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solmaris [256]
2 years ago
12

EVIDENCE NOTEBOOK

Chemistry
1 answer:
BARSIC [14]2 years ago
4 0

Answer:

Your body temperature would decrease

Explanation:

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What was the result of heating the mixture? All BUT ONE choice is correct.
snow_tiger [21]

Answer:

w gang alright

Explanation:

ay its b alright

4 0
3 years ago
12. A helium-filled weather balloon has a volume of 2.4 x 102 L at 99 kPa pressure and a
Luba_88 [7]

Answer:

The answer to your question is 8.74 g of He

Explanation:

Data

V = 2.4 x 10² L

P = 99 kPa

T = 0°C

mass = ?

Process

1.- Convert kPa to atm

P = 99 kPa = 99000 Pa

                   1 atm --------------- 101325 Pa

                    x       ---------------   99000 Pa

                   x = (99000 x 1) / 101325

                   x = 0.977 atm

2.- Convert temperature to °K

°K = 273 + 0

°K = 273

3.- Substitution

      PV = nRT

- Solve for n

      n = PV / RT

      n = (0.977)(2.4 x 10²) / (0.082)(273)

      n = 24.48 / 22.386

      n = 1.093 moles

4.- Calculate the grams of He

          8 g -------------------- 1 mol

           x    -------------------- 1.093 moles

           x = (1.093 x 8) / 1

           x = 8.74 g                      

6 0
3 years ago
Read 2 more answers
Why doesn’t vegetable oil stay dissolved in vinegar to make a salad dressing?
allsm [11]
In order for things to dissolve they must dissolve in a solvent similar to them (polar and polar or nonpolar and nonpolar) Oil is nonpolar and vinegar is polar therefore oil wont dissolve in vinegar
4 0
3 years ago
Read 2 more answers
Zzzzzzzzzzzzzzzzzzzzzzz
Korolek [52]

Answer:

Wake up wake up wake up wake up

Explanation:

wake up wake up wake up wake up

6 0
2 years ago
Iodine-131 decays with a half-life of 8.02
dlinn [17]
Radioactive material undergoes 1st order decay kinetics.

For 1st order decay, half life = 0.693/k

where k = rate constant

k = 0.693/half life = 0.693/8.02 = 0.0864 day-1

Now, for 1st order reaction,
k = \frac{2.303}{t} X log \frac{initial.conc}{final.conc}

Given: t = 6.01d, initial conc. = 5mg

∴0.0864 = \frac{2.303}{6.01} X log \frac{5}{final.conc}
∴ final conc. = 2.975 mg
3 0
3 years ago
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