Please it's due today
Please it's due today - 1

Answers

Answer 1
Answer:

Answer:

B

Explanation:

Newton's third law. states that:

Action and reaction are equal and opposite.


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How many valence electrons must a lithium atom lose to obtain a complete valence shell?A. one
B. two
C. three
D. four

Answers

Answer:

C. three

Explanation:

Final answer:

A lithium atom must lose one valence electron to achieve a full valence shell, resulting in a positively charged lithium cation (Lit) with a noble gas configuration similar to helium.

Explanation:

To achieve a full valence shell, a lithium atom must lose one electron. Lithium has an atomic number of 3, which means it has three electrons: two in the first shell and one in the second shell. Since the first shell (1s) is already full with two electrons, lithium has a single electron in the 2s subshell of the second shell. This single electron is the valence electron.

According to the Lewis diagram, lithium (Li) has only one valence electron in its second shell. By transferring this lone electron to another atom, lithium's electron configuration will resemble that of helium (He), with two electrons in its first shell, thus achieving a stable noble gas configuration. This transfer results in the formation of a lithium cation, denoted as Lit, with a charge of 1+.

It is important to note that when lithium becomes a cation, it does not necessarily mean it has a complete valence shell in terms of helium or neon. Instead, it has achieved stability by having a full inner shell, which mimics the noble gas configuration of helium.

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You have a stock solution of epinephrine at a concentration of 1 mg/mL. Knowing that the pipette you will use delivers 20 drops/mL, a. calculate the number of drops of the stock solution that must be added to a smooth muscle bath containing 25 mL of Locke’s solution so that the final concentration of epinephrine in the muscle bath will be 100 µg/mL, and

Answers

Answer : The number of drops pf the stock solution mist be added are, 50 drops.

Explanation :

As we are given that the concentration of stock solution 1 mg/mL and pipette delivers 20 drops/mL. That means,

1 mg of epinephrine = 1 mL = 20 drops

The final volume of Locke's solution = 25 mL

and the final concentration needed = 100\mu g/mL

As, 1 mL of solution contains = 100\mu g

So, 25 mL of solution contains = (25mL)/(1mL)* 100\mu g=2500\mu g=2.5mg

Conversion used :1\mu g=0.001mg

Now we have to determine the number of drops needed.

As, 1 mg of epinephrine contains 20 drops.

So, 2.5 mg of epinephrine contains 2.5 × 20 = 50 drops.

Therefore, the number of drops pf the stock solution mist be added are, 50 drops.

Final answer:

If you want to achieve a final concentration of 100 µg/mL of epinephrine in a 25 mL solution, when using a stock solution of 1 mg/mL and a pipette that delivers 20 drops/mL, you need to add 50 drops of your stock solution.

Explanation:

Since we are asked to find the number of drops of stock solution required to achieve a final concentration of 100 µg/mL in a 25 mL solution, the first step is to convert the concentration of the stock solution to the same units, µg/mL. Hence, 1 mg/mL is equal to 1000 µg/mL. Further, we know that 1 mL of the stock solution contains 1000 µg of epinephrine, and our pipette delivers 20 drops/mL, so 1 drop of stock solution contains 1000 µg / 20 drops = 50 µg. Thus, if we need a 100 µg/mL concentration in 25 mL, we need a total of 100 µg/mL * 25 mL = 2500 µg of epinephrine. Therefore, to achieve this, we must add 2500 µg / 50 µg/drop = 50 drops of our stock solution. Hence,

50 drops

of the stock solution should be added to achieve the desired concentration.

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In compliance with conservation of energy, Einstein explained that in the photoelectric effect, the energy of a photon (hv) absorbed by a metal is the sum of the work function (Φ), the minimum energy needed to dislodge an electron from the metal’s surface, and the kinetic energy (Ek) of the electron: hv = Φ + Ek. When light of wavelength 358.1 nm falls on the surface of potassium metal, the speed (u) of the dislodged electron is 6.40 x 10⁵ m7s. (a) What is Ek (½mu²) of the dislodged electron? (b) What is Φ (in J) of potassium?

Answers

Answer:

 a) 1.866 × 10 ⁻¹⁹ J      b)   3.685 × 10⁻¹⁹ J

Explanation:

the constants involved are

h ( Planck constant) = 6.626 × 10⁻³⁴ m² kg/s

Me of electron = 9.109 × 10 ⁻³¹ kg

speed of light = 3.0 × 10 ⁸ m/s

a) the Ek ( kinetic energy of the dislodged electron) = 0.5 mu²

Ek = 0.5 × 9.109 × 10⁻³¹ × ( 6.40 × 10⁵ )² = 1.866 × 10 ⁻¹⁹ J

b) Φ ( minimum energy needed to dislodge the electron ) can be calculated by this formula

hv =   Φ + Ek

where Ek = 1.866 × 10 ⁻¹⁹ J

v ( threshold frequency ) = c / λ where c is the speed of light and λ is the wavelength of light = 358.1 nm = 3.581 × 10⁻⁷ m

