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horrorfan [7]
3 years ago
15

Is it possible for a balloon with an initial internal pressure equal to 250.0 kPa

Chemistry
1 answer:
MrRa [10]3 years ago
6 0
No, don't try, it will explode  close to 187 kPa
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Which of the following changes occurs in the nucleus when a position particle is given off
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Explanation:

When the position particle is given off, the nucleus' atom's turn to a cluster and the wavelength becomes shorter. Hope this helps!

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3 years ago
What is the different between a mixture and a compound?
krek1111 [17]
A compound is 2 or more substances chemically combined, mixtures arent

4 0
3 years ago
Convert 7.8 moles of carbon tetrafluoride into grams.
Vesnalui [34]

Answer:

686.43363984 is the answer when 7.8 moles is converted.

6 0
3 years ago
The isotope Np-238 has a half life of 2.0 days if 96 grams of it were present on Monday how much will remain six days later
Kipish [7]

Answer:

12.02 g

Explanation:

From the question given above, the following data were obtained:

Half life (t½) = 2 days

Original amount (N₀) = 96 g

Time (t) = 6 days

Amount remaining (N) =..?

Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:

Half life (t½) = 2 days

Decay constant (K) =?

K = 0.693 / t½

K = 0.693 / 2

K = 0.3465 /day

Therefore, the rate of disintegration of the isotope is 0.3465 /day.

Finally, we shall determine the amount of the isotope remaining after 6 days as follow:

Original amount (N₀) = 96 g

Time (t) = 6 days

Decay constant (K) = 0.3465 /day.

Amount remaining (N) =.?

Log (N₀/N) = kt / 2.303

Log (96/N) = (0.3465 × 6) / 2.303

Log (96/N) = 2.079/2.303

Log (96/N) = 0.9027

Take the anti log of 0.9027

96/N = anti log (0.9027)

96/N = 7.99

Cross multiply

96 = N × 7.99

Divide both side by 7.99

N = 96 /7.99

N = 12.02 g

Therefore, the amount of the isotope remaining after 6 days is 12.02 g

3 0
3 years ago
The rate constant for a certain reaction is k = 5.40×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the
IceJOKER [234]

Answer : The concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

Explanation :

The expression for first order reaction is:

[C_t]=[C_o]e^{-kt}

where,

[C_t] = concentration at time 't'  (final) = ?

[C_o] = concentration at time '0' (initial) = 0.100 M

k = rate constant = 5.40\times 10^{-3}s^{-1}

t = time = 17.0 min = 1020 s (1 min = 60 s)

Now put all the given values in the above expression, we get:

[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}

[C_t]=4.05\times 10^{-4}M

Thus, the concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

4 0
3 years ago
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