries two positive charges, so it is necessary to determine the number of moles of Mg^2+ and then convert it to equivalents.

Given:

Number of moles of Mg^2+ = 2.75 mol

To calculate the equivalents, we use the relationship that one mole of Mg^2+ is equal to 2 equivalents of Mg^2+ (since each Mg^2+ ion carries two positive charges):

Equivalents of Mg^2+ = Number of moles of Mg^2+ * 2

Equivalents of Mg^2+ = 2.75 mol * 2

Equivalents of Mg^2+ = 5.50 equivalents

Therefore, in a solution containing 2.75 mol of Mg^2+, there are 5.50 equivalents of Mg^2+ present.

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Answer:

10.58 J/g-°C

Explanation:

To find the specific heat capacity of the metal, you need to know how much heat was lost when it reacted with water.

You know that there are 55.4 g of water, the initial temp. of water is 25°C, and the final temp. (the mixture's temp.) is 63.4°C.

You should also know that the specific heat capacity of water is 4.186 J/g-°C.

Plug this into the equation the q=mcΔT.

q = (55.4 g)(4.186 J/g-°C)(63.4°C - 25°C)

q = 8905.12896 J

If 8905.12896 J was gained by the water, then 8905.12896 J must have been lost from the metal.

You know that there are 23 g of the metal and that its initial temp. is 100°C.

Plug this information into q=mcΔT.

8905.12896 J = (23 g)(C)(63.4°C - 100°C)

C = 10.58 J/g-°C

*When you plug all of this into the calculator, it will result in a negative number but keep in mind that heat was LOST by the metal so 8905.12896 J  is essentially negative. So the negative cancels out.*

Answer:

i dont know i do this for points

Explanation:

Answer:

How to make scientific observations?

Observe something through your senses or record information using scientific tools and instruments and ask questions about your scientific observations (e.g., natural phenomena).

How to calculate data involving metric units?

To convert from one unit to another within the metric system usually means moving a decimal point (e.g., 1000000mm = 100000cm = 10000dm = 1000m = 100dkm = 10hm = 1km).

Answer:

5 mL

Explanation:

As this is a problem regarding dilutions, we can solve it using the following formula:

Where subscript 1 refers to the initial concentration and volume, while 2 refers to the final C and V. Meaning that in this case:

We input the data:

And solve for V₁:

Answer:

5 mL of 5.0 M H₂SO₄ (aq) are needed to prepare 100 mL of 0.25 M H₂SO₄ (aq).

Explanation:

In chemistry, dilution is the reduction of the concentration of a chemical in a solution.

Then, dilution consists of preparing a less concentrated solution from a more concentrated one, and it consists simply by adding more solvent to the same amount of solute. That is, the amount or mass of the solute is not changed, but the volume of the solvent varies: as more solvent is added, the concentration of the solute decreases, since the volume (and weight) of the solution increases.

A dilution is calculated by the expression:

Ci*Vi = Cf*Vf

where:

In this case, you know:

Replacing:

5 M*Vi = 0.25 M* 100 mL

Solving:

[tex]Vi= \frac{0.25 M*100 mL}{5 M}[/tex]

Vi= 5 mL

5 mL of 5.0 M H₂SO₄ (aq) are needed to prepare 100 mL of 0.25 M H₂SO₄ (aq).

Answer:

Chromatography

Distillation

Evaporation

Filtration

Explanation:

Chromatography involves solvent separation on a solid medium.

Distillation takes advantage of differences in boiling points.

Evaporation removes a liquid from a solution to leave a solid material.

Filtration separates solids of different sizes.

Mixtures can be physically separated by using methods that use differences in physical properties to separate the components of the mixture, such as evaporation, distillation, filtration and chromatography.

Based on the hypotheses discussed in Chapter 12 regarding melting points and the forces that hold materials together, we can make a prediction about the forces holding atoms of cesium and lithium together in their solid metal forms.

One hypothesis suggests that stronger forces between atoms result in higher melting points. Another hypothesis proposes that metals are held together by metallic bonds, where positively charged metal ions are surrounded by a sea of delocalized electrons.

Considering these hypotheses, we can infer that cesium atoms would be held together by stronger forces compared to lithium atoms in their solid metal forms. This is because cesium is located further down the periodic table, belonging to Group 1 (alkali metals), whereas lithium is in Group 2 (alkaline earth metals). As we move down a group in the periodic table, the atomic radius generally increases, leading to weaker forces of attraction between atoms.

Therefore, the larger atomic size of cesium compared to lithium would result in weaker interatomic forces, making cesium's solid metal form have a lower melting point compared to lithium.

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Explanation:

molarity = no. of moles of solute/solution in litres

molarity =0.202/7.98

=0.025 M

Lewis acid: Fe(NO3)3Lewis base: H2OThe Lewis acid is H+ and the Lewis base is NH3.

The Lewis concept of acids and bases, Lewis acids are those that have an incomplete valence shell and accept electrons from a Lewis base. A Lewis base has at least one electron pair available to form a covalent bond with a Lewis acid, filling its valence shell. The Lewis acid and base in each of the following reactions are given below:Fe(NO3)3(s) + 6H2O(l) ⟶ Fe(H2O)63+(aq) + 3NO3?(aq)The Lewis acid is Fe3+ and the Lewis base is H2O.NH3(g) + HCl(g) ⟶ NH4Cl(s)The Lewis acid is H+ and the Lewis base is NH3.Neither HCl nor NH3 is a Lewis acid or base in this particular reaction. Therefore, you made a mistake in your answer. The correct answers are as follows:Lewis acid: Fe(NO3)3Lewis base: H2OThe Lewis acid is H+ and the Lewis base is NH3.

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Answer: The percentage by mass of sulphur in [tex]Al_2(SO_4)_3[/tex] is 9.36%

Explanation:

Mass percent of an element is the ratio of mass of that element by the total mass expressed in terms of percentage.

[tex]{\text {Mass percentage}}=\frac{\text {mass of sulphur}}{\text {Total mass}}\times 100\%[/tex]

Given: mass of sulphur = 32 g/mol

mass of [tex]Al_2(SO_4)_3[/tex] = 342 g/mol

Putting in the values we get:

[tex]{\text {Mass percentage}}=\frac{32g/mol}{342g/mol}\times 100\%=9.36\%[/tex]

The percentage by mass of sulphur in [tex]Al_2(SO_4)_3[/tex] is 9.36%

The MW of rGFP in the band can be determined by comparing the mobility of the band with that of protein standards run on the same gel.

The Western blotting technique is utilized to identify and detect specific proteins in a sample of tissue or extract. If there is a band in the W2 lane of the Western blot, one can conclude that rGFP is present in that band. The physical protein structure of rGFP could not be inferred from the presence of a band in the W2 lane of the Western blot.  This requires additional analysis such as X-ray crystallography, nuclear magnetic resonance, or cryo-electron microscopy to analyze protein structure. MW is the molecular weight which can be determined using a molecular weight marker that runs in a parallel lane to the protein extract on the gel. In the Western blotting method, SDS-PAGE is typically used to separate proteins based on their size. The SDS-PAGE gel is calibrated with protein markers that have a known molecular weight. Hence, the MW of rGFP in the band can be determined by comparing the mobility of the band with that of protein standards run on the same gel.

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Answer:

The population of the long-legged birds decreases.

Explanation:

The population of the short-legged birds increases whereas the population of  long-legged birds decreases due to availability of food in that environment. The long-legged birds feed on fish whose population decreases due to drought conditions so the population of long-legged birds also decreases while on the other hand, the population of short-legged birds increases or remain the same because they feed on the insects and the insects are available in large amount and less affected by the drought conditions.

Sublimation is an effective technique for the isolation of pure caffeine because: The purified caffeine will vaporize and can be collected as a solid, leaving impurities behind.

So, the answer is B.

Sublimation is a chemical technique that is used to isolate the pure form of a substance from impure or mixed form. It is the phase transition of a solid directly to a gas without passing through a liquid phase.

It is an effective technique for the isolation of pure caffeine due to the following reasons:During sublimation, the purified caffeine will vaporize and can be collected as a solid, leaving impurities behind.

The caffeine is heated, and the heat causes the solid to vaporize directly from the solid phase to the gas phase, skipping the liquid phase. Thus, it separates the caffeine from impurities and provides pure caffeine. Therefore, option B is correct.

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A low vapor pressure indicates the presence of weak intermolecular forces in a liquid. The correct answer is: c.

The strength of intermolecular forces in a liquid determines the vapor pressure of the liquid. A liquid with strong intermolecular forces will have a low vapor pressure, while a liquid with weak intermolecular forces will have a high vapor pressure.

This is because the molecules in a liquid with weak intermolecular forces are more likely to escape from the surface of the liquid and enter the gas phase.

The other options are incorrect because they are all properties that indicate the presence of strong intermolecular forces in a liquid. A high boiling point indicates that a large amount of energy is required to overcome the intermolecular forces and vaporize the liquid.

A high surface tension indicates that the molecules in the liquid are strongly attracted to each other and to the surface of the liquid. A low heat of vaporization indicates that a small amount of energy is required to overcome the intermolecular forces and vaporize the liquid.

Therefore, the only property that indicates the presence of weak intermolecular forces in a liquid is a low vapor pressure.

Therefore, the correct option is C, a low vapor pressure.

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With a simple main effect, the results are analyzed as if you had separate experiments at each level of the other independent variable.

In the realm of factorial ANOVA, an enlightening endeavor known as a simple main effect analysis arises. Within this statistical examination, the intricate interplay of two or more independent variables upon a dependent variable is meticulously unraveled.

When a notable interplay between these independent variables materializes, the pursuit of comprehension beckons the astute pursuit of simple main effects analyses, illuminating the essence of the interaction.

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Answer:

The final balanced equation is : 2C2H2+5O2→4CO2+2H2O.

Answer: There is no eclipse. However, two or four times a year, the Moon travels through some portion of the Earth's penumbral or umbral shadows, resulting in one of the three types of eclipses mentioned above. When the Moon crosses between the Earth and the Sun, this occurs. This is only possible when the Moon is in its New Moon phase.

Explanation:

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Answer:

The amplitude of the speedboat waves were larger then the Breeze waves. The frequency of the speedboats waves were lower than the breeze waves. the speedboat waves had more of an effect on the boat, which tells us the speedboat waves had more energy.

Explanation:

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Answer:

speedboat waves artificail and waves by breezes natural

Explanation:

Answer:

Molarity is moles per liter. You have one mole in 0.750 liters

Explanation:

Answer:

3fe s )+ 4h2o G )= fe3o4 s )+ 4h2 G

Explanation: 3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g) (i) Iron metal is getting oxidised. (ii) Water is getting reduced. (iii) Water is acting as reducing agent.

Explanation:

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Answer:Divergent Boundaries(or Boundary)

Explanation:took the test and got it right

The balanced equation is:

(a) 2Mn^2+ + 2H2O2 + 4OH^- → 2MnO_2 + 4H2O.

(b)3SnO2^2- + 6OH^- + 2Bi(OH)3 → 3SnO3^2- + 2Bi + 9H2O. (c)14Cr2O7^2- + 7C2O4^2- + 22H2O → 4Cr^3+ + 14CO2 + 28H+ + 28e^-. (d)2ClO3^- + 16H^+ + 3Cl^- → 3Cl2 + 8H2O

(e)10BiO3^- + 60H^+ + 12Mn^2+ → 10Bi^3+ + 30H2O + 12MnO4^-

a. In the balanced redox equation Mn^2+ + H_2O_2 → MnO_2 + H_2O (in basic solution), the half-reactions are:

Reduction: Mn^2+ → MnO_2

Oxidation: H_2O_2 → H_2O

To balance the reduction half-reaction, we need to add four OH^- ions to the left side: Mn^2+ + 4OH^- → MnO_2 + 2H2O + 2e^-

To balance the oxidation half-reaction, we add four OH^- ions to the right side and water molecules to balance the oxygen atoms: H2O2 + 4OH^- → 2H2O + 2e^-

Now, multiply the reduction half-reaction by two and the oxidation half-reaction by one to equalize the electrons:

2(Mn^2+ + 4OH^- → MnO_2 + 2H2O + 2e^-)

H2O2 + 4OH^- → 2H2O + 2e^-

Finally, add the two half-reactions together and cancel out the common species: 2Mn^2+ + 2H2O2 + 4OH^- → 2MnO_2 + 4H2O

b. The balanced redox equation Bi(OH)3 + SnO2^2- → SnO3^2- + Bi (in basic solution) can be balanced by following these steps:

Reduction: SnO2^2- → SnO3^2-

Oxidation: Bi(OH)3 → Bi

To balance the reduction half-reaction, add two OH^- ions to the left side: SnO2^2- + 2OH^- → SnO3^2- + H2O + 2e^-

To balance the oxidation half-reaction, add six OH^- ions to the right side: Bi(OH)3 → Bi + 3H2O + 3e^-

Multiply the reduction half-reaction by three and the oxidation half-reaction by two to equalize the electrons:

3(SnO2^2- + 2OH^- → SnO3^2- + H2O + 2e^-)

2(Bi(OH)3 → Bi + 3H2O + 3e^-)

Combine the two half-reactions and cancel out the common species:

3SnO2^2- + 6OH^- + 2Bi(OH)3 → 3SnO3^2- + 2Bi + 9H2O

c. In the acidic solution, the balanced redox equation Cr2O7^2- + C2O4^2- → Cr^3+ + CO2 can be balanced as follows:

Reduction: Cr2O7^2- → Cr^3+

Oxidation: C2O4^2- → CO2

To balance the reduction half-reaction, add seven H2O molecules to the right side: Cr2O7^2- → 2Cr^3+ + 7H2O + 14e^-

To balance the oxidation half-reaction, add eight H^+ ions to the left side:

C2O4^2- + 2H2O → 2CO2 + 4H+ + 4e^-

Multiply the reduction half-reaction by two and the oxidation half-reaction by seven to equalize the electrons:

2(Cr2O7^2- → 2Cr^3+ + 7H2O + 14e^-)

7(C2O4^2- + 2H2O → 2CO2 + 4H+ + 4e^-)

Combine the two half-reactions and cancel out the common species:

14Cr2O7^2- + 7C2O4^2- + 22H2O → 4Cr^3+ + 14CO2 + 28H+ + 28e^-

d. In the acidic solution, the balanced redox equation ClO3^- + Cl^- → Cl2 + ClO2 can be balanced as follows:

Reduction: ClO3^- → ClO2

Oxidation: Cl^- → Cl2

To balance the reduction half-reaction, add two H^+ ions to the right side:

ClO3^- + 2H^+ → ClO2 + H2O + 2e^-

To balance the oxidation half-reaction, add two H2O molecules to the left side:

2Cl^- → Cl2 + 2e^-

Multiply the reduction half-reaction by two and the oxidation half-reaction by one to equalize the electrons:

2(ClO3^- + 2H^+ → ClO2 + H2O + 2e^-)

Cl^- → Cl2 + 2e^-

Combine the two half-reactions and cancel out the common species:

2ClO3^- + 16H^+ + 3Cl^- → 3Cl2 + 8H2O

e. In the acidic solution, the balanced redox equation Mn^2+ + BiO3^- → Bi^3+ + MnO4^- can be balanced as follows:

Reduction: BiO3^- → Bi^3+

Oxidation: Mn^2+ → MnO4^-

To balance the reduction half-reaction, add six H^+ ions to the left side:

BiO3^- + 6H^+ → Bi^3+ + 3H2O + 6e^-

To balance the oxidation half-reaction, add eight H^+ ions to the right side:

Mn^2+ → MnO4^- + 8H^+ + 5e^-

Multiply the reduction half-reaction by five and the oxidation half-reaction by two to equalize the electrons:

5(BiO3^- + 6H^+ → Bi^3+ + 3H2O + 6e^-)

2(Mn^2+ → MnO4^- + 8H^+ + 5e^-)

Combine the two half-reactions and cancel out the common species:

10BiO3^- + 60H^+ + 12Mn^2+ → 10Bi^3+ + 30H2O + 12MnO4^-

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