Which part of the electromagnetic spectrum is nearest to X-rays? microwaves infrared light gamma rays radio waves

Answers

Answer 1

Answer:

The correct answer to the following question will be Option A (Microwaves).

Explanation:

This tends to be a source of radiation, including some signals, infrared radiation, x-rays as well as gamma.  This appears to have a diverse variety of uses, involving communications, radar, and scheduling, but best known by most individual people. Sections of society used handheld devices, including tablets, as well as WiFi.

The other given choices are not related to the given circumstances. So that option A seems to be the appropriate one.

Answer 2

Answer:

its a

Explanation:


Related Questions

What biological sample do home pregnancy tests use to detect pregnancy?

Answers

Answer:

The biological sample that pregnancy tests use to detect pregnancy is urine

Answer:

yes

Explanation:

the urine in the the tester will find hormones to determine if pregnant or not

how many atoms are contained in 2.70g of aluminum provided that 32g of sulphur equals 6.02 × 10^(23)atoms​

Answers

Answer:

[tex]1.63 \times {10}^{24} [/tex]

one atom of an element = 6.02 \times {10}^{23} atom

The mass of one atom of sulphur = 32g

The mass of one atom of aluminium = 27g

so one atom of aluminium = 6.02 \times {10}^{23}

27g of AL = 6.02 \times {10}^{23} atom

2.70g of AL = X atoms

Then you cross multiply ........

and get the answer

.



Next, break down the equation shown into the skeletal half-reactions for oxidation and reduction. Which of
these pairs shows the two skeletal half-reactions with their correct assignments?
reduction half reaction: HNO, NO
oxidation half reaction SH,SO,
oxidation half reaction: HNO3 -> NO
reduction half reaction: SH2SO,
reduction half reaction: HNO3 -H,SO,
oxidation half reaction S -> H2SO4

Answers

Answer:

Its answer A

Explanation:

I just took the test

Answer:

A:

reduction half reaction: HNO3-> NO

oxidation half reaction S->H2SO4

Explanation:

a heating fan is supplied with 150 j of electricity which is nonverts to 113 j of thermal energy and 37 j of sound energy
How efficient is the heating fan?

Answers

Answer:

Heating Fan is 75.33% efficient.

Explanation:

As it is mentioned that the given fan is a HEATING FAN, it means the it is used for provided heat (thermal energy) to the surrounding. Hence, its efficiency depends upon how much heat/thermal energy is provides.

We can state that.

Total Energy provided = 150 J

Useful Energy = 113 J

Wasted Energy = 37 J

The efficiency of the heating fan can be given by following formula:

Efficiency = ( Useful Energy / Total Energy ) × 100

Efficiency = ( 113 J / 150 J) × 100

Efficiency = (0.7533)× 100

Efficiency = 75.33%

For each of the compounds, find the length of the longest carbon chain in the box provided.
CH,
CH3 -CH2-CH2-CH2-CH-CH2-CH2 - CH2 - CH3
CH,CH,CH,
HC
CH-CH2-CH3
CH,CH.CHCCH,CH,CH,
1
CH2
H.C CH.CH
CH3
<​

Answers

Answer:

In First compound the longest chain contains 7 C atoms.

In the Second compound the longest chain contains 9 C atoms.

In the Third compound the longest chin contains 7C atoms.

Explanation:

In first compound the longest chain contains 7 carbon atoms.In the second compound the longest chain contains 9 carbon atoms.In the third compound the longest chin contains 7 carbon atoms.

What is an atom?

An atom is defined as the smallest unit of matter which forms an element. Every form of matter whether solid,liquid , gas consists of atoms . Each atom has a nucleus which is composed of protons and neutrons and shells in which the electrons revolve.

The protons are positively charged and neutrons are neutral and hence the nucleus is positively charged. The electrons which revolve around the nucleus are negatively charged and hence the atom as a whole is neutral and stable due to presence of oppositely charged particles.

Atoms of the same element are similar as they have number of sub- atomic particles which on combination do not alter the chemical properties of the substances.

