As the [ht] in a solution decreases, what happens to the [OH???

Wuy Piolcululugu<br> Give the names of two organs in the chest.<br> 1. ...............<br> 2 m

Arada [10]

Answer:

The organs present inside the chest are :

1. The lungs

2. The heart

Explanation:

The chest cavity is also called as the thoracic cavity. It is the second largest hollow space of the body.In the bottom , it is enclosed by the diaphragm.

This cavity actually contain three space each round with mesothelium , pleural cavity and precardial cavity.

This contain the lungs , the tracheobronchial tree , the heart , the blood vessels which transport the blood between the heart and the lungs.

It also contain the esophagus .

Esophagus is the path through which the food passes from the mouth to the stomach.

8 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

How many controls or controlled variables can there be in an experiment

notsponge [240]

i would say 35 i think

3 0

3 years ago

One mole of acetyl chloride was mixed with one mole of dimethylamine. After the reaction is complete, what species can be found

fredd [130]

Answer:

N,N-dimethylacetamide is formed.

Explanation:

It is an example of a nucleophilic addition-elimination reaction. Here dimethylamine acts as a nucleophile.
In the first step, dimethyl amine gives nucleophilic addition reaction at carbonyl center of acetyl chloride.
In the second step, removal of Cl atoms occurs.
In the third step, deprotonation takes place from amino group to produce N,N-dimethylacetamide.
Full reaction mechanism has been shown below.

7 0

3 years ago

Calculate the percent by mass of a solution of Ca(NO3)2 that has 22.63 g dissolved in 896.92 g of water.

never [62]

Hey there!

To calculate the percent by mass of the Ca(NO₃)₂ we need to find the total mass first by adding.

896.92 + 22.63 = 919.55

In total, the solution is 919.55 grams.

To find the percent of Ca(NO₃)₂ in the solution, divide the mass of Ca(NO₃)₂ by the total mass and multiply by 100.

22.63 ÷ 919.55 = 0.0246

0.0246 x 100 = 2.46

Ca(NO₃)₂ makes up 2.46% of the solution.

Hope this helps!

6 0

3 years ago