v = ( 3.0 × 10 ⁸ m/s ) / (3.581 × 10⁻⁷ m ) = 8.378 × 10¹⁴ s⁻¹

hv = 6.626 × 10⁻³⁴ m² kg/s ×  8.378 × 10¹⁴ s⁻¹ = 5.551 × 10⁻¹⁹ J

5.551 × 10⁻¹⁹ J = 1.866 × 10 ⁻¹⁹ J + Φ

Φ = 5.551 × 10⁻¹⁹ J - 1.866 × 10 ⁻¹⁹ J = 3.685 × 10⁻¹⁹ J

How many liters of the antifreeze ethylene glycol [CH2(OH)CH2(OH)] would you add to a car radiator containing 6.50 L of water if the coldest winter temperature in your area is -10.°C? (The density of ethylene glycol is 1.11 g/mL. Assume the density of water at -10.°C is 1.00 g/mL.)

Answers

Answer:

Around 2.0 L of ethylene glycol needs to be added to the car radiator

Explanation:

The depression in freezing point ΔTf of a solution is directly proportional to its molality (m), i.e.

\Delta T_(f)= T_(f)^(0)-T_(f)=i*K_(f)*m

From the given information:

T_(f) = freezing pt of solution = -10.0 C

T_(f)^(0) = freezing pt of pure solvent = 0 C

Kf = freezing pt depression constant = 1.86 C/m

i = 1 for ethylene glycol antifreeze

[0-(-10.0)] C= 1*(1.86 C/m) *( m)\n\nm = 5.38

Molality = (moles\ of\ solute)/(kg\ solvent) \n\nTherefore,\ moles of antifreeze = molality* mass\ of\ water\n

Molality = (moles\ of\ solute)/(kg\ solvent) \n\nTherefore,\ moles of ethylene glycol = molality* mass\ of\ water\n

Volume of water = 6.50 L = 6500 ml

Density of water = 1.00 g/ml

Therefore mass of water = density * volume = 1.00g/ml*6500ml = 6500g = 6.50kg

moles\ of\ ethylene glycol= 5.38moles/kg*6.50kg = 34.9 moles

Molar mass of ethylene glycol = 62 g/mol

Mass of ethylene glycol needed = molar\ mass* moles = 62g/mol*34.9moles=2163.8g

Density of ethylene = 1.11 g/ml

Therefore, volume needed = (mass)/(density) =(2163.8g)/(1.11g/ml) =1949ml

 16 = m x 1.86 
m = 8.60 = moles solute / 6.50 Kg 

moles solute = 55.9 

mass solute = 55.9 x 62.068 g/mol=3470 g 

V = 3479/ 1.11 =3126 mL= 3.13 L 

delta T = 8.60 x 0.512 =4.40 
boling point = 104.4 °C

Hope this helps.

A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base (B) solution. Show all your work.pKb for the weak base = 4.82.

Molar mass of the weak base = 82.0343g/mole.

Note: pKa = -logKa

pKb = -logKb

pH + pOH = 14

[H+ ] [OH- ] = 10^-14

Answers

Answer:

11.39

Explanation:

Given that:

pK_(b)=4.82

K_(b)=10^(-4.82)=1.5136* 10^(-5)

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

moles = (Mass\ taken)/(Molar\ mass)

Thus,

Moles= (1.805\ g)/(82.0343\ g/mol)

Moles= 0.022\ moles

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)

Molarity=(0.022)/(0.055)

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

                                  B +   H₂O    ⇄     BH⁺ +        OH⁻

At t=0                        0.4                          -              -

At t =equilibrium     (0.4-x)                        x           x            

The expression for dissociation constant is:

K_(b)=\frac {\left [ BH^(+) \right ]\left [ {OH}^- \right ]}{[B]}

1.5136* 10^(-5)=\frac {x^2}{0.4-x}

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³  M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

pH = 14 - pOH = 14 - 2.61 = 11.39

It is possible for gases in the atmosphere to change the behavior of energy on earth. Agree or disagree

Answers

The green house gases in the atmosphere can change the behavior of energy on earth. Thus i agree with this statement.

What is green house effect?

The green house effect is defined as a process by which the radiations from the sun are absorbed by the green house gases like methane, Chloro fluoro carbons and not reflected back into the space. This makes the surface as insulator and prevents it from freezing.

Due to the increased level of green house gases, the temperature of the earth increases considerably and thereby causes global warming, depletion of ozone layer, smog and air pollution, etc.

It is the natural green house effect which affect the behaviour of heat energy radiated by the sun. The green house gases never let the radiations escape from the earth and increase the surface temperature of earth. This leads to global warming.

Thus green house gases affects the solar energy.

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Answer:

I agree

Explanation:

because the sunlight passes through the atmosphere and is absorbed by earth's surface.