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2 years ago

What is the molar mass of aluminum (Al)

Chemistry

2 answers:

Nookie1986 [14]2 years ago

8 0

Answer:

2.68573957

Explanation:

Papessa [141]2 years ago

6 0

Answer:

26.981539 u

Explanation:

One mole of Al atoms has a mass in grams that is numerically equivalent to the atomic mass of aluminum. The periodic table shows that the atomic mass (rounded to two decimal points) of Al is 26.98, so 1 mol of Al atoms has a mass of 26.98 g.

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Suppose that a 10-mL sample of a solution is to be tested for I− ion by addition of 1 drop (0.2 mL) of 0.13 M Pb(NO3)2.

Liono4ka [1.6K]

The minimum number of grams of I that must be present is ; 0.01836 * 10⁻³ mol

Given data :

Volume of solution to be tested for I-ion = 10 mL

Volume of Pb(NO₃)₂ = 0.2 mL

molarity of Pb(NO₃)₂ = 0.13 M

<h3>Determine the number of I that must be present </h3>

First step : calculate conc of PB²⁺ ions in the solution

conc of PB²⁺ ions = ( molarity of Pb(NO₃)₂ * volume of Pb(NO₃)₂ ) / ( total volume )

= ( 0.13 * 0.2 ) / ( 10 + 0.2 )

= ( 0.026 ) / ( 10.2 ) = 0.002549 M

Next step : determine the molarity of I

using the dissociation reaction of PbI₂

PbI₂(s) ---> Pb²⁺ (aq) + 2I (aq)

also; Ksp = [ Pb²⁺ ] [ I ]² ---- ( 1 )

From the question the given value of Ksp = 8.49 * 10⁻⁹

Therefore equation ( 1 ) becomes

8.49 * 10⁻⁹ = ( 0.002549 ) * [ I ]²

[ I ] = √ ( 8.49 * 10⁻⁹ ) / ( 0.002549 )

= 0.0018 M

Final step : Determine the minimum number of grams of I

moles of I = molarity of I * total volume

= 0.0018 M * 10.2 mL

= 0.01836 * 10⁻³ mol

Hence we can conclude that The minimum number of grams of I that must be present is ; 0.01836 * 10⁻³ mol

Learn more about Pb(NO₃)₂ : brainly.com/question/25071409

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2 years ago

What is part of every cell

Charra [1.4K]

Cytoplasm is an organelle in every cell on earth.

3 0

3 years ago

Read 2 more answers

In the laboratory a student combines 47.8 mL of a 0.321 M aluminum nitrate solution with 21.8 mL of a 0.366 M aluminum iodide so

alex41 [277]

Answer: The final concentration of aluminum cation is 0.335 M.

Explanation:

Given: $V_{1}$ = 47.8 mL (1 mL = 0.001 L) = 0.0478 L

$M_{1}$ = 0.321 M, $V_{2}$ = 21.8 mL = 0.0218 L, $M_{2}$ = 0.366 M

As concentration of a substance is the moles of solute divided by volume of solution.

Hence, concentration of aluminum cation is calculated as follows.

$[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}$

Substitute the values into above formula as follows.

$[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M$

Thus, we can conclude that the final concentration of aluminum cation is 0.335 M.

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3 years ago

Moles /Number of Particles Con

Olin [163]

(2.03x10^22)/(6.02x10^23) = .033721 mol Li

I hope this helps, if not, i am sorry

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3 years ago

The main reason that cs2 has a higher boiling point than co2 is that cs2

OlgaM077 [116]

Explanation:

Inspite of having similar intermolecular forces, CS2 has a higher boiling point than CO2, since it has a greater molar mass. The potential energy of molecules reduces until a certain level as they get closer to each other. Although the polarity of both CO2 and CS2 are cancelled because of their linear structure.

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3 years ago