Answer:
Time taken = 174 days
Explanation:
Half life is the time take for a subsrtance taken to decay to half of it's origial or initial concentration.
In this probel, the haf life is 87 days, this means that after evry 87 days, the concentration or mass of sulfur-35 decreases by half.
If the starting mass is 1.00g, then we have;
1.00g --> 0.5g (First Half life)
0.5g --> 0.25g (Second half life)
This means that sulphur-335 would undergo two half lives for 0.25g to remain.
Total time taken = Number of half lives * Half life
Time taken = 2 * 87
Time taken = 174 days
A Carbon-10 nucleus has 6 protons and 4 neutrons. Through radioactive beta decay, it turns into a Boron-10 nucleus, with 5 protons and 5 neutrons, plus another particle. What kind of additional particle, if any, is produced during this decay
Answer:
No additional particle was produced during the decay.
Explanation:
The equation of decay is given as;
¹⁰₆C + ⁰₋₁ e → ¹⁰₅B + x
To identify x, we have to calculate its atomic and mass number.
In the reactants side;
Atomic Number = 6 + (-1) = 5
Mass number = 10 + 0 = 10
In the products side;
Atomic Number = 5 + x
Mass Number = 10 + x
Generally, reactant = product
Atomic Number;
5 = 5 + x
x = 5 - 5 = 0
Mass Number;
10 = 10 + x
x = 10 - 10 = 0
This means no additional particle was produced during the decay.
Ethanol, , boils at 78.29 °C. How much energy, in joules, is required to raise the temperature of 2.00 kg of ethanol from 26.0 °C to the boiling point and then to change the liquid to vapor at that temperature? (The specific heat capacity of liquid ethanol is 2.44 J/g ∙ K, and its enthalpy of vaporization is 855 J/g.)
Answer:
THE HEAT REQUIRED TO CHANGE 2 KG OF ETHANOL FROM 26 °C TO THE BOILING POINT AND TO VAPOR AT THAT TEMPERATURE IS 1965.175 KJ.
Explanation:
Boiling point of ethanol = 78.29 °C = 78.29 + 273 K = 351.29 K
Mass = 2 kg = 2000 g
Final temp. = 26.0 °C = 26 + 273 K= 299 K
Change in temperature = (78.29 - 26) °C = 52.29 °C
1. Heat required to raise the temperature from 26 °C to the boiling point?
Heat = mass * specific heat * change in temperature
Heat = 2000 * 2.44 * 52.29
Heat = 255 175.2 J
2. Heat required to change the liquid to vapor at that temperature?
Heat = mass * enthalphy of vaporization
Heat = 2000 * 855
Heat =1 710000 J
The total heat required to raise the temperature of 2 kg of ethanol from 26 °C to the boiling point and then to change the liquid to vapor at that temperature will be:
Heat = mcT + m Lv
Heat = 255 175.2 J + 1710000 J
Heat = 1965175.2 J
Heat = 1965.175 kJ of heat.
Which diagram represents this molecule?
Answer:
C
Explanation:
The molecule has 8 carbon atoms joined by 7 C-C bonds.
The first two diagrams show 6 carbon atoms, not 8.
The last two diagrams show line segments representing C-C bonds. Only choice C shows 7 such segments.
The appropriate choice is C.
Answer:
C.
Explanation:
Acetonitrile, CH3CN, is a polar organic solvent that dissolves many solutes, including many salts. The density of a 1.80 M acetonitrile solution of LiBr is 0.826 g/mL. Calculate the concentration of the solution in units of (a) molality; (b) mole fraction of LiBr; (c) mass percentage of CH3CN.
Answer:
(a) [tex]m=2.69m[/tex]
(b) [tex]x_{LiBr}=0.099[/tex]
(c) [tex]\% LiBr=18.9\%[/tex]
Explanation:
Hello,
In this case, given the molality in mol/L, we can compute the required units of concentration assuming a 1-L solution of acetonitrile and lithium bromide that has 1.80 moles of lithium bromide:
(a) For the molality, we first compute the grams of lithium bromide in 1.80 moles by using its molar mass:
[tex]m_{LiBr}=1.80mol*\frac{86.845 g}{1mol}=156.32g[/tex]
Next, we compute the mass of the solution:
[tex]m_{solution}=1L*0.826\frac{g}{mL}*\frac{1000mL}{1L}=826g[/tex]
Then, the mass of the solvent (acetonitrile) in kg:
[tex]m_{solvent}=(826g-156.32g)*\frac{1kg}{1000g}=0.670kg[/tex]
Finally, the molality:
[tex]m=\frac{1.80mol}{0.670kg} \\\\m=2.69m[/tex]
(b) For the mole fraction, we first compute the moles of solvent (acetonitrile):
[tex]n_{solvent}=669.68g*\frac{1mol}{41.05 g} =16.31mol[/tex]
Then, the mole fraction of lithium bromide:
[tex]x_{LiBr}=\frac{1.80mol}{1.80mol+16.31mol}\\ \\x_{LiBr}=0.099[/tex]
(c) Finally, the mass percentage with the previously computed masses:
[tex]\% LiBr=\frac{156.32g}{826g}*100\%\\ \\\% LiBr=18.9\%[/tex]
Regards.
