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Shalnov [3]
3 years ago
15

What type of change produced the rust shown above? A. chemical, because a new substance with new properties was formed B. physic

al, because a new substance with new properties was formed C. physical, because the chemical makeup of the metal did not change D. chemical, because the chemical makeup of the metal did not change
Chemistry
1 answer:
Maru [420]3 years ago
6 0
Rust or oxidation is a chemical change because it creates a new substance.

Knowing that the correct answer should be (A.)

Hope this helps!
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A buffer solution contains 0.479 M NaHCO3 and 0.342 M Na2CO3. Determine the pH change when 0.091 mol HNO3 is added to 1.00 L of
pshichka [43]

Answer:

ΔpH = 0.20

Explanation:

The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2

HCO₃⁻ ⇄ H⁺ + CO₃²⁻

There are 0.479moles of NaHCO₃ and 0.342moles of Na₂CO₃.

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.342mol / 0.479mol

<em>pH = 10.05</em>

NaOH reacts with HCO₃⁻ producing CO₃²⁻, thus:

NaOH + HCO₃⁻ → CO₃²⁻ + H₂O + Na⁺

0.091 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:

CO₃²⁻: 0.342mol + 0.091mol: 0.433mol

HCO₃⁻: 0.479mol - 0.091 mol: 0.388mol

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.433mol / 0.388mol

pH = 10.25

That means change of pH, ΔpH is:

ΔpH = 10.25 - 10.05 = <em>0.20</em>

<em />

I hope it helps!

3 0
3 years ago
An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

7 0
3 years ago
If there is a loss of 5.63 x 10-7kg of mass in a nuclear reaction, how many Joules of energy would be released? Recall that c =
Fudgin [204]
Thank you for posting your question here at brainly. 

E = mc^2 
<span>where E is the energy in joules, </span>
<span>m is the mass in kilograms, </span>
<span>and c is the speed of light. </span>

<span>E = mc^2 </span>
<span>E = (5.63 x 10^-7 kg)(3 x 10^8 m/s)^2 </span>
<span>E = 5.07 x 10^10 J </span>


5 0
4 years ago
HARD QUESTION (really need help)
torisob [31]

Answer:

answer is A

hope this helps

5 0
3 years ago
In the following net ionic equation, identify each reactant as either a Bronsted-Lowry acid or a Bronsted-Lowry base.
gizmo_the_mogwai [7]

Answer:

Bronsted lowry base = Proton acceptor = H2O

Bronsted lowry acid = Proton donor = HF-

Explanation:

The equation is given as;

HF-(aq) + H2O(l) --> F-(aq) + H3O(aq)

A bronsted lowry base is any specie that can accept hydrogen ion (proton) from another molecule.

Basically a bronsted lowry base is  a proton acceptor while a bronsted lowry acid is a proton donor.

In the reaction above, upon comparing both the reactants and products;

Bronsted lowry base = Proton acceptor = H2O

Bronsted lowry acid = Proton donor = HF-

3 0
3 years ago
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