Categorize the following materials: bronze alloy, mouse growing an ear on its back, porcelain dentures

Dr. Franck cuts a bar of pure gold into smaller and smaller pieces. Will this action change the element that makes up the bar? E

Anvisha [2.4K]

Answer:

i believe this is a chemical or physical question? well your answer to that is no the element does not change because the gold is still gold it is still physical because you have just cut it into piece it is still gold

Explanation:

lmk if it was helpful :/

7 0

1 year ago

A student wants to form 2‑hexanol using acidic hydration. He finds four alkenes in the inventory cabinet that could be possible

julia-pushkina [17]

Answer:

Look on the picture.

Explanation:

He could find only 2 isomers of n-hexane alkenes for this reaction. Other two could be marked from other direction.

8 0

2 years ago

Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.0550 M Ba(OH)2 solution. What is the concentration of the exces

Shalnov [3]

Answer: The concentration of excess $[OH^-]$ in solution is 0.017 M.

Explanation:

1. $Molarity=\frac{moles}{\text {Volume in L}}$

moles of $HCl=Molarity\times {\text {Volume in L}}=0.250\times 0.075=0.019moles$

1 mole of $HCl$ give = 1 mole of $H^+$

Thus 0.019 moles of $HCl$ give = 0.019 mole of $H^+$

2. moles of $Ba(OH)_2=Molarity\times {\text {Volume in L}}=0.0550\times 0.225=0.012moles$

According to stoichiometry:

1 mole of $Ba(OH)_2$ gives = 2 moles of $OH^-$

Thus 0.012 moles of $Ba(OH)_2$ give = $2 \times 0.012=0.024$ moles of $OH^-$

$H^++OH^-\rightarrow H_2O$

As 1 mole of $H^+$ neutralize 1 mole of $OH^-$

0.019 mole of $H^+$ will neutralize 0.019 mole of $OH^-$

Thus (0.024-0.019)= 0.005 moles of $OH^-$ will be left.

$[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M$

Thus molarity of $[OH^-]$ in solution is 0.017 M.

4 0

1 year ago

Read 2 more answers

The dissociation of calcium carbonate has an equilibrium constant of Kp = 1.16 at 1073 K . CaCO3(s) ⇄ CaO(s) + CO2(g) If you pla

tino4ka555 [31]

Answer:

<h3> $Pressure of CO_2 in the container=1.6 atm$ </h3>

Explanation:

First balance the chemical equation:

$CaCO_3(s)$ ⇄ $CaO(s) + CO_2(g)$

two components are solid so these two will not exert any kind of pressure in the container so at equilibrium only CO2 will apply pressure on the container

Therefore only partial pressure of CO2 will be taken for the calculation of equilibrium pressure constant i.e. Kp

$K_p=[CO_2]$

$[CO_2]=p$

$K_p=p$

$p=K_p = 1.16atm$

$Pressure of CO_2 in the container=1.6 atm$

8 0

1 year ago

An aqueous feed solution of 1000 kg/h containing 23.5 wt % acetone and 76.5 wt % water is being extracted in a countercurrent mu

JulijaS [17]

Question:

The question is incomplete. What is required to calculate was not added.The equilibrium data was not also added. Below is the additional questions and the answers.

1. Calculate the minimum solvent that can be used.

2.Using a solvent rate of 1.5 times the minimum, calculate the number of

theoretical stages.

Answer:

1. Minimum solvent = 411.047

2. N = 5

Explanation:

See the attached files for explanations.

5 0

1 year ago