If the specific heat of Copper is 0.385 J/g °C and I add 10 kJ of heat to a 10kg block of copper, how much will the temperature of copper change?

a. 0.384 °C
b. 2.60 °C
c. 38.4 °C
d. 384 °C
e. more than 400°C

The specific heat of the metal object with a mass of 22.7g heated to to temperature of 97.0°C and then transferred to an insulated container containing 84.7 g of water at 20.5 ∘C is 0.815J/g°C

How to calculate specific heat?

The specific heat capacity of a metal can be calculated using the calorimetry equation as follows:

Q = mc∆T

Where;

Q = quantity of heat absorbed

m = mass of substance

c = specific heat capacity

∆T = change in temperature

mc∆T (water) = -mc∆T (metal)

84.7 × 4.18 × 3.8 = - (22.7 × c × -72.7)

1345.375 = 1650.29c

c = 0.815J/g°C

Therefore, the specific heat of the metal object with a mass of 22.7g heated to to temperature of 97.0°C and then transferred to an insulated container containing 84.7 g of water at 20.5 ∘C is 0.815J/g°C.

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Answer:

20 neutrons

Explanation:

(not really any just look at a periodic table)

Answer:$100

Explanation:Divide 60 by 12=5

5x200=$1000

Answer:

We can simply refer to the cation in the ionic compound as magnesium. Phosphorus, Pstart text, P, end text, is a group 15 element and therefore forms 3- anions.

Explanation:

Because it is an anion, we add the suffix -ide to its name to get phosphide as the name of the ion.

It is likely that the temperature of the steam is higher than the temperature of the water boiling in the kettle. The intense heat felt by Alka from the puff of steam supports this observation.

In general, the temperature of steam produced from boiling water is higher than the temperature of the water itself. When water boils, it undergoes a phase change from a liquid to a gas, forming steam.

During this phase change, the water absorbs heat energy from the heat source, such as a stove or electric kettle, and converts it into the latent heat of vaporization.

The boiling point of water is 100 degrees Celsius (212 degrees Fahrenheit) at standard atmospheric pressure. At this temperature, the water molecules have enough energy to overcome the intermolecular forces and transition into the gaseous state.

However, steam is hotter than the boiling point of water because it contains additional heat energy in the form of latent heat. The heat energy absorbed during vaporization is stored as latent heat within the steam. As the steam gushes out of the spout of the kettle, it releases this latent heat energy, which can be felt as intense heat.

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A dam constructed to produce tidal power does so by harnessing water flow to drive turbines and electric generators .

Tidal energy was employed in grain mills to crush grains mechanically, just like wind energy was. grain crushing Here, the tidal energy generated by the turbines was used. Hydroelectric dams, which serve as significant energy storage, also employ tidal energy to store energy.

Tidal power can harm marine life because tidal turbines' whirling blades can cause marine organisms to perish. Fish habitations in tidal power settings may be impacted by noise from the turbines' rotation. Tidal energy can also affect how sediment and water are processed.

Therefore, option B is correct.

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Answer: A warm front is the transition area where a mass of warm air moves to replace a mass of cold air. On a weather map, a warm front is usually drawn using a solid red line with half circles pointing in the direction of the cold air that will be replaced. Warm fronts usually move from southwest to northeast. A warm front can initially bring some rain, followed by clear skies and warm temperatures.

Explanation: I got this from a website called SciJinks.gov. I hope that it answers your question and that this helps. Good luck!

Answer:

In case anyone wants both

Explanation:

Cold fronts appear on weather maps in areas of ( large temperature and air pressure decrease )

Warm fronts appear on weather maps in areas of ( temperature and air pressure increase )

D. A change in a substance with no new substances being formed

There are 2 ways a substance can change, physical changes and chemical changes. In all changes, physical or chemical, mass is always conserved.

Physical vs. Chemical

The main difference between a physical and chemical change is the substance at the end.

New substances are only created by chemical changes. Additionally, only chemical changes break apart bonds. Physical changes maintain the same compounds and elements, so bonds are not broken.

Examples of Physical Changes

Physical changes do not change the actual substance, but they can change the form of the substance.

All of these change the state of matter or shape of the substance, but not the chemical makeup.

Multiplying by 1000 will allow you to convert 35 liters to milliliters

1 Liters = 1000 milliliters

Thus, multiply the value in liter by 1000 will convert it to milliters

With the above scale, we can convert 35 liters to milliliters. Details below:

We can convert 35 liters to milliliters as illustrated below:

1 Liters = 1000 milliliters

Therefore,

35 liters = (35 liters × 1000 milliliters) / 1 liters

35 liters = 35000 milliliters

Thus, 35 liters is equivalent to 35000 milliliters

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Answer:

33 cm/83.82 in

Explanation:

Area is length multiplied by width, so we get

10 × 3.30 = 33. To convert this to inches, multiply the answer by 2.54 (33 × 2.54) resulting in 83.82.

