I need helpppppppppp, please. What type of radioactive elements does Seabrook Station, New Hampshire's have?

Answer:

motion of molecules because we gonna see which one is in a solid state and what's so ever

Answer:

alpha decay

Explanation:

Nuclear reactions accompany nuclear decay and this type of decay is alpha decay.

There are two types of nuclear reactions which are nuclear fusion and nuclear fission .They involve the combination and disintegration of the element's nucleus respectively.

In nuclear fission, the nucleus of the atom is bombarded with electrons of low energy which splits the nucleus in to two parts .Large amount of energy is released in the process.It is used in nuclear power reactors as it produces large amount of energy.

In nuclear fusion,on the other hand, is a reaction which occurs when two or more atoms combine to form a heavy nucleus.Large amount of energy is released in the process which is greater than that of the energy which is released in nuclear fission process.

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Answer:

200 kPa

(it is actually 199.999, but rounding up it is 200)

hope this helps!

Answer:

1500 Torr is  about 200 kPa

Answer:

What are the pros of coal energy? ans :

Huge Global Reserves.

Not an Intermittent Energy Source.

Reliable Fuel.

Inexpensive Energy Source.

Minimal Wastes, Consumable Byproducts

Smokeless Alternatives

Compatible With Other Energy Sources.

Answer:

soil retains water and nutrients

Explanation:

abiotic factors

water

soil

pH

Answer: B

Explanation:

Answer:

Thus is because of the it ionic nature of compound the when NaCl is dissolve in the water the Na and CL ions are free to move in the solution because of this they can conduct electricity and serves as an electrolyte

Explanation:

Given the data from the question, the identity of the unknown metal having a of mass 133 g is Cobalt

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

Density = mass / volume

Density = 133 / 15

Density = 8.86 g/mL

Comparing the density of the unknown metal (i.e 8.86 g/mL) with those given in the chart in the question above, we can conclude that the unknown metal is Cobalt

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Answer:

D. Stores calcium and phosphorus

Explanation:

Correct me if I'm wrong

Answer:

B. Design homes that can be built on stilts where flooding is likely to occur.

C. Replace power lines that have been damaged by severe weather.

Explanation:

[tex]Hope[/tex] [tex]this[/tex] [tex]helps![/tex] [tex]:)[/tex]

The volume that would be necessary to decrease the pressure to 1.0 atm is mathematically given as

v2=6.4l

Question Parameter(s):

A 2.0 l container of oxygen had a pressure of 3.2 atm

Generally, the equation for the   is mathematically given as

P1v1=p2v2

Therefore

v2=piv1/p2

v2=2*3.2/1

v2=6.4l

In conclusion, the  volume is

v2=6.4l

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Answer:

Explanation:

Answer:

measure the final temperature of water

Answer:

0.5 N

Explanation:

The ball pushes back with the same force as Alice applies to it

Answer:

0.5 Newtons

Explanation:

did it on edge and got it right

Answer:

B h2

Explanation:

Answer:H2

Explanation:

Answer:

Intermolecular Forces: Review. Electronegativity is a measure of an atom's ability to attract the shared electrons of a covalent bond to itself. If atoms bonded together have the same electronegativity, the shared electrons will be equally shared.

Explanation:

hope this helped

Answer:

Kinetic energy, Kelvin temperature, direction, warmer, cooler, Kelvin, Absolute zero

Explanation:

THe answers are in order and when there is a comma move to the next ______

Answer:

Biological Functions: processes and structures by which organisms adjust to short-term or long-term changes in their environment

Answer:

[tex]2.31*10^3 L[/tex] of gas are in the container.

Explanation:

1. First, convert temperature to kelvins. Otherwise, you will get the wrong answer. T = 56 + 273.15 = 329.15 K.

2.Use the ideal gas law. [tex]PV = nRT[/tex]

3.Re-arrange for volume and use the gas contstant R = [tex]\frac{0.082(L*atm)}{K*mol}[/tex]

4.Insert values. [tex]PV = nRT = \frac{(7.7 moles)(0.082 L*atm/(K*mol))(329.15K)}{(0.090 atm)} = 2.31*10^3 L.[/tex]

Answer:

Approximately [tex]7.25[/tex].

