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uysha [10]
2 years ago
4

A gas occupies 3.5 L at standard pressure. Find the volume of the gas when the pressure is 1140 mm Hg.

Chemistry
1 answer:
telo118 [61]2 years ago
4 0

Answer: 2.33L

Explanation:

P1 = 760mmHg

V1 = 3.5L

P2 = 1140mmHg

V2 =?

P1V1 = P2V2

760 x 3.5 = 1140 x V2

2660 = 1140 V2

2660/1140 = V2

V2 = 2.33L

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How many moles of lithium chloride will be formed by the reaction of chlorine with 0.046 mol of lithium bromide in the following
Feliz [49]

Answer:

The answer is 0.046 mol.

Explanation:

By looking at the balanced equation, you can form a ratio of lithium chloride and lithium bromide using the coefficient value :

ratio of lithium bromide <u>(</u>2LiBr)

= 2

ratio of lithium chloride (2LiCl)

= 2

So the ratio is 2 : 2 then simplify into 1 : 1 . Which means that 1 mol of lithium bromide is equal to 1 mole of lithium chloride.

In this case, 0.046 mol of lithium bromide will form <u>0</u><u>.</u><u>0</u><u>4</u><u>6</u><u> </u><u>m</u><u>o</u><u>l</u> of lithium chloride.

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2 years ago
How many quarters fit in a 1.75 liter bottle?
WARRIOR [948]
1.85 quarts can fit into a 1.75 liter bottle
8 0
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A scientist stores a very large number of particles in a container in order to determine the number of particles the scientist m
frutty [35]

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2 years ago
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At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5
Annette [7]

Answer:

4600s

Explanation:

2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

-\frac{d[B]}{dt}=k[B] - - -  -\frac{d[B]}{[B]}=k*dt

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.  

-\frac{d[P(B)]}{P(B)}=k*dt  

-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt

Integrating we get:

\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt

-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})

Clearing for t2:

\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}

ln[P(N_{2}O_{5})]=ln(650)=6.4769

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t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s

4 0
2 years ago
Danny sails a boat downstream. The wind pushes the boat along at 21 km/hr. The current runs downstream at 15 km/hr. What is the
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Explanation:

For downstream motion of the boat, the actual velocity of the boat is the sum of velocity of the water current and the velocity of the boat due to pushed by wind.

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Velocity of the boat going downstream, u = 21 km/h

Actual velocity of the boat = v'

v' = v + u

⇒v' = 15 km/h + 21 km/h

⇒u = 21 km/h +15 km/h = 36.0 km/h downstream

Thus, the correct answer is option D.

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2 years ago
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