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Calculate the payback time if she insulates the loft with 300 mm insulation when: area of loft space = 100m^2 cost of roll of 100 mm thick loft insulation that covers 8.3m^2 = £20 cost of putting the insulation into the loft = £120 savings per year = £80 per 100 mm Payback Time = installation cost divided by annual savings

Answers

Answer: Payback time = 0.0075

Explanation: Since payback time is calculated as:

payback time = [tex]\frac{installation cost}{annual savings}[/tex]

First determine the installation cost:

100 mm thick insulation covers 8.3 m². Then 300 mm covers 24.9 m².

To cover 8.3m² costs £20. Then, the cost to cover 24.9 m² is:

cost = [tex]\frac{20*24.9}{8.3}[/tex]

cost = £60

The cost of putting the insulation is £120, so the total cost is:

total cost = £60 + £120

total cost = £180

Savings per year per 100 mm thick is £80. For 300 mm, the value of annual savings is:

savings = [tex]\frac{300*80}{100}[/tex]

savings = 240

payback time = [tex]\frac{installation cost}{annual savings}[/tex]

payback time = [tex]\frac{180}{240}[/tex]

payback time = 0.75

Primary succession is most likely caused by?

Answers

Answer:

volcanic eruption.

mark my answer as brainlest......

Which consists of only one type of atom?

Answers

Answer:

A chemical element

Explanation:

A chemical element consists of only one type of atom.

How does the government control scientific research

Answers

Answer:

The government allocates a budget for research every year. The spending of that money is determined by government priorities. Some of the money is spent directly, in government-funded research centers.

Other money is distributed to other research institutions.

Money spent by other institutions for research has no government oversight.

Explanation:

En la electrólisis del Cloruro de plomo se han depositado 2.6 gramos del metal en 20 minutos el peso atómico del plomo es 206 calculo la intensidad de la corriente eléctrica empleada AYUDENNNNN PLISSSS es para hoy!!!!!

Answers

Answer:

2.01 A

Explanation:

Para esta pregunta debemos empezar por la semi-reaccion del plomo:

[tex]Pb^+^2~_(_a_q_)~+~2e^-~->~Pb_(_s_)[/tex]

Se intercambian dos electrones en la semi-reacción del plomo. Si tenemos en cuenta la ecuación:

[tex]n=\frac{I*t}{z*F}[/tex]

Donde:

n= Moles depositados

I= Intensidad de corriente (en Amperios)

z= Numero de electrones intercambiados

F= Constante de faraday = 96484 C/mol

t=tiempo (en segundos)

Que conocemos de esta ecuación?

Los moles (se pueden calcular a partir de la masa atómica del plomo, 207.2 g/mol)

[tex]2.6~g~Pb\frac{1~mol~Pb}{207.2~g~Pb}=0.0125~mol~Pb[/tex]

El tiempo (hay que convertirlo a segundos):

[tex]20~min\frac{60~s}{1~min}=1200~s[/tex]

Z (Numero de electrones)

De acuerdo a la semi-reacción son intercambiados 2 electrones.

Por lo tanto podemos resolver para "I":

[tex]I=\frac{n*z*F}{t}=\frac{0.0125~mol*2*96484\frac{C}{mol}}{1200~s}[/tex]

[tex]I~=~2.01~\frac{C}{s}=2.01~A[/tex]

Espero que sea de ayuda!

A student weighs out a 6.64 g sample of , transfers it to a 500. mL volumetric flask, adds enough water to dissolve it and then adds water to the 500. mL tick mark. What is the molarity of cobalt(II) fluoride in the resulting solution

Answers

Answer:

the molarity of cobalt(II) fluoride in the resulting solution is = 0.137  M

Explanation:

Given that :

a student dissolves 6.64 g of CoF₂  into 500 mL of water

volume of the solution(water) = 500 mL = 0.50 L

The standard  molar mass of CoF₂ is 96.93 g/mol

number of moles of CoF₂ = mass of CoF₂/molar mass of CoF₂

number of moles of CoF₂ = 6.64 g/96.93 g/mol

number of moles of CoF₂ = 0.0685 mol

The molarity of any given substance is known to be as the number of moles of solute dissolved in one litre of solution.

Thus ;

Molarity of  cobalt(II) fluoride CoF₂ in the resulting solution is = number of moles / Volume in (L)

Molarity of  cobalt(II) fluoride CoF₂  = 0.0685 mol/ 0.50 L

= 0.137  M

Thus ; the molarity of cobalt(II) fluoride in the resulting solution is = 0.137  M

Which refers to the density of the medium’s particles at the compression of a sound wave? A.amplitude B.pitch C.volume D.wavelength

Answers

Answer:

A is the answer.

Explanation:

The sound wave is a longitudinal wave which travels in the form of compression and rarefaction . The point where the particles are closer to each other is called compression and the point where the particles are far apart from each other is called rarefaction.

Answer:

Amplitude

Explanation:

Amplitude describes the displacement of the medium's particles. The higher the amplitude, the more dense it is, and the lower the amplitude, the less dense it is.

Pitch is incorrect because it describes the sound

Volume is incorrect because it describes the sound.

Wavelength is incorrect because it describes the distance between a period of a wave.

Identify the correct statement regarding the strength of chemical bonds.
Weak bonds require more surface area to form than strong bonds.
A weak bond forms from a smaller change in energy compared to a strong bond.
Strong bonds occur with high temperature and weak bonds with low temperature.
Weak bonds require more heat to form than strong bonds.

Answers

Answer:

B.

Explanation:

A weak bond forms from a smaller change in energy compared to a strong bond.

name the bleaching agent for cloth​

Answers

Hydrogen peroxide is one of the most common bleaching agents. It is the primary bleaching agent in the textile industry, and is also used in pulp, paper, and home laundry applications.

bleaching agent is a material that lightens or whitens a substrate through chemical reaction. The bleaching reactions usually involve oxidative or reductive processes that degrade color systems. These processes may involve the destruction or modification of chromophoric groups in the substrate as well as the degradation of color bodies into smaller, more soluble units that are more easily removed in the bleaching process. The most common bleaching agents generally fall into two categories: chlorine and its related compounds (such as sodium hypochlorite) and the peroxygen bleaching agents, such as hydrogen peroxide and sodium perborate. Reducing bleaches represent another category. Enzymes are a new category of bleaching agents. They are used for textile, paper, and pulp bleaching as well as for home laundering. Chlorine‐containing bleaching agents are the most cost‐effective bleaching agents known. They are also effective disinfectants, and water disinfection is often the largest use of many chlorine‐containing bleaching agents. They may be divided into four classes: chlorine, hypochlorites, N‐chloro compounds, and chlorine dioxide. Except to bleach wood pulp and flour, chlorine itself is rarely used as a bleaching agent. The principal form of hypochlorite produced is sodium hypochlorite. Other hypochlorites include calcium hypochlorite and bleach liquor, bleaching powder and tropical bleach. The principal solid chlorine bleaching agents are the chlorinated isocyanurates, eg, sodium dichloroisocyanurate dihydrate. Other N‐chloro compounds include halogenated hydantoins, and sodium N‐chlorobenzenesulfonamide (chloramine B). Chlorine dioxide is a gas that is more hazardous than chlorine. Large amounts for pulp bleaching are made by several processes in which sodium chlorate is reduced with chloride, methanol, or sulfur dioxide in highly acidic solutions by complex reactions. Hydrogen peroxide is one of the most common bleaching agents. It is the primary bleaching agent in the textile industry, and is also used in pulp, paper, and home laundry applications. Hydrogen peroxide reacts with many compounds, such as borates, carbonates, pyrophosphates, sulfates, etc, to give peroxy compounds or peroxyhydrates. Peracids have superior cold water bleaching capability versus hydrogen peroxide because of the greater electrophilicity of the peracid peroxygen moiety. Lower wash temperatures and phosphate reductions or bans in detergent systems account for the recent utilization and vast literature of peracids in textile bleaching. The reducing agents generally used in bleaching include sulfur dioxide, sulfurous acid, bisulfites, sulfites, hydrosulfite (dithionites), sodium sulfoxylate formaldehyde, and sodium borohydride. These materials are used mainly in pulp and textile bleaching.

ANSWER:
Hydrogen peroxide is one of the bleaching agents that is most commonly used worldwide.
HOPE IT HELPS!!!

A 75 lb (34 kg) boy falls out of a tree from a height of 10 ft (3 m). i. What is the kinetic energy of the boy when he hits the ground? Round your answer to the nearest joule. ii. What is the speed of the boy when he hits the ground? Round your answer to two significant figures. iii. Using the conversion factors of 1 m = 1.094 yd and 1 mi = 1760 yd, calculate the speed of the boy in miles per hour when he hits the ground.