Explanation:

If the concentration of [tex]{\rm H^{+}}[/tex] ions in a solution is [tex]x\; {\rm M}[/tex], the [tex]{\rm pH}[/tex] of that solution would be [tex](-\log_{10}(x))[/tex].

Note that the base of the logarithm in this calculation should [tex]10[/tex]. One way to be sure is to state the base explicitly. Using the change of base rule of logarithms:

[tex]\begin{aligned}-\log_{10}(x) = -\frac{\ln(x)}{\ln(10)}\end{aligned}[/tex].

In this question, it is implied that the concentration of [tex]{\rm H^{+}}[/tex] in the given solution is [tex]5.6 \times 10^{-8}\; {\rm M}[/tex], such that [tex]x = 5.6 \times 10^{-8}[/tex]. Using the equations above:

[tex]\begin{aligned}& \text{pH of this solution} \\ =\; & -\log_{10}(x) \\ =\; & -\frac{\ln(x)}{\ln(10)} \\ =\; & -\frac{\ln(5.6 \times 10^{-8})}{\ln(10)} \\ \approx\; & 7.25\end{aligned}[/tex].

Since the bell weighs 190 N and is 45 m high, the bell’s potential energy is 8550 J

To answer the question, we need to know what potential energy is.

This is energy due to position. It is given by PE = mgh where

Also, W = mg = weight of object.

So, PE = Wh

Now, given that the bells weight 190 N and is 45 m high,

So, PE = Wh

= 190 N × 45 m

= 8550 Nm

= 8550 J

So, the bell’s potential energy is 8550 J

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The correct balanced equation of a beta decay undergone by Mercury-203 is as follows:

203 80 Hg → 203 81 + 0 -1 e

Beta decay in radioactivity refers to the release of a beta particle by a radioactive element.

Beta particle is characterized by the possession of mass number 0 and atomic number -1.

This means that if Mercury-203 undergoes beta decay, the atomic number of the resulting isotope will be +1 as follows:

203 80 Hg → 203 81 Tl + 0 -1 e

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The concentration of the hydronium ion in hydrochloric acid is 0.0045 M, and the pH of the solution is 2.34.

pH is the potential of the hydrogen or the hydronium ions in the aqueous solution.

As the solution contains [tex]4.5 \times 10^{-3} \;\rm M\;[/tex] HCl the concentration of the hydronium ion will be the same, [tex]4.5 \times 10^{-3} \;\rm M.[/tex]

The pH of the solution is calculated as:

[tex]\begin{aligned} \rm pH &= \rm -log[H^{+}]\\\\&= - \rm log (4.5 \times 10^{-3})\\\\&= 2.34\end{aligned}[/tex]

The concentration of the hydroxide ion is calculated from pH and hydronium ion as:

[tex]\begin{aligned} \rm [H_{3}O^{+}][OH^{-}] &= 10^{-14}\\\\&= \dfrac{1 \times 10^{-14}}{4.5 \times 10^{-3}}\\\\&= 2.2 \times 10^{12}\end{aligned}[/tex]

Now, for the calcium hydroxide solution, the calculations are shown as,

[tex]\begin{aligned} \rm (H_3}\rm O^{+}) &= \rm antilog (-pH)\\\\&= \rm antilog (-8)\\\\&= 10^{-8} \;\rm M\end{aligned}[/tex]

pOH is calculated as:

[tex]\begin{aligned} \rm pOH &= 14- 8 = 6\\\\\rm [OH^{-}] &= \rm antilog (-6)\\\\&= 10^{-6} \end{aligned}[/tex]

The concentration of calcium hydroxide is calculated as:

[tex]\begin{aligned} &= \dfrac{1}{2} \times \rm [OH^{-}]\\\\&= 5 \times 10^{-4} \;\rm M\end{aligned}[/tex]

Therefore, the pH and the pOH give the concentration of the hydrogen or the hydronium ion and the hydroxide ion.

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Answer: A heating curve graphically represents the phase transitions that a substance undergoes as heat is added to it.

The plateaus on the curve mark the phase changes. The temperature remains constant during these phase transitions.

Water has a high boiling point because of the strong hydrogen bonds between the water molecules; it is both a strong hydrogen bond donor and acceptor.

