How many grams of iron are there in a 65.2g sample of rust, Fe2O3?

Select the correct answer.

Fittoniya [83]

Answer:

subscribe to me on you-tube for brainliest custom link since u cant do a you-tube link :/

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Explanation:

7 0

3 years ago

If the pOH of a cesium hydroxide solution is known to be 4.00. what is the (OH)? *Please round your answer to the appropriate nu

Snowcat [4.5K]

Answer:

i going to be aniston i would say take a gess

4 0

3 years ago

If the half life of iridium-182 is 15 years, how much of a 3 gram sample is left after 2 half-lives?

padilas [110]

Answer:

D. 0.75 grams

Explanation:

The data given on the iridium 182 are;

The half life of the iridium 182, $t_{(1/2)}$ = 15 years

The mass of the sample of iridium, N₀ = 3 grams

The amount left, N(t) after two half lives is given as follows;

$N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}} }$

For two half lives, t = 2 × $t_{(1/2)}$

∴ t = 2 × 15 = 30

$\dfrac{t}{t_{(1/2)}} = \dfrac{30}{15} = 2$

$\therefore N(t) = 3 \times\left (\dfrac{1}{2} \right )^2 = 0.75$

∴ The amount left, N(t) = 0.75 grams

4 0

3 years ago

In a thin layer chromatography experiment, a plate of length 9.3 cm was used and a horizontal line was made at 1.45 cm above the

dezoksy [38]

Answer: The $R_f$ value is 0.664

Explanation:

Distance travelled by solvent front = (7.7-1.45)cm = 6.25 cm

Distance travelled by unknown = (5.6-1.45) cm = 4.15 cm

The retention factor or the $R_f$ value is defined as the ratio of distance traveled by the unknown to the distance traveled by the solvent front.

$R_f=\frac{\text {distance travelled by unknown}}{\text {distance travelled by solvent}}$

$R_f=\frac{4.15}{6.25}=0.664$

Thus the $R_f$ value is 0.664

6 0

3 years ago

Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, a

lesya692 [45]

Answer:

The process which releases most heat is E)

Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

Processes with temperatures < 0ºC : by using specific heat of ice (Sh ice) multiplied by the change in temperature (ΔT= Final Temperature - Initial Temperature)⇒ Sh ice x ΔT

Processes of ice melting (at 0ºC): by using heat of fusion of ice (ΔH fus) multiplied by a conversor factor (1 mol H20= 18 g)⇒ ΔHfus x 1mol/18g

Processes between 0ºC and 100ºC: by using specific heat of liquid water (Sh liq) multiplied by change in temperature ⇒ Sh liq x ΔT

Processes of water evaporation (at 100ºC): by using heat of vaporization (ΔH vap) multiplied by the conversor factor ⇒ ΔH vap x 1mol/18 g

Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

Heat= 2.01 J/ºC x (110ºC - 140ºC)

Heat= -60.3 J (heat is released)

E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

Heat= (2.01 J/ºC x (100ºC-106ºC)) - (40.7 x 10³ J/mol x 1mol/18 g) + (4.18 J/ºC x (88ºC -100ºC)

Heat= -2323.32 J (heat is released) <u>THIS IS THE RIGHT ANSWER</u> (the more negative= the more released)

7 0

3 years ago