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2 years ago

Which statement is true: Mg -> Mg2+ + 2e-

Chemistry

1 answer:

Damm [24]2 years ago

3 0

Answer:

(2) Mg²⁺ loses 2 electrons.

Explanation:

Let's consider the following half-reaction.

Mg ⇒ Mg²⁺ + 2 e⁻

Magnesium is a metal that loses 2 electrons to fulfill the octet rule. When magnesium loses electrons, the oxidation number increases from 0 to +2. This kind of half-reaction is known as oxidation, and has to be accompanied by a reduction of another species.

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The shape of hair is determined in part by the pattern of disulfide bonds in keratin, its major protein. From first to last, arr

UNO [17]

Answer:

I.A thiol containing reagent and Heating(gentle)

II. Disulfide bond Breaker for Keratin

III. curling the hair physically

IV.A good oxidizing agent application

V. Disulfide bonds reformer for keratin to hold the shape

Explanation:

The hair is mostly made of Keratin which is a structural proteins with about 15% cysteine

In order to change the hair appearance on the outside a reducing agent is required containing a thiol as to reduce the formation of the sulfide and disulfide bridges formation in the hair so the that any changes can be applied without the body naturally turning it back, i-e allowing easier manipulation after this hairs are to be curled physically in order to keep this a oxidizing agent is applied and then finally a keratin reformer is added to hold the shape

4 0

1 year ago

A 36 gram sample of water has an initial temperature of 22 degrees Celsius after the sample absorbs 1200joules of heat energy th

UkoKoshka [18]

We can use the heat equation,
Q = mcΔT

Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

Let's assume that the finale temperature is T.

Q = 1200 J

m = 36 g
c = 4.186 J/g °C

ΔT = (T - 22)

By applying the formula,

1200 J = 36 g x 4.186 J/g °C x (T - 22)

(T - 22) = 1200 J / (36 g x 4.186 J/g °C)

(T - 22) = 7.96 °C

T = (7.96 + 22) °C = 29.96 °C

T = 30 °C

Hence, the final temperature is 30 °C.

8 0

2 years ago

Read 2 more answers

Octane is a liquid component of gasoline. Given the following vapor pressures of octane at various temperatures, estimate the bo

Hitman42 [59]

Answer:

110.8 ºC

Explanation:

To solve this problem we will make use of the Clausius-Clayperon equation:

lnP = - ΔHºvap/RT + C

where P is the pressure, ΔHºvap is the enthalpy of vaporization, R is the gas constant, T is the temperature, and C is a constant of integration.

Now this equation has a form y = mx + b where

y = lnP

x = 1/T

m = -ΔHºvap/R

Now we have to assume that ΔHºvap remains constant which is a good asumption given the narrow range of temperatures in the data ( 104-125) ºC

Thus what we have to do is find the equation of the best fit for this data using a software as excel or your calculator.

T ( K) 1/T ln P

377 0.002653 5.9915

384 0.002604 6.2115

390 0.002564 6.3969

395 0.002532 6.5511

398 0.002513 6.6333

The best line has a fit:

y = -4609.5 x + 18.218

with R² = 0.9998

Now that we have the equation of the line, we simply will substitute for a pressure of 496 mm in Leadville.

ln(496) = -4609.5(1/Tb) + 18.218

6.2066 = -4609.5(1/Tb) +18.218

⇒ 1/Tb = (18.218 - 6.2066)/4609.5 = 0.00261

Tb = 383.76 K = (383.76 -273)K = 110.8 ºC

Notice we have touse up to 4 decimal places since rounding could lead to an erroneous answer ( i.e boiling temperature greater than 111, an impossibility given the data in the question). This is as a result of the value 496 mmHg so close to 500 mm Hg.

Perhaps that is the reason the question was flagged.

7 0

2 years ago

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by No

liraira [26]

Answer:

Molality = 0.066 m

Molar mass = 608.36 g/mol

Explanation:

It seems the question is incomplete. However a web search us shows this data:

" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "

The freezing-point depression is expressed by:

ΔT=Kf * m

We put the data given by the problem and solve for m:

2.63 °C = 40°C·kg/mol * m

m = 0.06575 m

For the calculation of the molar mass: Molality is defined as moles of solute per kilogram of solvent:

0.06575 m = Moles reserpine / kg camphor

25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor

We calculate moles of reserpine:

0.06575 m = Moles reserpine / 0.025 kg camphor

Moles reserpine = 1.64x10⁻³ mol

Finally we use the mass of reserpine and the moles to calculate the molar mass:

1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol

Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.

8 0

2 years ago

How much heat is required to raise the temperature of 670g of water from 25.7"C to 66,0°C? The specific heat

inna [77]

Answer:

Explanation:

q= mc theta

where,

Q = heat gained

m = mass of the substance = 670g

c = heat capacity of water= 4.1 J/g°C

theta =Change in temperature=( 66-25.7)

Now put all the given values in the above formula, we get the amount of heat needed.

q= mctheta

q=670*4.1*(66-25.7)

=670*4.1*40.3

=110704.1

8 0

2 years ago