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BARSIC [14]
3 years ago
7

write the balanced reaction for the complete combustion of C11(subscript) H24 by diatomic O2. show all work

Chemistry
1 answer:
shutvik [7]3 years ago
7 0
Abs are not getting mad annoying and piper are you doing ok I don’t know
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If iodine-131 has a half-life of 8 days, how much of a 64.0 g sample of iodine-131 will remain after 32 day?
Mariulka [41]

Answer:

4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.

Explanation:

Half life (t1/2) = 8 days

Original mass (No) = 64 g

Elapsed time (t) = 32 days

Mass remaining (Nt) = ?

Using the half life equation we can obtain the mass remaining (Nt)

Nt = No (1/2) ^t/t1/2

Substituting the values, we have;

Nt = 64 * ( 1/2 ) ^32/8

Nt = 64 * (1/2) ^4

Nt = 64 * 0.0625

Nt = 4 g

So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.

8 0
3 years ago
Increasing temperature can:
umka21 [38]

The answer would be A. You cannot decrease mass because of increasing temp so it can't be C and adding neutrons or protons would be changing the atom and therefore the element, and that is not possible through just an increase of temperature, so it can't be B or D. Hope this was helpful! ;)

3 0
3 years ago
Read 2 more answers
A reaction produced 37.5 L of oxygen gas at 307 K and 1.25 atm. How many moles of oxygen were produced?
MariettaO [177]

On the basis of the given data:

Volume (V) is 37.5 L

Temperature (T) is 307 K

Pressure (P) is 1.25 atm

In order to calculate the number of moles, the formula to be used is PV = nRT, Here R is 0.0821 Latm/mol/K

n = PV / RT

n = 1.25 atm × 37.5 L / 0.0821 Latm/mol/K × 307 K

n = 1.85 mol

3 0
3 years ago
The rate constant for this second‑order reaction is 0.380 M − 1 ⋅ s − 1 0.380 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶product
Sati [7]

Answer: 8.38 seconds

Explanation:

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

a_0 = initial concentartion = 0.860 M

a= concentration left after time t = 0.230 M

k = rate constant =0.380M^{-1}s^{-1}

\frac{1}{0.860}=0.380\times t+\frac{1}{0.230 }

t=8.38s

Thus it will take 8.38 seconds for the concentration of  A to decrease from 0.860 M to 0.230 M .

5 0
3 years ago
Substance A and substance B combine to form substance C. Three illustrations of 3-D atomic structures of substances: substance A
ANEK [815]

Answer:

To answer this question we need the illustration

Explanation:

You have left the question incomplete since we do not have the illustration making this impossible to answer

8 0
3 years ago
Read 2 more answers
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