The first change of phase is melting, during which the temperature stays the same while water melts. The second change of phase is boiling, as the temperature stays the same during the transition to gas

Explanation:

That's latent heat of vaporization.

Answer:

The answer should be I and III

Answer:

| and |||

Explanation:

Answer:

If a metal is less reactive than carbon, it can be extracted from its oxide by heating with carbon. The carbon displaces the metal from the compound, and removes the oxygen from the oxide. This leaves the metal.

Answer:

D  86.8%

Explanation:

Two bromines mole wt = 2 * 79.9 = 159.8 gm  (from periodic table)

Mg = 24.3  gm

159.8 / ( 159.8 + 24.3)  = .868 = 86.8 %

Answer:

exothermic

Explanation:

Diagrams shows reactants.....add a little energy to get the reaction going

 then energy is released and the final product has less energy ....so energy was released (heat) so this would be an exothermic reaction

The values of molarity and volumes for the given questions are  2,100 mL, 0.75 M, 0.1 M, 50,000 mL, 2.07 M and 1,200 mL respectively.

Relation between the molarity and volume of the solution will be shown by the following equation:
M₁V₁ = M₂V₂, where

M₁ & V₁ is the molarity and volume of one solution whereas M₂ & V₂ are the molarities and volume of the another solution.

V₂ = (750)(2.8) / (1) = 2,100 mL

M₂ = (0.15)(125) / (25) = 0.75 M

M₂ = (100)(0.15) / (150) = 0.1 M

V₂ = (10)(250) / (0.05) = 50,000 mL

M₂ = (1.5)(345) / (250) = 2.07 M

V₂ = (2.4)(500) / (1) = 1,200 mL

Hence required values of molarity and volume are 2,100 mL, 0.75 M, 0.1 M, 50,000 mL, 2.07 M and 1,200 mL respectively.

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Answer:

Many people have a qualitative idea of what is meant by concentration. Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. In chemistry, the concentration of a solution is the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for solution reactions. Chemists use many different methods to define concentrations, some of which are described in this section.

Molarity

The most common unit of concentration is molarity, which is also the most useful for calculations involving the stoichiometry of reactions in solution. The molarity (M) is defined as the number of moles of solute present in exactly 1 L of solution. It is, equivalently, the number of millimoles of solute present in exactly 1 mL of solution:

molarity=molesofsolutelitersofsolution=mmolesofsolutemillilitersofsolution(4.5.1)

(4.5.1)molarity=molesofsolutelitersofsolution=mmolesofsolutemillilitersofsolution

The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as MM . An aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the molar concentration of a solute. Therefore,

[sucrose]=1.00M(4.5.2)

(4.5.2)[sucrose]=1.00M

is read as “the concentration of sucrose is 1.00 molar.” The relationships between volume, molarity, and moles may be expressed as either

VLMmol/L=L(molL)=moles(4.5.3)

(4.5.3)VLMmol/L=L(molL)=moles

or

VmLMmmol/mL=mL(mmolmL)=mmoles(4.5.4)

(4.5.4)VmLMmmol/mL=mL(mmolmL)=mmoles

Figure 4.5.14.5.1 illustrates the use of Equations 4.5.34.5.3 and 4.5.44.5.4 .

alt

Figure 4.5.14.5.1 : Preparation of a Solution of Known Concentration Using a Solid Solute

Example 4.5.14.5.1 : Calculating Moles from Concentration of NaOH

Calculate the number of moles of sodium hydroxide (NaOH) in 2.50 L of 0.100 M NaOH.

Given: identity of solute and volume and molarity of solution

Asked for: amount of solute in moles

Strategy:

Use either Equation 4.5.34.5.3 or Equation 4.5.44.5.4 , depending on the units given in the problem.

Solution:

Because we are given the volume of the solution in liters and are asked for the number of moles of substance, Equation 4.5.34.5.3 is more useful:

molesNaOH=VLMmol/L=(2.50L)(0.100molL)=0.250molNaOHmolesNaOH=VLMmol/L=(2.50L)(0.100molL)=0.250molNaOH

Exercise 4.5.14.5.1 : Calculating Moles from Concentration of Alanine

Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine.