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I am Lyosha [343]
3 years ago
7

Krypton-79 has a half-life of 35 hours. how many half-lives have passed after 105 hours?

Chemistry
1 answer:
Nookie1986 [14]3 years ago
6 0
The number of half -lives that has passed after 105  hours  for krypton-79 that has  half-life of 35 hours is calculated as below

if 1 half life = 35 hours

what about 105 hours = ? half-lives

= (1 half life  x105  hours) /35 hours = 3 half-lives  has passed after 105 hours
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Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result
Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f)  12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only  the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) =  4.886

(d) mol HA reacted = 150 x 10⁻³ L  x  x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) =  5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH  = √(kb x (A⁻)  where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA,  200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒   Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0
3 years ago
The Nutrition Facts label for crackers states that one serving contains 19 g of carbohydrate, 4 g of fat, and 2 g of protein. Wh
sp2606 [1]

<u>Answer:</u> The amount of kilocalories contained in the given serving of crackes is 0.120 kCal

<u>Explanation:</u>

<u>As per the USDA:</u>

Carbohydrates provide 4 calories per gram, protein provides 4 calories per gram, and fat provides 9 calories per gram

We are given:

Mass of fat in the crackers = 4 g

Mass of carbohydrates in the crackers = 19 g

Mass of protein in the crackers = 2 g

Conversion factor used:  1 kCal = 1000 Cal

Applying unitary method:

  • <u>For Fat:</u>

1 gram of fat provides 9 calories

So, 4 gram of fat will provide = \frac{9}{1}\times 4=36Cal=0.036kCal

  • <u>For Carbohydrates:</u>

1 gram of carbohydrates provides 4 calories

So, 19 gram of carbohydrates will provide = \frac{4}{1}\times 19=76Cal=0.076kCal

  • <u>For Proteins:</u>

1 gram of proteins provides 4 calories

So, 2 gram of fat will provide = \frac{4}{1}\times 2=8Cal=0.008kCal

Total kilocalories per serving = [0.036 + 0.076 + 0.008] kCal = 0.120 kCal

Hence, the amount of kilocalories contained in the given serving of crackes is 0.120 kCal

8 0
3 years ago
Can anyone answer these ? 40 points !!
lesantik [10]

1st one is D

2nd one is A

4 0
3 years ago
Energy is released when the nucleus of an atom splits and two smaller atoms are formed. What is the name of this process?
Svetllana [295]

Explanation:

It's (D), nuclear fission................

5 0
2 years ago
How many molecules are contained in 0.800 mol o2?
Ksenya-84 [330]
Multiply .800 moles of O2 by Avagadro's number divided by 1 mole. This will get rid of the moles on the bottom and leave you with molecules. So technically .800 times 6.02x10^23.
6 0
3 years ago
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