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stellarik [79]
2 years ago
12

A particular radioactive isotope has a half-life of (2.50+a) hours. if you have (24.5+b) g of the isotope at 10:00 am, how much

will you have at 7:30pm? give your answer in grams (g)and with 3 significant figures.
Chemistry
1 answer:
gavmur [86]2 years ago
3 0
Its a not b and i dont know the reast

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A) How many moles of CO2 and H2O are formed from 3.85 mole of propane C3H8 (This calculation needs to be done twice-once fro CO2
4vir4ik [10]

Answer:

a) 11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

15.4 moles of water are formed from 3.85 mole of propane.

b)0.5176 moles of water are formed from 0.647 mole of oxygen gas.

0.1294 moles of propane are consumed.

Explanation:

C_3H_7+5O_2\rightarrow 3CO_2+4H_2O

a) Moles of propane = 3.85 moles

According to reaction, 1 mole of propane gives 3 moles of carbon dioxide gas.

Then 3.85 moles of propane will give:

\frac{3}{1}\times 3.85 mol=11.55 mol of carbon dioxide gas.

11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

According to reaction, 1 mole of propane gives 4 moles of water gas.

Then 3.85 moles of propane will give:

\frac{4}{1}\times 3.85 mol=15.4 mol of water .

15.4 moles of water are formed from 3.85 mole of propane.

b) Moles of oxygen gas = 0.647 moles

According to reaction, 5 mole of oxygen gas gives 4 moles of water.

Then 0.647 moles of oxygen will give:

\frac{4}{5}\times 0.647 mol=0.5176 mol of water.

0.5176 moles of water are formed from 0.647 mole of oxygen gas.

According to reaction, 5 mole of oxygen gas reacts with 1 mole of propane.

Then 0.647 moles of oxygen will give:

\frac{1}{5}\times 0.647 mol=0.1294 mol of propane.

0.1294 moles of propane are consumed.

5 0
2 years ago
A 520-gram sample of seawater contains 0.317 moles of NaCl. What is the percent composition of NaCl in the water?
Volgvan

Answer:

c

Explanation:

8 0
2 years ago
During this reaction, p4 + 5o2 → p4o10, 0.800 moles of product was made in 15.0 seconds. what is the rate of reaction? 56.8 g/mi
Eduardwww [97]
The solution for this problem would be:
The mass of P4O10 is computed by: 0.800 mol x 284 g/mol = 227g t = 15.0 s ( 1 min / 60 s) = 0.25 min 
So solving for the rate will be mass over t = m/t = 227/0.25 = 908 g/min would be the answer for this problem.
3 0
2 years ago
A sample of n2 gas occupies 2.40 l at 20°c. if the gas is in a container that can contract or expand at constant pressure, at wh
Bas_tet [7]
From Charle's law the volume of a fixed mass of a gas is directly proportional to the absolute temperature at constant pressure.
Therefore';
V1/T1=V2/T2
    Where; V1 = 2.40 l, T1 = 273 +20= 293 K, V2 = 4.80, and T2= ?
 2.4/293= 4.8/T2
 T2= (4.8×293)/2.4
       = 586 K or  313° C
     

4 0
2 years ago
Read 2 more answers
15. Using the information below, calculate ΔHf° for PbO(s)
Svet_ta [14]
ΔH(reaction) = ΔH(formation of products) - ΔH(formation of reactants) 
ΔH(reaction) = ( 1*ΔH(Pb(s)) + 1*ΔH(CO2(g)) ) - ( 1*ΔH(PbO(s)) + 1*ΔH(CO(g)) ) 
ΔH(reaction) = ( 0 + -393.5 ) - ( ΔH(PbO(s)) + -110.5 ) 
ΔH(reaction) = -283 - ΔH(PbO(s)) 
-131.4 = -283 -ΔH(PbO(s)) 
ΔH(PbO(s)) = -151.6 kJ

So, the best answer is A.
5 0
1 year ago
Read 2 more answers
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