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Tasya [4]
3 years ago
7

Which of the following is NOT a unit energy? (select all that applies)

Chemistry
2 answers:
enot [183]3 years ago
6 0

Joules, Kelvin and Calories

never [62]3 years ago
4 0

Answer:

Kelvin, Celsius, Fahrenheit

Explanation:

See my other answer for an explanation.

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Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. Classify each of the following me
Anika [276]

Answer:

a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

Explanation:

Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. The reactivity series of metals arranges metals based on decreasing order of reactivity. The more reactive metals are found higher up in the series while the least reactive metals are found at the lower ends of the series. Thus, metals above iron in the reactivity series can serve as sacrificial anodes by protecting against corrosion, while those lower than iron cannot.

Based on the reactivity series, the following metals can be classified as either a sacrificial anode for iron or not:

a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

6 0
3 years ago
A solution is prepared from 26.7 g of an unknown compound and 116.2 g acetone, C3H6O, at 313 K. At this temperature the vapor pr
Cloud [144]

Answer:

The molecular weight of the unknown compound is 267.7 g/mol

Explanation:

Lowering vapor pressure → Colligative property where the vapor pressure of solution is lower than vapor pressure of pure solvent

ΔP = P° . Xm

0.526 atm - 0.501 atm = 0.025atm

0.025 atm = 0.526 atm . Xm

Xm = 0.025 atm / 0.526 atm → 0.0475 (mole fraction)

Mole fraction = Moles of solute / Total moles (solute + solvent)

0.0475 = Moles of solute / Moles of solute + Moles of solvent

We determine the moles of solvent → 116.2 g . 1mol / 58 g = 2 moles

0.0475 = Moles of solute / Moles of solute + 2

0.0475 moles of solute + 0.095 = Moles of solute

0.095 = Moles of solute - 0.0475 moles of solute

0.095 / 0.9525 = Moles of solute → 0.0997 moles

Molas mass = g/mol → 26.7 g / 0.0997 mol = 267.7 g/mol

4 0
3 years ago
Please answer this question for meeeeeeeeeee pleeeeeeeeeease!
Anettt [7]

Answer:battery in a flashlight so A

Explanation:

6 0
3 years ago
A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in
LenKa [72]

Answer:

P=3.7atm

Explanation:

Hello,

In this case, it is possible to determine the pressures of both helium and neon as shown below:

n_{He}=\frac{P_{He}V_{He}}{RT}=\frac{5.6atm*3.0L}{0.082\frac{atm*L}{mol*K}*298.15K} =0.688molHe\\\\n_{Ne}=\frac{P_{Ne}V_{Ne}}{RT}=\frac{3.6atm*4.5L}{0.082\frac{atm*L}{mol*K}*298.15K}=0.663molNe

Now, one considers the total moles (addition between both neon's and helium's moles) and the total volume to compute the final pressure as shown below:

P=\frac{n_TRT}{V_T} =\frac{(0.688+0.663)mol*0.082\frac{atm*L}{mol*K}*298.15K}{9.0L}=3.7atm

Best regards.

8 0
3 years ago
What kind of bond is formed when lithium and fluorine combine to form lithium fluoride?
cupoosta [38]

Answer:

ionic bond

Explanation:

4 0
3 years ago
Read 2 more answers
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