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3 years ago

The atomic mass of Aluminum is 26.98154 g/mol. Is it possible to have 5.0*10-25g of Al

1 answer:

7 0

I believe the correct answer is no, it is impossible to have 5.0*10^-25g of Al. One atom of Al weighs 4.48*10^-23 g/atom Al which means it is impossible to have any weight of Al less than that. Hope this answers the question. Have a nice day.

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X(g)+4Y(g)→2Z(g), ΔH∘=−75.0 kJ Before the reaction, the volume of the gaseous mixture was 5.00 L. After the reaction, the volume

vlabodo [156]

Isobaric transition, first law: H=ΔU+w for a gas expansion: w=Pext∗ΔV to convert to joules, you need the gas constants. R = 0.08206 L atm/mol*K, R=8.314 J/mol*K w=Pext∗ΔV∗8.314 J/mol∗K0.08206 L atm/mol∗K ΔU=ΔH−[Pext∗ΔV∗8.314 J/mol∗K0.08206 L atm/mol∗K] ΔU=−75000 J−[(43.0atm)∗(2−5)L∗8.314 J0.08206 L atm] Then you need to convert to kJ. by the way U=E, internal energy.

7 0

2 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

2 years ago

List two detection (i.e. visualisation) techniques commonly used to visualise compounds in TLC.

olya-2409 [2.1K]

Answer:

The most common non-destructive visualization method for TLC plates is ultraviolet (UV) light. A UV lamp can be used to shine either short-waved (254nm) or long-waved (365nm) ultraviolet light on a TLC plate with the touch of a button

Explanation:

hope this helps

7 0

3 years ago

Part A of the lab involved adding 4 mL increments of distilled water to 5.00 mL of antimony trichloride solution. The antimony t

Mrrafil [7]

Answer:

0.0238M SbCl3, 1.07M H+, 1.14M Cl-

Explanation:

The total volume of the solution is:

4mL + 5.00mL + 12.0mL = 21mL

As the volume of the SbCl3 is 5.00mL, the dilution factor is:

21mL / 5.00mL = 4.2 times

The concentration of SbCl3 is:

0.10M SbCl3 / 4.2 times = 0.0238M SbCl3

The concentration of H+ = [HCl]:

4.5M / 4.2 times = 1.07M H+

The initial concentration of Cl- is:

3 times SbCl3 + HCl = 0.10M*3 + 4.5M =

3 times SbCl3 because 1 mole of SbCl3 contains 3 moles of Cl-

4.8M Cl- / 4.2 times = 1.14M Cl-

3 0

2 years ago

the gravitational force between two objects depends on the distance between the objects and each object?

Nina [5.8K]

Answer:

Gravitational force between two objects depends on: a. The weight of the objects c. The mass of each object b. The distance between the objects d. Both b and c Please select the best answer from the choices provided A B C D

✓d

7 0

2 years ago