what is seed reproduction​

Answer:

That means we do not need to use chemical reactions to separate them. ... The one liquid component in milk is water, and the other is fatty oil. ... Identify the type of substances (solid, liquid or gas) that are mixed in each

Explanation:

Answer:

\( \huge \sf \colorbox{pink}{hey \: there}\)

Explanation:

The mixture of clay or sand with water is muddy. The small clay particles become suspended in the water. This kind of mixture is called a suspension. Suspensions are opaque; that means they are cloudy and we cannot see through them very well.

Answer:

people ,shape ,behavior, grades ,friends

Answer:

c. nitrogen gas

Explanation:

Our air is a homogeneous mixture of many different gases and therefore qualifies as a solution. Approximately 78% of the atmosphere is nitrogen, making it the solvent for this solution.

The volume of the gas at STP would be 23.93 liters.

The volume of gas at STP (Standard Temperature and Pressure), we need to use the Ideal Gas Law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature. First, we need to calculate the number of moles of gas in the initial sample. We can use the formula n = PV/RT, where P is the initial pressure, V is the initial volume, R is the gas constant, and T is the initial temperature.
n = (2.31 atm) x (25.15 L) / [(0.0821 L atm/mol K) x (201 + 273.15 K)]
n = 1.067 moles

Now, we can use the molar volume of gas at STP, which is 22.4 L/mol, to calculate the volume of gas at STP.
V = n x 22.4 L/mol
V = 1.067 moles x 22.4 L/mol
V = 23.93 L
Therefore, the volume of the gas at STP would be 23.93 liters.

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Answer:

Earlier today, my friend got a call from the company where she had recently applied for a job. They informed her that she had been selected for an interview, which was scheduled for later this week. She was overjoyed and relieved to hear the news, as she had been searching for a job for several months and this was one of the positions she was most excited about. The call was definitely the highlight of her day, and it put her in a great mood for the rest of the afternoon! :) She immediately called her parents and friends to share the good news and celebrate!!

Answer:

a. Cl2 is evolved at the cathode

Explanation:

When an aqueous solution of NaCl (table salt) is electrolyzed, the positively charged ions, Na+, move to the anode while the negatively charged ions, Cl-, move to the cathode. At the anode, the Na+ ions lose electrons, resulting in the formation of Na atoms which then combine with other elements in the solution. At the cathode, the Cl- ions gain electrons, forming chlorine gas (Cl2) which is evolved as a gas.

The heat of reaction is -334.4 J.

Heat of reaction of AgCl is -66.88KJ/mol.

Precipitation of silver chloride is an exothermic process

Calculation:

We can calculate the heat using Calorimeter.

Formula of calorimeter is,

q = c.m.ΔT

Where,

q= heat of reaction, c= specific heat capacity

Now, mass of solution= 100g

ΔT = T2 - T1

Initial temperature (T1) = 22.60°c

Final temperature (T2) = 23.40°c

ΔT = 23.40°c - 22.60°C = 0.80°c

Now put all the value in the formula of calorimeter

q= c.m.ΔT

q= (4.18J/°c.g) × 100g × 0.80°c

q= 334.4 J

Heat will be released as the solution will get warmer

So, Heat of the reaction = -334.4 J

Let's consider the reaction between AgN\(O_{3}\)  and CaC\(l_{2}\)

2 AgN\(O_{3}\) (aq) + CaC\(l_{2}\)(aq) ⇄ 2AgCl + Ca(N\(O_{3}\)\()_{2}\)

From this equation we get,

2 moles of  AgN\(O_{3}\) produce 2 moles of AgCl

Means, 1 mole AgN\(O_{3}\)  gives 1 mole AgCl.

So, the mole of AgN\(O_{3}\) in solution,

AgN\(O_{3}\); n= molarity of  AgN\(O_{3}\) × volume of  AgN\(O_{3}\) in litres

= 0.100 mol/L × 0.0500 L

= 0.0050 mol

This is mole of  AgN\(O_{3}\) in solution = moles of AgCl formed = 0.0050 mol

To get heat of reaction in moles,

q= \(\frac{-334.4J}{0.0050 mol}\)

q = -66880 J/mol

q= -66.88KJ/mol

The heat of reaction for the AgCl formed

= -66.88KJ/mol

Precipitation of silver chloride is an exothermic process since the heat is released in this case.

What is an exothermic reaction?

The part of a solution that transforms into a different state while creating the solution or the part that isn't overly abundant; the substance that dissolves in another material.

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The glowing splint in the reaction of hydrogen gas and oxygen gas to form water provides the activation energy for the reaction.

