All categories

3 years ago

What relationship do you see between an electronic configuration of an element to its most common ion formed?

2 answers:

6 0

An ion is an atom that has lost or gained an electron so the relation ship you would see would be a negative or a positive one depending on if it gained or lost an electron.

otez555 [7]3 years ago

5 0

An ion is an atom that has gained or lost and electron for example carbon 14 will go to nitrogen

You might be interested in

scientific description about the pizza dough doubles its size when it is placed in a warm place . how does this happen

Marina CMI [18]

Answer is: because carbon(IV) oxide is released.
For example if baking soda is used for the pizza dough, than on heat carbon(IV) oxide is product of chemical reaction:
2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O.
If we use yeast (single-celled microorganism), also carbon(IV) oxide produced, because yeast turns sugars into carbon(IV) oxide.

6 0

3 years ago

Type the chemical name of the compound that contains 1 copper atom, 1 oxygen

Bas_tet [7]

Answer:

Copper (I) hydroxide

Explanation:

Copper (I) hydroxide is a chemical compound with the chemical formula of CuOH. One copper atom (Cu), one oxygen atom (O), and one hydrogen atom (H.)

4 0

3 years ago

Classify each of the four compounds as a conjugated, isolated, or cumulated diene. Compound A: Two alkenes are joined by a sigma

Lina20 [59]

Explanation:

Conjugated diene is the one that contains alternate double bonds in its structure. That means both the double bonds are separated by a single bond.

Cumulated diene is the one that contains two double bonds on a single atom. This means it has two double bonds continuously.

Isolated double-bonded compound has a single bond isolated by two to three single bonds.

Compound A: Two alkenes are joined by a sigma bond.

For example:

$-CH_2=CH-CH=CH2-$

It is a conjugated diene.

Compound B: Two alkenes are joined by a C H 2 group.

It is a cumulative diene.

Compound C: Two alkenes are joined by C H 2 C H 2.

Then it is an isolated alkene.

Compound D: A cyclohexene has a double bond between carbons 1 and 2. Carbon 3 is an sp 2 carbon that is bonded to another s p 2 carbon with an alkyl substituent.

Hence, compound D is a conjugated diene.

8 0

3 years ago

A sample of nitrogen is initially at a pressure of 1.7 kPa, a temperature of -10 C and a volume of 7.5 m3. Then the volume is de

zhannawk [14.2K]

Answer:

$\boxed{\text{2.6 kPa}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}$

Data:

p₁ = 1.7 kPa; V₁ = 7.5 m³; T₁ = -10 °C

p₂ = ?; V₂ = 3.8 m³; T₂ = 200 K

Calculations:

(a) Convert temperature to kelvins

T₁ = (-10 + 273.15) K = 263.15 K

(b) Calculate the pressure

$\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}$

7 0

3 years ago

Draw the product that valine forms when it reacts with excess CH3CH2OH and HCl followed by a wash with aqueous base.

-BARSIC- [3]

Answer:

Product: ethyl L-valinate

Explanation:

If we want to understand what it is the molecule produced we have to analyze the reagents. We have valine an amino acid, in this kind of compounds we have an amine group ( $NH_2$ ) and a carboxylic acid group ( $COOH$ ). Additionally, we have an alcohol ( $CH_3CH_2OH$ ) in the presence of HCl (a strong acid) in the first step, and a base ( $OH^-$ ).

When we have an acid and an alcohol in a vessel we will have an esterification reaction. In other words, an ester is produced. As the first step, the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the second step, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In step 3, a proton is transferred to produce a better leaving group ( $H_2O$ ). In step 4, a water molecule leaves the main structure to produce again the double bond C=O. Finally, a base ( $OH^-$ ) removes the hydrogen from the C=O bond to produce ethyl L-valinate

See figure 1

I hope it helps!

7 0

3 years ago