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gtnhenbr [62]
3 years ago
12

5. Rocks and sediment change their form as they move through the rock cycle. Which change

Chemistry
1 answer:
mina [271]3 years ago
8 0

Answer:

Rock to Magma

Explanation:

Lava to Rock is incorrect because its "Freezing" The Lava

Magma to Lava basically a name change. No type change.

Rock to Sediment is Erosion of the rock.

Rock to Magma Is the Rock being super-heated, melting it. Similar to Ice and water but more extreme.

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≡↔∛ππΔ≅×π⊄∴<br>khvjh j hjgvvvvvvvvvvvvvh jh
cupoosta [38]

its 40

I just took the test right now to get this answer

7 0
3 years ago
Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result
Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f)  12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only  the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) =  4.886

(d) mol HA reacted = 150 x 10⁻³ L  x  x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) =  5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH  = √(kb x (A⁻)  where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA,  200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒   Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0
3 years ago
That's how you can
kirza4 [7]
2)8/8-9_12-/78)022-/7/$
6 0
3 years ago
5.1169 mol of Ne is held at 0.9148 atm and 911 K. What is the volume of its container in liters?
sveticcg [70]

By applying the Boyle's equation and substituting our given data the volume of the container was found to be 418.14 Litres

<h3>Boyle's  Law</h3>

Given Data

  • number of moles of Ne = 5.1169 mol
  • Pressure = 0.9148 atm
  • Temperature = 911 K

We know that the relationship between pressure and temperature is given as

PV = nRT

R = 0.08206

Making the volume subject of formula we have

V= nRT/P

Substituting our given data to find the volume we have

V = 5.1169*0.08206*911/0.9148

V = 382.522353554/0.9148

V = 418.14 L

Learn more about Boyle's law here:

brainly.com/question/469270

5 0
2 years ago
What items could scientists use to date a layer of rock
Charra [1.4K]

By Using relative and radiometric dating methods hope this helps!!

4 0
3 years ago
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