A 2. 50g sample of zinc is heated, then placed in a calorimeter containing 60. 0g of water. The temperature of water increases from 20. 00 degrees C. The specific heat of Zinc is 0. 390J/g Degree C. What was the initial temperature of the zinc metal sample?

To determine the equation of the parabola that passes through the given points, we can use the standard form of a quadratic equation, which is y = ax² + bx + c.

Substituting the coordinates of the three points into this equation will give us a system of three equations in three variables, which can be solved to find the values of a, b, and c that make up the equation of the parabola. The process is as follows:Substituting (-12, 1333) into the equation, we get: 1333 = a(-12)² + b(-12) + cSimplifying, we get: 1333 = 144a - 12b + c(1)Substituting (4,77) into the equation, we get: 77 = a(4)² + b(4) + cSimplifying, we get: 77 = 16a + 4b + c(2)Substituting (8,403) into the equation, we get: 403 = a(8)² + b(8) + cSimplifying, we get: 403 = 64a + 8b + c(3)Solving equations (1), (2), and (3) simultaneously, we get: a = -3, b = 45, and c = 1427Substituting these values back into the standard form equation, we get: y = -3x² + 45x + 1427Therefore, the equation of the parabola that passes through the points (-12, 1333), (4,77), and (8,403) is y = -3x² + 45x + 1427.

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They are classified on the basis of their physical features- Shape and Appearance.

The four basic shapes of bacteria are Coccus (spherical or ovoid), bacillus (rod-like), vibrio (comma-shaped ), and spirilla (spiral or helical shaped).

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The overall enthalpy change i.e ΔH for the system is -1300 kJ.

Enthalpy change in a reaction is defined as the difference in the potential energy of the products and the potential energy of the reactants.

This is represented by ΔHreaction.

The reaction of enthalpy change is given as

ΔHreaction= ΔHproducts -  ΔHreactants

where ΔHproducts is the potential energy of the products

=[(-200) + (-300)]kJ

= -500kJ

And,  ΔHreactants is the potential energy of the reactants

=800KJ

Therefore, by putting these values ΔHreaction can be calculated as

ΔHreaction

= [-500-800] KJ

=-1300 KJ

Thus, the overall enthalpy change is -1300KJ

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Answer: I believe it would be using group input to discuss and prepare preliminary designs

Explanation:

The other substances utilized by enzymatic receptors and G protein-coupled receptors to relay the message within the cytoplasm are known as second messengers.

Second messengers are small molecules or ions that act as intermediaries in intracellular signal transduction pathways.

When an extracellular ligand binds to an enzymatic receptor or a G protein-coupled receptor on the cell surface, it initiates a cascade of events that ultimately leads to a cellular response. The activation of these receptors triggers the production or release of second messengers within the cell.

Common examples of second messengers include:

Cyclic adenosine monophosphate (cAMP): cAMP is generated by the activation of adenylate cyclase, an enzyme stimulated by G protein-coupled receptors. cAMP acts as a second messenger by binding to and activating protein kinase A (PKA), which phosphorylates target proteins and initiates specific cellular responses.

Inositol trisphosphate (IP3) and diacylglycerol (DAG): These second messengers are generated following the activation of phospholipase C (PLC) by G protein-coupled receptors. PLC cleaves phosphatidylinositol 4,5-bisphosphate (PIP2) into IP3 and DAG. IP3 triggers the release of calcium ions (Ca²⁺) from intracellular stores, while DAG activates protein kinase C (PKC), which phosphorylates target proteins and regulates cellular processes.

Calcium ions (Ca²⁺): Calcium ions act as versatile second messengers involved in numerous cellular signaling pathways. They can be released from intracellular stores or enter the cell through ion channels upon receptor activation. Calcium ions regulate various processes, including muscle contraction, neurotransmitter release, and gene expression.

These second messengers play vital roles in transmitting signals from the cell surface to the interior of the cell, amplifying and modulating the initial signal, and ultimately mediating a cellular response. Their regulation and precise control are essential for proper cellular functioning and maintaining homeostasis.

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Answer:

C. Theories have been confirmed through tests; hypotheses haven't.

Hope this helps!!

From Newton's second law,

F = ma

Where F is the force

m is the mass

and a is the acceleration

Q.1
Here, we are to determine the force acting on the box

m = 3 kg

a = 9 m/s²

Using the formula,

F = ma

F = 3 × 9

F = 27N

Hence, the force acting on the box is 27N

Q.2

Here, we are to determine the acceleration of the car

m = 780 kg

F = 235 N

Using the same formula,

F = ma

235 = 780 × a

a = 235 / 780

a = 0.30 m/s²

Hence, the acceleration of the car is 0.30 m/s²

Q.3

Here, we are to determine the object's mass

a = 5 m/s²

F = 10 N

Using the same formula,

F = ma

10 = m × 5

m = 10 / 5

m = 2 kg

Hence, the object's mass is 2 kg

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Answer: I believe the answer is B. Sorry if i'm wrong

Explanation:

Answer:

no H2O molecules are produced:)

Explanation:

Answer: 18 g of water occupies one mole and one mole contains Avagadro’s no. of molecules = 6.023*10^23 molecules .

Explanation:question says 72 g of water . 72 g of water occupies 4 moles.

So the no. of molecules in 72 g of water has 4*(6.023*10^23) molecules .

Answer:

The answer is C for sure!

Explanation:

In the picture we have the electrons given, so we have to determine the nucleons and protons.

Answer:

Both similes and metaphors are a poetic device because authors can be seen using them while writing their poem to give the reader a deeper understanding of the character and the setting. Metaphor is a figure of speech that talks about one this but refers to another. On the other hand, a simile is a figure of speech that compares two things together. Usually using the work "like" or "as".

Hoped it helped! Have a great day!

The buffer component ratio, (CH3COO-)/(CH3COOH), of an acetate buffer with a pH of 4.47 and a Ka of CH3COOH of 1.8 x 10^- is: 0.54

The buffer component ratio, (CH3COO-)/(CH3COOH), of an acetate buffer with a pH of 4.47 and a Ka of CH3COOH of 1.8 x 10^-5 can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([CH3COO-]/[CH3COOH])

So, the buffer component ratio (CH3COO-)/(CH3COOH) of the acetate buffer is approximately 0.54.

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Answer:

4 yards/second is the answer

Answer:

the answer is 4 yards per second

Explanation:

80 yards in 20 seconds

20+20+20+20=80

there are 4, 20s to make 80

in other words...

80÷4=20

. B) excoriation. An abrasion of the skin by scratching, trauma, or chemicals is termed as Excoriation.  Excoriation refers to a superficial injury to the skin caused by mechanical trauma.

such as scratching, rubbing, or scraping. This can occur due to a variety of reasons, including skin conditions such as eczema or psoriasis, insect bites, or contact with chemicals or other irritants. An abrasion of the skin by scratching, trauma, or chemicals is termed as Excoriation. Therefore, option B is the correct answer. Excoriation refers to a superficial injury to the skin caused by mechanical trauma. Excoriations can range from mild to severe and may be accompanied by pain, itching, or bleeding. Treatment depends on the underlying cause and may include topical medications, dressings, or other interventions to promote healing and prevent infection.

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Mutualism occurs when organisms need the same resource that is in limited supply .

Mutualism is the association of the organisms of two species in which both biologically interact with each other and are mutually benefited

Organisms participating in mutualism are known as mutualists .

Mutualism exist between plants , between animals and between plants and animals .

if similar interaction occur within a species, they are called as co-operation.

Types of mutualistic relationship :

they are resource-resource relationships, service-service relationships, and service-resource relationships.

Mycorrhizal association is an example of resource - resource relationship .

ants belonging to genus Acacia trees is the example of service-service relationship.

pollination by insects is an example of service-resource relationship.

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Answer:

C

Explanation:

i took the test

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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Answer:

1.25 moles NaNO

Explanation:

Molarity = moles/Volume (L)

moles = Molarity (M) x Volume (L)

M = 2.5 M

V = 500 mL = 0.5L

moles = M x V = 2.5 x 0.5 = 1.25 moles

Final chloride ion concentration is 1.03 mol / Cl- L when 34. g of zncl2 have been dissolved in 696. ml of water.

First we need to convert  zncl2 to moles:

34 g zncl2 x 1 mole zncl2/ 136.3 g = 0.24 mol zncl2

Given the molecular formula, we have three chlorine atoms for every mole of zncl2

0.24 mol zncl2 x 3 mole Cl-/ 1 mole  zncl2 =  0.72 mol Cl-

Now we will divide this by the volume of the solution as liters

696 ml x 1 L/ 1000 mL =  0.696 L

Concentration =  0.72 mol Cl- / 0.696 L = 1.03 mol / Cl- L

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Answer:

By making sense of the World around them

Explanation:

maybe?

The primary components of clean dry air are nitrogen (78%), oxygen (21%), and argon (0.9%).

Clean dry air is the basic requirement for a healthy living environment. Clean air is required for better respiratory health and better living. Pure air is free from water vapors and other particles. Nitrogen, Oxygen, and Argon are the primary components of clean, dry air. Nitrogen is the most abundant gas in the air with 78% of the volume. Oxygen is the second most abundant gas in the air, making up 21% of the volume.

Argon, the third most abundant gas in the air, makes up about 0.9% of the volume. Other trace elements include neon, helium, methane, hydrogen, carbon dioxide, and ozone, which together make up less than 0.1% of the atmosphere. Water vapor is present in the air, but it's not always pure. Air with a relative humidity of less than 50 percent is classified as dry air. Pure air is essential for respiratory health.

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Answer:

Iron will disolve in acid. The acid eats away at the iron. :)

Have an amazing day!!

Please rate and mark brainliest!!

This involves the transfer to Clive, one of its franchisees, of the right to use BHI's trademark, business model, and other proprietary information to operate a Burger Heaven restaurant.

Burger Heaven, Inc. (BHI) conducts a chain-style business operation type of franchise. Clive agrees to follow BHI's standards and procedures in running the restaurant, including using BHI's approved ingredients and recipes, maintaining a certain level of cleanliness and customer service, and paying fees and royalties to BHI.

In exchange, Clive benefits from the brand recognition and support of BHI's marketing and training programs, as well as the opportunity to profit from the popularity of the Burger Heaven concept. This type of franchise allows for consistency across multiple locations and provides a proven business model for aspiring entrepreneurs.

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Answer:

wind and humidity

Explanation:

ocean and tides are not charectiristics of describing the weather

Answer: A. Winds

C. Humidity

Explanation: because wind blows air and humidity is heat like warm fog or warm.

Answer:

number 3.

Explanation:

Answer:

3

Explanation:

No, the volume of water in a beaker is not an extensive property.

Extensive properties are those that depend on the amount or size of the substance being measured. In other words, they are properties that change with the quantity of the substance. Examples of extensive properties include mass, volume, and total energy.

In the given scenario, the volume of water in the beaker is 50 milliliters. This volume remains the same regardless of the quantity of water present. Whether it's 50 milliliters or 500 milliliters, the volume measurement does not change. Therefore, the volume of water in the beaker is an example of an intensive property.

Intensive properties are independent of the amount or size of the substance. They are characteristics that remain constant regardless of the quantity of the substance. Examples of intensive properties include temperature, density, and color.

It's important to note that the distinction between extensive and intensive properties depends on the specific property being considered. While volume is typically an extensive property for a bulk substance, in the case of a fixed volume of water in a beaker, it becomes an intensive property.

In summary, the volume of water in a beaker is not an extensive property but rather an intensive property because it does not change with the quantity of the substance.

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The flame test can be effectively used to identify a cation in a solution because when a cation is placed in a flame, it will emit a different color with

visibile light.

Ions are charged particles that form when atoms gain or lose electrons. When an atom gains electrons, it forms negatively charged ions called negative ions. On the other hand, positively charged ions, called cations, are formed when atoms lose electrons. These ions can be detected in solution using special tests. The flame test involves exposing a joint to a flame and identifying the joint by flame color. The color of the flame test is due to the fact that electrons in metal cations are excited and jump to higher energy levels.

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Answer:

The correct answer is - D C2H4.

Explanation:

Saturated hydrocarbons are hydrocarbons with single covalent C-C bonds. They are known as alkanes. The general formula for these hydrocarbons is  CnH2n+2

Unsaturated hydrocarbons the hydrocarbons with double or triple covalent C-C bonds. They are known as alkenes and alkynes respectively. The general formula for these hydrocarbons is  CnH2n and CnHn-2

For the given options:

Option D: C2H4, is the simplest alkene with a double bond so it is an unsaturated hydrocarbon.

The volume of 8.8g of carbon dioxide at STP is 4.38 L.

1 mole of any gas will take up 22.4 L of space at standard temperature and pressure (STP). A balanced chemical equation and the Ideal Gas Law can be used to determine the amount or mass of gas consumed or created in a chemical process.

n = m/M

where m is the molar mass of carbon dioxide and M is its mass in terms of molecules.

Considering that the molar mass of carbon dioxide is 44.01 g/mol:

n = 8.8 g / 44.01 g/mol

n = 0.1998 mol

Next, we can plug in the values of n, R, P, and T into the ideal gas law and solve for V:

V = (nRT)/P

V = (0.1998 mol x 0.0821 L·atm/(mol·K) x 273.15 K) / 1 atm

V = 4.38 L

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91.9 grams of potassium phosphate are produced from 72.9 grams of KOH.

The balanced chemical equation for the given reaction is:

3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O

From the balanced equation, we can see that the stoichiometric ratio of KOH to K₃PO₄ is 3:1. This means that for every 3 moles of KOH reacted, 1 mole of K₃PO₄ is produced.

First, we need to calculate the number of moles of KOH present in 72.9 g of KOH. The molar mass of KOH is 56.11 g/mol, so the number of moles of KOH is:

72.9 g KOH × (1 mol KOH/56.11 g KOH) = 1.30 mol KOH

Now, using the stoichiometric ratio from the balanced equation, we can calculate the number of moles of K₃PO₄ produced:

1.30 mol KOH × (1 mol K₃PO₄/3 mol KOH) = 0.433 mol K₃PO₄

Finally, we can use the molar mass of K₃PO₄ to convert from moles to grams:

0.433 mol K₃PO₄ × (212.27 g K₃PO₄/mol) = 91.9 g K₃PO₄

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