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lesya692 [45]
2 years ago
13

A certain liquid X has a normal boiling point of 111.20 celsius and a boiling point elevation constant Kb=0.95. calculate the bo

iling point of a solution made of 11. g of ammonium chloride dissolved in 400 g of X.
Chemistry
1 answer:
Goshia [24]2 years ago
8 0

Answer: the boiling point is = 137.325°C

Explanation:

From the formula: ∆Tb= Kb*m

From the question, Kb= 0.95, m= 27.5, T1= 111.2°C

Substitute into ∆Tb= Kb*m

∆Tb= 0.95*27.5= 26.125

∆Tb= T2-T1

Hence

T2- 111.2=26.125

T2= 26.125+ 111.2= 137.325°C

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5 0
1 year ago
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Explanation:

6 0
2 years ago
A piece of potassium metal is added to water in a beaker. The reaction that takes place is 2K(s) 2H20(/) ..... 2KOH(aq) H2(g) Pr
kakasveta [241]

Answer:

Explanation:

The given reaction is exothermic . So ΔH is negative .

Gas is evolving so work done by gas is positive or w is positive.

Change in internal energy that is ΔU is negative.

q = u - w

u is negative , w is positive so q is negative .

4 0
3 years ago
Tarnish on copper is the compound CuO. A tarnished copper plate is placed in an aluminum pan of boiling water. When enough salt
oksian1 [2.3K]

Answer:

The standard cell potential is 2.00 V

Explanation:

<u>Step 1:</u> Data given

Cu is cathode because of higher EP

Al3++3e−→Al       E∘=−1.66 V     anode

Cu2++2e−→Cu    E∘=0.340 V    cathode

<u>Step 2:</u> Balance both equations

2*(Al → Al3+-3e−)       E∘=1.66 V    

3*(Cu2++2e−→Cu)    E∘=0.340 V

<u>Step 3:</u> The netto equation

2 Al + 3Cu2+ +6e- → 2Al3+ + 3Cu -6e-

2 Al + 3Cu2+  → 2Al3+ + 3Cu

<u>Step 4:</u> Calculate the standard cell potential

E∘cell = E∘cathode - E∘anode

E∘cell = E∘ Cu2+/Cu - E∘ Al3+/Al

E∘cell =0.340 V - (-1.66) = 2.00 V

The standard cell potential is 2.00 V

4 0
2 years ago
A fossil was analyzed and determined to have a carbon-14 level that is 70 % that of living organisms. The half-life of C-14 is 5
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Expression for rate law for first order kinetics is given by:

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where,

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a - x = amount left after decay process = \frac{70}{100}\times 100=70

t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{70}

t=2948years

Thus the fossil is 2948 years old.

5 0
2 years ago
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