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What would indicate that a physical change takes place when copper is drawn into wire?

Lina20 [59]

Any change in which the composition of material does not change that is it retains its identity but changes its state or form is known as a physical change.

The properties of metal to draw them into wires is known as ductility. When a copper is drawn into wire the only change that occurs is change in its shape and size no change will take place into its composition that is the wires are still possessing the properties of copper metal. Thus, a physical change takes place when copper is drawn into wire.

8 0

3 years ago

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ICMP scanning involves in checking for the live systems, which can be done by sending the following ping scan request to a host.

abruzzese [7]

Answer:

ICMP Echo Request

Explanation:

ICMP Echo Request is a form of probe or message sent by a user to a destination system.

5 0

3 years ago

What happens when an electron in its lowest energy level or ground state when it is by absorbs energy?

tester [92]

Explanation:

atom changes from a ground state to an excited state by taking on energy from its surroundings in a process called absorption. The electron absorbs the energy and jumps to a higher energy level. In the reverse process, emission, the electron returns to the ground state by releasing the extra energy it absorbed

8 0

3 years ago

Calculate the molarity of a solution obtained dissolving 10.0 g of cobalt(Ⅱ) bromide tetrahydrate in enough water to make 450 mL

Vladimir [108]

Answer:

The molarity of the CoBr2•4H2O solution is 7.64 × 10-2 M

Explanation:

Cobalt (II) bromide tetrahydrate

• Cobalt - A transition metal with Roman numeral (II) → charge: +2 → Co2+

• Bromide - anion from group 7A → -1 charge → symbol: Br-

• Tetrahydrate- tetra- means 4 and hydrate is H2O

The chemical formula of the compound is: CoBr2•4H2O

We then need to determine the number of moles of CoBr2•4H2O since this is the only information missing for us to find molarity. Notice that the volume of the solution is already given.

We’re given the mass of CoBr2•4H2O. We can use the molar mass of CoBr2•4H2O4 to find the moles.

•The molar mass of CoBr2•4H2O is:

CoBr2•4H2O

1 Co x 58.93 g/mol Co = 58.93 g/mol

2 Br x 79.90 g/mol Br = 159.80 g/mol

8 H x 1.008 g/mol H = 8.064 g/mol

4 O x 16.00 g/mol O = 64.00 g/mol

________________________________________

Sum = 290.79 g/ mo

The moles of CoBr2•4H2O is:

= 10.0 g CoBr2•4H2O x $\frac{ 1 mol CoBr_2 . 4H_2O}{290.79 g CoBr_2 . 4H_2O}$

= 0.0344 mol CoBr2•4H

We know that the volume of the solution is 450 mL.

We can now calculate for molarity:

Convert mL to L → 1 mL = 10-3 L

Formula:

Molarity (M)= Mole of solute / Liters of solution

= 0.0344 mol CoBr2•4H / 450 mL x 1 ml / 10^ -3 L

= 0.0764

= 7.64 × 10-2 mol/L

8 0

3 years ago

What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i

UNO [17]

Answer:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

$Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-$

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

$Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)$

The net equation is then:

$Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)$

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

$Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-$

Actual cathode half-equation:

$Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)$

Actual net reaction:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

6 0

3 years ago