Sorry this is a bit long

b. The following reaction takes place in an acidic solution. Use the following steps to balance the equation. (5 points)
Zn(s) + NO3–(aq) Zn2+(aq) + NO(g)


i. Write the reduction and oxidation half-reactions (without electrons). (1 point)











ii. Balance the equations for atoms. Use H2O and H+ to balance O and H. (1 point)
















iii. Balance the equations for charge. (1 point)











iv. Equalize charges for the two half-reactions. (1 point)











v. Add the equations and simplify to get a balanced equation. (1 point)

Answer:

Explanation:

Driving a car (burning gas is a chemical change) and almost all the plastics we use are made by chemical reactions of different components.

Answer:

Applied Chemistry is the scientific field for understanding basic chemical properties of materials and for producing new materials with well-controlled functions. (:

Answer:

Explanation:

Applied Chemistry is the scientific field for understanding basic chemical properties of materials and for producing new materials with well-controlled functions. (:

Answer: I go search some information

Explanation:I come back with a answer

Human blood appears to be a red liquid to the naked eye, but under a microscope we can see that it contains four distinct elements:  

   IamSugarBee

The complete sentences are:

Page 3:

The effect of energy in phase transitions of matter is that it is required to break the intermolecular forces that hold the particles of a substance together. When energy is added to a substance, the particles move faster and the intermolecular forces are broken. This can cause the substance to change phase.

The interactive demonstration on the sample of water shows that energy is required to melt ice and boil water. When the ice is heated, the particles start to move faster and the ice melts. The temperature of the water stays constant at 0°C until all of the ice has melted. This is because the energy is being used to break the intermolecular forces in the ice. Once all of the ice has melted, the temperature of the water starts to rise again. When the water is boiled, the particles move so fast that they escape from the liquid state and become a gas. The temperature of the water stays constant at 100°C until all of the water has boiled. This is because the energy is being used to break the intermolecular forces in the water. Once all of the water has boiled, the temperature of the steam starts to rise again.

The complete sentences:

Page 4:

Heating curves show the temperature of a substance as it is heated. The curve has a horizontal line at the melting and boiling points, which indicates that the temperature does not change during these phase changes.

Cooling curves show the temperature of a substance as it is cooled. The curve has a horizontal line at the melting and boiling points, which indicates that the temperature does not change during these phase changes.

Both curves show that the temperature of a substance increases as it is heated and decreases as it is cooled.

A heating curve is more choppy than a cooling curve because there are more phase changes during heating than during cooling.

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Answer:

see below (I hope this helps!)

Explanation:

You are a student, therefore, the people who are the same "rank" as you are your fellow students, this makes them your peers.

Answer:

CO

Explanation:

the rest are elements. CO is made up of one carbon atom and one oxygen atom

Answer:

Fe

Iron – Ferrum (Fe)

Iron's Latin name, 'ferrum', gives it its symbol Fe; it simply means 'iron' or 'sword', and is possibly of Semitic origin.

The number of moles of calcium chloride ( CaCl₂) are contained in the given sample is 0.045 moles.

Given that :

The mass of the calcium chloride, CaCl₂ = 5 g

The molar mass of the calcium chloride, CaCl₂ = 110.98 g /mol

The number of moles can be calculated by the formula given below :

The number of moles = mass / molar mass

where,

Mass = 5 g

Molar mass = 119.98 g/mol

The number of moles = 5 g / 110.98 g /mol

The number of moles = 0.045 moles.

The moles of the calcium chloride ,  CaCl₂ is 0.045 mol.

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The false statement is applied pressure only refers to atmospheric pressure and the correct option is option 1.

It is the ratio of the force applied to the surface area over which the force is applied. Pressure can be defined as the force applied perpendicular to the surface of an object per unit area over which that force is distributed. Atmospheric pressure is the pressure exerted by the atmosphere on the earth.

Thus applied pressure not only represents atmospheric pressure.

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Based on formal charges and electronegativity, the more likely Lewis structure between the two is the RIGHT structure because oxygen is more electronegative than nitrogen.

Electronegativity is the ability of an atom to attract electrons in a chemical bond, and since oxygen has a higher electronegativity than nitrogen, it is more stable when it has a higher number of bonds in the structure. The right structure better fulfills this requirement, making it more likely.

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Answer:

PLA is very user friendly that is stronger and stiffer than other materials. It melts easily and doesn't warp very often. Although it is very brittle, it still is a very popular material choice.

Explanation:

Answer:

if i remember correctly it's B

Explanation:

The heat generated by the freezing of 2.5g of water is 2.5 g * 1 mol/18.02 g/mol * -285.83 kJ/mol.

The term freezing has to do with the change of water from liquid to solid. This is accompanied by an enthalpy change.

From the options that are presented in the question, the heat generated by the freezing of 2.5g of water is 2.5 g * 1 mol/18.02 g/mol * -285.83 kJ/mol.

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Answer:

ughh blue milk and oreos

Explanation:

An imbalance in the lonization energy leads to the distribution of charge in molecules. A polar molecule is one that has an unequal distribution of charges, causing it to have a positive end and a negative end.

In the case of an uneven distribution of charges, which molecules are present?

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Answer:

through water

Explanation:

I THINK 18 OR 21 I AM SORRY IF I AM WRONG BRO

The right response is b) APF is higher for a box filled with bronze balls.

The atomic packing factor (APF) is a measure of the amount of space that is occupied by atoms in a crystal structure. It is calculated by dividing the total volume of atoms in a unit cell by the volume of the unit cell. The higher the APF, the more closely packed the atoms are in the crystal structure.

In this case, since the boxes have the same dimensions, the unit cell for both boxes will be the same. However, the size and weight of the balls differ. The wooden balls are more massive than the hollow bronze balls, but they are smaller in size. This means that the wooden balls will have a higher density and will occupy less space than the hollow bronze balls.

Since the APF is calculated based on the volume of atoms in the unit cell divided by the volume of the unit cell, the size and density of the balls will affect the APF. In this case, the hollow bronze balls will occupy more space in the unit cell compared to the wooden balls, as they have a larger radius and lower density. Therefore, the box filled with hollow bronze balls will have a higher APF compared to the box filled with wooden balls.

Therefore, the correct answer is b) Box filled with bronze balls has higher APF.

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Answer:

C8H10

Explanation:

n (CH2) = 112

n (12 + 1 + 1) = 112

n (14) = 112

n = 8

Molecular Formula: C8H10

43.2 g of H₂0 would form when 4.8 g of Hydrogen reacts with 38.4 g of Oxygen.

Stoichiometry is defined as the relationship between reactants and products that are used to determine quantitative data.

It is the study of balanced chemical reactions.

To find the mass of the H20 form, we have to use stoichiometry.

First of all, we have to write a balanced chemical reaction

The reaction is as follows:

\(2H_2 + O_2 \rightarrow 2H_2O\)

The molar mass of O₂ = 32 g

The molar mass of H₂O = 18 g

According to the above reaction,

2 moles of Hydrogen react with 1 mole of Oxygen to form 2 moles of H2O

According to stoichiometry,

32 g of O₂ forms 36 g of H₂O

38.4 g of O₂ forms \(\frac{36 \times 38.4 } {32}\)of H₂O

                         = 43.2 g of H₂O

Thus, the mass of H₂Oform when 4.8 g of Hydrogen reacts with 38.4 g of Oxygen is 43.2 g

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1. The volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2.The appropriate initial condition for the ODE is S(0) = 0, as there is no salt initially in the tank.

3. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. The solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^{2/2} + 30t)} dt) / e^{(t^{2/2} + 30t)})\)

5.  the salt concentration in the tank as t→infinity is zero.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being added and drained. The tank is being filled at a rate of 2 liters per minute and drained at a rate of 1 liter per minute.

Since the tank starts with an initial volume of 30 liters, the volume of water in the tank at time t can be calculated using the equation:
Volume(t) = Initial volume + (Rate of filling - Rate of draining) * t

Volume(t) = 30 + (2 - 1) * t

So, the volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2. Let S(t) denote the amount of salt in the fish tank at time t in grams.