Answers

Answer:

Kinetic energy of boy just before hitting the ground is [tex]\approx[/tex]1000 J.

Speed of boy just before hitting the ground is 7.67 m/s

or 17.16 mi/hr.

Explanation:

Given that:

Mass of boy = 75lb = 34 kg

Height, h = 10ft = 3m

To find:

Kinetic energy of boy when he hits the ground.

As per law of conservation of energy The potential energy gets converted to kinetic energy.

[tex]\therefore[/tex] Kinetic energy at the time boy hits the ground = Initial potential energy of the boy when he was at the Height 'h'

The formula for potential energy is given as:

[tex]PE = mgh[/tex]

Where m is the mass

g is the acceleration due to gravity, g = 9.8 [tex]m/s^2[/tex]

h is the height of object

Putting all the values:

PE = [tex]34 \times 9.8 \times 3 \approx 1000\ J[/tex]

Hence, Kinetic energy is [tex]\approx[/tex]1000 J.

Formula for Kinetic energy is:

[tex]KE = \dfrac{1}{2}mv^2[/tex]

where m is the mass and

v is the speed

Putting the values and finding v:

[tex]1000 = \dfrac{1}{2}\times 34 \times v^2\\\Rightarrow v^2 = 58.82\\\Rightarrow v = 7.67\ m/s[/tex]

Given that:

1 m = 1.094 yd and 1 mi = 1760 yd

[tex]\Rightarrow 1609\ m = 1\ mi[/tex]

Converting 7.67 m/s to miles/hour:

[tex]\dfrac{7.67 \times 3600}{1609}=17.16\ mi/h[/tex]

Consider an electrochemical cell based on the spontaneous reaction 2AgCl(s) + Zn(s) → 2Ag(s) + 2Cl– + Zn2+. If the zinc ion concentration is kept constant at 1 M, and the chlorine ion concentration is decreased from 1 M to 0.001 M, the cell voltage should:

Answers

Answer:

there is an increase by 0.18 V in the cell voltage.

Explanation:

The given equation of the reaction can be well written as

[tex]2AgCl_{(s)} + Zn _{(s)} \to 2Ag_{(s)} + 2 Cl^- _{(aq)}+ Zn^{2+}_{(aq)}[/tex]

By application of Nernst Equation ; we have the expression

[tex]E_{cell} = E^0- \dfrac{0,059}{n}log (\dfrac{[product]}{[reactant]})[/tex]

here in the above equation;

n = number of electrons transferred in the equation of the reaction

n = 2

Also;

[tex]E^0 = E_{cathode} - E_{anode}[/tex]

[tex]E^0 = E_{Ag^+/Ag} - E_{Zn^+/Zn}[/tex]

[tex]E^0 = +(0.80 \ V) - (-0..76 \ V)[/tex]

[tex]E^0 = (0.80 \ V +0..76 \ V)[/tex]

[tex]E^0 = 1.56 \ V[/tex]

If the zinc ion concentration is kept constant at 1 M; we have:

[tex]E_{cell} = E^0- \dfrac{0.059}{n}log (\dfrac{[product]}{[reactant]})[/tex]

[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log ({[Zn^{2+} ]}{[Cl^{2-}]})[/tex]

[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log (1)[/tex]

Since log(1) = 0

Therefore;

[tex]E_{cell} = 1.56\ V[/tex]

When the chlorine ion concentration is decreased from 1 M to 0.001 M; we have;

[tex]E_{cell} = E^0- \dfrac{0.059}{n}log (\dfrac{[product]}{[reactant]})[/tex]

[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log ({[Zn^{2+} ]}{[Cl^{2-}]})[/tex]

[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log ({[1*0.001^2}]})[/tex]

[tex]E_{cell} = 1.56 - 0.0295 \ * \ log ({[1*10^{-6}}]})[/tex]

[tex]E_{cell} = + 1.737 \ V[/tex]

The change in voltage = [tex]E_{cell} - E^0[/tex]

=( 1.737 - 1.56 )V

= 0.177 V

≅ 0.18 V

Thus; from the following observation; there is an increase by 0.18 V in the cell voltage.

The voltage of the cell increased by 0.18 V.

