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4 years ago

10

The atom diagnostically associated with organic compounds is

1 answer:

Liono4ka [1.6K]4 years ago

8 0

Carbon would be the atom that is diagnostically associated with organic compounds.

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Jennifer is applying a constant force to swing a can on the end of a string, as shown here. If the can is moving around the circ

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Answer:

Explanation:

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3 years ago

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34 grams of carbon react with an unlimited amount of h2o. the reaction is: c + h2o → co + h2 the equation is balanced. the start

VladimirAG [237]

A periodic table is tabular arrangement of the chemical elements, ordered by their atomic number,electronic configuration and recurring chemical properties. carbon is in group 14 and period two. The atomic mass of carbon from periodic table is 12.01 g/mole

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4 years ago

Plz answerrrr I dunnoooo

aleksley [76]

Midnight is your answer.

plz mark as brainliest. i really need it ;-;.

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4 years ago

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Need help on this question asap pleasee

algol [13]

Answer:

9.0 moles of CaO

Explanation:

We have the reaction equation as follows;

Fe2O3 + Ca3(PO4)2 -------> 2FePO4 + 3CaO

Now we know from the equation that;

1 mole of iron III oxide yields 3 moles of CaO

Therefore;

3 moles of iron III oxide yields 3 * 3/1

= 9.0 moles of CaO

6 0

3 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