A chemical reaction involves the combination of two or more substances to yield a new substance which is different from the starting materials. It may also involve the break up a substance.

Since energy is required for a chemical reaction to take place, the glowing splint in the reaction of hydrogen gas and oxygen gas to form water provides the activation energy for the reaction.

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Answer:

pH = 4.5, concentration = 0.045 M.

pH = 6.5, concentration = 0.175 M.

Explanation:

The ka for the can be calculated by using the formula below;

Ka = 10^-pka = 10^-5.960 = 1.1 × 10^-6

The concentration of hydrogen ion at pH = 4.50 can be calculated as given below;

{H^+ } = 10^-4.50 = 3.2 × 10^-5 M.

(NB=> 10 in this regards means the inverse of log).

The next step is to determine the distribution coefficient which can be calculated by using the formula below;

distribution coefficient = (partition coefficient) × ka / ka + ( concentration of Hydrogen ion,H^+).

distribution coefficient =( 5 × 1.1 × 10^-6 ) / 1.1 × 10^-6 + 3.2 × 10^-5 M. = 5.5 × 10^-6/ 3.2 = 0.00000171875

The fraction remaining from the compound = 45.0 mL / 45.0 mL + (0.00000171875 × 125).

= 0.999995.

Thus, the concentration at pH = 4.5 = 0.999995 × 0.0450 M = 0.045 M

(B). pH=6.50, thus the concentration of Hydrogen ion = 10^-6.5 = 3.2 × 10^-7 M.

distribution coefficient = (partition coefficient) × ka / ka + ( concentration of Hydrogen ion,H^+).

distribution coefficient = (5 × 1.1 × 10^-6)/ 1.1 × 10^-6 + 3.2 × 10^-7 M).

distribution coefficient = 5.5 × 10^-6/ 1.42 × 10^-6 = 3.9.

Therefore, the concentration = 3.9 × 0.0450 M = 0.175 M.

Answer:

30 ml

Explanation:

Answer:

We can use the principle of conservation of energy to solve this problem. The heat lost by the metal is equal to the heat gained by the water:

Q metal = -Q water

where Q metal is the heat lost by the metal, and Q water is the heat gained by the water.

The heat lost by the metal can be calculated using the formula:

Q metal = m metal * c metal * ΔT metal

where m metal is the mass of the metal, c metal is its specific heat, and ΔT metal is the change in temperature of the metal.

The heat gained by the water can be calculated using the formula:

Q water = m water * c water * ΔT water

where m water is the mass of the water, c water is its specific heat, and ΔT water is the change in temperature of the water.

We know the values of all the variables except c metal, so we can solve for it. We can start by calculating the values of Q metal and Q water:

Q metal = -Q water

m metal * c metal * ΔT metal = -m water * c water * ΔT water

Substituting the given values, we get:

5.4 g * c metal * (100.0 °C - T) = -142 g * 4.18 J/(g*°C) * (T - 24.2 °C)

Simplifying and solving for c metal, we get:

c metal = [142 g * 4.18 J/(g*°C) * (T - 24.2 °C)] / [5.4 g * (100.0 °C - T)]

Multiplying out, we get:

c metal = [593.56 J/(°C) * (T - 24.2 °C)] / [5.4 g * (100.0 °C - T)]

To solve for c metal, we need to find the value of T that satisfies the equation. We can do this by substituting the given value of ΔT water = 0.9 °C into the equation and solving for T:

c metal = [593.56 J/(°C) * (T - 24.2 °C)] / [5.4 g * (100.0 °C - T)]

c metal = [593.56 J/(°C) * (T - 24.2 °C)] / [540 g - 5.4 g * T]

0.9 g * [593.56 J/(°C) * (T - 24.2 °C)] = [540 g - 5.4 g * T] * c metal

535.2044 J/(°C) * (T - 24.2 °C) = 540 g * c metal - 5.4 g * T * c metal

535.2044 J/(°C) * T - 12931.7808 J = 540 g * c metal - 5.4 g * c metal * T

5.4 g * c metal * T + 535.2044 J/(°C) * T = 540 g * c metal + 12931.7808 J

T * (5.4 g * c metal + 535.2044 J/(°C)) = 540 g * c metal + 12931.7808 J

T = [540 g * c metal + 12931.7808 J] / [5.4 g * c metal + 535.2044 J/(°C)]

Substituting the given values, we get:

T = [540 g * c metal + 12931.7808 J] / [5.4 g * c metal + 535.2044 J/(°C)]

T = [540 g * c metal + 12931.7808 J] / [5.4 g * c metal + 535.2044 J/(°C)]

T ≈ 23.3 °C

Therefore, the specific heat of the metal is:

c metal = [142 g * 4.18 J/(g°C) * (T - 24.2 °C)] / [5.4 g * (100.0 °C - T)]

c metal ≈ 0.39 J/(g°C)

So the specific heat of the metal is approximately 0.39 J/(g*°C).