The equation of the reaction is; 2AgCl(s) + Zn(s) → 2Ag(s) + 2Cl– + Zn2+

We know that;

E°cell = 1.36 - (-0.76) = 2.12 V

If the cells are both at 1M concentration the Ecell = E°cell = 2.12 V

When the concentration of Cl- decreased from 1 M to 0.001 M

Ecell =  E°cell - 0.0592/n log Q

Substituting values;

Ecell = 2.12 V -  0.0592/2 log (1 × (0.001)^2)

Ecell = 2.298 V

Increase in voltage = 2.298 V - 2.12 V = 0.18 V

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How many moles of silver are equivalent to 2.408 x 10^24 atoms

Answers

i found this i hope this helps you

The mole is used to measure small particles like atoms and molecules.  The 4 moles of silver is equivalent to [tex]\bold{2.408 x 10^2^4 }[/tex] atoms.

Given here,

The number of atoms

[tex]\bold{2.408 x 10^2^4 }[/tex]

Number of moles = ?

1 mol of substance = [tex]\bold{ 6.02 x10^2^3}[/tex]

Hence,

moles of silver,

[tex]\bold {= \dfrac {2.408 x 10^2^4 } { 6.02 x10^2^3} }}\\\\\bold {= 4 mol}[/tex]

Therefore, the 4 moles of silver is equivalent to [tex]\bold{2.408 x 10^2^4 }[/tex] atoms.

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what is the mass of 1.5 mol of aluminum?

Answers

Answer:

40.47 g Al

Explanation:

In 1 mol of Al, there are 26.98 grams of Al, so Al is 26.98 g/mol.

Simply multiply 1.5 by 26.98 to get your answer.

5.What is the chemibal formula for lead (II) iodide

Answers

Answer: The chemical formula for lead (II) iodide is [tex]PbI_2[/tex]

Explanation:

Lead (II) iodide is a ionic compound because it are formed by transference of electrons between metals and non metals.The bond formed between a metal and a non-metal is always ionic in nature.

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

The nomenclature of ionic compounds is given by:

1. Positive is written first followed by the oxidation state of metal in roman numerals in square brackets.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

The chemical formula for lead (II) iodide is [tex]PbI_2[/tex]

Phloem contains specialized cells that move sugars and other nutrients both
up and down in plants. Which object is the best model of one of these cells?
A. Water balloon

Answers

Answer:Drinking straw

Explanation: my quiz said so

The object that represents the best model of one of the phloem cells is the drinking straw.

What is Phloem?

Phloem may be defined as a type of vascular tissue in plants that significantly conducts the passage of sugars and other metabolic products from the leaves to the roots. This mediates the migration of sugar which is synthesized in the leaves during photosynthesis.

Apart from this, the phloem also regulates the movement of nutrients from the roots to the upper part of the body. This function of phloem cells shares homology with the object known as a drinking straw that mediates the transport of any fluid in both directions.

Therefore, a drinking straw is an object that represents the best model of one of the phloem cells.

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Your question seems incomplete. The most probable complete question is as follows:

Can openerWater balloonDrinking StrawKitchen strainer

At what angle(s) should investigators photograph a tool mark? A. parallel B. perpendicular C. 45 degrees D. varying

Answers

The answer is b. perpendicular

Answer:

B

Explanation:

which of the following us not a property of salt in the ionic and covalent properties lab

Answers

Explanation:

I think Low melting point

because NaCl (salt has mp 801 degC)

Water has mp 0 degC

what is meant by amoeba and what is the name of its parts​

Answers

Answer:

amoeba is the unicellular organism which can be seen by only microscope but not with our nacked eyes

Answer:

An amoeba often called an amoeboid, is a type of cell or unicellular organism which has the ability to alter its shape, primarily by extending and retracting pseudopods Amoebae do not form a single taxonomic group; instead, they are found in every major lineage of eukaryotic organisms. Amoeboid cells occur not only among the protozoa, but also in fungi, algae, and animals.

name of parts of ameoba

3 parts – the cytoplasm, plasma membrane and the nucleus. The cytoplasm can be differentiated into 2 layers – the outer ectoplasm and the inner endoplasm. The plasma membrane is a very thin, double-layered membrane composed of protein and lipid molecule

i hope this will help you

what is Lfusion in the equation Q=mLfusion

Answers

Answer:

[tex]L_{\text{fusion}[/tex] is latent heat of fusion.