Answer:

they are 10 but I'll be listing three

Explanation:

cirrus,cirrocumulus and cirrostratus

I hope this helps....good luck with the rest...again, I hope this helps

Answer:

The waters' temp increased

Explanation:

The temperature of the water in the swimming pool has increased due to the heat from the Sun. As a result, the particles in the water are moving faster and have a higher kinetic energy than in the morning.

Answer:

HBr+LiOH->H2O+LiBr

Explanation:

Hydrogen and Lithium ions have a charge of +1, Bromine and Hydroxide ions have a charge of -1. Therefore, the result.

Reagent bottles must be closed when not in use because they may contain volatile or sublime substances.

Reagent bottles are most commonly used to store chemical reagents, including acid and alkali solvents that can be safely stored due to anti-corrosion capabilities.

Reagent bottles, often known as graduated bottles, are glass, plastic, borosilicate, or related-substance containers with specific caps or stoppers. They are used in labs to store chemicals in liquid or powder form in cabinets or on shelves.

Some reagent bottles are tinted amber (actinic), brown or red to protect light-sensitive chemical compounds from visible light, ultraviolet and infrared radiation which may alter them; other bottles are tinted blue (cobalt glass) or uranium green for decorative purposes -mostly vintage apothecary sets.

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Answer:

its 4

Explanation:

Answer:

4 is the answer

Explanation:

2 fingers plus 2 fingers is 4 fingers, Not 1,250

Answer:

Neutrons - 5

Protons - 4

Electrons - 4

Explanation:

Beryllium (Be) is the 4th element on the Periodic Table.

It has 4 protons and 4 electrons (neutral charge)

Atomic Mass = # of Protons + # of Neutrons

9 = 4 + x

x = 5 Neutrons

The appropriate items to their respective bins are:

Covalent: Carbon dioxide

Ionic: Table salt

Metallic: Iron

Molecular: Sugar

What is bins?

Bins are containers used for storing and organizing items. They can be made from a variety of materials, including wood, metal, and plastic, and come in many shapes and sizes. Bins are often used in large warehouses, retail stores, and other business establishments to store and organize products, tools, and other items. In the home, bins can be used to store toys, shoes, and other items that need to be kept organized.

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ΔH₁ = ΔH₂ - ΔH₃

ΔH₁ = -635.1-(178.3) KJ

ΔH₁ = -813.4 KJ

The given reaction is a double displacement reaction where, Cl and NO₃ group displaces between and sodium and silver atoms. The complete reaction is written as follows:

\(\rm NaCl (aq) + AgNO_{3} (aq) \rightarrow AgCl (s) + NaNO_{3} (aq)\).

There are various kinds of reactions classified based on the way of regrouping of atoms in the reactants.  In a double displacement reaction, two species from the reactant side are displaced of each other.

The reaction between sodium chloride and silver nitrate gives sodium nitrate in aqueous form and the precipitate of silver chloride. Here the groups nitrate and chlorine displaces between sodium and silver. Hence it is a double displacement reaction.

The net reaction represents the formation of the silver chloride precipitate.

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Answer:

2.5 g

Explanation:

First we have to calculate the diffusion flux (J). To calculate, we use the formula:

\(J=-D\frac{dC}{dx}=-D\frac{C_2-C_1}{x_2-x_1}\\ \\Given\ that:\\\\x_2-x_1=0.1\ cm,D=110*10^{-8}cm^2/s,C_2=0.02g/cm^3,C_1=0.44g/cm^3\\\\Substituting:\\\\J=-110*10^{-8}(\frac{0.02-0.44}{0.1} )=4.62*10^{-6}g/cm^2s\\\\But, J=\frac{M}{At}\\ \\t=3\ hours=10800\ s,A=50\ cm^2\\\\Substituting:\\\\4.62*10^{-6}=\frac{M}{50*10800}\\ \\M=4.62*10^{-6}*50*10800\\\\M=2.5\ g\)

Answer: Ionic Compound

Explanation:

An ionic compound is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element or the metal and the element which accepts the electrons is known as electronegative element or non metal.

A covalent compound is formed when an element shares its valence electron with another element. This bond is formed between two non metals.

Mercury flouride contain ionic bonds as it is made up of mercury metal and fllourine non metal.

Answer:

A.) 1

Explanation:

Propane only exists in one conformation. It does not have enough carbons to form branches, and there are only hydrogens attached to each carbon. Furthermore, there is no way to twist the carbon or change its orientation (ex. cis- and trans-) to result in a different structure of propane. There is no other way to represent the molecule without drawing a different molecule.

Explanation:

from

de Broglie wavelength= h/ mv

substitute the values

joules= mass(kg) × acceleration (m/s^2) × distance(m)= kgm^2/s^2.

sorry I don't have my calc with me

According to De Broglie's Hypothesis, matter has both particle and wave properties. The wavelength of De Broglie is given as λ = h p,

wavelength (in m) of a proton =  11.09 x\(10^ {-55\) m.

λ= h m v.

= 6.63 x \(10^{-34}\) x  1.673 x \(10^{-24\)x 1000

= 11.09 x\(10^ {-55\) m

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Answer: When silver is exposed to sulfur-containing gases in the air, it discolours and then darkens as it reacts with the gas to form a surface layer of tarnish. This process is called tarnishing.

Hope this helped :)

When hydrogen bromide (HBr) is added to 2-methylbut-2-ene ((CH3)2CCHCH3), an electrophilic addition reaction takes place, where the π bond of the alkene is broken, and the hydrogen and bromine atoms are added to the resulting carbocation.

The reaction proceeds through a Markovnikov addition, where the hydrogen atom attaches to the carbon atom with the greater number of hydrogen atoms.

In this case, the initial addition of HBr to 2-methylbut-2-ene leads to the formation of a primary carbocation, as the positively charged carbon atom only has one alkyl group attached to it. The primary carbocation is relatively unstable, and it can undergo a rearrangement to form a more stable secondary carbocation.

The major product that is typically obtained is the 2-bromo-2-methylbutane. The hydrogen atom from HBr adds to the carbon with three hydrogen atoms (the more substituted carbon), resulting in the formation of a secondary carbocation.

On the other hand, a minor product is also formed, which is 3-bromo-2-methylbutane. This product arises from the addition of HBr to the primary carbocation, which is less stable. Although the primary carbocation is less favored, it can still be formed and lead to the formation of the minor product.

In summary, the addition of HBr to 2-methylbut-2-ene yields two products: the major product is 2-bromo-2-methylbutane, resulting from the addition of HBr to the more stable secondary carbocation, and the minor product is 3-bromo-2-methylbutane, originating from the less stable primary carbocation.

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Through tests and theoretical calculations, the cause of the high yield and the process of the accidental oxidation reaction were determined.

When two of the substances sodium and chloride combine to create sodium chloride, common table salt, the reaction is 2Na + Cl2 → 2NaCl. A kind of synthesis process in which metallic nanoparticles react with CO2 to produce the equivalent metallic carbonate is illustrated by the aforementioned interaction between calcium oxide and CO2.

Synthesis is the process of creating chemical compounds through the reaction of simpler constituents. A difficult and complex task that necessitates the ongoing creation of novel reactions, catalysts, and procedures is the building of sophisticated and defined new molecules.

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When working with dilutions we usually use this formula:

Vi * Mi = Vf * Mf

Where:

Vi = initial volume or the volume before the dilution

Mi = initial molarity or the molarity before the dilution

Vf = final volume or the volume after the dilution

Mf = final molarity or the molarity after the dilution

According to the problem we have:

Vi = ? mL

Mi = 11.65 M

Vf = 500 mL

Mf = 3.00 M

Since the initial volume is our unknown we will solve the formula for Vi:

Vi = Vf * Mf / Mi

Replacing:

Vi = 500 mL * 3.00 M / 11.65 M

Vi = 129 mL

We have to take 129 mL of the solution.

*

initial volume of solution is 128.75 ml

Here we are given with the initial and final molarity of the solution, and we have to find out the initial volume of the solution: -

Molarity of solution (M1) =11.65M

(Here M stands for molarity of the solution)

Let volume of solution = V1 ml

Final volume of the solution = 500 ml (V2)

Final molarity of the solution = 3.00 M (M2)

Using the formula: -

M1V1=M2V2

Substituting the values in the given expression from the above-mentioned values

11.65 x V1 = 500 x 3

V1= 128.75 ml

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Mangrove trees' robust roots aid in stabilizing the soil and preventing soil erosion. This is especially crucial in swampy coastal locations, where the soil is frequently loose and prone to wind and wave erosion. Mangrove trees work to stabilize the coastline and stop land erosion by securing the soil in place.

Mangrove trees also aid in removing contaminants from the water, enhancing the quality of the water in coastal areas while serving as an essential home for a variety of wildlife species. Mangrove trees contribute significantly to environmental protection and are an essential component of coastal ecosystems.

Mangrove trees are crucial for environmental preservation because they hold the soil in place and stop erosion in marshy coastal areas. They also provide crucial habitats for numerous wildlife species, stabilize the coastline, and stop the land from being lost to erosion while enhancing water quality by filtering pollutants from the ocean.

Mangrove trees are an essential part of coastal ecosystems overall, and preserving them is essential for preserving a healthy and sustainable environment.

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Answer:

C.Barometer

Explanation: