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3 years ago

How many aluminum atoms are in 2.26 g of aluminum?

Chemistry

1 answer:

ki77a [65]3 years ago

4 0

Answer:

Aluminum atoms are in 3.78 g of aluminum

02214076x10^23)8.4367659370640769254888x10^22 atoms, therefore there are 8.4367659370640769254888x10^22 atoms of Aluminum(Al) present in 3.78 grams of Aluminum(Al).

Explanation:

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How many helium nuclei fuse together when making carbon?

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3 years ago

Copper, a metal known since ancient times, exists in two stable isotopic forms, 6329Cu (69.09%) and 6529Cu (30.91%). Their atomi

Svetradugi [14.3K]

Answer:

Wt. Avg. Atomic Weight => 63.35457 amu

Explanation:

Given Isotopic %Abundance fractional Wt Avg

At. Mass (amu) abundance contribution

Cu-63 62.93 69.09 0.6909 43.4783

Cu-65 64.9278 20.0668 0.200668 20.0668

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= 43.4783 amu + 20.0668 amu = 63.35457 amu

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3 years ago

How many moles are in 2.04 x 10^8 atoms of calcium?

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Answer:

2.0 moles

Explanation:

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2 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago