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inn [45]
3 years ago
10

What do you think about this volcanoes? Its for my science project, so be honest or brutally honest as possible. And tell me wha

t I do I need to do to make it look better.

Chemistry
2 answers:
uranmaximum [27]3 years ago
6 0
It looks pretty good I can see a lot of effort. Just one thing: the lava shouldn't be so equally spread out. It usually flows a majority in one direction. Good luck!
Anestetic [448]3 years ago
4 0
Wow! That's a cool volcano. In elementary I had to make a volcano and I used clay to make it. But yours kind of looks like a pumpkin... omg sorry... but i was just being honest... haha. Good luck on that! 
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Will the vapors of acetic acid go to the floor or ceiling of a room
solong [7]
The vapor of acetic acid will go to the floor of the room.

This is because the density of acetic acid vapor is equal to 2.07 is higher (almost twice) than the density of air which is equal to only 1
8 0
3 years ago
Need help with this.
Jobisdone [24]

Answer:

18.2 g.

Explanation:

You need to first figure out how many moles of nitrogen gas and hydrogen (gas) you have. To do this, use the molar masses of nitrogen gas and hydrogen (gas) on the periodic table. You get the following:

0.535 g. N2 and 1.984 g. H2

Then find out which reactant is the limiting one. In this case, it's N2. The amount of ammonia, then, that would be produced is 2 times the amount of moles of N2. This gives you 1.07 mol, approximately. Then multiply this by the molar mass of ammonia to find your answer of 18.2 g.

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3 years ago
You’re beautiful *heart*
Georgia [21]
I’m not your just blind. I’m sorry
8 0
3 years ago
Decide which of the following statements are True and which are False about equilibrium systems:A large value of K means the equ
ivanzaharov [21]

Answer:

a. True

b. False

c. True

d.  False

e. False

Explanation:

A. (true) The equilibrium constant K is defined as

\frac{Products}{reagents}

In any case  

aA +Bb ⇌ Cd +dD

where K is:

K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}

A large value on K means that the concentration of products is bigger than the concentrations of reagents, so the forward reaction is favored, and the equilibrium lies to the right.

B. (False) When we work with gases, we use partial pressure to make calculations in the equilibrium, so we estimate Kp as:

Kp= \frac{(P_{C})^{c}(P_{D})^{d}}{(P_{A})^{a}(P_{B})^{b}}

Using the ideal gas law, we can get a relationship between K and Kp  

Pv=nRT where P=\frac{n}{v}*RT we know that \frac{n}{v} is the molar concentration. When we replace P in the expression for Kp we get:

Kp= \frac{[C]^{c}*(RT)^{c}[D]^{d}*(RT)^{d}}{[A]^{a}*(RT)^{a}[B]^{b}*(RT)^{b}}

Reorganizing the equation:

Kp= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}*\frac{(RT)^{c+d}}{(RT)^{a+b}}

We can see K in the expression  

Kp= K*(RT)^{c+d-a-b}

Delta n = c+d-a-b

Kp= K*(RT)^{delta n}

For the reaction  

H_{2}(g) + F_{2}(g)-- equilibrium---2HF(g)

Delta n = 2-1-1=0

Kp= K*(RT)^{0}

So Kp=K in this case.

C. (true) The value of K just depends on the temperature that’s why changing the among of products won’t have any effect on its value.  

D. (false) as we can see this reaction involve a heterogeneous system with solids and gases. For convention the concentration for solids and liquids can be considered constant during the reaction that’s why they’re not include in the calculation for the equilibrium constant. Taking this into account the expression for the equilibrium for this reaction is:

CaCO_{3}(s)---equilibrium----CaO(s) + CO_{2}(g)

K= [CO_{2}]

So we can see that [CaCO_{3}] is not include in the expression.  

E. (False) The equilibrium is defined as the point where the rate of the forward reaction is the same to the rate of the reverse reaction. The value of K is telling you which reaction is favored but the rate of both reactions is the same in this point. (see picture)  

3 0
3 years ago
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bija089 [108]
O le elo ma muli pipilo ma le kaoga e fai gi o meaoga
4 0
3 years ago
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