6th grade science i will give brainliest

Draw the mechanism arrows for both propagation steps for the radical addition of HBr to the alkene. When drawing single-headed r

Vitek1552 [10]

Answer:

See the attached file for the structure

Explanation:

See the attached file for the explanation

7 0

3 years ago

what is the final molarity of a dilute solution of NaOH that is prepared by pipetting 5.00 mL of 7.00 M NaOH into a 100mL volume

Angelina_Jolie [31]

Answer:

$M_{diluted}=0.35M$

Explanation:

Hello,

In this case, since in a dilution process the moles of the solute must remain unchanged, we use the volumes and molarities as shown below:

$M_{diluted}V_{diluted}=M_{concentrated}V_{concentrated}$

Clearly, the concentrated solution is 7.00M and the diluted solution is unknown, thus, the concentration of the diluted solution after the dilution to 100 mL is:

$M_{diluted}=\frac{M_{concentrated}V_{concentrated}}{V_{diluted}} =\frac{7.00M*5.00mL}{100mL}\\ \\M_{diluted}=0.35M$

Best regards.

4 0

4 years ago

If you have 74.4 liters Argon what is the mass in grams

Sloan [31]

1 liter is 1000 grams so 74.4*1000=74400 grams

7 0

3 years ago

Explain why some metals are extracted by heating their oxides with carbon, but some metals cannot be extracted in this way

Oxana [17]

Answer:

This happens because a metal is less reactive than carbon and it can be extracted from its oxides by heating with carbon. The carbon displaces metal from the compound and removes the oxygen from the oxide.

8 0

3 years ago

II. Ionic Equations

mario62 [17]

Answer:

Complete ionic: $\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Net ionic: $\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

Explanation:

Start by identifying species that exist as ions. In general, such species include:

Soluble salts.
Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ . These three salts will exist as ions:

Each $\rm AgNO_3\, (aq)$ formula unit will exist as one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion.
Each $\rm CaCl_2$ formula unit will exist as one $\rm Ca^{2+}$ ion and two $\rm Cl^{-}$ ions (note the subscript in the formula $\rm CaCl_2\!$ .)
Each $\rm Ca(NO_3)_2$ formula unit will exist as one $\rm Ca^{2+}$ and two $\rm {NO_3}^{-}$ ions.

On the other hand, $\rm AgCl$ is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each $\rm AgNO_3\, (aq)$ formula unit will exist as only one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion. However, because the coefficient of $\rm AgNO_3\, (aq)\!$ in the original equation is two, $\!\rm AgNO_3\, (aq)$ alone should correspond to two $\rm Ag^{+}\!$ ions and two $\rm {NO_3}^{-}\!$ ions.

Do not rewrite the salt $\rm AgCl$ because it is insoluble.

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of $\rm Ca^{2+}$ and two units of $\rm {NO_3}^{-}$ . Doing so will give:

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}$ .

Simplify the coefficients:

$\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

7 0

3 years ago