A 5.4 g sample of the metal is heated to the 100.0 °C and is placed in the beaker containing 142 g of the water at 24.2 °C. The specific heat of the metal is 1.322 J/ g °C.

The mass of the metal = 5.4 g

The final temperature = 25.1 °C

The initial temperature = 100 °C

The specific heat capacity of metal = x

The mass of the water = 142 g

The final temperature = 25.1 °C

The initial temperature = 24.2 °C

The specific heat capacity of water = 4.184 J/ g °C

Loss of Heat of Metal = Gain of Heat by Water

-q metal = + q metal

- 5.4 × x × ( 25.1 - 100 ) = 142 × 4.184 ( 25.1 - 24.2 )

404.46 x = 534.71

x = 1.322 J/ g °C

The specific heat capacity of metal is  1.322 J/ g °C.

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Answer:

C. aquifer

Explanation:

I just did a lesson on this

:)

We have that The  the mass change when the copper coin was made to look silver is an increase in mass

Correct option C

It increased.

It is important to note that the copper coin after its cutting into shape will have a specific mass or weight and the silver coating solution will also have a net value of mass or weight

Therefore

The  the mass change when the copper coin was made to look silver is an increase in mass

Correct option C

It increased.

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Answer:

44 g/mol

Explanation:

Types of luminous flames:

1. Yellow Luminous Flame

2. Smoky Luminous Flame

3. Orange Luminous Flame

4. Blue Luminous Flame

Luminous flames are characterized by their visible glow, which is caused by the incomplete combustion of fuel. The presence of soot particles in the flame causes the emission of light. There are different types of luminous flames, which can be classified based on their fuel composition and burning conditions. Here are some common types of luminous flames:

1. Yellow Luminous Flame: This is the most common type of luminous flame, often seen in open fires, candles, and gas stoves. It appears yellow due to the presence of soot particles in the flame. Yellow flames indicate incomplete combustion of hydrocarbon fuels, such as methane, propane, or natural gas. The high carbon content in these fuels leads to the formation of soot, which emits visible light.

2. Smoky Luminous Flame: This type of flame is characterized by a significant amount of black smoke and soot production. It is commonly observed in poorly adjusted or malfunctioning burners or engines. The excessive presence of unburned fuel in the flame results in incomplete combustion and the emission of dark smoke particles.

3. Orange Luminous Flame: An orange flame indicates a higher combustion temperature compared to a yellow flame. It is often seen in more efficient burners or when burning fuels with a higher carbon content, such as oil or diesel. The higher temperature helps in burning more of the carbon particles, reducing the amount of soot and making the flame appear less yellow.

4. Blue Luminous Flame: A blue flame is typically associated with complete combustion. It indicates efficient burning of fuel, resulting in minimal soot formation. Blue flames are commonly observed in gas burners or Bunsen burners. The blue color is a result of the combustion of gases, such as methane, in the presence of sufficient oxygen.

It's important to note that the luminosity of a flame can vary depending on factors such as fuel-air mixture, combustion temperature, and the presence of impurities. Achieving complete combustion and minimizing the production of soot is desirable for efficient and cleaner burning processes.

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Answer:

angle of incidence.

Answer: During aerobic respiration, catabolic reactions convert larger complex organic molecules into ATP, the chemical that drives most physiological processes in the body.

Explanation:

To know the element with eletronic configuration 2.8.6 we need to:

First:

do the eletronic distribution - 1s² 2s² 2p⁶ 3s² 3p⁶

Answer

2 L of hydrogen react with 1 L of oxygen

4 mol of hydrogen reacts with 2 moles of oxygen.

Explanation

In the equation: 2H2 + O2 → 2H2O

All that applies to the equation are:

2 L of hydrogen react with 1 L of oxygen

4 mol of hydrogen reacts with 2 moles of oxygen.

The SI unit of pressure is the Pascal (Pa).

The boiling point of water is lower on Mount McKinley than the boiling point of water in NYC.

What is Pressure?

Pressure is defined as the amount of force applied perpendicular to the surface of an object per unit area over which that force is distributed. In other words, it is the force per unit area that an object exerts on another object. Pressure can be measured in various units such as pascal (Pa), bar, pounds per square inch (psi), and atmospheres (atm), among others. It is an important concept in physics and is used to describe many phenomena, including fluid dynamics, weather patterns, and even the behavior of gases in space.

At lower elevations, atmospheric pressure is higher compared to higher elevations.

Standard atmosphere or standard atmospheric pressure is equal to 101325 Pa.

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The missing labels are:

It is a way to represent a chemical reaction.

Let's consider the following chemical equation.

CuCO₃(s) + H₂SO₄(aq) →  CuSO₄(aq) + H₂O(l) + CO₂(g)

The missing labels are:

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Answer:

\(M_{KOH}=0.3704M\)

Explanation:

Hello there!

In this case, since the titration of bases when using monoprotic acids like KHP, occurs in a 1:1 mole ratio, it is possible to use the following equation, because at the endpoint the moles of the KHP and KOH get equal:

\(n_{KHP}=n_{KOH}\)

In such a way, we first calculate the moles of KOH given the mass and molar mass of KHP:

\(n_{KHP}=n_{KOH}=3.458g*\frac{1mol}{204.22g}=0.0169mol\)

Next, since we have the volume of KOH, we first take it to liters (0.04571 L) to that we obtain the following concentration:

\(M_{KOH}=\frac{0.0169mol}{0.04571L}\\\\M_{KOH}=0.3704M\)

Regards!

Answer:

hydrogen ion dissolve in water is acid and hydroxide ion dissolve in water is basi

the recation in which oxidation and reduction goes side by side is redox redaction

Answer: for me everything ig

Explanation:

Answer:

solution

Explanation:

Subaquatic soil can be classified as sediment or soil based on its geological properties and formation processes.

Sediment refers to any material that is transported and deposited by water, wind, ice, or gravity. Sediments can be composed of various materials, such as minerals, rocks, organic matter, and even human-made debris.

Sediments can accumulate in different environments, such as rivers, lakes, oceans, and deserts, and can be deposited in layers over time.

Subaquatic soil can be classified as sediment or soil based on its geological properties and formation processes. If it has primarily formed through sediment deposition, it is more appropriate to classify it as sediment.

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The number of moles of gas, a balloon inflated to 7 L of volume at STP will be 0.31 moles.

The molar volume of a gas is the volume of one mole of a gas at STP.

At STP, one mole (6.02×10²³ particles) of any gas occupies a volume of 22.4 L

Therefore,

If 1 mole of any gas occupies 22.4 L of volume at STP

Then, X mole of gas occupies 7 L of Volume at STP

Now,

Let's equate both the above conditions ;

X / 7 L = 1 / 22.4 L

  X      = 1/ 22.4 L x 7 L  

  X      = 0.31 moles

Hence, the number of moles of gas, a balloon inflated to 7 L of volume at STP will be 0.31 moles.

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Answer:

Explanation:

IUPAC nomenclature is based on naming a molecule's longest chain of carbons connected by single bonds, whether in a continuous chain or in a ring. All deviations, either multiple bonds or atoms other than carbon and hydrogen, are indicated by prefixes or suffixes according to a specific set of priorities

Answer:

Electronegativities generally increase from left to right across a period

Therefore the answer is B.

The mass of calcium carbonate that must be decomposed to produce 25.0 L of carbon dioxide gas at 26 °C and 0.997 atm is 104.1 g

The volume of CO₂ produced with a 78% yield would be 19.7 L.

The mass of calcium carbonate that must be decomposed to produce 25.0 L of carbon dioxide gas at 26 °C and 0.997 atm is calculated from the equation of the reaction as follows:

The balanced equation of reaction: CaCO₃(s) → CaO(s) + CO₂(g)

The number of moles of CO₂ produced is determined from the ideal gas equation:

n = PV/RT

n = (0.997 atm) × (25.0 L) / [(0.0821 L·atm/(mol·K)) × (299 K)]

n = 1.04 mol CO₃

From the balanced equation;

1 mol of CaCO₃ produces 1 mol of CO₂.

Moles of CaCO₃₃required = 1.04 mol of CaCO3.

the mass of CaCO3 needed will be:

mass = 1.04 mol × 100.09 g/mol

mass = 104.1 g

If there is only a 78% yield, the actual moles of CO₂ produced would be:

moles = 0.78 × 1.04 mol

moles = 0.81 mol

solving for volume using the ideal gas equation:

V = nRT/P = (0.81 mol) × (0.0821 L·atm/(mol·K)) × (299 K) / (0.997 atm)

V  = 19.7 L

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