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3 years ago

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A 1.11 L Flask is filled with 1.15g of argon at 25*C. A Sample of Ethan vapor is added to the same flask until the total pressur

e is 1.450 atm. what is the partial pressure of ethane in the flask?

1 answer:

ycow [4]3 years ago

7 0

Answer:

I thik the pressure will be something about 3 or 4 percent

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What is the formula for tin(IV) sulfide?<br> A. Sn4S<br> B. SnS2<br> C. Sns<br> D. SnS4

defon

Answer:

$SnS_{2}$

Explanation:

The formula for tin(IV) sulfide is SnS $_{2}$

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3 years ago

BRAINLIEST PLEASE HELP!!!

Natalka [10]

Answer : first opinion and also last
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A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

$-w=\frac{dm}{dt}$

w is the mass flow

m is the mass of salt

$-v*C=\frac{dm}{dt}$

v is the volume flow

C is the concentration

$C=\frac{m}{V+(6-3)*L/min*t}$

$-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}$

$-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}$

$-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)$

$-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)$

$m=90kg*[2000L/(2000L+3*L/min*t)]$

a) Initially: t=0

$m=90kg*[2000L/(2000L+3*L/min*0)]=90kg$

b) t=210 min (3.5 hr)

$m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg$

c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

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3 years ago

The most common cooling mechanism for cloud formation is ________.

KatRina [158]

The answer is "Rising and expanding air or atmosphere cooling".
To form a cloud the air must be cooled to the temperature of dew point. when there is expansion of air, it cools to the dew point and thus the formation of cloud happens. This process is very common and used for the formation of clouds.

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3 years ago

For the following reaction, 5.04 grams of nitrogen gas are allowed to react with 8.98 grams of oxygen gas: nitrogen(g) + oxygen(

OLga [1]

Answer:

1. 10.8 g of NO

2. N₂ is the limting reagent

3. 3.2 g of O₂ does not react

Explanation:

We determine the reaction: N₂(g) + O₂(g) → 2NO(g)

We need to determine the limiting reactant, but first we need the moles of each:

5.04 g / 29 g/mol = 0.180 moles N₂

8.98 g / 32 g/mol = 0.280 moles O₂

Ratio is 1:1, so the limiting reactant is the N₂. For 0.280 moles of O₂ I need the same amount, but I only have 0.180 moles of N₂

Ratio is 1:2. 1 mol of N₂ can produce 2 moles of NO

Then, 0.180 moles of N₂ may produce (0.180 .2) / 1 = 0.360 moles NO

If we convert them to mass → 0.360 mol . 30 g/1 mol = 10.8 g

As ratio is 1:1, for 0.180 moles of N₂, I need 0.180 moles of O₂.

As I have 0.280 moles of O₂, (0.280 - 0.180 ) = 0.100 moles does not react.

0.1 moles . 32 g/mol = 3.2 g of O₂ remains after the reaction is complete.

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3 years ago

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