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Ugo [173]
3 years ago
10

Describe a physical change and a chemical change that an iron nail could undergo.

Chemistry
1 answer:
galina1969 [7]3 years ago
8 0
Physical change

iron nail have high melting point, thus, when it is heated, it will not melt very easily.

chemical change

iron nail can rust due to the exposure to the air(oxygen). iron in the nail and oxygen in the air will react together to form iron oxide(rust).
furthermore, the reaction is irreversible.
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Will calcium metal react with chromium (III) chloride? Explain your answer
iren [92.7K]

Calcium metal with chromium(III) chloride would react. Calcium metal is more reactive than chromium metal; therefore, putting calcium metal in chromium chloride would create a reaction, where as putting chromium metal in calcium chloride would not form any reaction.

5 0
3 years ago
Aluminium +hydrochloride acid ________+_________​
REY [17]

Answer:

Aluminium Chloride + Hydrogen

Please vote for Brainliest and I hope this helps!

4 0
2 years ago
Cu is in FCC structure. We know that even at close to melting temperature, vacancies are still sparsely distributed, i.e., it is
sveticcg [70]

Answer:

6

Explanation:

FCC is face centered cubic lattice. In FCC structure, there are eight atoms at the eight corner of the cubic unit cell and one atom centered in each of the faces. FCC unit cells consist of four atoms, (8/8) at the corners and (6/2) in the faces.

Given that, Cu has FCC structure and it contains a vacancy at origin (0, 0, 0). And there is no other vacancy directly adjacent to the vacancy at the origin. So, all the adjacent positions contain Cu atoms. Hence, the total number of adjacent atoms of the vacancy at origin can jump into this vacancy.

the above FCC unit cell clearly indicates that there are six adjacent atoms adjacent to the vacancy at origin

So, the total number of adjacent atoms of the vacancy at origin can jump into this vacancy is 6.

5 0
3 years ago
A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain
sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

-w=\frac{dm}{dt}

w is the mass flow

m is the mass of salt

-v*C=\frac{dm}{dt}

v is the volume flow

C is the concentration

C=\frac{m}{V+(6-3)*L/min*t}

-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}

-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}

-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)

-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)

m=90kg*[2000L/(2000L+3*L/min*t)]

a) Initially: t=0

m=90kg*[2000L/(2000L+3*L/min*0)]=90kg

b) t=210 min (3.5 hr)

m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg

c) If time trends to infinity the  division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0
3 years ago
• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL vo
garik1379 [7]

Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c)  0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

1 mole         3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b)  if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

10 ml            17.50 ml

(x) M              0.200 M

Molarity = \frac{0.2*17.5}{1000}

= 0.0035 mole

c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

= 0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

Molar Concentration =  \frac{mole \ \ of \ soulte }{ Volume \ of \ solution }

Molar Concentration = \frac{0.00166 \ mole \ of \  H_3PO_4 }{10}*1000

Molar Concentration = 0.166 M

∴  the molar concentration of phosphoric acid in the original stock solution = 0.166 M

6 0
3 years ago
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