To show that S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t),

we need to take the derivative of S(t) with respect to t and substitute it into the given ODE.

Taking the derivative of S(t), we have:

S'(t) = 0 - (1+0)S(t) + 0

S'(t) = -S(t)

Substituting this into the given ODE, we get:

-S(t) = 70 - (t+30)S(t)

Simplifying the equation, we have:

S'(t) = 70 - (t+30)S(t)

Therefore, S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t).

The appropriate initial condition for the ODE is S(0) = 0,

as there is no salt initially in the tank.

3. This ODE is a first-order linear ordinary differential equation. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. To solve the initial value problem for S(t) using the method of integrating factors, we first rewrite the ODE in standard form:

S'(t) + (t+30)S(t) = 70

The integrating factor is given by:
\(\mu(t) = e^{(\int (t+30) dt)} = e^{(t^2/2 + 30t)\)

Multiplying both sides of the equation by μ(t), we have:
\(e^{(t^2/2 + 30t)} * S'(t) + e^{(t^2/2 + 30t)} * (t+30)S(t) = 70 * e^{(t^2/2 + 30t)\)

Applying the product rule to the left side of the equation, we get:
\((e^{(t^{2/2} + 30t) * S(t))' = 70 * e^{(t^{2/2} + 30t)})\)

Integrating both sides of the equation with respect to t, we have:
\(\int (e^{(t^2/2 + 30t)} * S(t))' dt = \int (70 * e^{(t^2/2 + 30t))} dt\)

Using the fundamental theorem of calculus, the left side becomes:
\(e^{(t^2/2 + 30t)} * S(t) = \int (70 * e^{(t^2/2 + 30t))} dt\)

Simplifying the right side by integrating, we get:
\(e^{(t^2/2 + 30t)} * S(t) = 70 * \int e^{(t^2/2 + 30t)} dt\)

At this point, the integration of \(e^{(t^2/2 + 30t)\) becomes difficult to express in terms of elementary functions.

Hence, the solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^2/2 + 30t)} dt) / e^{(t^2/2 + 30t)\)

5. As t approaches infinity, the exponential term \(e^{(t^2/2 + 30t)\) becomes very large, causing the salt concentration S(t) to approach zero. Therefore, the salt concentration in the tank as t→infinity is zero.

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The salt concentration in the tank as t approaches infinity is 70/3.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being poured into the tank and the rate at which water is being drained from the bottom.

At a rate of 2 litres per minute, water is being poured into the tank. So after t minutes, the amount of water poured into the tank is 2t litres.

At a rate of 1 litre per minute, water is being drained from the tank. So after t minutes, the amount of water drained from the tank is t litres.

Since the tank was initially filled with 30 litres of fresh water, the volume of water in the tank at time t is given by:
Volume(t) = 30 + 2t - t
Volume(t) = 30 + t

2. Let S(t) denote the amount of salt in the fish tank at time t. To determine the ODE for S(t), we need to consider the salt being poured into the tank and the salt being drained from the tank.

The salt concentration in the water being poured into the tank is 35 grams per litre. So the amount of salt being poured into the tank per minute is 35 * 2 = 70 grams.

The amount of salt being drained from the tank per minute is S(t)/Volume(t) * 1.

Therefore, the ODE for S(t) is:
S'(t) = 70 - S(t)/Volume(t)

The initial condition for this ODE is S(0) = 0, since there was no salt in the tank initially.

3. The ODE S'(t) = 70 - S(t)/Volume(t) is a first-order linear ODE. It is not separable since the variables S(t) and Volume(t) are mixed together.

4. To solve the initial value problem for S(t), we can rewrite the ODE as:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)

This is a linear ODE of the form y'(t) + p(t)y(t) = g(t), where p(t) = 1/Volume(t) and g(t) = 70 * Volume(t).

To solve this type of ODE, we can multiply both sides by an integrating factor, which is the exponential of the integral of p(t).

The integrating factor is exp(integral of 1/Volume(t) dt) = exp(ln(Volume(t))) = Volume(t).

Multiplying both sides of the ODE by the integrating factor, we get:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)
Volume(t) * S'(t) + Volume(t) * S(t) = 70 * Volume(t)^2
( Volume(t) * S(t) )' = 70 * Volume(t)^2

Integrating both sides with respect to t, we get:
Volume(t) * S(t) = 70/3 * Volume(t)^3 + C

S(t) = 70/3 * Volume(t)^2 + C/Volume(t)

Using the initial condition S(0) = 0, we can solve for C:
0 = 70/3 * 30^2 + C/30
C = -70000

Therefore, the solution for S(t) is:
S(t) = 70/3 * Volume(t)^2 - 70000/Volume(t)

5. As t approaches infinity, the volume of water in the tank becomes very large. In this case, we can approximate the volume of the tank as t, since the rate at which water is being poured in is 2 litres per minute. So the salt concentration in the tank as t approaches infinity is given by:
S(t)/Volume(t) = (70/3 * t^2 - 70000/t) / t
As t approaches infinity, the second term (-70000/t) approaches 0, so the salt concentration in the tank as t approaches infinity is:
S(t)/Volume(t) = 70/3 * t^2 / t = 70/3 * t

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a) The maximum observable range of redshifts that produces Lyman-alpha line is 0 ≤ z ≤ 10.6

b) i) identifying the galaxy with a radio source, ii) looking for other Lyman lines, iii) a coincidence with a continuum break

c)The maximum redshift range over which galaxies can be found using this strategy is z = 7 to z = 15.5.

d)Earth's atmosphere absorbs the radiation, and the laboratory conditions are not the same as interstellar conditions.

a) Lyman-alpha line is produced by the hydrogen atoms that have electrons that are in the ground state being raised to the first excited state. Over a certain range of redshifts, the Lyman-alpha line is redshifted to longer wavelengths that are observable by an optical spectrograph. The maximum observable range of redshifts that produces Lyman-alpha line is 0 ≤ z ≤ 10.6 (depending on the exact details of the galaxy's emission profile).

b) Observing the galaxy in the infrared can help in the identification of the Lyman-alpha line as it is shifted to longer wavelengths. Three ways to increase confidence in the correct identification of the Lyman-alpha line are:

i) identifying the galaxy with a radio source, ii) looking for other Lyman lines, iii) a coincidence with a continuum break.

c) The strategy that needs to be adopted is to look for the Lyman limit, which is the point at which the spectrum is cut off by the absorption of all hydrogen in the galaxy. To be certain that the line you detect is the Lyman-alpha line, you need to look for a decrement in the flux of the galaxy at wavelengths shorter than the line and a decrement in the flux at wavelengths longer than the line. This is because the Lyman limit will be shifted to longer wavelengths at higher redshifts, so to maximize the redshift range over which galaxies can be found, you need to search for the Lyman limit at the longest wavelength possible. The maximum redshift range over which galaxies can be found using this strategy is z = 7 to z = 15.5.

d) The reason why such lines can be seen from interstellar gas but not the Earth's atmosphere or in the laboratory is that the Earth's atmosphere absorbs the radiation, and the laboratory conditions are not the same as interstellar conditions. The forbidden lines from the interstellar gas are not affected by dust absorption because they are produced in regions where dust is not present.

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The binder found in high-fiber foods that inhibits the absorption of iron and zinc is oxalic acid Choose raw fruits and vegetables instead of juice, and consume the peels. Alternative fiber sources include whole buckwheat, whole wheat couscous, quinoa, bulgur, wheat germ, chia water seeds, hemp seeds, lentil pasta, and edamame pasta.

Popcorn is a complete grain. Serve it low-fat and without butter for a healthy snack option. It is a white crystalline substance that dissolves colorlessly in water. It got its name from the fact that early researchers obtained oxalic acid from blooming plants of the genus Oxalis, sometimes known as wood-sorrels.

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The product at the negative electrode is not always a metal because it will only be produced if it is less reactive than hydrogen.

These are materials which donate electrons in a chemical reaction and have high heat and electrical conductivity.

Metals are more reactive than hydrogen which is why it is not always produced at the cathode.

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