ella is interested in combining substances to form a new material. what does she need to know before she combines the substances to create a specific material? a. whether the substances are synthetic or from natural resources b. that all materials react to combine and form new products accidentally c. the physical and chemical properties of the material that will be formed d. the physical and chemical properties of the substances and how they react chemically

The axon involved in the transmission of pain that is thinly myelinated and conducts the first feeling of pain that is often felt as coming on as a sharp, rapid feeling is called the A-delta fiber.

A-delta fibers are one of the two main types of sensory neurons involved in the transmission of pain signals to the brain. The other type is called C fibers, which are unmyelinated and transmit a slower, more diffuse and longer-lasting feeling of pain. A-delta fibers are responsible for the initial, sharp sensation of pain that is felt immediately after an injury.

In contrast, C fibers are responsible for the more persistent, dull, and aching sensation of pain that is felt after the initial sharp sensation has subsided. Both A-delta fibers and C fibers play important roles in the experience of pain, and both types of fibers are involved in the transmission of pain signals to the brain.

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Answer:

A. physical change

b. physical change

c. chemical change

d. chemical change

Answer:

The correct option is;

A. 0

Explanation:

The ammonia molecule has four pairs of electron whereby being located around the central nitrogen atom is tetrahedral in shape.

Of the four pairs of electrons, three are the bonds between nitrogen and hydrogen, such that there is only one lone pair of electron in the ammonia molecule that is located on the central nitrogen atom

The one electron on each hydrogen atoms and one of the electrons on the nitrogen atom form a bonding pair of electrons, that is the bonding pair of electrons are shared by the two nuclei, hydrogen and nitrogen

Since a lone pair are a pair of electrons attached to only one nucleus, the hydrogen in ammonia has no lone pair of electrons.

Answer:

Explanation:

The grass hopper feeds on the plants so the secondary consumer lacking its food source would decrease as the producer increases with not predator.

Answer:

Option B. 4 moles of the gaseous product

Explanation:

Data obtained from the question include:

Initial volume (V1) = V

Initial number of mole (n1) = 2 moles

Final volume (V2) = 2V

Final number of mole (n2) =..?

Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:

V1/n1 = V2/n2

V/2 = 2V/n2

Cross multiply

V x n2 = 2 x 2V

Divide both side by V

n2 = (2 x 2V)/V

n2 = 2 x 2

n2 = 4 moles

Therefore, 4 moles of the gaseous product were produced.

Answer:

We know that 18 g of H2O contains 1 mole (or 6.022 x 10^23) molecules of water because the molar mass of H2O is 18 g/mol.

So 1 g water will contain 1/18 moles of H2O molecules.

Now 15 g of water will contain 15/18 moles of H2O.

So molecules of H2O = (15/18) x 6.022 x 10^23 = 5 x 10^23 molecules.

Explanation:

I learned this in science hope this helps.

NH3 or ammonia is a basic substance. By taking a proton (H+) from an acid to create the ammonium ion (NH4+), it can function as a weak base. Ammonia will combine with water molecules in water to generate

Ammonia (NH3), an odourless gas with a wide range of industrial applications, is widely utilised. It is a crucial ingredient in the creation of plastics, cleaning supplies, and fertilisers. In addition to these uses, ammonia is also a refrigerant and is a component in the manufacturing of drugs and explosives. It can accept a proton (H+) from an acid to generate the ammonium ion (NH4+) since it is a basic molecule. Ammonia creates hydroxide and ammonium ions in water, which results in a basic solution. Yet, ammonia is also a hazardous gas that, if handled improperly, may be detrimental to both humans and animals.

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Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

\(k = \frac{1}{2} m {v}^{2} \\ \)

where

m is the mass

v is the velocity

From the question

m = 1.6 kg

v = 25 m/s

We have

\(k = \frac{1}{2} \times 1.6 \times {25}^{2} \\ = 0.8 \times 625 \\ = 500\)

We have the final answer as

Hope this helps you

Answer:

Explanation:

1. Test Your Pool Water Testing your water is always the first step when trying to correct your pool chemistry.

2. Add Muriatic Acid to Lower pH and Alkalinity.

3. Allow Time for the Acid to Circulate.

4. Aerate the Water to Raise the pH.

5. Allow Time for Aeration to Work.

Answer:

0.67 atm

Explanation:

From the question,

Applying Boyles law

PV = P'V'......................... Equation 1

Where P = Initial Pressure of the sample of gas, P' = Final pressure of the sample of gas, V = Initial volume of the sample of gas, V' = Final volume of the sample of gas.

make P' the subject of the equation above

P' = PV/V'.................... Equation 2

Given: P = 10.0 L, V = 1.0 atm, V' = 15.0 L

Substitute these values into equation 2

P' = (10.0×1.0)/15.0

P' = 0.67 atm

To determine which air mass polygon in the source regions folder would most likely be characterized as a continental tropical (ct) air mass, we need to consider the characteristics of this type of air mass.

Continental tropical air masses originate in hot and dry regions, typically over deserts or semi-arid areas. As they move away from their source region, they can become very warm and dry, and are often associated with clear skies and high temperatures.
Based on these characteristics, the air mass polygon in the source regions folder that is most likely to be characterized as a continental tropical air mass would be the one covering the southwestern United States and northern Mexico. This region is known for its hot and dry climate, with temperatures often reaching well above 100 degrees Fahrenheit in the summer months. The air mass that originates from this region would be very warm and dry, and would likely move northward into the central and eastern United States, bringing with it hot and dry conditions.
It is important to note that air masses can change and evolve as they move away from their source region, so it is possible for a continental tropical air mass to become modified or transformed as it moves into different areas. However, based on the characteristics of the source region and the expected behavior of a continental tropical air mass, the southwestern United States and northern Mexico would be the most likely source region for this type of air mass.

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Respuesta:

2.17 × 10⁻³ g/cm³

Explicación:

Paso 1: Información provista

Paso 2: Convertir 64 °C a Kelvin

Usaremos la siguiente fórmula.

K = °C + 273.15 = 64 °C + 273.15 = 337 K

Paso 3: Calcular V₂

Usaremos la ecuación combinada de los gases ideales.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

V₂ = P₁ × V₁ × T₂ / T₁ × P₂

V₂ = 96.4 kPa × 969 cm³ × 273.15 K / 337 K × 100 kPa = 757 cm³

Paso 4: Calcular la densidad del gas a  TPE

A TPE, 1.64 g del gas ocupan 757 cm³.

ρ = 1.64 g/757 cm³ = 2.17 × 10⁻³ g/cm³

Answer:

Its the third one (aa)

Explanation:

Answer:

Coefficient = 1.58

Exponent = - 5

Explanation:

pH = 2.95

Molar concentration = 0.0796M

Ka = [H+]^2 / [HA]

Ka = [H+]^2 / 0.0796

Therefore ;

[H+] = 10^-2.95

[H+] = 0.0011220 = 1.122 × 10^-3

Ka = [H+] / molar concentration

Ka = [1.122 × 10^-3]^2 / 0.0796

Ka = (1.258884 × 10^-6) / 0.0796

Ka = 15.815 × 10^-6

Ka = 1.58 × 10^-5

Coefficient = 1.58

Exponent = - 5

To convert 53.0 g of glucose to CO2 and H2O, we need to calculate the number of grams of oxygen required. The answer is approximately 159.0 g of oxygen.

The balanced equation given is:

\(\[ C_6H_{12}O_6(aq) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l) \]\)

From the equation, we can see that for every 1 molecule of glucose, we need 6 molecules of oxygen to convert it into carbon dioxide (\(CO_2\)) and water (\(H_2O\)).

To calculate the number of grams of oxygen required, we first convert the mass of glucose (53.0 g) into moles. The molar mass of glucose (\(C_6H_{12}O_6\)) is 180.16 g/mol.

\(\[ \text{Moles of glucose} = \frac{\text{Mass of glucose}}{\text{Molar mass of glucose}} = \frac{53.0\, \text{g}}{180.16\, \text{g/mol}} \approx 0.294\, \text{mol} \]\)

According to the balanced equation, the molar ratio between glucose and oxygen is 1:6. Therefore, the number of moles of oxygen needed is:

\(\[ \text{Moles of oxygen} = 0.294\, \text{mol} \times 6 = 1.764\, \text{mol} \]\)

Finally, we convert the moles of oxygen into grams using the molar mass of oxygen (32.00 g/mol):

\(\[ \text{Mass of oxygen} = \text{Moles of oxygen} \times \text{Molar mass of oxygen} = 1.764\, \text{mol} \times 32.00\, \text{g/mol} = 56.5\, \text{g} \]\)

Therefore, approximately 159.0 grams of oxygen are required to convert 53.0 grams of glucose to CO2 and H2O.

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The given reaction is the calcium carbide reaction of calcium hydroxide with water to produce acetylene.

The balanced equation is CaC2+2H2OCa(OH)2+C2H2.

Molar mass of acetylene, C2H2 = 2x12+2x1 = 26 g/mol

we get 64 g of CaC2 becomes 26 g of C2H2. 5.0g CaC2..

Molecular weight of Ca(OH)2=1(atomic weight of Ca)+2(atomic weight of O)+2(atomic weight of H)=40+2(16)+2(1)=40+32+2=74.

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Explanation:

It has to be C, because Carbon can bond with four other atoms and is like a four-hole wheel.

(Sorry if my elplanation was kinds choppy Oof, helpfully it makes sense?)

The colors of flames in experiments can vary based on several factors, including the temperature and composition of the burning material.

In general, the color of a flame is determined by the emission spectrum of the excited molecules and atoms in the flame. When a material is burned, the heat and energy generated excites the molecules and atoms, causing them to emit light. The specific colors that are emitted depend on the temperature of the flame, as well as the chemical composition of the burning material.

For example, in a very hot flame, such as the flame produced by a welder's torch, the temperature can be high enough to excite and ionize the atoms of the burning material, causing them to emit light across the entire visible spectrum and beyond. This results in a flame that appears white.

In cooler flames, such as those produced by a candle, the temperature is not high enough to ionize the atoms, but it is still high enough to excite them and cause them to emit light. The specific colors that are emitted in this case depend on the chemical composition of the burning material. For example, in a candle flame, the wax is primarily composed of hydrocarbons, which emit a yellow-orange light when burned. The blue color that is often seen in the center of the flame is due to the reaction between the hydrogen and carbon in the wax, which produces excited molecules that emit blue light.

In conclusion, the colors of flames in experiments can vary based on the temperature and composition of the burning material. The specific colors that are observed depend on the conditions within the flame and the chemical reactions that are taking place.

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Answer:

1/3p0

Explanation:

The combined gas law:

P1V1/T1 = P2V2/T2, where P, V and T are Pressure, Volume, and Temperature.  Temperature must always be in Kelvin.  The subscriopts 1 and 2 are for initial (1) and final (2) conditions.

In this case, temperature is constant (adiabatically). V1 = 2.0L and V2 = 6.0L.  I'll assume P1 = p0.

Rearrange the combined gas law to solve for final pressure, P2:

P1V1/T1 = P2V2/T2

P2 = P1*(V1/V2)*(T2/T1)  [Note how I've arranged the volume and temoperature terms - as ratios.  This helps us understand what the impact of raising or lowering one on the variables will do to the system].

No enter the data:

P2 = P1*(V1/V2)*(T2/T1):  [Since T2 = T1, the (T2/T1) terms cancels to 1.]

P2 = p0*(2.0L/6.0L)*(1)

P2 = (1/3)p0

The final pressure is 1/3 the initial pressure.

Given the following: \begin{tabular}{cc|c|c} T
xN
​
& 0.139N & −0.220N & −0.28N \\ \hlineT
yN
​
& 0.209N & 0.117N & 0N \end{tabular}Calculating the x and y components of the net force on the ring ΣF: For x components of ΣF:ΣF
x
​
= T
x
1
​
 + T
x
2
​
 + T
x
3
​
ΣF
x
​
= 0.139 N - 0.220 N - 0.28 N ΣF
x
​
= -0.361 NFor y components of ΣF:ΣF
y
​
= T
y
1
​
 + T
y
2
​
 + T
y
3
​
ΣF
y
​
= 0.209 N + 0.117 N + 0 N ΣF
y
​
= 0.326 NThus, the components of the net force are: ΣF
x
​
= -0.361 N, ΣF
y
​
= 0.326 N

Newton’s 1st law: Every body will continue in a state of rest or of uniform motion in a straight line unless compelled to change that state by forces impressed upon it. This law is obeyed since the sum of the forces on the ring is not zero. It would continue in its motion, if there were no net force acting upon it.

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Answer: a)Complete ionic equation:

2NH₄⁺ + S²⁻ + Fe²⁺ + SO₄²⁻ → 2NH₄⁺ + SO₄²⁻ + FeS

Net ionic equation:

Fe²⁺ + S²⁻ → FeS

b) Complete ionic equation:

2Na⁺ + SO₃²⁻ + Ca²⁺ + 2Cl⁻ → 2Na⁺ + 2Cl⁻ + CaSO₃

Net ionic equation:

SO₃²⁻ + Ca²⁺ → CaSO₃

c) Complete ionic equation:

Cu²⁺ + SO₄²⁻ + Ba²⁺ + 2Cl⁻ → Cu²⁺ + 2Cl⁻ + BaSO₄

Net ionic equation:

Ba²⁺ + SO₄²⁻ → BaSO₄

Explanation:

(a) Balanced formula equation:

(NH₄)₂S + FeSO₄ → (NH₄)₂SO₄ + FeS

Complete ionic equation:

2NH₄⁺ + S²⁻ + Fe²⁺ + SO₄²⁻ → 2NH₄⁺ + SO₄²⁻ + FeS

Net ionic equation:

Fe²⁺ + S²⁻ → FeS

(b) Balanced formula equation:

Na₂SO₃ + CaCl₂ → NaCl + CaSO₃

Complete ionic equation:

2Na⁺ + SO₃²⁻ + Ca²⁺ + 2Cl⁻ → 2Na⁺ + 2Cl⁻ + CaSO₃

Net ionic equation:

SO₃²⁻ + Ca²⁺ → CaSO₃

(c) Balanced formula equation:

CuSO₄ + BaCl₂ → CuCl₂ + BaSO₄

Complete ionic equation:

Cu²⁺ + SO₄²⁻ + Ba²⁺ + 2Cl⁻ → Cu²⁺ + 2Cl⁻ + BaSO₄

Net ionic equation:

Ba²⁺ + SO₄²⁻ → BaSO₄

Answer:

its A i put B and got it wrong

Explanation:

Answer:

CARBON-HYDROGEN BOND

Explanation:

The non-covalent interactions in the options are the hydrogen bond, the Van der Waal's bond, the dipole-dipole interaction. The odd one out of the options is the carbon - hydrogen bond.

The carbon-hydrogen bond is a covalent bond involving carbon and hydrogen atoms in which carbon shares its outer valence electrons with up to four hydrogen atoms either forming a complete shell with single bonds or complete shells with double or triple bonds. The type of bonding is found in hydrocarbons and other organic compounds. For example the alkanes, alkenes, alkynes and aromatic hydrocarbons.

The non- covalent bonds are the bonds that bind macro-molecules and do not involve the sharing of pairs of electrons. They are much more weaker than covalent bonds and are very important in the formation of proteins and nucleic acids. This type of bonding includes the electrostatic interactions, hydrogen bonds, hydrophobic bonds, Van der Waal's interactions. The different types of the non-covalent bonds differ in geometry, strength and specificity.

So therefore, carbon-hydrogen bond is the odd one of the non-covalent interactions listed in the options.

Diluting a solution does affect the pH, but not the way one might expect

Diluting a solution does affect the pH, but not the way one might expect. The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration [H+], so as the concentration of hydrogen ions changes, the pH changes as well. However, when a solution is diluted, the concentration of hydrogen ions remains the same, while the concentration of all other ions and molecules in the solution decreases proportionally. This means that the pH of the diluted solution remains the same as the original solution. For example, if a solution has a pH of 3 and is diluted by a factor of 10, the [H+] concentration will remain the same, but the concentration of all other species in the solution (e.g., buffer molecules) will decrease by a factor of 10. Therefore, the pH will remain at 3.

In summary, diluting a solution does not affect the pH, but rather, it changes the concentration of all species in the solution proportionally, while the hydrogen ion concentration and thus the pH remain constant

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Answer:

C

Explanation:

Von Helmont think that plants get their nourishment from soil because he concluded an experiment and watered the plant after which he came to know that.

Plant growth is the ability of a plant to develop its roots in length, sets in length, and breadth in girth. Plant growth occurs in four phases. These stages are as follows...

Plant seeding, germination, and seedling formation all increase.

John Baptista van Helmont (1580-1644) contributed to the understanding of photosynthesis. He planted a willow tree in a measured amount of dirt. After five years, he observed that the willow tree weighted around 74 kg more than it did when he first planted it. Van Helmont decided that plant development could not be solely attributable to soil minerals because the weight of the soil had scarcely altered. He assumed that the increased plant material had come only from the water.

For the first time, this invalidated the Ancient Greeks' earlier idea that plants grew bulk by absorbing minerals from the earth. Van Helmont was the first to put this theory to the test and discover it to be false.

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Explanation:

Eating and digesting cheese pizza is a Chemical Change because it is an irreversible reaction. It can't be undone.