Explanation:

Latent heat is basically termed as hidden heat. It is defined as the amount of heat required to change 1 g of a substance at the temperature of its melting point from the solid to the liquid state at constant temperature.The formula that relates heat and latent heat of fusion is given by :

[tex]Q=mL_{\text{fusion}}[/tex]

Here,

m is mass of substance

[tex]L_{\text{fusion}[/tex] is latent heat of fusion.

what is the molar mass of magnesium tartrate

Answers

Answer:

172.385 g/mol

Explanation:

Magnesium Tartrate is C4H4MgO6

C - 12.01 g/mol

H - 1.01 g/mol

Mg - 24.305 g/mol

O - 16.00 g/mol

12.01(4) + 1.01(4) + 24.305 + 16(6) = 172.385 g/mol

Answer:

172.38

Explanation:

[tex]C_4H_4MgO_6\\C=12.01\\H=1.01\\Mg=24.30\\O=16.00\\\\4(12.01)+4(1.01)+24.30+6(16.00)\\48.04+4.04+24.30+96\\=172.38[/tex]

C = 12.01

H=1.01

Mg=24.30

O =16.00

4(12.01)+4(1.01)+24.30+6(16.00)

48.04 +4.04+24.30+96

=172.38

Which statement correctly describes a feature of the rock cycle?
PA
Rocks are preserved through the process.
Rocks change from one type to another.
Different rock groups are not related to one another.
Rocks change from one type to another in a specific order.

Answers

Answer:

It’s D

Explanation:

Answer:

D Rocks change from one type to another in a specific order.

Explanation:

Which sequence represents the relationship between pressure and volume of an ideal gas as explained by the kinetic-molecular theory? more gas particles Right arrow. More collisions Right arrow. Higher pressure smaller volume Right arrow. Crowded particles Right arrow. Less collisions Right arrow. Lower pressure smaller volume Right arrow. Crowded particles Right arrow. More collisions Right arrow. Higher pressure more gas particles Right arrow. More kinetic energy Right arrow. More volume Right arrow. Higher pressure

Answers

Answer:

smaller volume ⇒ Crowded particles ⇒ More collisions  ⇒ Higher pressure

Explanation:

Smaller the volume , more crowed the particles . Then the particles will have rapid collisions so the free mean path is decreased , hence the pressure will be increased as follows

[tex]P=\frac{K_bT}{\sqrt{2}\pi d^2\lambda }[/tex]

where λ is mean free path , P is pressure .

The sequence the represent the relationship between pressure and volume of an ideal gas is: smaller volume right arrow Crowded particles right arrow More collisions right arrow Higher pressure

The kinetic molecular theory made five postulates which are used to explain the behaviour of gases.

From the postulates, he uses the kinetic molecular theory to explain Boyle's Law because the majority of a gas's volume in space is usually empty and may be compressed.

So, when a gas is compressed without affecting its temperature, the average kinetic energy of the gas particles remains constant. The particles continue to flow at the same rate, but the container has reduced.

As a result, the particles go from one end of the container to another in less time. This suggests they're hitting the barriers (collision) more frequently. Each and every increase in the frequency of collisions with the walls, thus, results in an increase in the gas's pressure.

Hence, as the volume of a gas decreases, the pressure of the gas increases.

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After 273 m3 of ethylene oxide at 748 kPa and 525 K is cooled to 293 K, it is allowed to expand to 1100. m3. The new pressure is _____kPa

Answers

Answer:

[tex]\large \boxed{\text{104 kPa}}[/tex]

Explanation:

To solve this problem, we can use the Combined Gas Laws:

[tex]\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}[/tex]

Data:

p₁ = 748 kPa;  V₁ =   273 m³;  n₁ = n₁;  T₁ =  525 K

p₂ = ?;             V₂ = 1100. m³; n₂ =  n₁; T₂ =  293 K

Calculations:

[tex]\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{748 kPa}\times \text{273 m}^{3}}{n _{1}\times \text{525 K}} & = &\dfrac{p_{2}\times \text{1100. m}^{3}}{n _{1}\times \text{293 K}}\\\\\text{390.0 kPa} & = &3.754{p_{2}}\\p_{2} & = & \dfrac{\text{390.0 kPa}}{3.754}\\\\ & = & \textbf{104 kPa} \\\end{array}\\\text{The new pressure is $\large \boxed{\textbf{104 kPa}}$}[/tex]

Other Questions
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