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vladimir2022 [97]
3 years ago
5

1.

Chemistry
2 answers:
astraxan [27]3 years ago
7 0
1) electric energy

2) toaster

3) music heard from a radio
Fudgin [204]3 years ago
7 0

Answer:

1.Electrical energy

2. Toaster

3. Music heard from a radio

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How many of the zeros in the measurement 0.050060 are significant
Feliz [49]
I think 3 of them are its been 1 half years since ive done this i dont take chemistry anymore
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3 years ago
Read 2 more answers
An unknown aqueous metal analysis yielded a detector response of 0.255. When 1.00 mL of a solution containing 100.0 ppm of the m
AVprozaik [17]

Answer:

1.022ppm is the unknown concentration of the metal

Explanation:

Based on Lambert-Beer law, the increasing in signal of a detector is directly proportional to its concentration.

The unknown concentration (X) produces a signal of 0.255

99mL * X + 1mL * 100ppm / 100mL produces a signal of 0.502

0.99X + 1ppm produce 0.502, thus, X is:

0.255 * (0.99X + 1 / 0.502) =

X = 0.503X + 0.508

0.497X = 0.508

X =

1.022ppm is the unknown concentration of the metal

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3 years ago
How do you draw a CS2 lewis structure
kicyunya [14]

Answer: LOL. I think thats how-

6 0
2 years ago
An engine operates on a Carnot cycle that uses 1mole of an ideal gas as the
sattari [20]

Answer:

Step 1;

q = w = -0.52571 kJ, ΔS = 0.876 J/K

Step 2

q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ

Explanation:

The given parameters are;

P_i = 100 N·m

T_i = 327 K

P_f = 90 N·m

Step 1

For isothermal expansion, we have;

ΔU = ΔH = 0

w = n·R·T·ln(P_f/P_i) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71

w ≈<em> -0.52571</em> kJ

At state 1, q = w = -0.52571 kJ

ΔS = -n·R·ln(P_f/P_i) = -1 × 8.314 × ln(90/100) ≈ 0.876

ΔS ≈ 0.876 J/K

Step 2

q = 0 for adiabatic process

ΔU = 25×(27 - 327) = -7,500

w = ΔU = <em>-7.5 kJ</em>

ΔH = ΔU + n·R·ΔT

ΔH = -7,500 + 8.3142 × 300 = -5,005.74

ΔH = ΔU = <em>-5.00574 kJ</em>

6 0
2 years ago
The temperature of a 500 ml sample of gas increases from 150 k to 300 k. what is the final volume of the sample of gas, if the p
seropon [69]
Upon a constant pressure (P), volume (V) of a gas will vary in direct proportion to changes in temperature (T). So V1/T1 = V2/T2
V2 = V1T2/T1 = (500)(300)/150
V2 = 150000/150 = 1000 mL
5 0
3 years ago
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